Is Distance/space Lorentz contracted?

[Saw]

I think what I said agrees with all of your points and PeterDonis's too.

Ah, then I misunderstood. Thanks for the comment.

 

[PeterDonis]

Two frames cannot agree on simultaneity regarding any two events
On #1 alone,, #2 is an automatic conclusion.

Ok, good.

Just looking at it mentally, it seems to me , we have R now behind ,so its clock is running behind ,exactly opposite the normal situation. F is ahead, with clocks running ahead.

I'm not sure what you mean by "opposite the normal situation". What is the "normal" situation I'm supposed to compare this one to, that comes out the opposite way?

But in any case, we haven't gotten to whose clock is running "behind" or "ahead" yet; all we've established so far is that events Rn and Fn+1 aren't simultaneous to either R or F, for n > 1. Now we take the next step: call the proper time elapsed on R's clock at event Rn, T-Rn. Call the proper time elapsed on F's clock at event Fn+1, T-Fn+1. These two proper times are equal, but let's keep separate symbols for them. Then we have the following:

* In R's MCIF at event Rn, the time coordinate of event Fn+1 is *less* than T-Rn.

* In F's MCIF at event Fn+1, the time coordinate of event Rn is *greater* than T-Fn+1.

This is a precise statement of what I think you were trying to get at by saying that R's clock is running behind and F's clock is running ahead. However, a I think a more natural interpretation for R and F to put on these observations would be:

* R, at event Rn, sees F pulling ahead of him, because in R's MCIF at event Rn, F has already passed event Fn+1--i.e., F has already passed his corresponding station.

* F, at event Fn+1, sees R falling behind him, because in F's MCIF at event Fn+1, R has not yet reached event Rn--i.e., R has not yet reached his corresponding station.

These interpretations are simply manifestations of the fact that the spatial separation of R and F, in the frame in which they start out--the station frame--*must* inevitably turn partially into time separation in any other frame in relative motion to the station frame. And the time separation must inevitably lead to "speed separation" in each observer's MCIF at any point along their worldlines.

Furthermore, all of the above will lead to a change in the distance between R and F, as measured in either of their MCIFs. For example, consider R's MCIF at event Rn (n > 1). What is the distance between R and F in that frame? Well, the "distance" between any two worldlines in a given frame is the spacelike interval between points on each of the worldlines that have the same time coordinate. But we've already seen that, whatever event on F's worldline has the same time coordinate as Rn in R's MCIF at Rn, it is *not* Fn+1--it is some event further along F's worldline than Fn+1. So the interval between events Rn and Fn+1 is irrelevant to answering the question, "how far away will F be from R when R passes station #n?"

And similarly, to answer the question, how far away R will be from F when F passes station n+1, we will have to find the event on R's worldline with the same time coordinate as Fn+1, in F's MCIF at Fn+1, and that event will not be as far along R's worldline as Rn. I won't do the math explicitly here (I can in a follow-up post if desired), but when you do all the calculations, it turns out that both distances, that between R and F at Rn and between F and R at Fn+1, are *larger* than the spatial separation between R1 and F2 in the station frame (which is the original separation between R and F, since those are the events where they start out). And of course this is consistent with R's observation that F is pulling away and F's observation that R is falling behind.

All this is also consistent with what is observed in the station frame. In that frame, the spatial separation between the worldlines remains constant. But in that frame, the ships are moving (for any time t > 0), so the separation between them in the station frame is Lorentz contracted, compared to the separation between them in either ship's MCIF (in which one ship is momentarily at rest).

Now, let's consider your suggestion:

They turn off the engines and resynch their clocks so that R is running ahead and F running behind and we have an inertial frame like any other ,except that it is exactly the same size as it was at rest in its original frame as observed in that original frame.

You failed to specify the most important thing: at precisely which events on their worldlines do R and F turn off their engines? It makes a difference.

Let's suppose that R turns off his engine at event Rn and F turns off his engine at event Fn+1 (i.e., they turn off their engines when they each pass their respective stations n and n+1, with the exact value of n being prearranged). This is what I think you meant to specify. Then, they resynchronize their clocks; R emits a light pulse towards F from event Rn, and F emits a light pulse towards R at event Fn+1.

What will happen? Of course their clocks will be out of sync. But to re-sync them, they have to know the distance between the ships, in what will now become their common (inertial) rest frame. What is that distance? They will find (and they can check, of course, by timing round-trip light pulses) that the distance is *not* the same as the original separation between them in the station frame; it is *larger*--in fact, just enough larger so that its Lorentz-contracted value, as it appears in the station frame, is the same as the original separation in the station frame. So it is *not* true that the separation between R and F, in this new inertial frame, will be "exactly the same size as it was at rest in its original frame". (It *is* true that the separation between R and F, in the *station frame*, will be the same as it always was--and this may have been what you meant by your statement of "exactly the same size" above, I'm not sure--as long as R and F each stop their engines at events which are simultaneous in the station frame, as I specified they did, above).

Armed with this (larger) distance value, R and F can each, by observing the "time stamps" on light pulses giving the time of emission, correcting those for light-travel time, and comparing those with the time of reception, correct their clock readings so that they are in sync. How they correct depends on what they want the new "origin" of time in their new common rest frame to be, but in general, R will have to advance his clock reading, or F will have to move back his clock reading, or some combination of the two, in order to get them in sync. (Btw, at the end of the synchronization, both clocks will have a common "origin of time"--R's clock will not be running ahead or F's running behind.)

As this is assumed to apply to a single ship as well , this presents an extraordinary picture.

No, it is *not* assumed to apply to a single ship as well--at least not by me. Everything I've said above applies only to the case of separate ships, with no physical connection between them, who each experience equal, constant proper acceleration, and therefore travel along the worldlines I've labeled R and F.

To apply any of what I've said to a single ship, you would have to set up a scenario such that the front end of the ship traveled along the same worldline as the "front" ship F in the scenario I've been discussing, and the rear end of the ship traveled along the same worldline as the "rear" ship R in that scenario. That can be done, of course, but it's going to look like a very unusual scenario. For example, here's one way of doing it:

Let there be a line of stations, numbered #1, #2, #3, etc., all moving inertially, and all at rest relative to one another. Let there be a long ship whose rear end, R, is located next to station #1, and whose front end, F, is located at station #2, at time t = 0 in the station frame. Let there be rocket engines at each end of the ship (R *and* F), and let them be identical in construction and performance, so that each imparts an equal, constant, proper acceleration to its end of the ship. At time t = 0 in the station frame, both rocket engines turn on, and remain on continuously thereafter.

In the scenario I've just stated, the worldlines of the rear and front ends of the ship *would* be the same as those, R and F, that I've been discussing. And in this scenario, the ship would gradually be stretched, and eventually, when the tension caused by the two rocket engines, each pushing its end of the ship, exceeded the tensile strength of the ship's hull, the ship would break apart.

But notice that to achieve that result, I had to put rocket engines at each end of the ship--not a typical design. :-) Suppose instead that I adopted a more normal rocket design principle:

Let there be a line of stations, numbered #1, #2, #3, etc., all moving inertially, and all at rest relative to one another. Let there be a long ship whose rear end, R, is located next to station #1, and whose front end, F, is located at station #2, at time t = 0 in the station frame. Let there be a rocket engine at the rear of the ship that, when turned on, imparts a constant proper acceleration to its end of the ship. At time t = 0 in the station frame, the rocket engine turns on, and remains on continuously thereafter.

In *this* scenario, the worldline of the *rear* end of the ship will be the same as the one, R, that I've been discussing. But as it stands, the scenario tells us *nothing* about how the *front* end of the ship will move! To draw any conclusions about that, we need to bring in some sort of physical assumption about how the force of the rocket engine at the rear of the ship gets transmitted through the ship's structure to the front of the ship. Different physical assumptions will result in different worldlines for the front of the ship, and that, in turn, will result in different predictions about what the ship will "look like" in its MCIF, and in the station frame--whether it will appear "Lorentz contracted" and by how much, and so on. All of that gets into what I called question #2 in my last post, and I haven't been discussing that up to now.

(Btw, one thing I haven't mentioned in any of the above: I've been quoting above how things would be observed in R's and F's MCIFs, without addressing the point of how the "accelerated frame" relates to the MCIFs--in fact I've simply been assuming that the view from the "accelerated frame" at a given event is the same as the view from the MCIF at that event. Do I need to go into how that works? If so, I will.)

And the real questions behind the query: Does the Lorentz contraction apply simply to material bodies or does it apply to extended frames and the space between bodies?
Is the phenomenon a purely physical one of forces, with implied stresses etc. or is it simply a stressfree result of the geometry of spacetime?
Is it fundamentally a temporal effect that is perceived as a spatial effect??

The question you're asking here is certainly a valid question, and hopefully will be somewhat clarified by what I'm saying. But the way you ask it may not be the best way to ask it. That's why I went to the trouble of separating the issue of "kinematics" from the issue of physical assumptions. Here's what I think is a better way to get at the points you're interested in.

(1) I would use the term "Lorentz contraction" *only* to refer to how a spacelike interval between a specific, defined pair of events appears in different frames which are in relative motion. In other words, the term should be used only for the "kinematic" aspects--the changes in "how things look" that are due *solely* to relative motion, *not* to any other physical assumptions.

(2) This means that the question of Lorentz contraction doesn't even come into play until you've already determined all the worldlines and events of interest. Lorentz contraction, defined as I recommend, is a "result of the geometry of spacetime", and that is all it is. It applies to any spacelike interval, once that interval is specified, regardless of whether the interval is occupied by empty space or by a material body (in the latter case, of course, the worldlines of all the points in the material body must already have been specified).

(3) All the "physical" questions about forces, stresses, etc. should be construed as questions about *how to specify the worldlines and events of interest*. In other words, they are questions about how to figure out which specific points and curves in the geometry of spacetime are the ones we are dealing with. This is where all the stuff about Born rigidity and so forth, and how reasonable various physical assumptions are about how bodies respond to stress, comes into play.

 

[PeterDonis]

If you want to contest this, you can do it in the thread Elegant Universe Example, where this subject is already advanced.

Ok, I'll look in that thread.

 

[PeterDonis]

Saw:

Ok, I'll look in that thread.

I looked and it looks like the thread has died. But I don't disagree with the position you were taking there: you were basically saying that (1) the "solution" (i.e., the "right" specification of who wins the race, for example) has to involve quantities that everyone can agree on, so there's no ambiguity, and (2) which quantities you choose has to be guided by the "spirit" of the problem. I agree with both of those statements. I was focusing on (1) in my comments, but I didn't mean to imply that (2) isn't important too.

 

[Austin0]

=PeterDonis;2358975] Originally Posted by Austin0
Two frames cannot agree on simultaneity regarding any two events
On #1 alone,, #2 is an automatic conclusion.

Ok, good.


Originally Posted by Austin0
Just looking at it mentally, it seems to me , we have R now behind ,so its clock is running behind ,exactly opposite the normal situation. F is ahead, with clocks running ahead.

I'm not sure what you mean by "opposite the normal situation". What is the "normal" situation I'm supposed to compare this one to, that comes out the opposite way?

I am well aware that I have to improve my grasp of accepted terminology. I am working on this. In this case the normal situation would be that relative to a another inertial frame the clocks in the rear would be running ahead of the clocks positioned forward in the direction of motion. As you point out here, that is exactly what we assume would transpire when they synch their clocks by the convention. The rear ship would set its clock ahead or the front ship would set its clock behind.

But in any case, we haven't gotten to whose clock is running "behind" or "ahead" yet; all we've established so far is that events Rn and Fn+1 aren't simultaneous to either R or F, for n > 1. Now we take the next step: call the proper time elapsed on R's clock at event Rn, T-Rn. Call the proper time elapsed on F's clock at event Fn+1, T-Fn+1. These two proper times are equal, but let's keep separate symbols for them. Then we have the following:

* In R's MCIF at event Rn, the time coordinate of event Fn+1 is *less* than T-Rn.

Fine or alternately; at event Rn the simultaneous location of F is Fn+1+? with T-F > T-Rn

* In F's MCIF at event Fn+1, the time coordinate of event Rn is *greater* than T-Fn+1.

OK or at Fn+1 the position of R is Rn minus (?) and the T-Rn-? is less than T-Fn+1

This is a precise statement of what I think you were trying to get at by saying that R's clock is running behind and F's clock is running ahead. However, a I think a more natural interpretation for R and F to put on these observations would be:

I assumed that a precise statement was not necessary and in fact you do seem to have figured out what I was saying in my casual way.

* R, at event Rn, sees F pulling ahead of him, because in R's MCIF at event Rn, F has already passed event Fn+1--i.e., F has already passed his corresponding station.

* F, at event Fn+1, sees R falling behind him, because in F's MCIF at event Fn+1, R has not yet reached event Rn--i.e., R has not yet reached his corresponding station.

Agreed. I was not assuming anything from the perspective of the ships, but only from our perspective of looking at the diagram and applying the assumption of simultaneity from that perspective, to spacetime positions of the ships.

These interpretations are simply manifestations of the fact that the spatial separation of R and F, in the frame in which they start out--the station frame--*must* inevitably turn partially into time separation in any other frame in relative motion to the station frame. And the time separation must inevitably lead to "speed separation" in each observer's MCIF at any point along their worldlines.

Agreed of course. From the point where it became a matter of simply applying the transformation and its implications to the ships, there is no longer any question. You have already answered any question of physics because the transformation itself predicts the complete spatial-temporal coordinate outcome. When I said I was sure your mathematical analysis was correct I did not mean that personally. As I see it, from this point it is purely the logic of mathematics. That any theorem or results validly derived from allowable arguments and axioms is bound to be true. I don't need to see or do the math to be "convinced" that it is going to be totally consistent . I already assume so.
All the rest; the spatial separation etc. is all immediately obvious as soon as you adopt the assumption of that simultaneity. If R is behind and F is ahead of course the distance in that perspective is going to be greater. Etc,etc

You failed to specify the most important thing: at precisely which events on their worldlines do R and F turn off their engines? It makes a difference.

Of course you are correct , it makes a difference. But I didnt see how the difference is relevant to the actual question here so I didnt specify. So if F cuts its engine at Fn+1 it will be before Rn and therefore R will continue acceleration for a longer period thus equalizing the velocities, I would assume. There are interesting detaila here I am sure , figuring out the relative time dilation between the ships etc. But I assume they also would work out consistently and don't add anything to these questions.

How they correct depends on what they want the new "origin" of time in their new common rest frame to be, but in general, R will have to advance his clock reading, or F will have to move back his clock reading, or some combination of the two, in order to get them in sync. (Btw, at the end of the synchronization, both clocks will have a common "origin of time"--R's clock will not be running ahead or F's running behind.)

Of course. That would only apply as observed from an inertial frame moving in the direction from F to R

No, it is *not* assumed to apply to a single ship as well--at least not by me. Everything I've said above applies only to the case of separate ships, with no physical connection between them, who each experience equal, constant proper acceleration, and therefore travel along the worldlines I've labeled R and F.

So I may have been taking JesseM's statement too literally.

In the scenario I've just stated, the worldlines of the rear and front ends of the ship *would* be the same as those, R and F, that I've been discussing. And in this scenario, the ship would gradually be stretched, and eventually, when the tension caused by the two rocket engines, each pushing its end of the ship, exceeded the tensile strength of the ship's hull, the ship would break apart.

AH HA! Finally ;-) From this it seems you are saying that the assumption of stretching does not come from the physics of acceleration re: propagation of momentum etc. but is derived from a conception of simultaneity. That this is a temporal translation but in this context is given a very physical interpretation.
I am not questioning this. My own "intuition" is that simultaneity is at the core of a lot of questions and could very well have physical implications , or at least as "physical" as any other phenomena.
I would guess then that you expect a steel rod , uniformly electromagnetically accelerated to stretch apart?

But notice that to achieve that result, I had to put rocket engines at each end of the ship--not a typical design. :-) Suppose instead that I adopted a more normal rocket design principle:

It would appear that Born rigid acceleration is an ideal abstraction impossible to even approximate in the real world. SO it would seem that any propulsion from the rear , a more normal rocket design as you put it , would automatically be as close to Born acceleration as we could get ?

Let there be a line of stations, numbered #1, #2, #3, etc., all moving inertially, and all at rest relative to one another. Let there be a long ship whose rear end, R, is located next to station #1, and whose front end, F, is located at station #2, at time t = 0 in the station frame. Let there be a rocket engine at the rear of the ship that, when turned on, imparts a constant proper acceleration to its end of the ship. At time t = 0 in the station frame, the rocket engine turns on, and remains on continuously thereafter.

In *this* scenario, the worldline of the *rear* end of the ship will be the same as the one, R, that I've been discussing. But as it stands, the scenario tells us *nothing* about how the *front* end of the ship will move! To draw any conclusions about that, we need to bring in some sort of physical assumption about how the force of the rocket engine at the rear of the ship gets transmitted through the ship's structure to the front of the ship. Different physical assumptions will result in different worldlines for the front of the ship, and that, in turn, will result in different predictions about what the ship will "look like" in its MCIF, and in the station frame--whether it will appear "Lorentz contracted" and by how much, and so on. All of that gets into what I called question #2 in my last post, and I haven't been discussing that up to now.


The question you're asking here is certainly a valid question, and hopefully will be somewhat clarified by what I'm saying. But the way you ask it may not be the best way to ask it. That's why I went to the trouble of separating the issue of "kinematics" from the issue of physical assumptions. Here's what I think is a better way to get at the points you're interested in.

The course of this thread is ample demonstration of the truth of that

(1) I would use the term "Lorentz contraction" *only* to refer to how a spacelike interval between a specific, defined pair of events appears in different frames which are in relative motion. In other words, the term should be used only for the "kinematic" aspects--the changes in "how things look" that are due *solely* to relative motion, *not* to any other physical assumptions.

(2) This means that the question of Lorentz contraction doesn't even come into play until you've already determined all the worldlines and events of interest. Lorentz contraction, defined as I recommend, is a "result of the geometry of spacetime", and that is all it is. It applies to any spacelike interval, once that interval is specified, regardless of whether the interval is occupied by empty space or by a material body (in the latter case, of course, the worldlines of all the points in the material body must already have been specified).

(3) All the "physical" questions about forces, stresses, etc. should be construed as questions about *how to specify the worldlines and events of interest*. In other words, they are questions about how to figure out which specific points and curves in the geometry of spacetime are the ones we are dealing with. This is where all the stuff about Born rigidity and so forth, and how reasonable various physical assumptions are about how bodies respond to stress, comes into play.

I understand what you are saying and certainly agree on the effects being the "result of the geometry of spacetime" but don't see the distinction between kinematics and physics you are implying.
On one hand you seem to be saying that contraction has no physical implications but then adopt assumptions of simultaneity from kinematics that then seem to dictate both contraction and the physical forces involved in acceleration.
Perhaps you could elaborate on this??
Thanks

 

[PeterDonis]

AH HA! Finally ;-) From this it seems you are saying that the assumption of stretching does not come from the physics of acceleration re: propagation of momentum etc. but is derived from a conception of simultaneity. That this is a temporal translation but in this context is given a very physical interpretation.

The stretching isn't an assumption; it's a logical consequence of specifying the worldlines R and F the way we've specified them (that both ends of the ship, R and F, experience equal constant proper acceleration). Once you've specified that, the stretching has to happen, by the laws of SR; you don't have to assume it.

I would guess then that you expect a steel rod , uniformly electromagnetically accelerated to stretch apart?

If the rod has a rocket engine at each end, with both engines imparting equal, constant proper accelerations to their respective ends, as with the ship I specified in my last post, then yes, it will stretch apart. I'm not sure what "electromagnetically accelerated" means here, but if it means that some electromagnetic force is applied all along the rod, as with the scenario I describe below for Born rigid acceleration, then no, the rod won't stretch apart.

It would appear that Born rigid acceleration is an ideal abstraction impossible to even approximate in the real world. SO it would seem that any propulsion from the rear , a more normal rocket design as you put it , would automatically be as close to Born acceleration as we could get ?

Neither of the scenarios I specified corresponds to Born rigid acceleration. You're right that Born rigid acceleration is an ideal abstraction, but it's certainly possible to construct a scenario that would realize it, which is no more physically impossible than the others we've been considering. :-) Here it is:

Let there be a line of stations, numbered #1, #2, #3, etc., all moving inertially, and all at rest relative to one another. Let there be a long ship whose rear end, R, is located next to station #1, and whose front end, F, is located at station #2, at time t = 0 in the station frame. Let the ship be divided into very short segments, running from R to F, and let each segment have its own rocket engine; each rocket engine is programmed to impart to its segment an acceleration equal to \frac{c^2}{x}, where x is the distance of the segment from the origin in the station frame at time t = 0. This means that, as we move from the rear end R to the front end F of the ship, the acceleration imparted to each segment gradually gets smaller, in inverse proportion to distance. At time t = 0 in the station frame, all of the rocket engines turn on, and they all remain on continuously thereafter.

In the above scenario, the ship as a whole will undergo Born rigid acceleration. Born rigid acceleration is a "nice" way to accelerate the ship because (1) in the MCIF of any given segment of the ship, at any point on that segment's worldline, the proper length of the ship appears unchanged from its initial value in the station frame at time t = 0; (2) because of this, the internal stresses within the ship stay the same as it moves. But as you can see, you need to arrange things very precisely in order to achieve this--and of course no real rocket ship would be designed to do this. (If you see a reference to Born rigid acceleration requiring a "conspiracy of forces", that's what it refers to.)

On one hand you seem to be saying that contraction has no physical implications but then adopt assumptions of simultaneity from kinematics that then seem to dictate both contraction and the physical forces involved in acceleration.
Perhaps you could elaborate on this??

I'm not saying that Lorentz contraction has no physical implications. For one thing, kinematics is a part of physics. I'm just trying to separate two stages of "solving" a problem that's posed. First, you analyze the specification of the problem to determine all the worldlines and events of interest. Then, you use relativistic kinematics to determine how those worldlines and events will appear in whatever frames you like. You could call both stages "physical" if you want to.

Also, I'm not "assuming" the kinematics, or the simultaneity relations, or the physical forces involved in acceleration, or anything of that sort, except insofar as I'm assuming that SR is correct. The laws of SR *require* all these things to be related in certain specific ways, once you've determined the worldlines and events involved. That's all I'm trying to emphasize when I talk about kinematics.

 

[Austin0]

Re: Is Distance/space Lorentz contracted?

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Originally Posted by Austin0
From this it seems you are saying that the assumption of stretching does not come from the physics of acceleration re: propagation of momentum etc. but is derived from a conception of simultaneity.

The stretching isn't an assumption; it's a logical consequence of specifying the worldlines R and F the way we've specified them #1(that both ends of the ship, R and F, experience equal constant proper acceleration). Once you've specified that, the stretching has to happen, by the laws of SR; you don't have to assume it.

Hi PeterDonis In this case [i.e. a single ship] we have not yet specified the world lines have we?

#1 This appears to make assumptions:
A) That simple application of force from one end would not result in equal acceleration at both ends.
B) That length contraction would not intrinsically occur as the result of instananeous velocities.[clock hypothesis]
C) It seems to be taking kinetic conclusions arrived at through consideration of two ships and then applying them to a different situation. Simply assigning world lines on that basis.
But I may be misinterpreting your statements so further clarification might help.

Originally Posted by Austin0
I would guess then that you expect a steel rod , uniformly electromagnetically accelerated to stretch apart?

If the rod has a rocket engine at each end, with both engines imparting equal, constant proper accelerations to their respective ends, as with the ship I specified in my last post, then yes, it will stretch apart. I'm not sure what "electromagnetically accelerated" means here, but if it means that some electromagnetic force is applied all along the rod, as with the scenario I describe below for Born rigid acceleration, then no, the rod won't stretch apart.

My knowledge of electromagnetic acceleration is minimal derived mostly from reading about railgun technology, current and hypothetical. From what I understand there are various methods of application depending on the timing and directions of fields. It seems that it is possible to, in effect, pull a rod , or push-pull or apply uniform fields , so I imagine it is also possible to control the force in the specific format required by Born acceleration.
So in this context do you still think a uniformly accelerated rod would deform?

Originally Posted by Austin0
It would appear that Born rigid acceleration is an ideal abstraction impossible to even approximate in the real world. SO it would seem that any propulsion from the rear , a more normal rocket design as you put it , would automatically be as close to Born acceleration as we could get ?

Let there be a long ship whose rear end, R, is located next to station #1, and whose front end, F, is located at station #2, at time t = 0 in the station frame. Let the ship be divided into very short segments, running from R to F, and let each segment have its own rocket engine; each rocket engine is programmed to impart to its segment an acceleration equal to , where x is the distance of the segment from the origin in the station frame at time t = 0. This means that, as we move from the rear end R to the front end F of the ship, the acceleration imparted to each segment gradually gets smaller, in inverse proportion to distance. At time t = 0 in the station frame, all of the rocket engines turn on, and they all remain on continuously thereafter.

In the above scenario, the ship as a whole will undergo Born rigid acceleration. Born rigid acceleration is a "nice" way to accelerate the ship because (1) in the MCIF of any given segment of the ship, at any point on that segment's worldline, the proper length of the ship appears unchanged from its initial value in the station frame at time t = 0; (2) because of this, the internal stresses within the ship stay the same as it moves. But as you can see, you need to arrange things very precisely in order to achieve this--and of course no real rocket ship would be designed to do this.

This raises a lot of interesting questions.
It would seem that application of force to the rear would represent maximal resistence to expansion. Now if compression was the issue, then it would make sense to apply additional forward force throughout the system. But that is not the object in this case, so I don't see how this calibrated application has any effect regarding the reduction of expansion. ?

If this does apply then what is the expected result for the contents of this vehicle??
Does this mean that a steel ruler hanging from the wall of Einsteins elevator would stretch like silly putty??
What of a ruler that has been accelerated while positioned transverse to the motion and is then rotated longitudinally?

Suddenly subjected to the cumulative acceleration of the total course?

I'm not saying that Lorentz contraction has no physical implications. For one thing, kinematics is a part of physics. I'm just trying to separate two stages of "solving" a problem that's posed. First, you analyze the specification of the problem to determine all the worldlines and events of interest. Then, you use relativistic kinematics to determine how those worldlines and events will appear in whatever frames you like. You could call both stages "physical" if you want to.

Also, I'm not "assuming" the kinematics, or the simultaneity relations, or the physical forces involved in acceleration, or anything of that sort, except insofar as I'm assuming that SR is correct. The laws of SR *require* all these things to be related in certain specific ways, once you've determined the worldlines and events involved. That's all I'm trying to emphasize when I talk about kinematics.

Agreed kinematics is a part of physics of course. It seems to me this question enters an area of a certain ambiguity.
For a second consider Newtonian kinematics. I am sure you would agree that the Galillean transforms had no physical implications or assumptions beyond the invariance of Newtonian kinematics. Given a set of coordinate events in one frame, there were no additional considerations of physics, per se, required to derive accurate events in another frame.
I think we agree that this is equally applicable to the Lorentzian transforms as applied to events between inertial frames.
But regarding the transitional state between inertial frames is a different situation.
To get coordinate events in any frame then requires consideration of physics, assumptions about, not only the physics of acceleration, but also about the meaning and interpretation of the Lorentzian effects, which did not need to be considered at all when dealing with inertial frames.
So IMHO it is not just a question of whether a particular set of assumptions is consistent with the fundamental constraints of SR, but whether or not it is the only set of assumptions that is consistent or the set that is the most consistent with the physics of acceleration as we know it from empirical testing to this point. I do not presume to have an answer to this, but it does appear to me that there are other possibilites.
Thanks. BTW this is starting to get more and more interesting

 

[PeterDonis]

In this case [i.e. a single ship] we have not yet specified the world lines have we?

In each of the single ship cases I specified, the worldlines of any point on the ship where I put a rocket engine were specified (because the accelerations that the rocket engines imparted to their locations on the ship were specified). But those were the *only* points whose worldlines were specified.

So, in the first case, with a rocket engine at the rear and front end of the ship, the worldlines of the rear and front end were specified, but the worldlines of intermediate parts of the ship were not. The front and rear end worldlines being specified were sufficient to conclude that the ship would stretch, but the exact worldlines of intermediate points would depend on the exact details of the stretching--i.e., how each segment responded to the tensile stress caused by the rear and front ends getting further apart.

In the second case, with a rocket engine only at the rear, as I said then, only the rear end's worldline was specified; the worldlines of all the other parts of the ship would depend on the details of how the force imparted at the rear end was transmitted through the ship. More on that below.

In the third case, with rocket engines all along the ship (Born rigid acceleration), the worldlines of all parts of the ship *were* specified (at least to the accuracy of the size of each "segment" of the ship that was assigned an engine). I had to do that because it's the only way to realize Born rigid acceleration--in effect, specifying that an object undergoes Born rigid acceleration *does* specify the worldlines of every part of the object.

A) That simple application of force from one end would not result in equal acceleration at both ends.

That's correct. It can't possibly, because it takes time for the information that the force has been applied to travel from one end of the ship to the other.

That information can't travel faster than the speed of light, so, for example, if the ship is one light-second long, then the front end can't possibly begin accelerating sooner than 1 second after the rear end does (as seen in the "station frame", the frame in which the ship is initially at rest). For any actual known material, it will take a lot longer than that for the applied force to propagate through the ship (because the speed of sound in all known materials, which is the speed at which applied forces propagate through the material, is many orders of magnitude less than the speed of light).

B) That length contraction would not intrinsically occur as the result of instananeous velocities.(clock hypothesis)

Once again, don't confuse the kinematic effect of Lorentz contraction, which requires that you have already specified the worldlines and events you're dealing with, with the dynamic effects like how objects respond to applied forces, which are what you use to *figure out* which worldlines and events you're dealing with.

The kinematic effect is simply this: if I know the length of a given object in a given frame, I also know its length in any other frame, just by applying the Lorentz transformations. But in order to know its length in any frame, I have to know the worldlines of both its ends. If I don't know those worldlines yet, I don't know its length. If I know only one particular pair of events on the worldlines of the object's ends--for example, I know the spatial position of its rear and front ends at time t = 0 in the station frame--then I only know its length in the frame in which that pair of events is simultaneous, at the particular time of those events--in this case, the station frame at time t = 0. I don't have enough information in that case to even apply the Lorentz transformation, because those two events won't be simultaneous in any frame that's moving relative to the station frame, so I can't use them to determine the object's length in any other frame. I have to find a pair of events on the worldlines of the object's ends that are simultaneous in the *moving* frame to determine its length in that frame, and if I don't know the worldlines of the ends, I can't do that.

Discussions of the Lorentz transformations in textbooks usually gloss over all this, by just specifying (often implicitly) all the worldlines that will be needed to solve whatever problems are posed, including performing whatever Lorentz transformations are required. But if you look "under the hood", you'll see that all the above stuff still applies; it's just hidden by the way the problem is stated. In other words: the kinematic effects, as I've defined them above, *do* "intrinsically occur"--but you have to know a lot to calculate what they are. Gaining that knowledge requires figuring out the dynamic effects first, so you can determine the worldlines and events you're dealing with, and the dynamic effects are *not* "intrinsic"--they depend on the physical assumptions you make.

C) It seems to be taking kinetic conclusions arrived at through consideration of two ships and then applying them to a different situation.

Nope--the only conclusions I'm drawing for each case are based on the worldlines that I know are specified in each case (I discussed which ones those were for each case above).

My knowledge of electromagnetic acceleration is minimal derived mostly from reading about railgun technology, current and hypothetical. From what I understand there are various methods of application depending on the timing and directions of fields. It seems that it is possible to, in effect, pull a rod , or push-pull or apply uniform fields , so I imagine it is also possible to control the force in the specific format required by Born acceleration. So in this context do you still think a uniformly accelerated rod would deform?

Once again, if the force is controlled the way I specified for Born rigid acceleration, then no, the rod would not deform. (I would think, as you do, that this is possible, but I don't know if any actual electromagnetic acceleration technologies do it in practice.) If the force is applied some other way, then the deformation of the rod would depend on how the force was applied.

It would seem that application of force to the rear would represent maximal resistence to expansion. Now if compression was the issue, then it would make sense to apply additional forward force throughout the system. But that is not the object in this case, so I don't see how this calibrated application has any effect regarding the reduction of expansion. ?

In the case of Born rigid acceleration, we *are* applying additional forward force throughout the system. At each point along the ship, we're applying just enough force to keep it "ahead" of the rear end of the ship by just the right amount, so that the total length of the ship remains constant (and internal stresses likewise remain constant). If we applied more force--for example, if we applied just as much force at the front end of the ship as at the rear--the ship would stretch, as in the first "single ship" case I specified. If we applied less force, the ship would compress, which is what would happen, at least initially, in the second "single ship" case I specified, where there was only a rocket engine at the rear end of the ship.

(How do I know the ship would initially compress in this case? Because, as I said above, the information that the force has been applied at the rear can't travel through the ship faster than the speed of light, so there will be an unavoidable minimum time delay before each part of the ship forward of the rear end starts accelerating. That means the ship will initially compress, as each segment, when it first starts accelerating, moves closer to the segment in front of it, that hasn't received the information to start accelerating yet.)

Does this mean that a steel ruler hanging from the wall of Einsteins elevator would stretch like silly putty??

If the elevator were accelerated hard enough, yes. Just as a steel ruler hanging from the ceiling in a room that was stationary in a strong enough gravitational field would stretch. Remember that in many of these cases, the implied accelerations we've been talking about are in the billions or trillions of g's.

What of a ruler that has been accelerated while positioned transverse to the motion and is then rotated longitudinally?

Accelerated how? Does the acceleration stop before the ruler is rotated?

Suddenly subjected to the cumulative acceleration of the total course?

I'm not sure I understand what you're asking here.

To get coordinate events in any frame then requires consideration of physics, assumptions about, not only the physics of acceleration, but also about the meaning and interpretation of the Lorentzian effects, which did not need to be considered at all when dealing with inertial frames.

The last statement here is incorrect; the meaning and interpretation of Lorentz contraction and time dilation is an integral part of figuring out how SR works for inertial frames. If Einstein had not been able to give each of these effects a direct physical interpretation in inertial frames in relative motion, I doubt he would have made much headway in getting people to consider SR. There is no fundamental difference here between inertial and accelerated motion, except that an object on an accelerated worldline "changes its inertial frame" as it moves--you have to use its MCIF at any given event to determine how things will look to it at that event. But as long as you know the object's worldline, its MCIF is fully determined at each event.

So IMHO it is not just a question of whether a particular set of assumptions is consistent with the fundamental constraints of SR, but whether or not it is the only set of assumptions that is consistent or the set that is the most consistent with the physics of acceleration as we know it from empirical testing to this point. I do not presume to have an answer to this, but it does appear to me that there are other possibilites.

Once more, I think it's important to keep emphasizing the distinction between kinematic and dynamic effects. As far as the kinematic effects are concerned (i.e., how things look in different frames, once you've already determined the specific worldlines and events you're dealing with), there is only *one* way to deal with them that is consistent with the fundamental constraints of SR (basically, the principle of relativity and the fact that the speed of light is the same in all inertial frames, plus the clock postulate). Once you have a specific set of worldlines and events, the way things look in any frame, inertial or accelerated, is fully determined by the postulates of SR (the three I just listed above) and the specifications of the worldlines and events. The only restriction is that spacetime has to be flat, at least to a good enough approximation for the problem you're dealing with.

However, that still leaves quite a bit of freedom in dealing with accelerated objects, because you still have to somehow determine the specific worldlines of the various parts of those objects, and that will depend on what assumptions you make about how forces applied to one part of the object are transmitted to other parts--the dynamic effects. There are *lots* of different assumptions you could make here that are consistent with the laws of SR, and each set of assumptions will result in a different set of worldlines for the various parts of the object. So in that sense, yes, there are certainly lots of different ways that an accelerated object could "look" that are all consistent with SR.

BTW this is starting to get more and more interesting

Good. :-)

 

[Austin0]

Originally Posted by Austin0
In this case [i.e. a single ship] we have not yet specified the world lines have we?

In each of the single ship cases I specified, the worldlines of any point on the ship where I put a rocket engine were specified (because the accelerations that the rocket engines imparted to their locations on the ship were specified). But those were the *only* points whose worldlines were specified.

So, in the first case, with a rocket engine at the rear and front end of the ship, the worldlines of the rear and front end were specified, but the worldlines of intermediate parts of the ship were not. The front and rear end worldlines being specified were sufficient to conclude that the ship would stretch, but the exact worldlines of intermediate points would depend on the exact details of the stretching--i.e., how each segment responded to the tensile stress caused by the rear and front ends getting further apart.

It seems to me that the above conclusions are justified only if:

A) It is justifiable to completely disregard the physics and existence of the intermediate body. I.e. That the motion and acceleration of the actual points of propulsion are completely independant of the connection with the total system.

B) You can apply an interpretation of acceleration as applied force and disregard the interpretation of acceleration as a change in the motion of a system.

C) That the Lorentz contraction can be completely disregarded as a factor in the physical situation regarding the relevant worldlines ,but can then, simply be applied as a kinematic result, ex post facto


In the second case, with a rocket engine only at the rear, as I said then, only the rear end's worldline was specified; the worldlines of all the other parts of the ship would depend on the details of how the force imparted at the rear end was transmitted through the ship. More on that below.

In the third case, with rocket engines all along the ship (Born rigid acceleration), the worldlines of all parts of the ship *were* specified (at least to the accuracy of the size of each "segment" of the ship that was assigned an engine). I had to do that because it's the only way to realize Born rigid acceleration--in effect, specifying that an object undergoes Born rigid acceleration *does* specify the worldlines of every part of the object.

Originally Posted by Austin0
A) That simple application of force from one end would not result in equal acceleration at both ends.

That's correct. It can't possibly, because it takes time for the information that the force has been applied to travel from one end of the ship to the other.

That information can't travel faster than the speed of light, so, for example, if the ship is one light-second long, then the front end can't possibly begin accelerating sooner than 1 second after the rear end does (as seen in the "station frame", the frame in which the ship is initially at rest). For any actual known material, it will take a lot longer than that for the applied force to propagate through the ship (because the speed of sound in all known materials, which is the speed at which applied forces propagate through the material, is many orders of magnitude less than the speed of light).

We are in complete agreement here regarding the propagation time. I would propose for the purposes of this discussion that Mach 1 be considered the practical limit. If the force applied is sufficient to make the speed of light relevant, I would say it is beyond the bounds of realism and billions of g's would imply inertial forces far beyond the ability of an addition of a few extra engines to counter.
Would you agree that as far as the principles involved are concerned ,the magnitude of the acceleration rate is not important?
WHile we agree on the importance of propagation time, we seem to have different ideas concerning the implications of this fact.
For discussion I will give a quick outline of my understanding and you or someone else with greater knowledge can point out any areas where I diverge from known physics.
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal occilation. At initial application ,there is no actual motion at the point of contact, the energy is simply passed along , with no net motion of either molecules or pressure.
It is not until the entire system has received enough energy that there is actual coordinate movement of the system as a whole.
As the magnitude is increased and exceeds the ability to transmit it fast enough, there is a build up , a delay at the point of introduction which could result in actual compression until the whole system reaches the point of motion. The initial compression you mentioned.
But is there any reason to expect this to remain once the system is in motion and the momentum is uniform throughout the system and additional applied energy is constant?
Or to assume that the intermolecular tensile forces would not reach some kind of equilibrium with the applied force??
Even possibly a sustained occilation?
Even more important, is there any reason to expect ,not only a sustained compression but an incrementally increasing compression or expansion over time?
My own thought is that this would only be reasonable if it was a dynamic rate of acceleration.
Obviously the situation becomes more complex with more elastic subjects.
Say a spring. Where the vector of applied force would not be aligned with the path of propagation and the path length would be longer relative to the mass and system length.
Torsion effects etc etc. would seem to lead to more intitial compression or expansion but not neccessarily any more of a sustained or cumulative effect.


Originally Posted by Austin0
B) That length contraction would not intrinsically occur as the result of instananeous velocities.(clock hypothesis)

Once again, don't confuse the kinematic effect of Lorentz contraction, which requires that you have already specified the worldlines and events you're dealing with, with the dynamic effects like how objects respond to applied forces, which are what you use to *figure out* which worldlines and events you're dealing with.

The kinematic effect is simply this: if I know the length of a given object in a given frame, I also know its length in any other frame, just by applying the Lorentz transformations.

QUOTE] I don't have enough information in that case to even apply the Lorentz transformation, because those two events won't be simultaneous in any frame that's moving relative to the station frame, so I can't use them to determine the object's length in any other frame.[/QUOTE]

In other words: the kinematic effects, as I've defined them above, *do* "intrinsically occur"--but you have to know a lot to calculate what they are.

Gaining that knowledge requires figuring out the dynamic effects first, so you can determine the worldlines and events you're dealing with, and the dynamic effects are *not* "intrinsic"--they depend on the physical assumptions you make.

I am not completely sure what the implications are of all the above quotes as some of them seem to be in agreement with what I have been saying while others seem to disagree.
But one thing seems sure, taken all together they appear ro agree with me when I said that the situation is not as simple as it might appear.


Originally Posted by Austin0
It would seem that application of force to the rear would represent maximal resistence to expansion. Now if compression was the issue, then it would make sense to apply additional forward force throughout the system. But that is not the object in this case, so I don't see how this calibrated application has any effect regarding the reduction of expansion. ?

In the case of Born rigid acceleration, we *are* applying additional forward force throughout the system. At each point along the ship, we're applying just enough force to keep it "ahead" of the rear end of the ship by just the right amount, so that the total length of the ship remains constant (and internal stresses likewise remain constant).

You seem to have missed the point of what I was saying.
In this case the proposition is: Single point thrust from the rear of the system would result in system expansion. Expansion is the "problem"
Logically it would seem to follow that a physical solution to this problem would be a counter thrust applied to the front. A compressive thrust against the direction of motion.
Do you disagree with this logic?
It would also seem to follow that the addition of forward propulsion would not only not reduce expansion it would reduce the compressive effect of a single rear drive.
It would make sense if you wanted to reduce compression.
DO you disagree with this logic??

If

we applied more force--for example, if we applied just as much force at the front end of the ship as at the rear--the ship would stretch, as in the first "single ship" case I specified. If we applied less force, the ship would compress, which is what would happen, at least initially, in the second "single ship" case I specified, where there was only a rocket engine at the rear end of the ship.

(How do I know the ship would initially compress in this case? Because, as I said above, the information that the force has been applied at the rear can't travel through the ship faster than the speed of light, so there will be an unavoidable minimum time delay before each part of the ship forward of the rear end starts accelerating. That means the ship will initially compress, as each segment, when it first starts accelerating, moves closer to the segment in front of it, that hasn't received the information to start accelerating yet.)

Originally Posted by Austin0
What of a ruler that has been accelerated while positioned transverse to the motion and is then rotated longitudinally?

Accelerated how? Does the acceleration stop before the ruler is rotated?

Consider either case.
A ruler that has not been Born accelerated longitudinally during the course of a long acceleration but is then reoriented during acceleration.
And then the same ruler a second after the engines are turned off.
What do expect the result to be in either case?

Originally Posted by Austin0
Suddenly subjected to the cumulative acceleration of the total course?

I'm not sure I understand what you're asking here.

In the course of a prolonged acceleration according to the hypothesis , the system has undergone coordinate contraction as well as resisted assumed proper expansion due to the action of a specific physical acceleration , directional force. A ruler that has been translated while oriented transversely has not experienced this directed force along its length.
If rotated so that that length is aligned with the acquired motion [cumulative acceleration]
it should not be contracted according to the hypothesis unless you assume that it instantly
undergoes an equivalent acceleration through rotation, in which case I guess you could expect it to stretch and break unless it had Born rockets attached throughout to control the rotation?
Or other?

Originally Posted by Austin0
To get coordinate events in any frame then requires consideration of physics, assumptions about, not only the physics of acceleration, but also about the meaning and interpretation of the Lorentzian effects, which did not need to be considered at all when dealing with inertial frames.

The last statement here is incorrect; the meaning and interpretation of Lorentz contraction and time dilation is an integral part of figuring out how SR works for inertial frames. If Einstein had not been able to give each of these effects a direct physical interpretation in inertial frames in relative motion, I doubt he would have made much headway in getting people to consider SR. There is no fundamental difference here between inertial and accelerated motion, except that an object on an accelerated worldline "changes its inertial frame" as it moves--you have to use its MCIF at any given event to determine how things will look to it at that event.

Out of context again.
So here is a more complete context.
PeterDonis

That's why I went to the trouble of separating the issue of "kinematics" from the issue of physical assumptions.

The kinematic effect is simply this: if I know the length of a given object in a given frame, I also know its length in any other frame, just by applying the Lorentz transformations
Original Austin0
For a second consider Newtonian kinematics. I am sure you would agree that the Galillean transforms had no physical implications or assumptions beyond the invariance of Newtonian kinematics. Given a set of coordinate events in one frame, there were no additional considerations of physics, per se, required to derive accurate events in another frame.
I think we agree that this is equally applicable to the Lorentzian transforms as applied to events between inertial frames.
_______________________________________________________________________

SO I was actually agreeing with you and somehow this was reinterpreted into implying that I thought there was no physical interpretation or meaning to Lorentz effects in SR.
In fact you seem to be contradicting yourself, since you have said more than once that physical assumptions were irrelevant to the application of the coordinate transforms.

However, that still leaves quite a bit of freedom in dealing with accelerated objects, because you still have to somehow determine the specific worldlines of the various parts of those objects, and that will depend on what assumptions you make about how forces applied to one part of the object are transmitted to other parts--the dynamic effects. There are *lots* of different assumptions you could make here that are consistent with the laws of SR, and each set of assumptions will result in a different set of worldlines for the various parts of the object. So in that sense, yes, there are certainly lots of different ways that an accelerated object could "look" that are all consistent with SR.

I am happy to see an area of at least partial agreement ;-)
We haven't yet touched on the basis for the Born hypothesis. I hope you don;t get tired of this topic before we do as I am sure you are a lot more knowledgeable and I am very curious.
Thanks

 

[PeterDonis]

A) It is justifiable to completely disregard the physics and existence of the intermediate body. I.e. That the motion and acceleration of the actual points of propulsion are completely independant of the connection with the total system.

You're correct that, in specifying that the rocket engines at the front and rear end of the ship each imparted the same constant proper acceleration to their ends of the ship, I was assuming that the engines could do that independently of the forces exerted on the front and rear ends by other parts of the ship. In any real case, that would mean the engines would have to be controlled very precisely, to maintain the same constant proper acceleration; for example, as the ship started to stretch, the segment of the ship just to the rear of the front end would start pulling back on the front end, so the rocket engine at the front end would have to increase its rate of fuel burn *just* enough to compensate for this extra pull, in order to maintain the same constant proper acceleration. Since that's not physically impossible (however unlikely it might be in practice), it doesn't invalidate my specification of the scenario, since it's just a thought experiment.

B) You can apply an interpretation of acceleration as applied force and disregard the interpretation of acceleration as a change in the motion of a system.

This isn't a matter of interpretation; the two have to be connected, because of conservation of momentum. The applied force is

F = \frac{d P}{d \tau}

where P is the energy-momentum 4-vector of the small segment of the ship that's being accelerated, and \tau is the proper time of that same small segment. But we have

P = m U

where m is the rest mass of the segment and U is its 4-velocity; so we have

F = m \frac{d U}{d \tau} = m A

where A is the proper acceleration (i.e., the rate of change of 4-velocity with respect to proper time). This is the relativistic version of Newton's second law, and if momentum is conserved, it *has* to hold.

C) That the Lorentz contraction can be completely disregarded as a factor in the physical situation regarding the relevant worldlines ,but can then, simply be applied as a kinematic result, ex post facto

Again, this isn't an assumption, it's a *definition* I've made in order to make it easier to avoid confusion between what I've been calling kinematic and dynamic effects. You can define "Lorentz contraction" to include those dynamic effects, if you want to, but then you'll have to come up with a different name for the kinematic ones, because they'll still be separate, conceptually, from the dynamic ones.

I would propose for the purposes of this discussion that Mach 1 be considered the practical limit.

Mach 1 *is* the limit--the speed of sound in a given substance is the limit of how fast applied forces can be propagated in that substance. But the speed of sound varies widely according to the substance. The speed of sound in air, which is what people usually refer to as "Mach 1", is pretty slow as sound speeds go, about 340 m/s, or 10^-6 x the speed of light. In water the sound speed is about 5 times that, about 1500 m/s, and in steel it's about 20 times that in air, about 6,000 m/s.

Sound speeds in plasmas (such as, for example, the Sun) can actually get up to a few percent of the speed of light. For example, using the "ion sound velocity" formula from http://en.wikipedia.org/wiki/Plasma_parameters#Velocities", with parameters typical of the Sun's core (\gamma = 5/3 for a non-relativistic plasma, T_e = 1,500 eV, Z = 1 and \mu = 1 because the Sun is mostly hydrogen), yields a sound speed of about 5 x 10^7 cm/s, or about 2/10 of a percent of the speed of light; and the Sun is actually a fairly cool plasma--larger stars have significantly hotter cores, and therefore higher sound speeds.

If the force applied is sufficient to make the speed of light relevant, I would say it is beyond the bounds of realism and billions of g's would imply inertial forces far beyond the ability of an addition of a few extra engines to counter.

The muons that were trapped and their lifetimes measured to check the clock postulate experienced accelerations up to 10^18 g. There's nothing physically impossible about such accelerations, and someday we may learn how to make macroscopic objects that are able to withstand them; the laws of physics don't prohibit that. Anyway:

Would you agree that as far as the principles involved are concerned ,the magnitude of the acceleration rate is not important?

Yes.

Given that the magnitude of force is within the materials ability to transmit it fast enough,

The speed of transmission of forces through an object (i.e., the sound speed) doesn't depend on the magnitude of the force. Small forces don't get transmitted any faster than large ones.

*How* the force gets transmitted does depend greatly on the physical constitution of the object--see next item.

At initial application ,there is no actual motion at the point of contact, the energy is simply passed along , with no net motion of either molecules or pressure.

Unless the point of application of the force is physically restrained in some way so that it can't move (i.e., if there's no counterbalancing force to work against the applied force), it *will* move upon application of the force. That is, the individual atoms or molecules of the object will move; at the atomic/molecular level, of course, the applied force is transmitted by interactions between the atoms/molecules--more precisely, by the electrical repulsion between the electrons in the atoms/molecules.

As the first "layer" of atoms/molecules in the object move in response to the applied force, they will, of course, move closer to the next "layer" in the object, and will thus exert a force on that next layer. In other words, the applied force starts a "wave" of force moving through the object--more precisely, a longitudinal wave (i.e., a sound wave) of alternating compression and expansion, as each layer pushes against the next, which then responds by moving away and pushing against the next in turn.

If we assume that the applied force at one end of the object continues to be applied, then once the initial wave has traveled the length of the object, it will start to be damped out. This is because as each layer pushes against the next, not all of the energy in the push gets converted into motion of the next layer; some of the energy goes into the internal degrees of freedom in the atoms/molecules--i.e., it gets converted into heat. If the material is very stiff (e.g., very high tensile steel--or even better, carbon nanotubes), the wave will be damped out quickly; if the material is soft, it will take longer for the wave to be damped out. While the wave is damping, the object will be oscillating, expanding and compressing with gradually decreasing amplitude. Once the wave is damped out, the object will be in a state (assuming the force continues to be applied at one end) in which there is a compressive stress all along its length, and in which its material is somewhat hotter than it was before the force was applied (because some of the applied energy got converted into heat during the damping).

In this case the proposition is: Single point thrust from the rear of the system would result in system expansion.

As the above analysis shows, it wouldn't; it would result in compression.

To get coordinate events in any frame then requires consideration of physics, assumptions about, not only the physics of acceleration, but also about the meaning and interpretation of the Lorentzian effects, which did not need to be considered at all when dealing with inertial frames.

...

For a second consider Newtonian kinematics. I am sure you would agree that the Galillean transforms had no physical implications or assumptions beyond the invariance of Newtonian kinematics. Given a set of coordinate events in one frame, there were no additional considerations of physics, per se, required to derive accurate events in another frame.
I think we agree that this is equally applicable to the Lorentzian transforms as applied to events between inertial frames.
__________________________________________________ _____________________

SO I was actually agreeing with you and somehow this was reinterpreted into implying that I thought there was no physical interpretation or meaning to Lorentz effects in SR.
In fact you seem to be contradicting yourself, since you have said more than once that physical assumptions were irrelevant to the application of the coordinate transforms.

I know that we agree that converting coordinates from one *inertial* frame to another is just "kinematics"; but you appeared to be saying that that was *not* true of accelerating frames. I'm saying that, since an "accelerating frame" at any given event on an accelerated worldline is just the MCIF at that event, what applies to inertial frames applies equally to "accelerating frames". Converting coordinates from an inertial frame to an "accelerated frame" is just converting from the inertial frame to the MCIF, which is another inertial frame. So it's just "kinematics" in both cases. All the physics is in how you determine *which* accelerated worldlines you're dealing with. As long as we agree on that, I'm good.

We haven't yet touched on the basis for the Born hypothesis. I hope you don;t get tired of this topic before we do as I am sure you are a lot more knowledgeable and I am very curious.

I'm certainly not tired of the topic. If you have a question about the Born hypothesis, please ask.

 

[Austin0]

So, in the first case, with a rocket engine at the rear and front end of the ship, the worldlines of the rear and front end were specified, but the worldlines of intermediate parts of the ship were not. The front and rear end worldlines being specified were sufficient to conclude that the ship would stretch, but the exact worldlines of intermediate points would depend on the exact details of the stretching--i.e., how each segment responded to the tensile stress caused by the rear and front ends getting further apart.

=PeterDonis;2367574]You're correct that, in specifying that the rocket engines at the front and rear end of the ship each imparted the same constant proper acceleration to their ends of the ship, I was assuming that the engines could do that independently of the forces exerted on the front and rear ends by other parts of the ship. In any real case, that would mean the engines would have to be controlled very precisely, to maintain the same constant proper acceleration; for example, as the ship started to stretch, the segment of the ship just to the rear of the front end would start pulling back on the front end, so the rocket engine at the front end would have to increase its rate of fuel burn *just* enough to compensate for this extra pull, in order to maintain the same constant proper acceleration. Since that's not physically impossible (however unlikely it might be in practice), it doesn't invalidate my specification of the scenario, since it's just a thought experiment.

Here you are basing conclusions of both physics and resultant worldlines on an already arrived at assumption of stretching. Well, we know the stretching occurs, just look at the worldlines. Well we know these would be the worldlines because we know that stretching must occur.


Originally Posted by Austin0
B) You can apply an interpretation of acceleration as applied force and disregard the interpretation of acceleration as a change in the motion of a system.

This isn't a matter of interpretation; the two have to be connected, because of conservation of momentum. The applied force is

F = \frac{d P}{d \tau}

where P is the energy-momentum 4-vector of the small segment of the ship that's being accelerated, and \tau is the proper time of that same small segment. But we have

P = m U

where m is the rest mass of the segment and U is its 4-velocity; so we have

F = m \frac{d U}{d \tau} = m A

where A is the proper acceleration (i.e., the rate of change of 4-velocity with respect to proper time). This is the relativistic version of Newton's second law, and if momentum is conserved, it *has* to hold.

I was not suggesting that conservation of momentum wouldn't hold. I would assume that the overall acceleration of the system would be the sum of force applied at both ends.
I was talking about the apparent implication of what you were saying : that the only points of equal proper acceleration were the points of actual application of the force. Disregarding the possibility that the overall system change in velocity could be equivalent throughout , which would mean equal proper acceleration if that were the case . Would you disagree? That if accelerometer readings were equivalent at all parts of the ship, this would mean equal proper acceleration.

That information can't travel faster than the speed of light, so, for example, if the ship is one light-second long, then the front end can't possibly begin accelerating sooner than 1 second after the rear end does

Mach 1 *is* the limit--the speed of sound in a given substance is the limit of how fast applied forces can be propagated in that substance. But the speed of sound varies widely according to the substance. The speed of sound in air, which is what people usually refer to as "Mach 1", is pretty slow as sound speeds go, about 340 m/s, or 10^-6 x the speed of light. In water the sound speed is about 5 times that, about 1500 m/s, and in steel it's about 20 times that in air, about 6,000 m/s.

Well I completely disagree with your statement that the speed of sound is the limit of propagation of momentum. Apparently you do too as you were the one who brought up the actual upper limit of c.
BTW Do you seriously think I might be unaware that the speed of sound is dependant on the properties of the medium? :-(

Original Austin0
I would say it is beyond the bounds of realism and billions of g's would imply inertial forces far beyond the ability of an addition of a few extra engines to counter.

The muons that were trapped and their lifetimes measured to check the clock postulate experienced accelerations up to 10^18 g. There's nothing physically impossible about such accelerations, and someday we may learn how to make macroscopic objects that are able to withstand them; the laws of physics don't prohibit that. Anyway:

I made no statement regarding the possibility of greater accelerations but simply commented that at those accelerations a Born rigid ship wouldn't be either rigid or survive. That it was unrealistic to think so.

Original AUstin0
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.

The speed of transmission of forces through an object (i.e., the sound speed) doesn't depend on the magnitude of the force. Small forces don't get transmitted any faster than large ones.

Do you think I was suggesting that loud noises move faster than soft ones??
As I understand it, the magnitude of force or momentum is a function of both acceleration and mass in one case, and velocity and mass in the other. So propagation is not just dependant on the speed of transmission which is of course constant, but also volume so to speak..
As an extreme case : a planetoid rear ends the ship at .3c. In this case wouldn't you agree that the applied momentum would be beyond the materials ability to transmit it fast enough and the resultant acceleration would reach the front of the ship as motion much faster than the speed of sound?
Or at the other end. If the momentum is being imparted by a number of very small particles
we would assume that with few particles, the momentum would simply be dissapated as incoherent heat. NO motion. Increasing the number would reach a point where the energy would reach the entire system as oscillation which would build up into infintesimal actual system motion. Like ideally slow adiabatic expansion. At some magnitude the energy input [mass*v] is greater than can be simply transmitted and actual motion starts at the point of introduction, resulting in stress and compression. Further increases go up the scale to torquing of the system or actual disruption

Unless the point of application of the force is physically restrained in some way so that it can't move (i.e., if there's no counterbalancing force to work against the applied force), it *will* move upon application of the force. That is, the individual atoms or molecules of the object will move; at the atomic/molecular level, of course, the applied force is transmitted by interactions between the atoms/molecules--more precisely, by the electrical repulsion between the electrons in the atoms/molecules.

Of course the atoms will move but not neccessarily net motion.
You may be right about the repulsion between electrons but my assumption would be that the internal electrostatic and nuclear forces between the nucleus and the electron shells, ionization due to nucleus displacement, repulsion between the nucleus and electrons, etc
would be equally relevant if not actually predominant, given that the majority of inertial masss resides in the nucleus

As the first "layer" of atoms/molecules in the object move in response to the applied force, they will, of course, move closer to the next "layer" in the object, and will thus exert a force on that next layer. In other words, the applied force starts a "wave" of force moving through the object--more precisely, a longitudinal wave (i.e., a sound wave) of alternating compression and expansion, as each layer pushes against the next, which then responds by moving away and pushing against the next in turn.

This is of course what I was referring to by reciprocal oscillation. But two points.
In the transmission of sound there is no net motion of the molecules . Is there real motion of the pressure wave? IMHO, No. It is a back and forth motion.
The internal atomic forces do work at the speed of light, as far as I know, so the reciprocal reactions of these forces would be happening almost instantly compared to the slow propagation rate.So right from the beginning there would be oscillation inherent in the propagation.
The only thing actually moving is a net translation of energy , momentum, until it is sufficient to actually move the system.

If we assume that the applied force at one end of the object continues to be applied, then once the initial wave has traveled the length of the object, it will start to be damped out. This is because as each layer pushes against the next, not all of the energy in the push gets converted into motion of the next layer; some of the energy goes into the internal degrees of freedom in the atoms/molecules--i.e., it gets converted into heat. If the material is very stiff (e.g., very high tensile steel--or even better, carbon nanotubes), the wave will be damped out quickly; if the material is soft, it will take longer for the wave to be damped out.

You may be right here also but my assumption would be that the softer, the more degrees of freedom that a material has, the faster the coherent vibrations would be damped.
I could easily be wrong :-)
BTW after I sent that post I realized i hadn't mentioned heat dissapation and almost went back to edit as I knew you were going to jump on it, but naaH

While the wave is damping, the object will be oscillating, expanding and compressing with gradually decreasing amplitude. Once the wave is damped out, the object will be in a state (assuming the force continues to be applied at one end) in which there is a compressive stress all along its length, and in which its material is somewhat hotter than it was before the force was applied (because some of the applied energy got converted into heat during the damping).

Of course the dissapation of energy would be taking place from the beginning throughout the system .
There is no disagreement that there would be a certain energy loss but would it be significant??
There is also no disagreement that there would be a certain compressive stress,
after all what is an accelerometer? In its most primitive form it is simply a spring scale
and as such is just measuring inertial force through compression. But this hypothesis says that over time, the compression would increase. That the stretching or compression would continue to incrementally increase and that this would be sustained after acceleration was terminated.That is what I am questioning.
If we use Einsteins elevator as an example how significant is the compression factor??
How significant is it here, when raisng a rigid rod from horizontal to vertical??
You never did answer my question about what you thought would happen on a Born ship raising a rod in this manner.

My understanding of the basis of Born acceleration is that it is founded on hyperbolic geometry and an interpretation of accelerated lines of simultaneity. Are there other important principles involved?
I don't remember any analysis along the lines of this thread , did I just miss that part??
Thanks

 

[PeterDonis]

Austin0:

Here you are basing conclusions of both physics and resultant worldlines on an already arrived at assumption of stretching. Well, we know the stretching occurs, just look at the worldlines. Well we know these would be the worldlines because we know that stretching must occur.

No--I'm basing the *conclusion* that there is stretching on the *stipulation* that both the front and the rear ends of the ship experience the same constant proper acceleration. That stipulation is physically possible (though not very practical, as I said), so I'm allowed to make it in a thought experiment.

I was not suggesting that conservation of momentum wouldn't hold. I would assume that the overall acceleration of the system would be the sum of force applied at both ends.

Remember how I specified the scenario: I specified that each of the two rocket engines, the rear one and the front one, imparted *equal, constant proper acceleration* to its segment of the ship. As I said in my last post, to do that in practice would require fantastically precise control of the rocket engines, in order to adjust their actual thrust to compensate for the forces exerted by other parts of the rocket. I only gave an example for the front end of the rocket last time, but the same would apply to the rear as well.

In other words, I was specifying the acceleration of each end of the ship, which in turn specifies the *net* force on each end of the ship--but I did *not* go into any details about how that net force results from the combination of the force exerted by the rocket engines and the internal forces exerted by parts of the ship on each other. I agree this is not a very practical scenario, but as I said, it's not physically impossible.

That if accelerometer readings were equivalent at all parts of the ship, this would mean equal proper acceleration.

Yes, that's true. In the scenario I specified, with rocket engines at the front and rear ends of the ship, accelerometers at the front and rear ends would each give the same readings. I didn't discuss what accelerometers at intermediate points on the ship would read, but if they all also gave the same readings, then proper accelerations would be equal at all those intermediate points as well as at the front and rear ends. And if that were the case, the ship would gradually stretch until it broke apart when the stretching exceeded the tensile strength of its hull.

Well I completely disagree with your statement that the speed of sound is the limit of propagation of momentum. Apparently you do too as you were the one who brought up the actual upper limit of c.

I brought up the upper limit of c as an upper limit to the *possible sound speed* in a material; in other words, it's not possible for the sound speed in a material to be faster than c. I never said that propagation of force in a particular material could be faster than the sound speed in that material. It can't.

As an extreme case : a planetoid rear ends the ship at .3c. In this case wouldn't you agree that the applied momentum would be beyond the materials ability to transmit it fast enough and the resultant acceleration would reach the front of the ship as motion much faster than the speed of sound?

If the sound speed in the material of the ship were slower than .3c (which of course it would be), then the ship would be crushed against the planetoid. If that counts as the acceleration reaching the front of the ship faster than the sound speed in the ship, then yes, it would. Most physicists would probably say that the concept of "speed of sound" wouldn't apply in a case like this, where the structure of the material was destroyed. (I'm assuming here that the planetoid is much larger than the ship.)

Or at the other end. If the momentum is being imparted by a number of very small particles we would assume that with few particles, the momentum would simply be dissapated as incoherent heat. NO motion.

This would violate conservation of momentum. The particles impacting the rear of the ship have *some* forward momentum, and that has to get translated into forward momentum of the ship itself, however small it is. I'm assuming that the ship is out in empty space, with no other forces acting.

You may be right about the repulsion between electrons but my assumption would be that the internal electrostatic and nuclear forces between the nucleus and the electron shells, ionization due to nucleus displacement, repulsion between the nucleus and electrons, etc would be equally relevant if not actually predominant, given that the majority of inertial mass resides in the nucleus

You're correct that I was ignoring the internal details of forces within each atom, and just treating the atoms as single objects. That's a pretty good approximation for the case we were considering, but of course you're right that the forces have to be transmitted internally between the electron shells and the nucleus or the atom as a whole won't move.

In the transmission of sound there is no net motion of the molecules . Is there real motion of the pressure wave? IMHO, No. It is a back and forth motion.

That depends on the driving force. For ordinary sound in, for example, air, the driving force is oscillatory--something, like a speaker membrane or a person's vocal cords, is vibrating back and forth, with no net motion, so there's no net motion in the sound wave either. In the case of the ship, the driving force is not oscillating; it's constant in a particular direction. In that case, the driving force does result in net motion immediately at the end of the ship where the force is applied, and the transmission of the sound wave through the ship transmits the net motion.

You may be right here also but my assumption would be that the softer, the more degrees of freedom that a material has, the faster the coherent vibrations would be damped.

I would agree that more degrees of freedom in the material ought to mean more places for the energy of vibration to go, hence faster damping, but I'm not sure softer materials have more degrees of freedom. I'd have to refresh my memory about the details of how bulk properties like the stiffness of the material are derived from microscopic properties of the atoms or molecules.

But this hypothesis says that over time, the compression would increase. That the stretching or compression would continue to incrementally increase and that this would be sustained after acceleration was terminated. That is what I am questioning.

In the case I was discussing here--a rocket engine only at the rear end of the ship--I didn't mean to imply that the compression would continue to increase. It wouldn't; it would reach a steady state and then not increase any further. In that steady state, there would be a compressive *stress* along the ship's length, but it would be opposed by the inter-atomic or inter-molecular forces, which would have increased because the atoms or molecules were now closer together. So there would be a new equilibrium and no further compression.

Also, I didn't mean to imply that the compression would remain if the acceleration was terminated; I didn't discuss that case (I was only discussing what would happen while the force continued to be applied). If the acceleration was terminated, of course the material would expand again, by the reverse process to that which compressed it: the information that the acceleration had ceased would take time to propagate from the rear to the front of the ship, so segments of the ship more towards the front would continue to accelerate longer, so a wave of initial expansion would travel throughout the ship. It would oscillate between expansion and compression again, but with no force being applied at the rear, the oscillations would eventually damp out and the ship would be at the same length (in its rest frame) as it was originally.

(By the way, all of the above assumes that the stresses induced in the ship don't exceed the elastic limit of its material. If they do, the ship's structure might be permanently deformed so it wouldn't return to its original length when the applied force was removed.)

In the case where there was a rocket engine at the front *and* rear of the ship, the stretching would continue to increase for as long as the rocket engines remained on--or until the ship broke apart. If the engines were turned off while the ship was still intact, the ship would still be stretched, but if the elastic limit of the material was not exceeded, it would shrink back to its original length (in its rest frame).

If we use Einsteins elevator as an example how significant is the compression factor??

I haven't calculated it specifically; in general it would be proportional to the acceleration and inversely proportional to the strength of the material. If you want a more specific formula I can try to dig one up.

How significant is it here, when raisng a rigid rod from horizontal to vertical??
You never did answer my question about what you thought would happen on a Born ship raising a rod in this manner.

I haven't considered this situation yet because adding the transverse dimension adds complications that make things more difficult to analyze, and I'm not sure it would add anything to the basic picture we already have. In general, I don't think the compression *ratio* (the ratio of original length to compressed length) would change, but of course once the rod is raised from horizontal to vertical the compression will be along the longer dimension of the rod. So in general, when the rod is horizontal it will be compressed along its shorter dimension and the longer dimension will be unchanged; once it is raised to vertical, and everything has time to equilibrate again, it will be compressed along its longer dimension and the shorter dimension will be what it originally was (i.e., before any force was applied at all).

Also, all of the above applies to the case where the force is applied at the *rear* only. This case is *not* the same as Born rigid acceleration. Born rigid acceleration would make the case of the rod rotating from horizontal to vertical *much* more complex to analyze exactly, because every atom in the rod would need to travel along a very precise path. I *think* the end result would be that, both horizontally and vertically, the rod's dimensions would remain the same as its unaccelerated dimensions; but that's only based on the general property of Born rigid acceleration that it's supposed to do that. I haven't actually calculated the precise accelerations that would have to be imparted to each little piece of the rod to make that come about.

My understanding of the basis of Born acceleration is that it is founded on hyperbolic geometry and an interpretation of accelerated lines of simultaneity. Are there other important principles involved?
I don't remember any analysis along the lines of this thread , did I just miss that part??

I don't think there's been a detailed discussion of Born acceleration or Born rigidity yet in this thread. There's a discussion of it http://www.mathpages.com/home/kmath422/kmath422.htm" that covers the main points. The original reason why Born proposed the idea was to find some kind of relativistic version of the concept of a "rigid body". The property he wanted to try and preserve was that each atom in the body would maintain the same spatial distance from its neighboring atoms (so that the acceleration would be "stress-free"); given that requirement, relativistic kinematics was enough to determine the worldlines of each part of the body, as shown in the page I linked to above. As I've said, it would be practically impossible to realize Born rigid acceleration for a real object, since it would require very precise application of force to each little piece of the object.

 

[Austin0]

Austin0:

No--I'm basing the *conclusion* that there is stretching on the *stipulation* that both the front and the rear ends of the ship experience the same constant proper acceleration. That stipulation is physically possible (though not very practical, as I said), so I'm allowed to make it in a thought experiment.

I have no quibbles about how you want set up your parameters or small details of practicallity. But given your condition of same constant proper acceleration at each end, I missed how this leads to inevitably increasing stretching. :-)




Originally Posted by Austin0
Or at the other end. If the momentum is being imparted by a number of very small particles we would assume that with few particles, the momentum would simply be dissapated as incoherent heat. NO motion.

This would violate conservation of momentum. The particles impacting the rear of the ship have *some* forward momentum, and that has to get translated into forward momentum of the ship itself, however small it is. I'm assuming that the ship is out in empty space, with no other forces acting.

I also am assuming in empty space. But I do not see how there would be any violation of conservation of momentum.
My assumption is that the momentum from a few particles would be converted from linear motion into molecular vibration [heat] and ultimately radiated into space as infrared photons. That in losing coherence it would also lose any unified momentum vector and would be nullified in that regard. Individual motions pointing in all directions.
ANything wrong with this picture?
____________________________________________________________________________-
Originally Posted by Austin0
In the transmission of sound there is no net motion of the molecules . Is there real motion of the pressure wave? IMHO, No. It is a back and forth motion.

That depends on the driving force. For ordinary sound in, for example, air, the driving force is oscillatory--something, like a speaker membrane or a person's vocal cords, is vibrating back and forth, with no net motion, so there's no net motion in the sound wave either. In the case of the ship, the driving force is not oscillating; it's constant in a particular direction. In that case, the driving force does result in net motion immediately at the end of the ship where the force is applied, and the transmission of the sound wave through the ship transmits the net motion.

What you are saying about coherent sound production is obviously relevant to the pattern of transmitted sound [momentum] in air.
But it seems to me that the reciprocal oscillatory nature of momentum propogation is independant of the driving force and stems directly from the 3rd law of motion.
That this does not just apply to the propulsion itself, but at every intermediate transference of momentum from particle to partcle. Obviously in a molecular lattice structure the structure itself will impose certain periodicities and larger coordinations ,harmonics etc.,
but will not detract from this fundamental action-reaction inherent reciprocity.
A thought picture:
Take the classic series of suspended identical metal balls. One end ball is elevated and released gaining momentum until impact with the first ball in the line. The momentum passes from the moving ball into the next ball in line and then through the series until encountering the last ball. It enters that ball and the momentum is then translated into motion.
It passed through the intermediate balls without resultant motion. It propagated throughout the last ball without coordinate motion until the whole amount of energy was absorbed and then the whole ball moved as a whole system. It seems like we can logically infer that there was no motion at the point of introduction because as soon as that point of contact actually moved there could be no further transfer of momentum. Yet we know that the total amount of momentum in the last ball is equal to the amount in the first ball minus the small amount of loss through dissapation.
SO it would seem there could be no net motion until the transfer was complete.
Is there something here I am not seeing?

A space launch: the engines introduce thrust for a considerable period before actual motion occurs. My assumption is that a lot of momentum has to propagate through the system to overcome the opposite acceleration of gravity and then additional momentum has to go into and throughout the system to actually overcome inertia and result in system motion.
Am I seeing this wrong??.

I would agree that more degrees of freedom in the material ought to mean more places for the energy of vibration to go, hence faster damping, but I'm not sure softer materials have more degrees of freedom. I'd have to refresh my memory about the details of how bulk properties like the stiffness of the material are derived from microscopic properties of the atoms or molecules.

You're probably right about there being some weird polymers that are both soft and resonant but i think in general bronze bells ring longer than paper ones ?

In the case I was discussing here--a rocket engine only at the rear end of the ship--I didn't mean to imply that the compression would continue to increase It wouldn't; it would reach a steady state and then not increase any further..

Can we consider this initial compression/contraction and consider it is not really what we are talking about with this question and so disregard it from this point?

Also, I didn't mean to imply that the compression would remain if the acceleration was terminated; I didn't discuss that case (I was only discussing what would happen while the force continued to be applied). If the acceleration was terminated, of course the material would expand again, by the reverse process to that which compressed it:

It seems that this question involves the concept of permanent change.
Fundamental to SR is the concept that contraction does not occur within a system in relative motion but is a condition observed from other frames.
Practically speaking that means no perceptible change in internal geometry or metric.
It seems to me that the Born hypothesis is based on the idea that to preserve this condition through the transition between relative velocities [ acceleration], that specific structured force must be applied.
From what I gotten from you, it seems like the difference between Born acceleration and simple acceleration from the rear, is that the rear propulsion would produce some slight initial
compression that would disappear after reaching a new velocity. Wouldn't this also conform to the expectations of SR , with the small difference that, during the transition there would be a small temporary difference in internal geometry. SR already considers accelerated systems as being radically different from inertial ones and Born acceleration in any case is not realistically achievable , so the question would seem to be what is the internal basis of the hypothesis that would make it inevitable as a reality?

I don't think there's been a detailed discussion of Born acceleration or Born rigidity yet in this thread. There's a discussion of it http://www.mathpages.com/home/kmath422/kmath422.htm" that covers the main points. The original reason why Born proposed the idea was to find some kind of relativistic version of the concept of a "rigid body". The property he wanted to try and preserve was that each atom in the body would maintain the same spatial distance from its neighboring atoms (so that the acceleration would be "stress-free"); given that requirement, relativistic kinematics was enough to determine the worldlines of each part of the body, as shown in the page I linked to above. As I've said, it would be practically impossible to realize Born rigid acceleration for a real object, since it would require very precise application of force to each little piece of the object

No there has not been a discussion of the hypothesis. I meant did any of the reasoning regarding worldlines etc that have come up in this thread , enter into the original formulation of the hypothesis. I have seen the link before and have questions regarding the interpretation of accelerated lines of simultaneity that is portrayed there.
More later Thanks

 

[PeterDonis]

Austin0:

But given your condition of same constant proper acceleration at each end, I missed how this leads to inevitably increasing stretching.

Because that condition means that the front and rear ends of the ship, R and F, follow exactly the same worldlines as the separate ships, R and F, in our previous scenario, where we agreed that the proper distance between the ships, in either ship's MCIF, would increase--ship F would "pull away" from ship R, and ship R would "fall behind" ship F (these are just two different ways of saying the same thing). If R and F are the front and rear ends of a single ship, but follow the same worldlines as before, then the proper distance between them will still increase--which means that the ship as a whole must stretch.

The only difference when R and F are two ends of a single ship is that the intermediate parts of the ship exert a force on the front and rear ends, so the condition of "same constant proper acceleration" does *not* imply "same constant rocket thrust", as it did in the case of the separate ships. With a single ship, as I've said before, the rocket thrusts at the ends will need to vary to compensate for the internal forces exerted by the parts of the ship on each other, so that the *net* force at the front and rear ends remains equal and constant, since it's the *net* force that determines the proper acceleration--i.e., what would be measured by an accelerometer.

My assumption is that the momentum from a few particles would be converted from linear motion into molecular vibration [heat] and ultimately radiated into space as infrared photons. That in losing coherence it would also lose any unified momentum vector and would be nullified in that regard. Individual motions pointing in all directions.
ANything wrong with this picture?

Yes. A "unified momentum vector", by which I assume you mean a net momentum vector pointing in a particular direction, *cannot* be "nullified"--it *cannot* be converted into individual motions pointing in all directions, with no net bias in any particular direction. That violates conservation of momentum.

Take the classic series of suspended identical metal balls. One end ball is elevated and released gaining momentum until impact with the first ball in the line. The momentum passes from the moving ball into the next ball in line and then through the series until encountering the last ball. It enters that ball and the momentum is then translated into motion.

Here the driving force is a single initial impulse, not a continuous applied force. That's why the behavior is different. For your thought picture to match the situation of the continuously accelerating rocket, you would have to have a continuous push on the ball at one end of the series. That would impart a net motion to the whole set of balls--at least until you've pushed them so far that they break away from their suspension points.

A space launch: the engines introduce thrust for a considerable period before actual motion occurs.

That's because the rocket is held down by clamps to the launch pad. As soon as the clamps are released, the rocket starts moving. The clamps are there because it's not safe to let the rocket move when the thrust to weight ratio is too low; there's too much chance that it would topple. So the rocket is held down until the main engines are all at full thrust. (On the space shuttle, the SRBs ramp up to full thrust fast enough that this delay isn't necessary.)

Check out http://science.ksc.nasa.gov/shuttle/countdown/count.html" . Note the "SRB HOLDDOWN RELEASE COMMAND" at exactly T minus 00 minutes 00 seconds, and liftoff at that same moment.

Can we consider this initial compression/contraction and consider it is not really what we are talking about with this question and so disregard it from this point?

Yes, that's fine; the steady-state condition is really what we're interested in.

Fundamental to SR is the concept that contraction does not occur within a system in relative motion but is a condition observed from other frames.

Only if you're considering "contraction" to be the *kinematic* effect--i.e., you already know all the worldlines and events, and you're just figuring out how they would look from different frames. But if you start mixing in the dynamic effects--the internal forces that parts of an object exert on other parts, which can *change* the worldlines that the various parts travel on--then you're mixing together two different effects, and you have to be very precise in distinguishing which observations you're making are due to the fact that the parts of an object are traveling along different worldlines, and which are due to the way those worldlines look in different frames.

From what I gotten from you, it seems like the difference between Born acceleration and simple acceleration from the rear, is that the rear propulsion would produce some slight initial
compression that would disappear after reaching a new velocity. Wouldn't this also conform to the expectations of SR , with the small difference that, during the transition there would be a small temporary difference in internal geometry.

Everything I've said conforms to the expectations of SR; indeed, SR is the theory I'm using to determine all the effects I'm talking about.

With regard to the quote above, are you now talking about scenarios where objects are accelerated only for a limited period, and the acceleration is then removed? In any case like that, generally speaking, assuming that the elastic limit of the material is not exceeded (which can be a big "if"--see below), an object that is accelerated to a given velocity (as seen from its original rest frame) and then left to move inertially thereafter, will have the same dimensions in its new inertial rest frame as it had in its original rest frame, regardless of how it was accelerated--provided only that we allow enough time for dynamic effects such as waves of compression and expansion in the material to propagate and damp themselves out. The details of the acceleration will be important only if you want to know the *precise* worldlines that each part of the object will follow in getting from its original state of inertial motion to its final one.

Note that qualification I gave, that the elastic limit of the material is not exceeded. That means that the general prediction I just made--that objects which are accelerated will return to their original dimensions (in their new rest frames) when the acceleration is removed--is *not* just a prediction of SR alone. It also requires knowledge of the physics of elastic materials. Non-elastic materials will have different behavior, even though the same laws of SR can be used to help predict the behavior in both cases. That's why I've been so insistent on maintaining the distinction between "kinematic" effects (which are *purely* due to the laws of SR) and "dynamic effects" (which depend on other physical theories in addition to SR).

 

[Austin0]

Originally Posted by Austin0
. If the momentum is being imparted by a number of very small particles we would assume that with few particles, the momentum would simply be dissapated as incoherent heat. NO motion.

This would violate conservation of momentum. The particles impacting the rear of the ship have *some* forward momentum, and that has to get translated into forward momentum of the ship itself, however small it is. I'm assuming that the ship is out in empty space, with no other forces acting.

I also am assuming in empty space. But I do not see how there would be any violation of conservation of momentum.
My assumption is that the momentum from a few particles would be converted from linear motion into molecular vibration [heat] and ultimately radiated into space as infrared photons. That in losing coherence it would also lose any unified momentum vector and would be nullified in that regard. Individual motions pointing in all directions.
ANything wrong with this picture?


Y

es. A "unified momentum vector", by which I assume you mean a net momentum vector pointing in a particular direction, *cannot* be "nullified"--it *cannot* be converted into individual motions pointing in all directions, with no net bias in any particular direction. That violates conservation of momentum.

Hi PeterDonis It would appear that either my understanding of conservation of momentum is incomplete or there is a miscommunication here and we are talking about different things.
Would you agree that the statements I made above are a description of certain aspects of both entropy and damping , which in itself is a specialized case of entropy??

That when we were talking about degrees of freedom we were talking about entropic dispersal possibilities, microstates?

Considering a single particle:
It transfer its momentum to the system as a discrete definable vector. But inside the lattice, that definite locality and direction is immediately distributed. Initially having a generalized forward direction, as it propagates it would spread over a wider front and be redirected through degrees of freedon in other directions. Besides this, parts of the energy would be continually being localized as molecular vibration, heat. Translational motion halted.
Entropy.

Would you disagree?

a net momentum vector pointing in a particular direction, *cannot* be "nullified"--it *cannot* be converted into individual motions pointing in all directions, with no net bias in any particular direction.

Isn't this simply a question of boundary conditions?
Using your extreme example of a rod light seconds long. If one end is subjected to an impact [momentum] that is small relative to the overall mass, would you expect that momentum to ever reach the other end at all??
Or would it be expected that it would be dissipated as internal vibration and heat long before reaching the other end??
If this magnitude of momentum was sustained would it be expected to change this basic entropic resistance?
Would you disagree that it seems likely that there would be a threshold of minimum input of momentum /time required to overcome this inherent entropic resistance??
WOuldn't this apply to any system it just being a quantitative question??

____________________________________________________________________________



Take the classic series of suspended identical metal balls. One end ball is elevated and released gaining momentum until impact with the first ball in the line. The momentum passes from the moving ball into the next ball in line and then through the series until encountering the last ball. It enters that ball and the momentum is then translated into motion.
It passed through the intermediate balls without resultant motion. It propagated throughout the last ball without coordinate motion until the whole amount of energy was absorbed and then the whole ball moved as a whole system. It seems like we can logically infer that there was no motion at the point of introduction because as soon as that point of contact actually moved there could be no further transfer of momentum. Yet we know that the total amount of momentum in the last ball is equal to the amount in the first ball minus the small amount of loss through dissapation.

Here the driving force is a single initial impulse, not a continuous applied force. That's why the behavior is different. For your thought picture to match the situation of the continuously accelerating rocket, you would have to have a continuous push on the ball at one end of the series. That would impart a net motion to the whole set of balls--at least until you've pushed them so far that they break away from their suspension points.

Would you disagree with the following view?
The driving force in this case is a continuous applied force , albeit of very short duration. The momentum passes from the first ball to the second not as a single instantaneous pulse but as a sustained transference through the point of contact.

Couldn't we in principle extend this time interval greatly up to some limit, simply by increasing the size of the spheres??

AS far as the principles involved are concerned the duration is not really important is it?

Couldn't we, from a frame at rest wrt the intermediate spheres, plot a clear picture of the complete sequence of events??
Couldn't we also graph the momentum itself, independant from the masses involved?
Give it both a spatial location as well as a defined velocity for any given instant? [with a certain statistical uncertainty regarding location].
So at T= (diameter/speed of sound) the momentum would be located, essentially , within sphere #2 etc, etc.
Culminating in its location within the final sphere before any coordinate motion of that sphere??

Now I may be seriously mistaken but I see several apparently fundamental priciples of momentum and inertia demonstrated here.
That, disregarding entropy, momentum simply passes through mass as long as there is an open pathway.
That inertia resists actual system motion until the momentum has propagated completely through that system.
That if the momentum of the final sphere is equal to the momentum of the initial sphere then the conservation of momentum would seem to dictate that none of it remains in the intermediate spheres as motion other than residual vibration.

It is understood that these are special conditions : identical mass and composition, perfectly symmetric transfer paths etc. That changing these parameters , for instance increasing the initial momentum significantly etc. would result in interactions that would not be so neat , but do you see any reason why these basic priciples wouldn't fundamentally apply as generalizations?

Or some basic understanding that I am missing?

If I have errors of understanding here I definitely want to get them right.

That's because the rocket is held down by clamps to the launch pad. As soon as the clamps are released, the rocket starts moving. The clamps are there because it's not safe to let the rocket move when the thrust to weight ratio is too low; there's too much chance that it would topple. So the rocket is held down until the main engines are all at full thrust. (On the space shuttle, the SRBs ramp up to full thrust fast enough that this delay isn't necessary.)

Check out http://science.ksc.nasa.gov/shuttle/countdown/count.html" . Note the "SRB HOLDDOWN RELEASE COMMAND" at exactly T minus 00 minutes 00 seconds, and liftoff at that same moment.

Yes I was lacking in knowledge of the mechanics and thought that the gantry supports were basically passive restraints against torquing SO BAD analogy.

Thanks for your patience

 

[PeterDonis]

Considering a single particle:
It transfer its momentum to the system as a discrete definable vector. But inside the lattice, that definite locality and direction is immediately distributed. Initially having a generalized forward direction, as it propagates it would spread over a wider front and be redirected through degrees of freedon in other directions. Besides this, parts of the energy would be continually being localized as molecular vibration, heat. Translational motion halted.
Entropy.

Would you disagree?

Yes. *That would violate conservation of momentum.* You *cannot* take a "discrete definable vector" of momentum and make it disappear. You can't dissipate it as heat. You can't redistribute it into other internal degrees of freedom. The total momentum of the system--"system" meaning the single particle and the larger object that it hits--*must* be the same after the collision as before. That's what conservation of momentum means. If the total momentum was a "discrete definable vector" before the collision, it must still be a "discrete definable vector" after the collision.

This is such a basic point that I want to make sure it's clear in this simple example before discussing any of the other examples in your post. Let me write down explicitly what the momentum looks like before and after the collision, in the frame in which the large object is initially at rest:

Before collision: Single particle with mass m and moving to the right (positive x-direction) with velocity v. Large object with mass M at rest.

Total momentum before collision: m v. Note that this is a vector in the positive x-direction (the same direction as v). Also note that this is a non-relativistic expression; if you want to look at the relativistic case, we can, but it doesn't change anything essential to what we're discussing right now.

After collision: Object with mass m + M moving to the right (positive x-direction) with velocity w.

Total momentum after collision: \left( m + M \right) w. Again, this is a vector in the positive x-direction (the same direction as w, which is the same direction as v).

Equating the momentum before and after the collision, as required by conservation of momentum, we have:

m v = \left( m + M \right) w

or, rearranging terms,

w = \frac{m}{m + M} v

So w will be much smaller than v, but it will *not* be zero, and since the object is now alone in empty space with no other objects to collide with, its velocity will remain w; there's nowhere for it to "dissipate" to, because interactions internal to an object can't change its total momentum.

Before going any further, please let me know your thoughts on the above.

 

[Austin0]

Yes. *That would violate conservation of momentum.* You *cannot* take a "discrete definable vector" of momentum and make it disappear. You can't dissipate it as heat. You can't redistribute it into other internal degrees of freedom. The total momentum of the system--"system" meaning the single particle and the larger object that it hits--*must* be the same after the collision as before. That's what conservation of momentum means. If the total momentum was a "discrete definable vector" before the collision, it must still be a "discrete definable vector" after the collision.

This is such a basic point that I want to make sure it's clear in this simple example before discussing any of the other examples in your post. Let me write down explicitly what the momentum looks like before and after the collision, in the frame in which the large object is initially at rest:

Before collision: Single particle with mass m and moving to the right (positive x-direction) with velocity v. Large object with mass M at rest.

Total momentum before collision: m v. Note that this is a vector in the positive x-direction (the same direction as v). Also note that this is a non-relativistic expression; if you want to look at the relativistic case, we can, but it doesn't change anything essential to what we're discussing right now.

After collision: Object with mass m + M moving to the right (positive x-direction) with velocity w.

Total momentum after collision: \left( m + M \right) w. Again, this is a vector in the positive x-direction (the same direction as w, which is the same direction as v).

Equating the momentum before and after the collision, as required by conservation of momentum, we have:

m v = \left( m + M \right) w

or, rearranging terms,

w = \frac{m}{m + M} v

So w will be much smaller than v, but it will *not* be zero, and since the object is now alone in empty space with no other objects to collide with, its velocity will remain w; there's nowhere for it to "dissipate" to, because interactions internal to an object can't change its total momentum.

Before going any further, please let me know your thoughts on the above.

There is no disagreement on the abslolute application of the fundamental conservation of mass and energy.
There is no disagreement on the truth of inertial motion or the conservation of momentum as applied to inertial motion.
there is no disagreement on the validity of the formula for force and momentum as applied to macrosystems in general.
There is no disagreement that principles and mathematical descriptions of physics do not exist if they are not validated to a high dgree of accuracy.

But you have presented this question in the context of absolutes.
Of abstract principles divorced from the real world.
So given a complex system regarded as an abstract point particle. GIven a system ideally isolated from the world with no photon transmission possible.
Then you are abslutely right.
Whether any principle, no matter how valid, can be quantitatively applied to real world systems in this absolute way is another question.

But this is not the real point of disagreement .
Viewed in this absolute way , the mathematical expression, having been derived from consideration of interactions between point particles or ideal elastic bodies defines an instantaneous change in momentum in reaction to force.
You seem to apply this to a complex system as a local reality. I.e. a ship with multiple points of applied momentum can be viewed as a set of separate systems. The momentum applied locally results in different velocities throughout the overall system because motion starts immediately at the source . Or at least this is what it seems to mean to me you have been saying .
My view is that the global condition is predominant and energy, momentum, enters the system and is propagated throughout the system which accelerates as a whole.
Do you have any other information that would help clarify this question?

 

[PeterDonis]

You seem to apply this to a complex system as a local reality. I.e. a ship with multiple points of applied momentum can be viewed as a set of separate systems. The momentum applied locally results in different velocities throughout the overall system because motion starts immediately at the source . Or at least this is what it seems to mean to me you have been saying .
My view is that the global condition is predominant and energy, momentum, enters the system and is propagated throughout the system which accelerates as a whole.
Do you have any other information that would help clarify this question?

Both of the views you have stated above are true. If you view the system as a single system, sitting in empty space, with no interactions with other systems other than the specific ones you've defined in the statement of the problem (for example, 1 rocket engine pushing on the rear of a spaceship, or a small particle hitting a larger object), then as a single system (e.g., the spaceship or the larger object), conservation of momentum applies just as I have stated it.

If, on the other hand, you view the system as a composite of subsystems (e.g., the spaceship as a composite of little segments, each with its own motion), then conservation of momentum applies to each individual subsystem, *and* to the system that's the composite of all the subsystems. The law of conservation of momentum for each individual system applies as I have stated it, but you have to obtain the *net* force on each individual system (e.g., each individual piece of the spaceship) to apply the law of conservation of momentum to it.

The law of conservation of momentum for the composite system (e.g., the whole spaceship) is then the sum of all the laws of conservation of momentum for the individual systems. What you will find when you do the sum is that all of the internal forces between parts of the composite system cancel out, and you're left with just the net external force on the whole system (e.g., the rocket engine pushing on the spaceship, or the small particle hitting the large object), which must balance with the net change in its total momentum, according to the law as I have stated it.

So *both* versions of conservation of momentum--the version that applies to the whole system, and the version that applies to each individual subsystem--are valid, and consistent with each other.

 

[Austin0]

=PeterDonis;2380008] Both of the views you have stated above are true.
So *both* versions of conservation of momentum--the version that applies to the whole system, and the version that applies to each individual subsystem--are valid, and consistent with each other

As far as I can see they are somewhat mutually exclusive. Certainly the expected end results appear to be.

Given: A ship with thrust applied to the front and back, with greater thrust at one end.

#1 The momentum from the front propagates through the ship to the back, while the momentum from the back propagates through to the front and then results in system motion . So the differential is equalized and the net acceleration is also equal at the front and the back.

#2 The momentum propagates locally from the source as motion and results in greater velocity at the end with greater thrust.

__________________________________________________________________________

If we assume that the applied force at one end of the object continues to be applied, then once the initial wave has traveled the length of the object, it will start to be damped out. This is because as each layer pushes against the next, not all of the energy in the push gets converted into motion of the next layer; some of the energy goes into the internal degrees of freedom in the atoms/molecules--i.e., it gets converted into heat. . While the wave is damping, the object will be oscillating, expanding and compressing with gradually decreasing amplitude. Once the wave is damped out, the object will be in a state (assuming the force continues to be applied at one end) in which there is a compressive stress all along its length, and in which its material is somewhat hotter than it was before the force was applied (because some of the applied energy got converted into heat during the damping).

Where did this energy come from?

Would it completely stop after initial adjustment to acceleration??

Would the system be getting continually hotter??

__________________________________________________________________________--
THE SPHERES

Here the driving force is a single initial impulse, not a continuous applied force. That's why the behavior is different. For your thought picture to match the situation of the continuously accelerating rocket, you would have to have a continuous push on the ball at one end of the series. That would impart a net motion to the whole set of balls--at least until you've pushed them so far that they break away from their suspension points..

Would you disagree with the following view?
The driving force in this case is a continuous applied force , albeit of very short duration. The momentum passes from the first ball to the second not as a single instantaneous pulse but as a sustained transference through the point of contactCouldn't we in principle extend this time interval greatly up to some limit, simply by increasing the size of the spheres??

AS far as the principles involved are concerned the duration is not really important, is it?

Couldn't we, from a frame at rest wrt the intermediate spheres, plot a clear picture of the complete sequence of events??
Couldn't we also graph the momentum itself, independant from the masses involved?
Give it both a spatial location as well as a defined velocity for any given instant? [with a certain statistical uncertainty regarding location].

So at T= (diameter/speed of sound) the momentum would be located, essentially , within sphere #2 etc, etc.
Culminating in its location within the final sphere before any coordinate motion of that sphere??

Now I may be seriously mistaken but I see several apparently fundamental priciples of momentum and inertia demonstrated here:

That, disregarding entropy, momentum simply passes through mass as long as there is an open pathway.

That inertia resists actual system motion until the momentum has propagated completely through that system.

That if the momentum of the final sphere is equal to the momentum of the initial sphere then the conservation of momentum would seem to dictate that none of it remains in the intermediate spheres as motion other than residual vibration.

It is understood that these are special conditions : identical mass and composition, perfectly symmetric transfer paths etc. That changing these parameters , for instance increasing the initial momentum significantly etc. would result in interactions that would not be so neat , but do you see any reason why these basic priciples wouldn't fundamentally apply as generalizations?

Or some basic understanding that I am missing?

 

[Austin0]

Yes. *That would violate conservation of momentum.* You *cannot* take a "discrete definable vector" of momentum and make it disappear. You can't dissipate it as heat. You can't redistribute it into other internal degrees of freedom. The total momentum of the system--"system" meaning the single particle and the larger object that it hits--*must* be the same after the collision as before. That's what conservation of momentum means. If the total momentum was a "discrete definable vector" before the collision, it must still be a "discrete definable vector" after the collision.

This is such a basic point that I want to make sure it's clear in this simple example before discussing any of the other examples in your post.

So w will be much smaller than v, but it will *not* be zero, and since the object is now alone in empty space with no other objects to collide with, its velocity will remain w; there's nowhere for it to "dissipate" to, because interactions internal to an object can't change its total momentum.

Before going any further, please let me know your thoughts on the above.

AM I incorrect in my understanding, that kinetic enrgy in its many forms including heat,
results in an increase in inertial mass??
If this is the case, how does this fit into the momentum conservation equation?
It would appear that any part of the directed applied momentum that was internally transformed into kinetic energy, would then contribute to the overall system conservation of momentum as an increase in inertial mass rather than, neccessarily, an increase in system velocity.
Is there some principle I am unaware of that would negate this concept??
Are we going to discuss any of the other examples in my post??

 

[PeterDonis]

Austin0: I'm going to respond to parts of both your posts here.

As far as I can see they are somewhat mutually exclusive. Certainly the expected end results appear to be.

The expected end results should be the same for both versions of conservation of momentum. If you get different results by applying the two different versions then there's a mistake somewhere.

Given: A ship with thrust applied to the front and back, with greater thrust at one end.

Which end? It makes a difference. Also, does "thrust" mean the actual force applied (presumably by a rocket engine), or the resultant acceleration? They're not necessarily the same, as I've noted before. See next comment.

#1 The momentum from the front propagates through the ship to the back, while the momentum from the back propagates through to the front and then results in system motion . So the differential is equalized and the net acceleration is also equal at the front and the back.

#2 The momentum propagates locally from the source as motion and results in greater velocity at the end with greater thrust.

I'm not exactly sure what you mean by the above. Rather than try to parse it, let me instead describe how a situation like this one would be analyzed with the tools I'm used to using.

First of all, as I noted above, it makes a difference whether "thrust" means the applied force (which might be combined with other forces, internal to the ship, to obtain the resultant net force--and acceleration--of a given small piece of the ship) or the resultant acceleration (which is a result of the net force, not the rocket thrust alone). I'm going to assume below that "thrust" means the applied force, since that's the usual meaning of that term.

Let's first write down in general terms what both versions of conservation of momentum will look like (applied to the ship as a whole, and applied to each piece):

(A) Treat the ship as a single object. The total momentum transferred to the ship per unit time will be equal to the sum of the thrusts on the two ends. Since we're treating the ship as a single object, we don't care how that force is propagated through the ship, or what it does to the internal stresses or distances within the ship. All we care about is that the momentum transferred to the ship per unit time must be equal to the total applied thrust (sum of both ends). In other words, we have

F = T_R + T_F = \frac{d P}{d \tau}

for the ship as a whole, where T_R and T_F are the externally applied thrusts at the front and rear ends. That's all there is to it.

(B) Treat the ship as an extended object. For simplicity, we'll treat it here as composed of three segments: R, the rear segment, where the rear thrust is applied; M, the middle segment, where no thrust is applied (but which can exchange internal forces with the other segments), and F, the front segment, where the front thrust is applied.

Then conservation of momentum for the segments individually looks like this:

F_R = T_R + S_{MR} = \frac{d P_R}{d \tau_R}

F_M = S_{RM} + S_{FM} = \frac{d P_M}{d \tau_M}

F_F = T_F + S_{MF} = \frac{d P_F}{d \tau_F}

where we've used T for the thrusts (rear and front) and S for the internal forces between each segment, with the order of subscripts indicating the direction of the internal force (for example, S_{MR} is the internal force exerted by the middle segment on the rear segment). In order to check for consistency with (A) above, we calculate the total momentum gained by the ship per unit time by adding the three forces above:

F = F_R + F_M + F_F

and check to see that the sum equals the sum given in (A) above, i.e., the sum of the externally applied thrusts.

Let's now suppose that the ship starts out at rest, with all forces zero. Then, at time t = 0 in the initial rest frame, the front and rear thrusts are applied. The internal forces at that moment will be zero, because there has been no time for the ship's internal structure to respond to the applied forces. So we will have

F_R = T_R

F_M = 0

F_F = T_F

when the thrusts are initially applied. This means that, initially, the rear and front segments will gain momentum, but the middle one will not. And you can see from the above that, intially, the sum of the momentum gains by the rear and front segments will be equal to the total momentum gained by the ship, which is equal to the sum of the applied thrusts at front and rear. So momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

In order to predict in detail how the internal forces will develop, we would need a detailed model of the material properties of the ship's structure, which I don't want to discuss here. But in general terms, you can see from the above that, assuming the front and rear thrusts are in the same direction, the initial effect will be that the rear segment will move closer to the middle segment, while the front segment will move further from it. So, after some small time has elapsed, we will have something like this:

F_R = T_R - C

F_M = C + S

F_F = T_F - S

where C is an internal pressure between the rear and middle segments, which arises because the segments are moving closer together and the material is assumed to resist compression (note the signs of the C terms in each force), and S is an internal tension between the middle and front segments, which arises because the segments are moving further apart, and the material is assumed to resist stretching (again, note the signs of the S terms in each force).

Once again, the above equations automatically conserve momentum for each individual segment. But if we add up the total momentum gained by all the segments, we find that, once again, it's just the sum of the external thrusts applied to front and rear; all the internal forces cancel out (because momentum gained by one segment due to each internal force is exactly balanced by momentum lost by the other segment due to the same force). So again, momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

All this will continue to be true as the internal forces vary with time; for example, the above suggests that after some more time has elapsed, the middle segment will move away from the rear segment and towards the front segment, as part of the oscillation we were discussing in earlier posts. Then we will have:

F_R = T_R + S

F_M = - S - C

F_F = T_F + C

where now S is a tension between the rear and middle segments and C is a pressure between the middle and front segments. The same analysis as above still applies: the internal forces are different than before, but they still cancel out when the total momentum is added up. So at each individual instant of time, momentum is conserved individually for each ship segment, and total momentum is conserved for the ship as a whole. Everything is consistent.

I can discuss more specific cases if you want (meaning specific relative values of the front and rear thrusts), but you can already see from the above that, regardless of whether the front or the rear thrust is larger, or whether they're both the same, conservation of momentum will hold as I've stated it. The only difference the specific relative values of the thrusts makes is to determine what kind of final state will be reached--i.e., will the ship end up stretched or compressed, and so forth--but that question is independent of the question of conservation of momentum.

I should note, since I'm going to address your questions about damping of the oscillations below, that the above model doesn't really include damping, except implicitly in whatever model you come up with for determining the internal forces between the segments. However, damping does not affect the correctness of what I've said above; all it does is add extra equations (dealing with the internal energies of each segment) that still will be consistent with the equations I've written above. If you want me to go into more detail about how that could work, let me know and I'll do so in a future post.

Where did this energy come from?

Would it completely stop after initial adjustment to acceleration??

Would the system be getting continually hotter??

The energy that heats the ship's structure as it oscillates in response to an externally applied thrust comes from the energy that fuels the source of the thrust--the rocket engine or whatever. The specific mechanism of how that would occur depends on how you model the internal forces between segments of the ship, as I noted above; but in general, the energy is converted into heat as a result of the relative motion between the internal parts of the ship.

If we assume that the ship reaches some steady-state equilibrium while under acceleration (i.e., a state where the internal parts of the ship are no longer moving relative to one another), then the heating of the ship's structure should stop once that steady-state equilibrium is reached, since at that point there is no longer any relative motion between the ship's parts.

AM I incorrect in my understanding, that kinetic enrgy in its many forms including heat,
results in an increase in inertial mass??
If this is the case, how does this fit into the momentum conservation equation?
It would appear that any part of the directed applied momentum that was internally transformed into kinetic energy, would then contribute to the overall system conservation of momentum as an increase in inertial mass rather than, neccessarily, an increase in system velocity.
Is there some principle I am unaware of that would negate this concept??

It seems like you're trying to separate out momentum from energy, and kinetic energy from other kinds of energy. That's probably not a good way to think of momentum and energy. Relativistically, the important object is the 4-momentum, which includes both momentum and energy and does not distinguish between types of energy--all of them appear in the 4-momentum and they aren't separated out. Trying to separate things out usually causes more problems than it solves.

That said, the quick answers to your questions are:

* Yes, kinetic energy, heat, etc. can all affect inertial mass. See my discussion starting in the paragraph after next.

* The momentum conservation equation deals with 4-momentum, which includes everything, as I noted above. Nothing is left out.

* As you apply constant force to an object, you change its 4-momentum. That's the relativistically invariant way to describe what happens.

In the simplest case, the object's rest mass is constant, and that means that (since 4-momentum is rest mass times 4-velocity) all of the change in the object's 4-momentum, in response to applied force, will be manifested in the object's 4-velocity. Viewed from a specific frame of reference, part of the change in 4-velocity will appear as a change in the object's inertial mass, and part will appear as a change in the object's velocity. The split between the two depends on your frame of reference. If you keep viewing things from the same frame of reference (for example, the initial rest frame of the object to which force is being applied), then as a constant force (thrust) is applied to an object, over time, more and more of the change in 4-velocity will appear as a change in inertial mass, and less and less as a change in velocity (as the object's observed velocity gets closer and closer to the speed of light).

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.

Are we going to discuss any of the other examples in my post??

I'd like your feedback on the above before addressing the spheres scenario or responding to your other questions. But I haven't forgotten the other questions.

 

[Austin0]

.

=PeterDonis;2384418]

The expected end results should be the same for both versions of conservation of momentum. If you get different results by applying the two different versions then there's a mistake somewhere.

Does this mean that both versions result in equal acceleration throughout the system??

Which end? It makes a difference. Also, does "thrust" mean the actual force applied (presumably by a rocket engine), or the resultant acceleration? They're not necessarily the same, as I've noted before. See next comment.

AS you point out later in this post it doesn't really make any difference which end or degrees of magnitude as far as the principles involved.
Yes I am using thrust as meaning force applied, not resultant acceleration.


I hope you will bear with me if I clarify an earlier concept I presented . For my own education and to check my understanding against your and other's knowledge.

Original AUstin0
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.
______________________________________________________________________
As a rough analogy;
assume a hydro-dynamic system within which the liquid velocity is magically limited to a range that is dependant simply on pressure- density.
Given: a pipe of a certain diameter and a variable input (volume/unit time) that starts at zero and incrementally increases.
Would it not be reasonable to suppose that the liquid would flow without increase in pressure up to some threshold of input??
That exceeding that threshold , since the velocity cannot increase , the pressure would .
This increased pressure would then increase the possible velocity, resulting in this high pressure/velocity itself progressing throughout the pipe until equilibrium was re-established throughout the entire system?
That increasing the diameter of the pipe would inevitably raise the threshold of stress/pressureless flow??
Taken into the context of propagation of momentum:
The pipe diameter would be mass distribution.
Given two rods of equal mass but different shape: Two sylinders , one short with a large cross section and one long and thin. Disregarding the system length and propagation time.
With the same force applied to their ends as an incrementally increasing magnitude starting at zero;
wouldn't it be reasonable to expect that the short rod transmitting this force ,distributed through a larger number of individual particles, would have a higher threshold of stressfree propagation than the rod with a smaller cross section ?
Is there some principle that would make this view untenable?

_____________________________________________________________

(A) Treat the ship as a single object. The total momentum transferred to the ship per unit time will be equal to the sum of the thrusts on the two ends. Since we're treating the ship as a single object, we don't care how that force is propagated through the ship, or what it does to the internal stresses or distances within the ship. All we care about is that the momentum transferred to the ship per unit time must be equal to the total applied thrust (sum of both ends). In other words, we have

F = T_R + T_F = \frac{d P}{d \tau}

for the ship as a whole, where T_R and T_F are the externally applied thrusts at the front and rear ends. That's all there is to it.

Complete agreement here as a generalization.

Original post austin0 #61
I would assume that the overall acceleration of the system would be the sum of force applied at both ends.

(B) Treat the ship as an extended object. For simplicity, we'll treat it here as composed of three segments: R, the rear segment, where the rear thrust is applied; M, the middle segment, where no thrust is applied (but which can exchange internal forces with the other segments), and F, the front segment, where the front thrust is applied.

Then conservation of momentum for the segments individually looks like this:

F_R = T_R + S_{MR} = \frac{d P_R}{d \tau_R}

F_M = S_{RM} + S_{FM} = \frac{d P_M}{d \tau_M}

F_F = T_F + S_{MF} = \frac{d P_F}{d \tau_F}

where we've used T for the thrusts (rear and front) and S for the internal forces between each segment, with the order of subscripts indicating the direction of the internal force (for example, S_{MR} is the internal force exerted by the middle segment on the rear segment). In order to check for consistency with (A) above, we calculate the total momentum gained by the ship per unit time by adding the three forces above:

F = F_R + F_M + F_F

and check to see that the sum equals the sum given in (A) above, i.e., the sum of the externally applied thrusts.

Let's now suppose that the ship starts out at rest, with all forces zero. Then, at time t = 0 in the initial rest frame, the front and rear thrusts are applied. The internal forces at that moment will be zero, because there has been no time for the ship's internal structure to respond to the applied forces. So we will have

F_R = T_R

F_M = 0

F_F = T_F

when the thrusts are initially applied. This means that, initially, the rear and front segments will gain momentum, but the middle one will not. And you can see from the above that, intially, the sum of the momentum gains by the rear and front segments will be equal to the total momentum gained by the ship, which is equal to the sum of the applied thrusts at front and rear. So momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.
Would you agree that compression is dependant on constraint?
That force applied to one end of a spring requires the other end to be fixed in order for compression to take place? That in this context the restraint is simply system inertia.
Isn't it possible that if the force is transmitted as fast as it is applied it then has no constraint ? It becomes a reciprocal motion with no net motion or compression?

Suppose we have a cyclindrical rod with the force applied to both ends but in opposite directions. What would happen?

A) [Local] The rod is stretched locally from the ends. The cross section decreasing and the length increasing at both ends with the middle retaining its original cross section or at least greater than the ends. Eventually breaking at one end or the other.

B) [Global] The rod stretches initially equally throughout its length, but because the force is applied to the ends they would tend to retain the cross section, so the middle of the rod would be the area of maximal stretching and eventual breakdown.

A or B ?

In order to predict in detail how the internal forces will develop, we would need a detailed model of the material properties of the ship's structure, which I don't want to discuss here.

I completely agree. There is nothing to be gained by trying to create a detailed model.
If we can't come to some conclusions based on basic principles we probably won't at all.

But in general terms, you can see from the above that, assuming the front and rear thrusts are in the same direction, the initial effect will be that the rear segment will move closer to the middle segment, while the front segment will move further from it. So, after some small time has elapsed, we will have something like this:

F_R = T_R - C

F_M = C + S

F_F = T_F - S

where C is an internal pressure between the rear and middle segments, which arises because the segments are moving closer together and the material is assumed to resist compression (note the signs of the C terms in each force), and S is an internal tension between the middle and front segments, which arises because the segments are moving further apart, and the material is assumed to resist stretching (again, note the signs of the S terms in each force).

Once again, the above equations automatically conserve momentum for each individual segment. But if we add up the total momentum gained by all the segments, we find that, once again, it's just the sum of the external thrusts applied to front and rear; all the internal forces cancel out (because momentum gained by one segment due to each internal force is exactly balanced by momentum lost by the other segment due to the same force). So again, momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

I agree with all of this. That overall momentum would be equivelant and that the internal stresses would cancel out.
I think there might be other ways they would cancel out though.

If we assume a large central mass with the propulsion units connected by springs at the front and back then I think your description would be accurate. Compression at the back and expansion at the front.
But given a uniform system isn't it possible that they both progress throughout the system
as waves and the established equilibrium is also essentially uniform throughout?

That given a greater force at one end, the total system acceleration would be the sum, but the internal stress would depend on the difference between the two forces, given that they were in the same direction??
That it would be equivalent to that difference being the only force , applied from one end??

All this will continue to be true as the internal forces vary with time; for example, the above suggests that after some more time has elapsed, the middle segment will move away from the rear segment and towards the front segment, as part of the oscillation we were discussing in earlier posts. Then we will have:

F_R = T_R + S

F_M = - S - C

F_F = T_F + C

where now S is a tension between the rear and middle segments and C is a pressure between the middle and front segments. The same analysis as above still applies: the internal forces are different than before, but they still cancel out when the total momentum is added up. So at each individual instant of time, momentum is conserved individually for each ship segment, and total momentum is conserved for the ship as a whole. Everything is consistent.

I can discuss more specific cases if you want (meaning specific relative values of the front and rear thrusts), but you can already see from the above that, regardless of whether the front or the rear thrust is larger, or whether they're both the same, conservation of momentum will hold as I've stated it. The only difference the specific relative values of the thrusts makes is to determine what kind of final state will be reached--i.e., will the ship end up stretched or compressed, and so forth--but that question is independent of the question of conservation of momentum.

Agreed, there is no gain from going into more specifics as far as the principles involved.


One point I am unclear on : given this equivalence of methods and balance of forces does this mean that the resultant velocity would then be the same at both ends??

If we assume that the applied force at one end of the object continues to be applied, then once the initial wave has traveled the length of the object, it will start to be damped out. This is because as each layer pushes against the next, not all of the energy in the push gets converted into motion of the next layer; some of the energy goes into the internal degrees of freedom in the atoms/molecules--i.e., it gets converted into heat. . While the wave is damping, the object will be oscillating, expanding and compressing with gradually decreasing amplitude. Once the wave is damped out, the object will be in a state (assuming the force continues to be applied at one end) in which there is a compressive stress all along its length, and in which its material is somewhat hotter than it was before the force was applied (because some of the applied energy got converted into heat during the damping

_____________________________________________________

The energy that heats the ship's structure as it oscillates in response to an externally applied thrust comes from the energy that fuels the source of the thrust--the rocket engine or whatever. The specific mechanism of how that would occur depends on how you model the internal forces between segments of the ship, as I noted above; but in general, the energy is converted into heat as a result of the relative motion between the internal parts of the ship.

Agreed some of the heat will result from the propulsion mechanism itself and the impossibility of 100% efficiency. I assume we agree this is not relevant and simply assumes some infrared radiative dissapation or the ship will eventually cook.
Also I assume we would agree that even if we posit an external force, an ion beam from an inertial frame or a light sail that the principles would still apply but aren't important for our enquiry?

If we assume that the ship reaches some steady-state equilibrium while under acceleration (i.e., a state where the internal parts of the ship are no longer moving relative to one another), then the heating of the ship's structure should stop once that steady-state equilibrium is reached, since at that point there is no longer any relative motion between the ship's parts.

Can we agree that the 2nd law of TD will still operate but that the results would be negligable and irrelevant to the real questions we are addressing??

It seems like you're trying to separate out momentum from energy, and kinetic energy from other kinds of energy. That's probably not a good way to think of momentum and energy. Relativistically, the important object is the 4-momentum, which includes both momentum and energy and does not distinguish between types of energy--all of them appear in the 4-momentum and they aren't separated out. Trying to separate things out usually causes more problems than it solves.

I think in actuality I am trying to integrate momentum and kinetic energy .
That there are three fundamental principles involved here.
Conservation of mass and energy.
Conservation of momentum.
2nd law of thermodynamics.
Correct me if I am wrong :
Newtonian conservation of momentum was based on a conception of instantaneous force and completely invariant mass.
SR evolved this concept in two fundamental ways.
1) Finite propagation speed of momentum [energy]
2) A variable mass in two distinct senses.
A) The relativistic inertial mass increase with velocity.
B) The equivalence and interchangability of energy and mass. This means that not only kinetic energy but the energy of internal stress etc results in increased inertial mass.
I trust we agree that the Lorentz mass increase is not relevant to our discussion.
As I understand the 4 momentum and the equations for conservation of momentum you have presented so far they are basically unchanged from the Newtonian form with regard to internal propagation of momentum and the equivalence of mass and energy.
The 4 momentum differs in its coordinate time component but still defines instantaneous change. Mass/energy is included in the concept of rest mass, invariant mass etc. but a possible change in inertial mass through applied force is not included in the basic formulations. It is still based on a concept of ideal elastic interactions. An assumption that both momentum and kinetic energy will be conserved. I think we would agree that in the real world there are no ideally rigid or elastic bodies.
So how does this get handled with complex systems?
I hope that we agree that the magnitude of effects involved are slight and we can proceed without regarding them further??

That said, the quick answers to your questions are:

* Yes, kinetic energy, heat, etc. can all affect inertial mass. See my discussion starting in the paragraph after next.

* The momentum conservation equation deals with 4-momentum, which includes everything, as I noted above. Nothing is left out.

* As you apply constant force to an object, you change its 4-momentum. That's the relativistically invariant way to describe what happens.

In the simplest case, the object's rest mass is constant, and that means that (since 4-momentum is rest mass times 4-velocity) all of the change in the object's 4-momentum, in response to applied force, will be manifested in the object's 4-velocity. Viewed from a specific frame of reference, part of the change in 4-velocity will appear as a change in the object's inertial mass, and part will appear as a change in the object's velocity. The split between the two depends on your frame of reference. If you keep viewing things from the same frame of reference (for example, the initial rest frame of the object to which force is being applied), then as a constant force (thrust) is applied to an object, over time, more and more of the change in 4-velocity will appear as a change in inertial mass, and less and less as a change in velocity (as the object's observed velocity gets closer and closer to the speed of light).

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.



I'd like your feedback on the above before addressing the spheres scenario or responding to your other questions. But I haven't forgotten the other questions

I hope not ;-) it would be good to get on to the really interesting stuff.
Thanks

 

[PeterDonis]

Austin0: Let me start with this from your post because I think it's the most basic point that needs to be clarified:

I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.

This is false. *Any* gain in momentum must result in *some* coordinate translation. More precisely: if, viewed in a given frame, an object gains some 4-momentum, and the 4-momentum gained has a spatial component (i.e., momentum as well as energy) in that frame, then the object must gain *some* coordinate velocity in that frame.

For example, consider a simpler version of the example in my last post. Let's consider a ship with a rocket engine only at the rear, and model it as having just two segments, the rear segment R with the rocket engine, and the middle segment M with no engine. Suppose we turn on the rocket engine at time t = 0 in the ship's initial rest frame. Then, at that instant, we have (same notation as my last post):

F_R = T_R

F_M = 0

So after a very small time dt, the rear segment will have acquired 4-momentum T_R dt. Since the force is in a particular direction (the "forward" direction of the ship), the 4-force T_R will have a spatial component, so the rear segment will have *some* coordinate velocity in the original rest frame. If we assume that the time dt is short enough that the 4-momentum acquired is small compared to the rear segment's rest mass M_R, then the 4-force will have *only* a spatial component, and the coordinate velocity will be approximately the non-relativistic value

v = \frac{T_R dt}{M_R}

This observation affects a number of the scenarios you've suggested: basically, *any* scenario where an application of a force in a particular direction does not immediately result in the object acquiring a momentum in that direction *cannot* be correct. The momentum may reside in only one part of the object (as above, only the rear segment has a momentum on the initial application of the force), but *some* part of the object must move immediately on application of a force. Given that observation, I'm not going to comment on scenarios you've suggested that appear to me to violate this rule. For example, this:

Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.

doesn't really capture what's going on. To see how the oscillation comes about, you have to model the object as a bunch of individual segments that can move independently (but may exert forces on adjacent segments), as we did with the ship; then you can see that *some* segment is always moving in response to the applied force, and the sum of the motions of all the segments always equals the total momentum gained from the applied force, so that conservation of momentum is satisfied. If this is what you mean by "reciprocal oscillation", that's fine, but it's not something that happens instead of the system moving as a whole; the system *is* moving as a whole at the same time that the oscillation is propagating through it.

One point I am unclear on : given this equivalence of methods and balance of forces does this mean that the resultant velocity would then be the same at both ends??

Let's go back to the full model we considered in my last post, where we have three segments of the ship, with thrusts applied at the front and rear segments and no thrust in the middle segment. For your question quoted just above to be meaningful, the ship must reach some sort of steady-state equilibrium, in which there is no relative motion between the parts of the ship--otherwise, as we found last time, there will still be oscillations that haven't been damped out. (This does make a fairly general assumption about the material structure of the ship--that it is adequately modeled by the general scheme of tension and compression forces we used last time.)

But remember the definition of Born rigid acceleration: the whole object must accelerate such that the distance between individual parts of the object, in each part's rest frame, remains constant. That means that any steady-state equilibrium of an accelerating object *must* be a state of Born rigid acceleration! In other words, Born rigid acceleration, while extremely unlikely to happen at the *start* of an object's acceleration (because the acceleration of each part of the object would have to be controlled so precisely), is a likely *end state* of an object's acceleration in response to constant continuous applied forces.

This is very helpful because it tells us how the net forces on each segment of the ship must be related (since the acceleration of each segment is determined by the net force on that segment). The net force must be highest at the rear end of the ship, and must decrease linearly with distance towards the front of the ship ("distance" as measured in the rest frame of the ship). That means that, if the applied thrusts at each segment are different than that relationship, then internal forces will develop within the ship to "correct" the total net force on each segment to what it "should" be to meet the Born rigid condition. (The forces may oscillate for a while until they settle into this equilibrium state, but assuming that there is at least some damping, they will eventually settle in.)

This also answers your question above: no matter *what* the starting thrusts are, *if* there is a steady-state equilibrium reached, then, since that equilibrium is a state of Born rigid acceleration, the acceleration at the front end of the ship must be *less* than at the rear end. The relative *velocities* of the end will depend on what frame we measure them in; but in the original rest frame of the ship, the velocity of the front end will be *less* than that of the rear end, while the ship continues to accelerate.

Let's now consider some examples to see what the final states look like; this will get back to your question about whether the ship will be stretched or compressed. Because the final state is one of Born rigid acceleration, in all cases, we must have

F_R > F_M > F_F

for the relationship between the forces on each segment (for a more exact relationship we would need to know the length of each segment in the ship's rest frame). Also, the net force on each segment must be positive, since the ship as a whole is moving in the positive x-direction.

(Case 1) The only thrust is on the rear segment. Then the final equilibrium state will look something like this:

F_R = T - C_R

F_M = C_R - C_F

F_F = C_F

where C_R is the pressure force between the rear and middle segments, and C_F is the pressure force between the middle and front segments. Note the signs of the forces; they *must* be as they are for the inequality F_R > F_M > F_F to hold. Thus, in this case, the ship will end up compressed.

(Case 2) There are equal thrusts on the front and rear segments. Then the final equilibrium state will look something like this:

F_R = T + S_R

F_M = S_F - S_R

F_F = T - S_F

where now we have stretching forces between the segments. Note, once again, the signs of the forces; if the thrusts on the front and rear segments are equal, then the forces *must* be stretching forces for the inequality F_R > F_M > F_F to hold. Thus, in this case, the ship will end up stretched.

Obviously, if the thrust on the front segment is *larger* than that on the rear, the ship will still end up stretched, just by a larger amount. If the thrust on the front segment is *smaller* than that on the rear, there should be a point at which the ship, in its final equilibrium state, will be neither stretched nor compressed (since if there is no force on the front end, the ship will end up compressed). However, that does *not* mean that the individual segments will be equidistant! In fact, for the overall length of the ship to remain the same even though there are net forces on each segment (which there must be for the segments to be accelerating), the forces must look like this:

F_R = T_R - C_R

F_M = C_R + S_F

F_F = T_F - S_F

In other words, there must be compression between the rear and middle segments, and stretching between the middle and front segments, with the amounts just canceling each other out, so that the overall length of the ship is the same as its original rest length.

A) [Local] The rod is stretched locally from the ends. The cross section decreasing and the length increasing at both ends with the middle retaining its original cross section or at least greater than the ends. Eventually breaking at one end or the other.

B) [Global] The rod stretches initially equally throughout its length, but because the force is applied to the ends they would tend to retain the cross section, so the middle of the rod would be the area of maximal stretching and eventual breakdown.

A or B ?

There are three issues with the above:

First, I'm not sure how you're deriving these two different predictions from a local vs. a global picture. The local picture is the one I'm going to expound below; the global picture would be something pretty simple, like this: the rod stretches until it breaks. *Where* it breaks is a local question, not a global one. :-) (I put the smiley face there, but I'm actually serious: as soon as you ask a question like "where does the rod break?", you're implicitly modeling the rod as an extended system with internal parts, not a single object, which means that you need a local model that looks at the internal parts of the rod and their interactions.)

Second, if you want to try and model how the cross section of the rod will respond to applied stresses, you will need to introduce additional assumptions about the material properties of the rod. I don't want to introduce that additional complication, so I'm just going to ignore the rod's cross section and talk about the internal energy stored in each segment, without discussing specifics about how it's stored.

Third, the models we've discussed so far don't include any internal energy stored in any of the segments, so they aren't adequate to model something like a rod breaking, since the general condition for a material to break under stress is that the stored internal energy in the material due to the stress exceeds the material's breaking strength. In order to add internal energy to our model, we would need, once again, additional assumptions about the material properties of the rod, to calculate how much internal energy is stored as a result of a given set of stresses. I'll give a very simple assumption of this sort below, but it will be very simple and crude, just to see in general how such a scheme might work; any real material would require more complicated models that, once again, I don't want to discuss here.

Let's model the rod the same way we did the ship above, with three segments, but now we have the force on the "rear" segment in the opposite direction from that on the front. The forces on the segments will then look like this:

F_R = S_R - T

F_M = S_F - S_R

F_F = T - S_F

where we have assumed that the force (T) on the front and rear ends is the same magnitude (but in opposite directions). If we further assume that the response of the material to stretching is uniform, then we should have S_F = S_R[/tex], and we have<br /> <br /> F_R = S - T<br /> <br /> F_M = S - S = 0<br /> <br /> F_F = T - S<br /> <br /> so that the middle of the rod will not move at all (but it will still be pulled on from each side). In other words, the rod will gain *no* net momentum in its original rest frame (its center of mass remains motionless). This is pretty much the simplest situation we can model.<br /> <br /> Now, as I noted above, in order to determine where the rod will break, we need to make some assumption about how the rod will store internal energy due to applied stress. Here&#039;s a simple assumption of that sort: each segment stores an amount of internal energy equal to<br /> <br /> U = K S<br /> <br /> where K is a constant that we assume to be an innate property of the material, and S is the &quot;balanced&quot; stress on the segment--in other words, the amount of stress that is &quot;equalized&quot; by forces in opposite directions, so that it doesn&#039;t actually result in any motion of the segment. For each segment, this will be the stretching force S between the segments, since that is the amount of stress that is &quot;balanced&quot; on both sides--for the middle segment, this is obvious, and for the rear and front segments, we can see that, for the rod to stretch at all, we must have T &amp;gt; S, so that S will be the stress that is &quot;balanced&quot; by equal forces in opposite directions, and +/- \left( T - S \right) will be the stress that is &quot;unbalanced&quot; and therefore results in motion of the segment.<br /> <br /> With this simple assumption about internal energy stored, we can see that the internal energy will be the same for each segment, since the balanced stress on each segment is the same. Therefore, under this model, the rod is equally likely to break *anywhere* along its length. If we had a material with a more complicated relation of internal energy to stress, we might find that the internal energy was greater in one segment than the others, so that one would be the one most likely to break. But again, I don&#039;t want to get into those complications here. Someone who has, in considerably more detail than I do here, is Greg Egan, at http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html&quot; .<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Agreed some of the heat will result from the propulsion mechanism itself and the impossibility of 100% efficiency. I assume we agree this is not relevant and simply assumes some infrared radiative dissapation or the ship will eventually cook. </div> </div> </blockquote><br /> Yes. In any real problem we&#039;d have to include heat dissipation, but that&#039;s a complication I didn&#039;t want to introduce for these simple models.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Can we agree that the 2nd law of TD will still operate but that the results would be negligable and irrelevant to the real questions we are addressing?? </div> </div> </blockquote><br /> I think we don&#039;t need to worry about the *details* of how dissipation happens, but we need to recognize that it has to happen somehow in order for the oscillations to damp out and the ship to reach an equilibrium. An idealized system with zero dissipation would continue to oscillate forever.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I think in actuality I am trying to integrate momentum and kinetic energy. </div> </div> </blockquote><br /> &quot;Kinetic energy&quot; isn&#039;t really a useful concept in relativistic problems. The 4-momentum and its components, total energy and total momentum, are the useful concepts. I&#039;ve never seen a relativistic problem where separating out the kinetic energy helped in the solution.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> That there are three fundamental principles involved here.<br /> Conservation of mass and energy.<br /> Conservation of momentum.<br /> 2nd law of thermodynamics. </div> </div> </blockquote><br /> The first two are aspects of a single law in relativity, conservation of energy-momentum. Or, when you get into more complicated problems that require using a stress-energy tensor instead of just a 4-momentum vector, the single law is conservation of the stress-energy tensor, in the form:<br /> <br /> \frac{D T^{ab}}{D x^b} = 0<br /> <br /> which says that the covariant divergence of the stress-energy tensor is zero.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I trust we agree that the Lorentz mass increase is not relevant to our discussion. </div> </div> </blockquote><br /> Yes, if by this you mean mass increase solely due to the effects of changing reference frames.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The 4 momentum differs in its coordinate time component but still defines instantaneous change. </div> </div> </blockquote><br /> &quot;Instantaneous&quot; in a relative sense, yes. Relativity includes the concept of &quot;events&quot;, which are taken to be points in spacetime, with no extension in any dimension. We&#039;ve been modeling the continuous application of forces, such as the thrust of a rocket engine on a segment of a ship, as a continuous succession of events along the worldline of the ship segment. Since the events have no extension, what happens at each event happens instantaneously, to the level of precision of our model; essentially we are modeling each ship segment as a spatial point. If we wanted a more accurate model, we could increase the resolution so that we were looking at individual atoms instead of macroscopic segments of the ship; then we would be able to see the &quot;travel time&quot; of forces internal to each ship segment, but the forces applied to individual atoms would be modeled as a succession of instantaneous events.<br /> <br /> <blockquote data-attributes="" data-quote="Austin0" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Austin0 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Mass/energy is included in the concept of rest mass, invariant mass etc. but a possible change in inertial mass through applied force is not included in the basic formulations. </div> </div> </blockquote><br /> Not in the ones in my previous posts, but it is (in a very simple fashion) in the model earlier in this post. Such models do require assumptions beyond the basic postulates of relativity, about the properties of materials, as I&#039;ve said, and we may not want to get into those complexities. For many problems it isn&#039;t necessary to do so.

 

[Austin0]

=PeterDonis;2389087]Austin0: Let me start with this from your post because I think it's the most basic point that needs to be clarified:

Originally Posted by Austin0
I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.

This is false. *Any* gain in momentum must result in *some* coordinate translation. More precisely: if, viewed in a given frame, an object gains some 4-momentum, and the 4-momentum gained has a spatial component (i.e., momentum as well as energy) in that frame, then the object must gain *some* coordinate velocity in that frame.

We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

I think we both agree that the energy component of 4-momentum , the scalar product of mass and speed, is both instantaneous and absolute upon application to a system. ?

SO the question is regarding the time it is internally propagating and whether that results in instantaneous motion of the back of the system,, not whether it results in internal motion at some internal point

So a question; assuming a purely x direction of the initial vector do you think that this component of the 4-vector is quantitatively conserved?? That there is zero propagated motion in y or z ??

Assume a thin slice ,one molecule thick in a 1000 x 1000 atom matrix.
Assume a single particle impacting one atom squarely in the x direction.
This is then propagated generally in the x dir. but inevitably there will be a yand z component also , yes?

Initially the impacted atom will translate in the x dir. but by the time the momentum has propagated to the middle, that original atom will have rebounded back into its original position in the matrix and will be oscillating at a decreasing amplitude and probably in additional directions due to the tensile forces from neighboring atoms.

We agree, no motion at this point in the front. So do you think there is net motion at the back??
For there to be would seem to neccessitate a coordinated pressure sustained all along the back. For system coordinate motion at the back at this point could only be a result of overall pressure and contraction along the whole face ,not simply the motion initially imparted to the first atom which was exceedingly localized and is now an oscillation..

Now at the point when the momentum reaches the front, what is the state of the back of the system?

Isn't it essentially in an inertial state with a slightly increased kinetic energy distributed as vibration??

Wouldn't the same consideration that applied at the beginning. I.e that the system couldn't move at the front until the energy reached there , also apply in this case?

That reaching the front would effect some degree of actual local motion of atoms but the inertia of the rest of the system would constrain that to an oscillation, which would then propagate backward through the system.

You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.

Doesn't it follow that, in principle, given a large enough system ,that a single particles momentum could be totally internally converted without ever reaching the opposite boundary??

Resulting in increased heat and inertial mass but no velocity ?

Conforming to the overall conservation of energy and mass and the conservation of energy-momentum also??
The math for the application of Force treats a complex system as a black box labled M
, no explicit consideration of internal structure etc.

What you put into one side of the box is what you get coming out of the box divided by overall mass, instantly.
I have agreed that within a certain macroscopic range this is operationally sound. With two cannonballs any other considerations are so neglible as to make the effort calculating them pointless.
But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.
Part of me would just like to agree to anything you say and get on to other aspects of the original enquiry but some of these questions are central to the larger issue so I can only hope we can reach some grounds of agreement sufficient to carry on.

This observation affects a number of the scenarios you've suggested: basically, *any* scenario where an application of a force in a particular direction does not immediately result in the object acquiring a momentum in that direction *cannot* be correct. The momentum may reside in only one part of the object (as above, only the rear segment has a momentum on the initial application of the force), but *some* part of the object must move immediately on application of a force. Given that observation, I'm not going to comment on scenarios you've suggested that appear to me to violate this rule. For example, this:

Originally Posted by Austin0
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.

doesn't really capture what's going on. To see how the oscillation comes about, you have to model the object as a bunch of individual segments that can move independently (but may exert forces on adjacent segments), as we did with the ship; then you can see that *some* segment is always moving in response to the applied force, and the sum of the motions of all the segments always equals the total momentum gained from the applied force, so that conservation of momentum is satisfied. If this is what you mean by "reciprocal oscillation", that's fine, but it's not something that happens instead of the system moving as a whole; the system *is* moving as a whole at the same time that the oscillation is propagating through it.

We have agreed that no motion can take place at the front until momentum has reached it.
If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .
Behind it how can there be general displacement?
Atomic reactions are exceedingly fast compared to the velocity of sound right??

If the energy of momentum is located at this internal front, conservation would seem to mean that it could not reside in actual increased momentum as velocity ,of all the particles behind it wouldn't it?

If this applies then wouldn't it seem to follow that a continuous application would operate on the same principle. Overall system velocity increase results from propagation throughout the system not locally except as an initial pressure adjustment?

We already agreed earlier that in a sound wave there is always a local motion someplace but no overall displacement of the medium. You suggested that sustained applied momentum was different
I thought that had been addressed but maybe not.
DO you think there is a fundamental difference in propagation between coherent sound and white sound??

DO you think there is any significant difference between white sound and the applied momentum of a large number of random particles??

Isn't each individual collision equivalent to microscopically ringing a tiny atomic resonant bell?

Hi PeterDonis I am out of time so will get back for the rest of your post.
Hopefully we can resolve these issues soon, Thanks

 

[Austin0]

=PeterDonis;2389087]
This also answers your question above: no matter *what* the starting thrusts are, *if* there is a steady-state equilibrium reached, then, since that equilibrium is a state of Born rigid acceleration, the acceleration at the front end of the ship must be *less* than at the rear end. The relative *velocities* of the end will depend on what frame we measure them in; but in the original rest frame of the ship, the velocity of the front end will be *less* than that of the rear end, while the ship continues to accelerate.

This appears to be entering a desired end result , the conclusion , directly into the deductive chain.
ANd then simply determining the physics and intermediate arguments in retrospect to conform to this conclusion.

Even going with the premise that unequal force applied at different locations would result in different stresses after equilibrium is established; If the force applied is constant
and the slight difference due to contraction or expansion is stabalized........ what would lead to continued relative motion between different parts of the ship?
Are you talking about a differing dynamic thrust increase at different parts of the ship?

(Case 1) The only thrust is on the rear segment. Then the final equilibrium state will look something like this:

F_R &gt; F_M &gt; F_F to hold. Thus, in this case, the ship will end up compressed.

(Case 2) There are equal thrusts on the front and rear segments. Then the final equilibrium state will look something like this:

F_R = T + S_R

F_M = S_F - S_R

F_F = T - S_F

where now we have stretching forces between the segments. Note, once again, the signs of the forces; if the thrusts on the front and rear segments are equal, then the forces *must* be stretching forces for the inequality F_R &gt; F_M &gt; F_F to hold. Thus, in this case, the ship will end up stretched.

I don't understand where any net stretching would occur if the forces are balanced at each end?
Or why you think that the inequality would hold??

Obviously, if the thrust on the front segment is *larger* than that on the rear, the ship will still end up stretched, just by a larger amount.

How does this equate. If the force is greater at the front then the force at the rear in the same direction would seem to have to decrease the inertial resistence from the rear and thus lessen the stretching forces from the front or am I missing something here?

If the thrust on the front segment is *smaller* than that on the rear, there should be a point at which the ship, in its final equilibrium state, will be neither stretched nor compressed (since if there is no force on the front end, the ship will end up compressed). However, that does *not* mean that the individual segments will be equidistant! In fact, for the overall length of the ship to remain the same even though there are net forces on each segment (which there must be for the segments to be accelerating), the forces must look like this:

F_R = T_R - C_R

F_M = C_R + S_F

F_F = T_F - S_F

In other words, there must be compression between the rear and middle segments, and stretching between the middle and front segments, with the amounts just canceling each other out, so that the overall length of the ship is the same as its original rest length.

Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??


Let's model the rod the same way we did the ship above, with three segments, but now we have the force on the "rear" segment in the opposite direction from that on the front. The forces on the segments will then look like this:

F_R = S_R - T

F_M = S_F - S_R

F_F = T - S_F

where we have assumed that the force (T) on the front and rear ends is the same magnitude (but in opposite directions). If we further assume that the response of the material to stretching is uniform, then we should have S_F = S_R[/tex], and we have<br /> <br /> F_R = S - T<br /> <br /> F_M = S - S = 0<br /> <br /> F_F = T - S<br /> <br /> so that the middle of the rod will not move at all (but it will still be pulled on from each side). In other words, the rod will gain *no* net momentum in its original rest frame (its center of mass remains motionless). This is pretty much the simplest situation we can model.<br /> <br /> Now, as I noted above, in order to determine where the rod will break, we need to make some assumption about how the rod will store internal energy due to applied stress. Here&#039;s a simple assumption of that sort: each segment stores an amount of internal energy equal to<br /> <br /> U = K S<br /> <br /> where K is a constant that we assume to be an innate property of the material, and S is the &quot;balanced&quot; stress on the segment--in other words, the amount of stress that is &quot;equalized&quot; by forces in opposite directions, so that it doesn&#039;t actually result in any motion of the segment. For each segment, this will be the stretching force S between the segments, since that is the amount of stress that is &quot;balanced&quot; on both sides--for the middle segment, this is obvious, and for the rear and front segments, we can see that, for the rod to stretch at all, we must have T &amp;gt; S, so that S will be the stress that is &quot;balanced&quot; by equal forces in opposite directions, and +/- \left( T - S \right) will be the stress that is &quot;unbalanced&quot; and therefore results in motion of the segment.<br /> <br /> With this simple assumption about internal energy stored, we can see that the internal energy will be the same for each segment, since the balanced stress on each segment is the same. Therefore, under this model, the rod is equally likely to break *anywhere* along its length. If we had a material with a more complicated relation of internal energy to stress, we might find that the internal energy was greater in one segment than the others, so that one would be the one most likely to break. But again, I don&#039;t want to get into those complications here. Someone who has, in considerably more detail than I do here, is Greg Egan, at http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html&quot; .<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> &quot;Kinetic energy&quot; isn&#039;t really a useful concept in relativistic problems. The 4-momentum and its components, total energy and total momentum, are the useful concepts. I&#039;ve never seen a relativistic problem where separating out the kinetic energy helped in the solution. </div> </div> </blockquote><br /> But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?<br /> Can you apply OHM&#039;s law, in principle, across its range of applicability without considering temperature at some levels of that range?<br /> Thanks<br /> <br /> <br /> .<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> &quot;Instantaneous&quot; in a relative sense, yes. Relativity includes the concept of &quot;events&quot;, which are taken to be points in spacetime, with no extension in any dimension. We&#039;ve been modeling the continuous application of forces, such as the thrust of a rocket engine on a segment of a ship, as a continuous succession of events along the worldline of the ship segment. Since the events have no extension, what happens at each event happens instantaneously, to the level of precision of our model; essentially we are modeling each ship segment as a spatial point. If we wanted a more accurate model, we could increase the resolution so that we were looking at individual atoms instead of macroscopic segments of the ship; then we would be able to see the &quot;travel time&quot; of forces internal to each ship segment,<b> but the forces applied to individual atoms would be modeled as a succession of instantaneous events.[/</b>QUOTE]<br /> Agreed entirely. Everything you have said. In fact we have both agreed that any input of momentum would result in immediate motion at the local atomic level.<br /> But the math does not make this distinction. A complex system appears in the equations as a discrete entity. Simply Mass <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Not in the ones in my previous posts, but it is (in a very simple fashion) in the model earlier in this post. Such models do require assumptions beyond the basic postulates of relativity, about the properties of materials, as I&#039;ve said, and we may not want to get into those complexities. For many problems it isn&#039;t necessary to do so </div> </div> </blockquote> Agreed we don&#039;t want to get sidetracked into specific consideration of materials etc.<br /> <br /> One last quickie. If there was a long line of spheres as before with an impacting sphere:<br /> A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.<br /> or <br /> B) No matter how many spheres were added to the line the last sphere would undergo &quot;some&quot; translation. <br /> So A) or B) ? </div> </div> </blockquote>

 

[PeterDonis]

Austin0: First, a general comment. You keep on throwing different, more complicated scenarios at me when we apparently haven't reached complete agreement on the simple one I've been discussing. Let's please hold off on more complicated scenarios (throwing in thousands of atoms, y and z dimensions of space, requiring assumptions about the tensile strength of materials and how they respond to stresses, etc.) until we've got complete agreement on the simple one.

We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

No, according to the definition of system motion that corresponds to the global version of conservation of momentum I gave several posts ago. According to that definition, the "system motion" is just the sum of the motions of all the individual parts of the system. If that sum results in a net motion, then the system as a whole is moving. So, for example, if we have a rocket just at the rear segment of a ship, and we turn on its engine, initially only the rear segment is moving, but since that equates to net motion when all the segments' motions are summed (since no other segment is moving), the system as a whole is moving--it has net momentum in a particular direction.

If this definition bothers you, then forget the term "system motion" and let's just talk about the individual motions of each part of the system. We'll get the same answers anyway when we try to predict actual measurements.

You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

I have agreed that dissipation can result in increased inertial mass (rest mass). I have *not* agreed that there is no necessary increase in coordinate velocity if the external applied forces to the system are in a particular direction. The only way to increase a system's mass (e.g., by heating it) without changing its coordinate velocity is to add energy to it in such a way that the net momentum cancels out--for example, by heating it with two laser beams from opposite directions, with the beam intensities exactly equal, so that their net momentum cancels out. Internal dissipation in response to an applied force in a particular direction won't do that.

Remember the rule I gave last time: *any* applied force in a particular direction *must* result in *some* coordinate velocity in that direction.

What you put into one side of the box is what you get coming out of the box divided by overall mass, instantly.

At the level of accuracy we've been modeling, yes. You're correct that how we see the momentum propagate (this isn't quite the term I would use) depends on the level of accuracy of our model. But that means that this:

But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.

is not quite accurate. The mechanisms in the models we've been discussing for propagating momentum are anything but fundamental. We've been modeling what are basically contact forces between macroscopic segments of objects; in this type of model, momentum transfer is basically instantaneous between adjacent objects. But those contact forces aren't fundamental; they're due to the underlying physics of atoms and the internals of atoms, which are ultimately due to quantum mechanics and mechanisms of momentum transfer that are quite different.

We've been discussing how we would deal relativistically with the motion of macroscopic objects that might have internal parts, but for which all the forces we're interested in can be modeled in the simple ways we've been modeling them. That let's us concentrate on the aspects that are due to relativity itself, rather than to other things like the material properties of objects. But that means we won't be able to answer some questions that are natural to ask, and still remain within those boundaries.

If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .

Once again, let's keep to the simple model before introducing complexities. Let's take the ship with three segments, but now, instead of applying a steady thrust at the rear, we'll apply the thrust only for a very short time--essentially, long enough to get the rear segment to move, but not long enough for its movement to develop any internal forces with other segments. Then, at time t = 0 in the original rest frame, when we apply the force at the rear, we have:

F_R = T

F_M = 0

F_F = 0

After a very short time dt, the rear segment will have moved slightly towards the middle one, and it will have momentum equal to T dt. But by then the thrust will be removed, so the forces will be:

F_R = - C

F_M = C

F_F = 0

Now let's suppose, just to make things simple, that the compressive force C just happens to be equal to the original thrust T. Then, after another short time dt, the rear segment will have stopped moving, and the middle one will be moving forward. Thus, the rear segment, in the original rest frame, will have zero momentum, and the middle one will now have momentum T dt, so that momentum is conserved. Then the forces will be:

F_F = 0

F_M = - T

F_F = T

Now wait one more short interval of time dt. Then the middle segment will have stopped moving, and the front segment will have moved forward, and will have momentum T dt, so that momentum is still conserved. At that point, the forces will be zero on all segments. But has the ship stopped moving? No--the front segment is still moving forward. That means that, if we wait another short interval of time dt, we will have:

F_F = 0

F_M = T

F_F = - T

as the front segment now moves further forward and stretches the ship. So the ship will keep moving forward. If we add some damping, to be more realistic, the oscillations will eventually damp out, and when they do, the final forward momentum of the ship will still be T dt, but of course in a steady state it will be equally distributed among the three segments, so the final velocity of the ship as a whole, as seen from its original rest frame, will be (assuming each segment has equal mass, and that the heating from the damping is negligible) one-third of the velocity that the original impulse force imparted to the rear segment.

This general picture would be the same if we took a bulk material and looked at the motion of its atoms; the final velocity of the object as a whole would still be the original momentum imparted by the impulse force, divided by the total mass of the object. If the original impulse force were applied to a single atom, then of course the initial velocity of that atom would be much higher than the final velocity--but the initial *momentum* would be the same as the final momentum.

We already agreed earlier that in a sound wave there is always a local motion someplace but no overall displacement of the medium. You suggested that sustained applied momentum was different.

The question whether sound waves are propagating in an object is independent of the question whether the object is undergoing overall motion. The phenomena may be related in a specific case (for example, the sound waves were caused by a very short impulse force applied to the object, which also induced overall motion because it was applied in a specific direction), but they are independent. The easiest way to separate them is as follows: overall motion of the object is motion of its center of mass, relative to some reference position we're interested in. Sound waves are waves of compression and expansion, viewed from the center of mass frame of the object (in other words, the frame in which its center of mass is at rest).

This appears to be entering a desired end result , the conclusion , directly into the deductive chain.

Then you haven't followed the argument correctly. Here's the deductive chain:

(1) For an object to undergo steady-state, equilibrium motion, in which all internal oscillations have been damped out, there must be no relative motion between the internal parts of the object, as viewed from the MCIF of each part at any event along the part's worldline.

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

(3) Therefore, any state of motion of an accelerated object which is a steady-state equilibrium, with no internal oscillations, as described above, must be a state of Born rigid acceleration.

The reason I introduced the condition of steady-state equilibrium is that that was the kind of state of motion we were interested in, and it's only that condition that leads to Born rigid acceleration as the final state. That doesn't mean that *every* state of motion of an accelerated object must be a state of Born rigid acceleration; there are at least two categories of states for which that isn't true:

(i) States in which the object is still undergoing internal oscillations that haven't yet been damped out--or, of course, oscillations that will never damp out, because we've modeled a perfectly elastic object with no damping;

(ii) States in which the object is not tending towards a steady-state equilibrium--for example, the scenario I discussed a number of posts ago, in which the acceleration of the front of the ship is such that the ship keeps stretching until it breaks. I'm sure we could come up with other such scenarios.

How does this equate. If the force is greater at the front then the force at the rear in the same direction would seem to have to decrease the inertial resistence from the rear and thus lessen the stretching forces from the front or am I missing something here?

You're missing what I was comparing it to. If the thrust on the front is greater than the thrust on the rear, the ship will be stretched by a greater amount than it would be if the thrust on the front was *equal* to the thrust on the rear.

Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??

There would be a gradient of compression; it would be greater at the bottom of the spring and less at the top, just as the water pressure in the ocean is greater the deeper you go, or air pressure at sea level is greater than at a high altitude.

But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?

My point is just that in relativity, when you calculate total energy, you're already including kinetic energy; you don't have to worry about it separately. And in virtually all problems where relativistic effects are significant, total energy is both easier and more useful to calculate.

But the math does not make this distinction. A complex system appears in the equations as a discrete entity. Simply Mass

At the level of accuracy we've been modeling, yes. You can always increase the accuracy of your model, at the expense of greater complexity and difficulty in calculation.

One last quickie. If there was a long line of spheres as before with an impacting sphere:
A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.
or
B) No matter how many spheres were added to the line the last sphere would undergo "some" translation.
So A) or B) ?

Good, we've gotten to the spheres, which was what I wanted to discuss next. :-)

The quick answer is B. Can you see why from what I've posted already?

 

[Austin0]

Austin0: First, a general comment. You keep on throwing different, more complicated scenarios at me when we apparently haven't reached complete agreement on the simple one I've been discussing. Let's please hold off on more complicated scenarios (throwing in thousands of atoms, y and z dimensions of space, requiring assumptions about the tensile strength of materials and how they respond to stresses, etc.) until we've got complete agreement on the simple one.

I am sorry you think my matrix scenario was unduly complex. We have agreed there is no point in complicating things with specific tensile strengths etc. I was not suggesting we do so. But within the context of basic principles I fail to see how it is essentially more complex that any of the scenarios we are considering. As it relates directly to the current question I am curious as to why it is simply not addressed.
It seems to me that part of the problem we are having is a basic difference in mindset.
I am viewing the question of applied momentum as a complete continuum, starting from zero and progressing through the range of magnitudes and conditions with different responses varying through that range.
You view it as an absolute linear response regardless of magnitude or internal considerations. ALthough I would agree that the conservation of mass and energy can be viewed in this way, I do not see how this could possibly apply to momentum unless the 2nd law of thermodynamics was not valid.
Or the fundamental conservation of mass and energy for that matter.
For the math , and your interpretation from the math, posits that a specific amount of energy could enter a system and increase internal kinetic energy and at the same time be 100% translated into system translational momentum without consideration of the internal characteristics and composition of that system.
I just don't see how that could work.

Originally Posted by Austin0
We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

No, according to the definition of system motion that corresponds to the global version of conservation of momentum I gave several posts ago. According to that definition, the "system motion" is just the sum of the motions of all the individual parts of the system. If that sum results in a net motion, then the system as a whole is moving. So, for example, if we have a rocket just at the rear segment of a ship, and we turn on its engine, initially only the rear segment is moving, but since that equates to net motion when all the segments' motions are summed (since no other segment is moving), the system as a whole is moving--it has net momentum in a particular direction.

This quote does not relate to my post at all but is addressed to a different stage where the momentum has progressed farther. See your quote following.

overall motion of the object is motion of its center of mass, relative to some reference position we're interested in.

Originally Posted by Austin0
You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

I

have agreed that dissipation can result in increased inertial mass (rest mass). I have *not* agreed that there is no necessary increase in coordinate velocity if the external applied forces to the system are in a particular direction. The only way to increase a system's mass (e.g., by heating it) without changing its coordinate velocity is to add energy to it in such a way that the net momentum cancels out--for example, by heating it with two laser beams from opposite directions, with the beam intensities exactly equal, so that their net momentum cancels out. Internal dissipation in response to an applied force in a particular direction won't do that.

Are you now saying that directed momentum entering a system cannot result in internal dissipation?


Originally Posted by Austin0
But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.

is not quite accurate. The mechanisms in the models we've been discussing for propagating momentum are anything but fundamental. We've been modeling what are basically contact forces between macroscopic segments of objects; in this type of model, momentum transfer is basically instantaneous between adjacent objects. But those contact forces aren't fundamental; they're due to the underlying physics of atoms and the internals of atoms, which are ultimately due to quantum mechanics and mechanisms of momentum transfer that are quite different.

Would you agree that the only actual instantaneous transfer of momentum does happen on the level of atoms not between macroscopic segments which each require a finite propagation time interval ?
Doesn't any discussion of the propagation of momentum neccessitate going down to the phonon level? Not in specifics but in principles.

We've been discussing how we would deal relativistically with the motion of macroscopic objects that might have internal parts, but for which all the forces we're interested in can be modeled in the simple ways we've been modeling them. That let's us concentrate on the aspects that are due to relativity itself, rather than to other things like the material properties of objects. But that means we won't be able to answer some questions that are natural to ask, and still remain within those boundaries.

We really have several different questions going which itself leads to confusion.
1) The fundamental propagation and how it relates to the initial stage of application and attaining equilibrium.
2) the actually more important question of the sustained period after the intial stage and how they evolve. For this question I agree that simpler models will suffice.
3) A peripheral question of the absolute application of the conservation of momentum which doesn't neccessarily add anything significant to the central enquiry but has come and up begs resolution.

Originally Posted by Austin0
If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .


This general picture would be the same if we took a bulk material and looked at the motion of its atoms; the final velocity of the object as a whole would still be the original momentum imparted by the impulse force, divided by the total mass of the object. If the original impulse force were applied to a single atom, then of course the initial velocity of that atom would be much higher than the final velocity--but the initial *momentum* would be the same as the final momentum.



Originally Posted by Austin0
This appears to be entering a desired end result , the conclusion , directly into the deductive chain.

Then you haven't followed the argument correctly. Here's the deductive chain:

(1) For an object to undergo steady-state, equilibrium motion, in which all internal oscillations have been damped out, there must be no relative motion between the internal parts of the object, as viewed from the MCIF of each part at any event along the part's worldline.

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

(3) Therefore, any state of motion of an accelerated object which is a steady-state equilibrium, with no internal oscillations, as described above, must be a state of Born rigid acceleration.

Right here in number 2 is the conclusion stated as an argument.

The reason I introduced the condition of steady-state equilibrium is that that was the kind of state of motion we were interested in, and it's only that condition that leads to Born rigid acceleration as the final state. That doesn't mean that *every* state of motion of an accelerated object must be a state of Born rigid acceleration; there are at least two categories of states for which that isn't true:

Here again

(

ii) States in which the object is not tending towards a steady-state equilibrium--for example, the scenario I discussed a number of posts ago, in which the acceleration of the front of the ship is such that the ship keeps stretching until it breaks. I'm sure we could come up with other such scenarios

.

Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.



Originally Posted by Austin0
Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??

There would be a gradient of compression; it would be greater at the bottom of the spring and less at the top, just as the water pressure in the ocean is greater the deeper you go, or air pressure at sea level is greater than at a high altitude.

You are quite right. I was failing to consider the weight of the spring itself. There would neccessarily be some gradient even in a vertical spring here on Earth without any weight.

Originally Posted by Austin0
But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?

My point is just that in relativity, when you calculate total energy, you're already including kinetic energy; you don't have to worry about it separately. And in virtually all problems where relativistic effects are significant, total energy is both easier and more useful to calculate.

I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.


Originally Posted by Austin0
One last quickie. If there was a long line of spheres as before with an impacting sphere:
A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.
or
B) No matter how many spheres were added to the line the last sphere would undergo "some" translation.
So A) or B) ?

Good, we've gotten to the spheres, which was what I wanted to discuss next. :-)

The quick answer is B. Can you see why from what I've posted already?

Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.

Getting on to the previous spheres would be fantastic. :-) Thanks

 

[PeterDonis]

Austin0:

I am viewing the question of applied momentum as a complete continuum, starting from zero and progressing through the range of magnitudes and conditions with different responses varying through that range.
You view it as an absolute linear response regardless of magnitude or internal considerations. ALthough I would agree that the conservation of mass and energy can be viewed in this way, I do not see how this could possibly apply to momentum unless the 2nd law of thermodynamics was not valid.
Or the fundamental conservation of mass and energy for that matter.
For the math , and your interpretation from the math, posits that a specific amount of energy could enter a system and increase internal kinetic energy and at the same time be 100% translated into system translational momentum without consideration of the internal characteristics and composition of that system.
I just don't see how that could work.

I think the above is really asking the same question as the following:

Are you now saying that directed momentum entering a system cannot result in internal dissipation?

No; I'm saying that directed momentum must result in directed momentum. :-)

Suppose I start with an object of rest mass M, at rest. I add some 4-momentum P to it; to keep it simple, say that in the object's original rest frame, P has only a spatial component in the x-direction, p_x. Then there are two possible results:

(1) No dissipation; then the object's final rest mass will be the same as its initial rest mass, M, and its final velocity will be (if the momentum p_x is small enough that the non-relativistic approximation is valid) v = p_x / M.

(2) With dissipation; then the object's final rest mass will be M' > M, and its final velocity will be (in the non-relativistic approximation) v' = p_x / M' < v. In other words, it will have the same final momentum, but its final velocity will be smaller because some of the applied energy went into its rest mass instead.

In either case, the object will have *some* coordinate velocity in the x-direction. There is no way to add a directed momentum p_x to the object and *not* have it end up with *some* coordinate velocity in the x-direction. Of course the "no dissipation" case is an idealization; no real object will have zero dissipation. But there are real cases where the dissipation is small enough that it can be ignored for the accuracy required for the specific problem. Also, of course, I'm ignoring the fact that a real object that was heated up by dissipation would then radiate that heat, which would further change its 4-momentum (since it would lose rest mass)--but then we would also have to account for the 4-momentum carried away by the radiation, in order to balance the books properly.

Does this address what you were thinking in the longer quote above? If not, please say what else isn't clear.

This quote does not relate to my post at all but is addressed to a different stage where the momentum has progressed farther. See your quote following.

The quote following was my comment that overall motion of a system is motion of its center of mass. But if any part of a system is moving, then its center of mass must be moving. The center of mass is not a physical object; it's an abstraction which is determined by the average location of all parts of the system. Saying that the center of mass is moving is *not* the same thing as saying that the particular physical segment of the object which happened to be located at the center of mass before it started moving, is moving. The latter seems to be what you are talking about, but it's not what I was talking about. Once again, though, if you're not comfortable with my definitions of "global" terms, we can just leave them out and talk about the motions of each individual part of the system. We'll end up with the same answers.

Would you agree that the only actual instantaneous transfer of momentum does happen on the level of atoms not between macroscopic segments which each require a finite propagation time interval ?

No; I would say the opposite. If you have a model that includes instantaneous transfer of momentum, it's because your model is not fundamental; it's at a level of detail that's not sufficient to capture the finite propagation times that are always present at the level of fundamental particles. It's at the level of fundamental particles and interactions that propagation times will always appear in one form or another.

Doesn't any discussion of the propagation of momentum neccessitate going down to the phonon level? Not in specifics but in principles.

In principle, if you want to understand how momentum propagates at a fundamental level, you have to look at quantum fields and interactions. (Phonons are quanta of vibration, but they're not really fundamental the way that, say, electrons and photons are.) At that level, momentum is transferred between real particles by the exchange of virtual particles; for example, the nucleus of an atom "holds" the electrons in certain orbitals by exchanging virtual photons with them in order to keep their momentum within certain limits. (Even this picture is arguably not fully fundamental, since it depends on perturbation theory, which is only an approximate solution to the equations of quantum field theory--but I don't want to get into that here.) This virtual photon exchange looks to us, at a macroscopic level, like an electromagnetic force between the nucleus and the electrons. Virtual particle exchanges take some amount of time--roughly speaking, that time is determined by the uncertainty principle combined with the magnitude of the momentum and energy being exchanged. But that time is so short that it can be ignored for many problems, and the force between the nucleus and the electrons can be treated as instantaneous, so that the atom acts like a single object.

Similarly, at the quantum level, interactions between atoms involve exchanges of virtual photons between the electrons in the atoms' outer shells, which take time. It's only in macroscopic models, when we don't care about the details of the interatomic forces, and we're working on time scales much larger than the interaction time for virtual particle exchanges, that we idealize the interatomic forces as instantaneous "contact forces" between atoms that have a definite size and collide like little billiard balls.

Right here in number 2 is the conclusion stated as an argument.

My premise 2 is a proposition that's already been proved by other means; it's not an assumption. You're correct that it's the crucial premise that leads to my conclusion, but that doesn't mean I'm arguing in a circle; it means my conclusion is correct.

Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.

Well, I specified exactly the scenario that would lead to it--that the *acceleration* experienced at the front and rear ends of the ship was the same. I explained how that was a *different* specification than saying that the *rocket thrust* at each end of the ship was constant and equal; the acceleration experienced at either end of the ship depends on the *net* force at that end, which is the resultant of the rocket thrust and any internal forces exerted on the end by other parts of the ship. I admitted that the specification of constant acceleration, rather than constant rocket thrust, was extremely unlikely to be realized by any real rocket (since it would require continually increasing rocket thrust at the front end of the ship), but unlikely is not the same as physically impossible.

It is true that I didn't make any specification of *how much* the ship would have to stretch before it broke, but I don't have to to know that it would eventually have to break with the specification I gave (constant acceleration of each end). For the ship not to break eventually, it would have to have infinite breaking strength, and that *is* physically impossible--not because values tending to infinity are impossible (after all, the scenario posits rocket thrust at the front end of the ship that must tend to infinity) but because relativity places finite limits on the breaking strength of materials. This condition is called the "weak energy condition", and it says that the stress in a material can't exceed its rest energy density (its rest energy divided by its volume in its rest frame). Maybe this is the additional physical principle you were looking for.

In any case, that specific scenario is not very important for our discussion; I brought it in only to show that it is possible to construct scenarios that do not tend to a steady-state equilibrium. Such scenarios may be extremely unlikely; that's fine.

I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.

Exactly; there is. Simple example: an object of rest mass m moving with velocity \beta (in units such that the speed of light = 1). The total energy of the object is E = \gamma m, where \gamma = 1 / \sqrt{1 - \beta^2}. Simple and direct. There is no simple and direct formula for the relativistic kinetic energy of the object; to obtain that, I have to subtract its rest mass m from its total energy E and call what's left over "kinetic energy". Take any other relativistic problem and you'll find the same thing: there will be a simple and direct formula for the total energy, but to get the kinetic energy, you'll have to subtract the rest energy from the total energy; there won't be a simple and direct formula for the kinetic energy alone.

Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.

I'm not sure either of these is relevant to the specific example where I gave the quick answer B. The reason B was the obvious quick answer is simple: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

Getting on to the previous spheres would be fantastic. :-)

OK, let's suppose now that spheres #1 through #N are in contact, each with its adjacent neighbors, and they all lie along the x-direction. We then hit sphere #1, from the left, with another sphere, sphere #0, which starts out with a certain momentum directed to the right--i.e., in the positive x-direction. (I'm leaving out the part about the spheres being suspended, which was in your original specification--I'm assuming the spheres to start out at rest, far out in empty space, with no other objects near enough to affect their motion.) Macroscopically, it looks like the impulse passes all the way through the row of spheres with no motion, until it finally "comes out" as motion at the other end, with sphere #N taking off while the rest of the spheres stay put. However, if we were to photograph the process with high-speed cameras, with enough spatial resolution to see deformations in the individual spheres, we would see this:

(1) Sphere #0 hits the left side of sphere #1. As a result, sphere #1 deforms--its left side is pushed inward, so there is now a net motion to the right of sphere #1. We'll assume, for simplicity, that sphere #0's motion stops completely as a result of its hitting sphere #1. Thus, sphere #1 now contains the momentum that was contained before in sphere #0. This will cause the center of mass of sphere #1 to shift to the right.

(2) Since sphere #1 is in contact with sphere #2, as soon as the wave of deformation has time to pass across sphere #1, and hit sphere #2, sphere #2 deforms, with its left side being pushed inward. If everything is tuned just right, sphere #2 deforming will coincide with sphere #1 "un-deforming"--the right side of sphere #1 will now push back on the rest of sphere #1, and sphere #1 will restore itself to its original shape, and its center of mass will shift back to the left, to where it was at the start of the process. Sphere #2 then contains the momentum transferred by the initial impulse.

(N) This process continues down the line until sphere #N is reached. Since this sphere has no neighbor to the right, once the wave of deformation starts at its left side (and sphere #(N-1) is pushed back into its original shape and position), there is nothing to keep all of sphere #N from continuing the motion. So now we see, macroscopically, that all the momentum of the initial impulse appears as motion of sphere #N to the right.

In other words, each sphere in succession does move to the right, but all of the motions except for the last one are reversed in the course of propagating the wave of deformation down the row of spheres. In a real case, things wouldn't be as "clean" as I've presented them, meaning that the momentum of the intial impulse would probably be distributed among two or more of the spheres during the process, though it would all come out contained in sphere #N at the end (if everything is tuned just right). But at any instant of time, viewed from the original rest frame of the spheres, the total momentum of the system as a whole--the sum of the net momentum contained in all the spheres, which must be directed to the right--will be equal to the initial impulse.

 

[Austin0]

The quote following was my comment that overall motion of a system is motion of its center of mass. But if any part of a system is moving, then its center of mass must be moving. The center of mass is not a physical object; it's an abstraction which is determined by the average location of all parts of the system. Saying that the center of mass is moving is *not* the same thing as saying that the particular physical segment of the object which happened to be located at the center of mass before it started moving, is moving. The latter seems to be what you are talking about, but it's not what I was talking about. Once again, though, if you're not comfortable with my definitions of "global" terms, we can just leave them out and talk about the motions of each individual part of the system. We'll end up with the same answers.

Of course the center of mass is an abstraction. A coordinate location.
But it is frame independant yes?? ANd it does correspond to a specific physical location within a system.
So saying the center of mass is moving, has a specific meaning of motion wrt an outside inertial frame. The center does not move relative to a comoving frame true?

So you have agreed that there can be no motion at a point in a system until momentum has reached it. YOu have proposed that there is actual instantaneous coordinate motion at the origin of applied momentum.
Now you are saying "But if any part of a system is moving, then its center of mass must be moving."
How is this consistent?

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

My premise 2 is a proposition that's already been proved by other means; it's not an assumption. You're correct that it's the crucial premise that leads to my conclusion, but that doesn't mean I'm arguing in a circle; it means my conclusion is correct.

Certainly it has not been mentioned in this discussion let alone proved.
At this point we are talking about the basic physics of acceleration as applied to some models of propulsion.

Originally Posted by Austin0
Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.

Well, I specified exactly the scenario that would lead to it--that the *acceleration* experienced at the front and rear ends of the ship was the same.

equal acceleration does not, in itself, constitute either an explanation or a proof of expansion to the point of disruption.

I admitted that the specification of constant acceleration, rather than constant rocket thrust, was extremely unlikely to be realized by any real rocket (since it would require continually increasing rocket thrust at the front end of the ship)

The requirement of increasing thrust at the front to maintain constant acceleration is also not demonstrated at this point.
ANd you never did answer when I asked if Born acceleration requires increasing thrust at all points.

It is true that I didn't make any specification of *how much* the ship would have to stretch before it broke

,
We have agreed that specific quantitative details may be disregarded for the purposes of our inquiry.

Originally Posted by Austin0
I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.

Exactly; there is. Simple example: an object of rest mass m moving with velocity \beta (in units such that the speed of light = 1). The total energy of the object is E = \gamma m, where \gamma = 1 / \sqrt{1 - \beta^2}. Simple and direct. There is no simple and direct formula for the relativistic kinetic energy of the object; to obtain that, I have to subtract its rest mass m from its total energy E and call what's left over "kinetic energy". Take any other relativistic problem and you'll find the same thing: there will be a simple and direct formula for the total energy, but to get the kinetic energy, you'll have to subtract the rest energy from the total energy; there won't be a simple and direct formula for the kinetic energy alone.

As far as I can see using this formula there will always be zero kinetic energy outside of the acquired momentum from acceleration. Of course there is no simple way of calculating internal kinetic energy. But we have agreed that for our purposes there is no need for actual quantitative values.
But an actual calculation of either rest mass or E in the foregoing equation would seem to neccessitate some non simple and direct calculation in order to include internal kinetic energy from dissipation.

I'm not sure either of these is relevant to the specific example where I gave the quick answer B. The reason B was the obvious quick answer is simple: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

On collision between sp/#1 and #2 there is an internal sound. A literal coherent sound wave the propagates both ways into #1 and #2 We know that these waves will resonate and reflect throughout the spheres. We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally. SO (sph #1) p > (#2) p or conversely
(n+1) p < (n) p So unless you propose some kind of Zenoesque infinite divisibility there will be an infintesimal but steady attenuation of the magnitude of momentum until eventual total diminishment yes?

OK, let's suppose now that spheres #1 through #N are in contact, each with its adjacent neighbors, and they all lie along the x-direction. We then hit sphere #1, from the left, with another sphere, sphere #0, which starts out with a certain momentum directed to the right--i.e., in the positive x-direction. (I'm leaving out the part about the spheres being suspended, which was in your original specification--I'm assuming the spheres to start out at rest, far out in empty space, with no other objects near enough to affect their motion.) Macroscopically, it looks like the impulse passes all the way through the row of spheres with no motion, until it finally "comes out" as motion at the other end, with sphere #N taking off while the rest of the spheres stay put. However, if we were to photograph the process with high-speed cameras, with enough spatial resolution to see deformations in the individual spheres, we would see this:

(1) Sphere #0 hits the left side of sphere #1. As a result, sphere #1 deforms--its left side is pushed inward, so there is now a net motion to the right of sphere #1. We'll assume, for simplicity, that sphere #0's motion stops completely as a result of its hitting sphere #1. Thus, sphere #1 now contains the momentum that was contained before in sphere #0. This will cause the center of mass of sphere #1 to shift to the right.

(2) Since sphere #1 is in contact with sphere #2, as soon as the wave of deformation has time to pass across sphere #1, and hit sphere #2, sphere #2 deforms, with its left side being pushed inward. If everything is tuned just right, sphere #2 deforming will coincide with sphere #1 "un-deforming"--the right side of sphere #1 will now push back on the rest of sphere #1, and sphere #1 will restore itself to its original shape, and its center of mass will shift back to the left, to where it was at the start of the process. Sphere #2 then contains the momentum transferred by the initial impulse.

(N) This process continues down the line until sphere #N is reached. Since this sphere has no neighbor to the right, once the wave of deformation starts at its left side (and sphere #(N-1) is pushed back into its original shape and position), there is nothing to keep all of sphere #N from continuing the motion. So now we see, macroscopically, that all the momentum of the initial impulse appears as motion of sphere #N to the right.

In other words, each sphere in succession does move to the right, but all of the motions except for the last one are reversed in the course of propagating the wave of deformation down the row of spheres. In a real case, things wouldn't be as "clean" as I've presented them, meaning that the momentum of the intial impulse would probably be distributed among two or more of the spheres during the process, though it would all come out contained in sphere #N at the end (if everything is tuned just right). But at any instant of time, viewed from the original rest frame of the spheres, the total momentum of the system as a whole--the sum of the net momentum contained in all the spheres, which must be directed to the right--will be equal to the initial impulse.[/QUOTE]

How could the energy be spread out through two or more of the spheres when it is clear that at the instant it reaches the end of the final sphere, that sphere will definitely move and break contact ending the transfer of momentum??
SO do you think that if we assume ideal crystal spheres at 0 deg. K , as close to rigid as theoretically possible that there would be deformation or that anything significant would be different as far as the propagation of momentum?
YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

 

[PeterDonis]

Of course the center of mass is an abstraction. A coordinate location.
But it is frame independant yes?? ANd it does correspond to a specific physical location within a system.
So saying the center of mass is moving, has a specific meaning of motion wrt an outside inertial frame. The center does not move relative to a comoving frame true?

The center of mass (CoM, to save typing) corresponds to a *coordinate* location within a system. Which particular physical part of the system, if any, is located at that coordinate location can change. The CoM is not "anchored" to any particular physical part of the system. It's just wherever the "average" location of all the parts of the object happens to be.

Whether or not the CoM is moving is frame dependent. You're correct that the CoM does not move in a comoving frame, if "comoving" is defined properly; but a "comoving" frame in this sense will not be an inertial frame, if there are external forces acting on the object.

The Wikipedia pages on http://en.wikipedia.org/wiki/Center_of_mass" have good (if brief) discussions of this stuff. However, they don't discuss what happens if the object is being acted on by an external force and so is being accelerated.

So you have agreed that there can be no motion at a point in a system until momentum has reached it. YOu have proposed that there is actual instantaneous coordinate motion at the origin of applied momentum.
Now you are saying "But if any part of a system is moving, then its center of mass must be moving."
How is this consistent?

As above, the CoM isn't "anchored" to any particular part of the object. It's just the "average" location of all the parts. It's not necessary that all the parts of the system are moving for the "average" location of the parts to be moving; just one part moving is enough.

Certainly it has not been mentioned in this discussion let alone proved.

The fact that Born rigid acceleration is the only state of motion of an accelerated object for which the internal distances between the parts remain constant has been mentioned several times in this thread, not just by me. We haven't proven it here, but there are plenty of discussions of this fact and how it comes about online and in textbooks. It's not a new idea; Born came up with the concept in, I believe, the 1920's. For a good quick reference to start with, try these online discussions of Rindler coordinates: http://en.wikipedia.org/wiki/Rindler_coordinates" (I think I've already posted these links earlier in this thread).

equal acceleration does not, in itself, constitute either an explanation or a proof of expansion to the point of disruption.

Once again, this has already been discussed in this thread. We discussed the motion of two separate spaceships with equal, constant acceleration, as in the "[URL spaceship paradox[/URL], and confirmed that their separation would continuously increase. Then I specified the motions of two ends of a single rocket ship in such a way that the worldlines of the two ends would be identical to those of the two separate spaceships. I admitted that my specification was extremely unlikely to occur in practice, but that doesn't make it physically impossible.

ANd you never did answer when I asked if Born acceleration requires increasing thrust at all points.

Not directly, but I did implicitly when I described the scenario that would produce Born rigid acceleration. I'll describe it again: Let there be a rocket ship that starts at rest along the x-axis of an inertial frame. At time t = 0 in that frame, let each individual segment of the ship start accelerating, with an acceleration equal to c^2 / x_s, where x_s is the x-coordinate of that segment in the original inertial frame. Then the ship as a whole will undergo Born rigid acceleration.

This specification, in itself, doesn't tell exactly what thrust will be required at each individual segment to produce the specified acceleration. However, if we assume that there are no internal stresses to start with in the ship, then, since Born rigid acceleration preserves the distances between adjacent segments, it should not produce any internal stresses. So the thrust required for each segment should simply be the rest mass of the segment times its specified acceleration. This does depend, though, on our assumption about the material properties of the ship being correct.

Also, the above specification is what would be required for an object to *start out* with Born rigid acceleration, and of course, as has been said before in this thread, it's extremely unlikely that it would happen that way in any real case. However, suppose an object starts out with some other "force profile" (for example, thrust only at the rear end), but then, as a result of the development of internal forces, reaches a state such that the net force on each segment of the object is exactly what it needs to be to cause all the segments to accelerate according to the same profile I described above. At that point, the object as a whole would have settled into a state of Born rigid acceleration; it's this kind of scenario I was talking about in previous posts. Of course in this case the internal stresses in the object would be non-zero, since the internal forces would be supplying part of the net force on each segment to make the force profile just right. But the internal stresses would not *change* any further once this state was reached; it would still be a stable steady-state equilibrium, just as an object resting on the surface of the Earth is in a stable steady-state equilibrium even though the stresses inside it are non-zero because of gravity.

We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally.

No, we do *not* know this, at least I don't. One more time: **this would violate conservation of momentum**. Conservation of momentum requires that the directed momentum before the collision must be equal to the directed momentum after the collision. So we must have (for the case where only sphere #1 is moving before the collision but both spheres #1 and #2 may be moving after the collision):

p_{1-before} = p_{1-after} + p_{2-after}

From this, the rest of what I said follows. Directed momentum simply cannot be "dissipated internally" like you're saying.

How could the energy be spread out through two or more of the spheres when it is clear that at the instant it reaches the end of the final sphere, that sphere will definitely move and break contact ending the transfer of momentum??

It could be spread between two or more spheres *during the process*; I'm just allowing for this possibility since I haven't done a detailed calculation to show microscopically how the momentum propagates. After the process ends, it will all be in the final sphere.

SO do you think that if we assume ideal crystal spheres at 0 deg. K , as close to rigid as theoretically possible that there would be deformation or that anything significant would be different as far as the propagation of momentum?

Assuming that the material properties of the spheres remain intact during the process (i.e., none of the spheres shatter, melt, fragment, etc.), and that those properties support elastic deformation (i.e., spheres made of clay or something like that wouldn't work), then there would be deformation as I've described it. The temperature of the spheres doesn't matter (as long as it's below the melting point).

YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

I'm proposing no such thing. I've said several times that any real object would have dissipation, and dissipation means that some of the *energy* added to the object gets turned into heat--i.e., it appears as an increase in the object's rest mass instead of an increase in the object's velocity. That is the sort of thing the 2nd law of TD deals with. But none of that changes the fact that *directed momentum* is conserved. I've already shown how that can happen, but one more time: consider an object of mass M, initially at rest, which then has some linear momentum p_x added to it:

(1) In the idealized case where there's no dissipation, the object's rest mass is unchanged, its linear momentum is p_x, and its final velocity (in the non-relativistic limit) will be v = p_x / M.

(2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x.

 

[PeterDonis]

Oops, forgot to include a comment on this:

As far as I can see using this formula there will always be zero kinetic energy outside of the acquired momentum from acceleration. Of course there is no simple way of calculating internal kinetic energy. But we have agreed that for our purposes there is no need for actual quantitative values.
But an actual calculation of either rest mass or E in the foregoing equation would seem to neccessitate some non simple and direct calculation in order to include internal kinetic energy from dissipation.

You call it "internal kinetic energy" but if it appears in the object's rest mass, it's not "kinetic energy" as far as that level of modeling is concerned. Of course all sorts of things that contribute to rest mass in a macroscopic model, like internal vibrations of molecules, will be treated as kinetic energy in a microscopic model. But if molecular vibrations, for example, are treated as kinetic energy in the microscopic model, they don't appear in the rest mass of the molecules or atoms; the model will explicitly include the motion of the molecules or atoms, and will track the explicit coordinates and velocities of those molecules or atoms, so that the energy in the vibrations appears as "external" kinetic energy, just like in the formula I gave.

For a macroscopic model, none of this is necessary; all you need is some kind of coefficient to tell you how much dissipation there is--i.e., how much of the energy applied to the object goes into increasing its rest mass. The viscosity of a fluid is an example of such a coefficient. In macroscopic models these coefficients are just intrinsic properties of materials, measured empirically; the underlying microscopic details are simply ignored.

 

[Austin0]

=PeterDonis;2395420]The center of mass (CoM, to save typing) corresponds to a *coordinate* location within a system. Which particular physical part of the system, if any, is located at that coordinate location can change. The CoM is not "anchored" to any particular physical part of the system. It's just wherever the "average" location of all the parts of the object happens to be.

If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

As above, the CoM isn't "anchored" to any particular part of the object. It's just the "average" location of all the parts. It's not necessary that all the parts of the system are moving for the "average" location of the parts to be moving; just one part moving is enough.

As above

The fact that Born rigid acceleration is the only state of motion of an accelerated object for which the internal distances between the parts remain constant has been mentioned several times in this thread, not just by me. We haven't proven it here, but there are plenty of discussions of this fact and how it comes about online and in textbooks.

I am somewhat familiar with the available info on the web regarding both Born and Rindler.
I have encountered descriptions in both cases but nothing in the way of explicit proof. I am also unaware of any empirical tests validating either ,,, have I missed something?

Once again, this has already been discussed in this thread. We discussed the motion of two separate spaceships with equal, constant acceleration, as in the "[URL spaceship paradox[/URL], and confirmed that their separation would continuously increase. Then I specified the motions of two ends of a single rocket ship in such a way that the worldlines of the two ends would be identical to those of the two separate spaceships

.

You specified. Does this mean that the physics neccessarily conforms to your specifications?
That a single ship would neccessarily conform to the identical world lines??
That this justifies an automatic assumption that a single extended ship is absolutely the same as two separate ships and there is no consequence derived from the matter connecting the ends?

Not directly, but I did implicitly when I described the scenario that would produce Born rigid acceleration. I'll describe it again: Let there be a rocket ship that starts at rest along the x-axis of an inertial frame. At time t = 0 in that frame, let each individual segment of the ship start accelerating, with an acceleration equal to c^2 / x_s, where x_s is the x-coordinate of that segment in the original inertial frame. Then the ship as a whole will undergo Born rigid acceleration.

This specification, in itself, doesn't tell exactly what thrust will be required at each individual segment to produce the specified acceleration. However, if we assume that there are no internal stresses to start with in the ship, then, since Born rigid acceleration preserves the distances between adjacent segments, it should not produce any internal stresses. So the thrust required for each segment should simply be the rest mass of the segment times its specified acceleration. This does depend, though, on our assumption about the material properties of the ship being correct.

Also, the above specification is what would be required for an object to *start out* with Born rigid acceleration, and of course, as has been said before in this thread, it's extremely unlikely that it would happen that way in any real case. However, suppose an object starts out with some other "force profile" (for example, thrust only at the rear end), but then, as a result of the development of internal forces, reaches a state such that the net force on each segment of the object is exactly what it needs to be to cause all the segments to accelerate according to the same profile I described above. At that point, the object as a whole would have settled into a state of Born rigid acceleration; it's this kind of scenario I was talking about in previous posts. Of course in this case the internal stresses in the object would be non-zero, since the internal forces would be supplying part of the net force on each segment to make the force profile just right. But the internal stresses would not *change* any further once this state was reached; it would still be a stable steady-state equilibrium, just as an object resting on the surface of the Earth is in a stable steady-state equilibrium even though the stresses inside it are non-zero because of gravity.

From this I understand that you are saying the the thrust would be different at different locales but would remain constant and not increase over time with Born acceleration. Correct?
It also appears you are saying it is not impossible that simple rear thrust could result in essentially Born acceleration or am I misreading you here?

: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

Originally Posted by Austin0
Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.



Originally Posted by Austin0
SO (sph #1) p > (#2) p or conversely
(n+1) p < (n) p So unless you propose some kind of Zenoesque infinite divisibility there will be an infintesimal but steady attenuation of the magnitude of momentum until eventual total diminishment yes?



Originally Posted by Austin0
We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally.

No, we do *not* know this, at least I don't. One more time: **this would violate conservation of momentum**. Conservation of momentum requires that the directed momentum before the collision must be equal to the directed momentum after the collision. So we must have (for the case where only sphere #1 is moving before the collision but both spheres #1 and #2 may be moving after the collision):

p_{1-before} = p_{1-after} + p_{2-after}-

From this, the rest of what I said follows. Directed momentum simply cannot be "dissipated internally" like you're saying.

Total agreement here- p_{1-before} = p_{1-after} + p_{2-after}-as I have explicitly stated before. ANd this continues to hold for

p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}

I have a couple of responses to this.

#1

Original PeterDonis (2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x

From this it follows that: v_{1-before} &gt; v_{2-after}-

ANd through extrapolation v_{1-before} &gt; v_{2-after}> v_{3-after}<br /> <br /> &gt; v_{4-after}> v_{5-after} &gt; v_{6-after}....>v_{N-after}= 0<br />

Unless there is some kind of infinite divisibility

COnsidering Sphere N with a v = 0 p_{1-before} = p_{N-after}

only if the increase in inertial mass of N was equivalent to the energy of v_{1-before}

or equivalent to the total increase in inertial mass distributed among all the previous spheres.

Clearly this could not be true in either case so logically
so it would seem to follow that:

p_{1-before} &gt; p_{N-after} And as there is an infintesimal incremental continuum between p_{1-before} and p_{N-after}

it follows that :

p_{1-before} &gt; p_{2-after} &gt; p_{3-after}&gt;,,,,,,,,,,&gt; p_{N-after}

ANd that there is no conflict between A) and B)

A) p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}

B) p_{1-before} &gt; p_{2-after} &gt; p_{3-after}&gt;,,,,,,,,,,&gt; p_{N-after}

That in both descriptions the total mass/energy content of the initial momentum is simply distributed among a large number of spheres with a steady decrease in velocity.





Originally Posted by Austin0
YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

I'm proposing no such thing. I've said several times that any real object would have dissipation, and dissipation means that some of the *energy* added to the object gets turned into heat--i.e., it appears as an increase in the object's rest mass instead of an increase in the object's velocity. That is the sort of thing the 2nd law of TD deals with. But none of that changes the fact that *directed momentum* is conserved. I've already shown how that can happen, but one more time: consider an object of mass M, initially at rest, which then has some linear momentum p_x added to it:

(1) In the idealized case where there's no dissipation, the object's rest mass is unchanged, its linear momentum is p_x, and its final velocity (in the non-relativistic limit) will be v = p_x / M.

(2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x.

Is there really any meaningful difference between the cases we are considering?

Propagation between a series of spatially separated interactions,
Propagation between a series of contact interactions
Propagation between atoms in a matrix
Or propagation of sound , which in the final analysis is just momentum.
SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??

Thanks

 

[PeterDonis]

Austin0:

If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

Why would motion be irrelevant unless it moved mass from one side of the line to the other? That condition doesn't correspond to any meaningful physical quantity that I'm aware of. For example, I can move parts of the object around without violating your condition and still change the location of the object's CoM, or the object's moment of inertia about an axis, etc. What meaningful physical invariant corresponds to the condition you're specifying?

I am somewhat familiar with the available info on the web regarding both Born and Rindler.
I have encountered descriptions in both cases but nothing in the way of explicit proof. I am also unaware of any empirical tests validating either ,,, have I missed something?

The specifications for Born rigid acceleration and Rindler coordinates are constructed in such a way that the distances between internal parts of the object remain constant *by construction*. If the references you've read don't explicitly prove, from the construction, that the distances remain constant, that's probably because the writers considered it sufficiently obvious that they could leave it to the reader to fill in the details. But if you want me to explicate the proof, here it is:

Let there be a ship which starts out at rest in an inertial frame (x, t), lying along the x-axis; each segment of the ship starts out at a given x-coordinate x_s. At time t = 0 in this original inertial frame, each segment starts accelerating as I specified previously for Born rigid acceleration. Then the worldline of each segment of the ship, in the coordinates of the original rest frame, will be

x^2 - t^2 = x_s^2

where I'm using units such that the speed of light is 1. (Rindler coordinates are constructed so that these worldlines, which look like hyperbolas in ordinary Minkowski coordinates, look like straight lines, but the invariant properties of the worldlines themselves are the same either way.) Let's re-write this equation in terms of x as a function of t:

x = \sqrt{t^2 + x_s^2}

Then the coordinate velocity of the segment, at a given event (t, x), will be

v = \frac{dx}{dt} = \frac{t}{\sqrt{t^2 + x_s^2}} = \frac{t}{x}

And the relativistic gamma factor will be

\gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - \frac{t^2}{x^2}}} = \sqrt{\frac{x^2}{x^2 - t^2}} = \frac{x}{x_s}

Now, the equation for coordinate velocity tells us that any line through the origin of the (x, t) frame, x = 0, t = 0 (these lines are the lines of constant q in Greg Egan's diagram), will be a line of simultaneity for any segment, at the point where the line intersects that segment's worldline. Thus, spacelike intervals along such a line will correspond to distances as measured by observers moving with each segment. In particular, given a point (x, t) along a segment's worldline, the proper distance from that point to the point x = 0, t = 0 is given by the Lorentz transformation:

x&#039; = \gamma \left( x - v t \right)

Substituting for v and gamma, we have

x&#039; = \frac{x}{x_s} \left( x - \frac{t^2}{x} \right) = \frac{1}{x_s} \left( x^2 - t^2 \right) = \frac{x_s^2}{x_s} = x_s

In other words, as Greg Egan notes in his discussion, each segment maintains a constant proper distance from the event x = 0, t = 0. That means that the proper distance between any two segments must also remain constant, since each maintains a constant proper distance from a fixed point.

I don't know that anyone has done a direct experiment where an object was put under Born rigid acceleration; such an experiment would be extremely difficult in practice. However, since everything I've said about Born rigid acceleration is a simple consequence of the laws of SR, and the laws of SR have been experimentally verified, I'm not too worried about my statements about Born rigid acceleration being correct.

You specified. Does this mean that the physics neccessarily conforms to your specifications?
That a single ship would neccessarily conform to the identical world lines??
That this justifies an automatic assumption that a single extended ship is absolutely the same as two separate ships and there is no consequence derived from the matter connecting the ends?

My specification is physically possible, as I've said several times. That's all that's required to consider its implications. In other words: there's nothing in the laws of physics that *prevents* the front and rear ends of a ship from following the worldlines I've specified, at least until the ship breaks from the stretching. It might be highly unlikely in practice for the front and rear ends of a ship to follow the worldlines I've specified; I've already said several times that for it to happen in practice, the rocket thrust at the front end would have to continually increase, to compensate for the increasing internal force of the rest of the ship pulling back on the front end because of the stretching. (Conversely, the rocket thrust at the rear end would have to continually decrease, because the increasing internal force from the rest of the ship due to the stretching would be pulling the rear end forward.) But the laws of physics permit the rocket thrust to behave this way, so my scenario is physically possible.

If it makes you feel any better, consider my scenario as a hypothetical. In other words, all I'm saying is that *if* the front and rear ends of the ship followed the worldlines I specified, the ship would continually stretch until it broke. It's physically possible for the front and rear ends to follow those worldlines, so my hypothetical is not ruled out by the laws of physics; but it's still only a hypothetical. I'm certainly not saying that *any* rocket ship with engines at the front and rear ends *has* to move this way; I'm only exploring the consequences *if* one did so.

From this I understand that you are saying the the thrust would be different at different locales but would remain constant and not increase over time with Born acceleration. Correct?

Correct.

It also appears you are saying it is not impossible that simple rear thrust could result in essentially Born acceleration or am I misreading you here?

You're correct here too, in the sense that simple rear thrust could lead to a steady-state equilibrium in which the object as a whole was undergoing Born rigid acceleration; but in that equilibrium state, there would be internal stresses in all parts of the ship, in addition to the rear thrust.

From this it follows that: v_{1-before} &gt; v_{2-after}

ANd through extrapolation v_{1-before} &gt; ... &gt; v_{N-after} = 0

Unless there is some kind of infinite divisibility

The last "= 0" here is not correct. The velocity will never go to zero; it can't, because that would imply zero momentum in that direction. This does not imply infinite divisibility; it implies that there are a finite number of objects among which the momentum can be shared. In this example, there were N spheres, a finite number; in the limit, we would be counting atoms, or subatomic particles. There are a finite number of those too, and even when the momentum has been shared among every last one of them, the velocity will still be nonzero.

ANd that there is no conflict between A) and B)

A) p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}

B) p_{1-before} &gt; p_{2-after} &gt; p_{3-after}&gt;,,,,,,,,,,&gt; p_{N-after}

That in both descriptions the total mass/energy content of the initial momentum is simply distributed among a large number of spheres with a steady decrease in velocity.

I have no problem with A), since it's just conservation of momentum. I'm not entirely sure about B). I would agree that, in general, p_{1-before} &gt; p_{N-after}, since some of the original momentum is likely to be distributed among other spheres besides sphere #N. But I'm not sure that this implies B). There's no reason why each of the momenta of the intermediate spheres *has* to fit into the "infinitesimal incremental continuum" between p_{1-before} and p_{N-after}; they could equally well be less than p_{N-after}.

Is there really any meaningful difference between the cases we are considering?

Propagation between a series of spatially separated interactions,
Propagation between a series of contact interactions
Propagation between atoms in a matrix
Or propagation of sound , which in the final analysis is just momentum.
SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??

I'm not sure what you mean here by "propagate without limit", but certainly all of the interactions you've described must obey the laws of conservation of momentum and energy. In that sense there's no meaningful difference between them. But the specifics of how the interactions are modeled in physics can vary quite a bit.

 

[Austin0]

Austin0;2397188]If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

=PeterDonis;2398074]Austin0:
Why would motion be irrelevant unless it moved mass from one side of the line to the other? That condition doesn't correspond to any meaningful physical quantity that I'm aware of. For example, I can move parts of the object around without violating your condition and still change the location of the object's CoM, or the object's moment of inertia about an axis, etc. What meaningful physical invariant corresponds to the condition you're specifying?

We are talking about systems accelerating along a linear path. No rotation. No systems moving at angles relative to the direction of motion.
Limited to this context do you think moving mass around before or behind the line would shift the CoM relative to the direction of travel?

The specifications for Born rigid acceleration and Rindler coordinates are constructed in such a way that the distances between internal parts of the object remain constant *by construction*. If the references you've read don't explicitly prove, from the construction, that the distances remain constant, that's probably because the writers considered it sufficiently obvious that they could leave it to the reader to fill in the details. But if you want me to explicate the proof, here it is:

Let there be a ship which starts out at rest in an inertial frame (x, t), lying along the x-axis; each segment of the ship starts out at a given x-coordinate x_s. At time t = 0 in this original inertial frame, each segment starts accelerating as I specified previously for Born rigid acceleration. Then the worldline of each segment of the ship, in the coordinates of the original rest frame, will be

x^2 - t^2 = x_s^2

where I'm using units such that the speed of light is 1. (Rindler coordinates are constructed so that these worldlines, which look like hyperbolas in ordinary Minkowski coordinates, look like straight lines, but the invariant properties of the worldlines themselves are the same either way.) Let's re-write this equation in terms of x as a function of t:

x = \sqrt{t^2 + x_s^2}

Then the coordinate velocity of the segment, at a given event (t, x), will be

v = \frac{dx}{dt} = \frac{t}{\sqrt{t^2 + x_s^2}} = \frac{t}{x}

And the relativistic gamma factor will be

\gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - \frac{t^2}{x^2}}} = \sqrt{\frac{x^2}{x^2 - t^2}} = \frac{x}{x_s}

Now, the equation for coordinate velocity tells us that any line through the origin of the (x, t) frame, x = 0, t = 0 (these lines are the lines of constant q in Greg Egan's diagram), will be a line of simultaneity for any segment, at the point where the line intersects that segment's worldline. Thus, spacelike intervals along such a line will correspond to distances as measured by observers moving with each segment. In particular, given a point (x, t) along a segment's worldline, the proper distance from that point to the point x = 0, t = 0 is given by the Lorentz transformation:

x&#039; = \gamma \left( x - v t \right)

Substituting for v and gamma, we have

x&#039; = \frac{x}{x_s} \left( x - \frac{t^2}{x} \right) = \frac{1}{x_s} \left( x^2 - t^2 \right) = \frac{x_s^2}{x_s} = x_s

In other words, as Greg Egan notes in his discussion, each segment maintains a constant proper distance from the event x = 0, t = 0. That means that the proper distance between any two segments must also remain constant, since each maintains a constant proper distance from a fixed point.

I don't know that anyone has done a direct experiment where an object was put under Born rigid acceleration; such an experiment would be extremely difficult in practice. However, since everything I've said about Born rigid acceleration is a simple consequence of the laws of SR, and the laws of SR have been experimentally verified, I'm not too worried about my statements about Born rigid acceleration being correct.

You have just given a description of Born acceleration. WHich is itself basically a definition of coordinate events.
The events described are exactly those of the essential Lorentz contraction and lead to exactly the same world lines.
SPECIFICALLY: the fundamental postulates propose that internal proper distances remain the same in inertial frames. ANd that the Lorentz contraction is regarded as a purely kinematic effect.

SO on the basis of this and the clock hypothesis, you could plot the worldline of the back of a normally propulsed ship, simply on the basis of constant proper acceleration.
You could then plot the front of the ship on the basis of instantaneous Lorentz contraction
and would then have a pair of worldlines that were indistinguishable from Born worldlines
Correct?

Based on these worldlines it would inevitably appear as if there had to be unequal acceleration at the front and the back. But this does not neccessarily have any physical meaning. It is simply a result of the conventions of Minkowski space and will have to be this way even if there is NO PHYSICAL cause of the contraction at all, but it is simply a kinematic effect taking place purely in the relative observation frame.
So aside from the description, the Born hypothesis is making a statement about the nature of the Lorentz contraction and is proposing that it will not take place through equal acceleration throughout a system , which will result in the opposite result . Catastrophic expansion.
It also seems to place a meaning to lines of simultaneity that I don't quite understand.

My specification is physically possible, as I've said several times. That's all that's required to consider its implications. In other words: there's nothing in the laws of physics that *prevents* the front and rear ends of a ship from following the worldlines I've specified, at least until the ship breaks from the stretching. It might be highly unlikely in practice for the front and rear ends of a ship to follow the worldlines I've specified; I've already said several times that for it to happen in practice, the rocket thrust at the front end would have to continually increase, to compensate for the increasing internal force of the rest of the ship pulling back on the front end because of the stretching. (Conversely, the rocket thrust at the rear end would have to continually decrease, because the increasing internal force from the rest of the ship due to the stretching would be pulling the rear end forward.) But the laws of physics permit the rocket thrust to behave this way, so my scenario is physically possible.

If it makes you feel any better, consider my scenario as a hypothetical. In other words, all I'm saying is that *if* the front and rear ends of the ship followed the worldlines I specified, the ship would continually stretch until it broke. It's physically possible for the front and rear ends to follow those worldlines, so my hypothetical is not ruled out by the laws of physics; but it's still only a hypothetical. I'm certainly not saying that *any* rocket ship with engines at the front and rear ends *has* to move this way; I'm only exploring the consequences *if* one did so.

WE need to go into this a little more perhaps. We talked about it in regard to two separate ships but haven't actually covered it at all with a single extended ship.


.


v_{1-before} &gt; v_{2-after}> v_{3-after}<br /> <br /> &gt; v_{4-after}> v_{5-after} &gt; v_{6-after}....>v_{N-after}= 0<br />

The last "= 0" here is not correct. The velocity will never go to zero; it can't, because that would imply zero momentum in that direction. This does not imply infinite divisibility; it implies that there are a finite number of objects among which the momentum can be shared. In this example, there were N spheres, a finite number; in the limit, we would be counting atoms, or subatomic particles. There are a finite number of those too, and even when the momentum has been shared among every last one of them, the velocity will still be nonzero.

Perhaps you could explain this . You say there are a finite number of objects among which the momentum can be shared but the momentum can never reach zero in that direction. Well that would seem to neccessitate an infinite number of objects or infinite division of the quantity of momentum . I.e. smaller and smaller momenta for each succeeding object but never reaching 0 ,,exactly like Achilles' smaller and smaller divisions of the distance from the tortoise which he never reaches.

I have no problem with A), since it's just conservation of momentum. I'm not entirely sure about B). I would agree that, in general, p_{1-before} &gt; p_{N-after}, since some of the original momentum is likely to be distributed among other spheres besides sphere #N. But I'm not sure that this implies B). There's no reason why each of the momenta of the intermediate spheres *has* to fit into the "infinitesimal incremental continuum" between p_{1-before} and p_{N-after}; they could equally well be less than p_{N-after}[/itex].

<br /> <br /> Well if any of them were less than a later sphere then there were seem to be a definite violation of conservation of momentum. Energy out of nowhere<br /> <br /> ____________________________________________________________________<br /> <br /> Originally Posted by Austin0 <br /> YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD<br /> <br /> <br /> <br /> Originally Posted by Austin0 <br /> Is there really any meaningful difference between the cases we are considering?<br /> <br /> Propagation between a series of spatially separated interactions,<br /> Propagation between a series of contact interactions<br /> Propagation between atoms in a matrix <br /> Or propagation of sound , which in the final analysis is just momentum.<br /> SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I&#039;m not sure what you mean here by <b>&quot;propagate without limit</b>&quot;, but certainly all of the interactions you&#039;ve described must obey the laws of conservation of momentum and energy. In that sense there&#039;s no meaningful difference between them. But the specifics of how the interactions are modeled in physics can vary quite a bit </div> </div> </blockquote><br /> Without limit. Velocity and momentum never reaching zero.<br /> <br /> Had a bit of flu ,, more later. Thanks

 

[PeterDonis]

Austin0:

We are talking about systems accelerating along a linear path. No rotation. No systems moving at angles relative to the direction of motion.
Limited to this context do you think moving mass around before or behind the line would shift the CoM relative to the direction of travel?

Yes. Suppose, for example, that our system is composed of two subsystems, a front segment and a rear segment. The segments are initially at rest, and are identical in length and mass in the original rest frame. That means your imaginary line would be drawn precisely at the boundary between the two segments.

Now apply a forward impulse to the rear edge of the rear segment. The rear segment will compress, but until a time equal to the length of the rear segment divided by the speed of light has elapsed, the front edge of the rear segment will not move. So we have the rear segment compressed, meaning the CoM of the rear segment (and hence of the whole system) moves forward, but no mass crosses your imaginary line.

You have just given a description of Born acceleration. WHich is itself basically a definition of coordinate events.
The events described are exactly those of the essential Lorentz contraction and lead to exactly the same world lines.
SPECIFICALLY: the fundamental postulates propose that internal proper distances remain the same in inertial frames. ANd that the Lorentz contraction is regarded as a purely kinematic effect.

No. The worldlines of each segment of the ship are accelerated worldlines, not inertial worldlines. The MCIF at each event along those worldlines is an inertial frame, but the ship segments are at rest in each MCIF only momentarily. So the fact that "internal proper distances remain the same in inertial frames" is irrelevant, since that only applies to objects moving on inertial worldlines.

You are correct that the distance between any two adjacent segments (i.e., between appropriate events on their worldlines), which remains constant in each MCIF along the worldlines, will appear to be increasingly Lorentz contracted in the "initial" frame, the one in which Born rigid acceleration begins at time t = 0. This is indeed a "kinematic" effect. However, the physics that determines which worldlines each segment follows is *not* just "kinematics"; it requires knowledge of forces and accelerations, which are genuine physical quantities. See further comments below.

SO on the basis of this and the clock hypothesis, you could plot the worldline of the back of a normally propulsed ship, simply on the basis of constant proper acceleration.
You could then plot the front of the ship on the basis of instantaneous Lorentz contraction
and would then have a pair of worldlines that were indistinguishable from Born worldlines
Correct?

Not really. You are ignoring the internal forces between the parts of the ship. Remember how I answered your question about simple rear thrust and Born acceleration:

You're correct here too, in the sense that simple rear thrust could lead to a steady-state equilibrium in which the object as a whole was undergoing Born rigid acceleration; but in that equilibrium state, there would be internal stresses in all parts of the ship, in addition to the rear thrust.

So the object would only be undergoing Born rigid acceleration *after* all the oscillations of internal forces had damped out, and the ship had settled into a steady-state equilibrium. In that equilibrium state, the *net* force on the rear end of the ship would be *less* than the rocket thrust applied there, because there would also be a compression force from the next segment forward that would act towards the rear, in the opposite direction from the thrust. So the *final* proper acceleration of the rear segment would be *less* that the proper acceleration you would calculate just from the rocket thrust alone.

When I gave the proof that Born rigid acceleration preserves internal distances, I ignored the complications above. To make my specification consistent with those complications, we would simply define the "initial" reference frame, the one in which each segment starts accelerating at time t = 0 in accordance with the precise profile required for Born rigid acceleration, to be the MCIF of the ship at the instant when all the internal oscillations are gone and the steady-state equilibrium has just been reached. The ship as a whole would be momentarily at rest in that frame, and the accelerations on each segment of the ship (due to internal forces, plus the rocket thrust at the rear end only) would be in just the right profile to produce Born rigid acceleration from then on. In essence, we would be ignoring the details of how the ship got to the steady-state equilibrium and just focusing on the equilbrium state itself.

If we do focus just on the equilibrium state, then you're basically correct that we can calculate what all the worldlines have to look like for Born rigid acceleration based on relativistic kinematics. However, doing that can lead to confusion if we forget that we were originally led to calculate those worldlines by specific *physical* assumptions, so our calculations are only relevant in specific physical situations where those assumptions are applicable.

Based on these worldlines it would inevitably appear as if there had to be unequal acceleration at the front and the back. But this does not neccessarily have any physical meaning.

Wrong. The acceleration of a worldline is simply the derivative of its 4-velocity with respect to proper time:

A = \frac{dU}{d \tau}

This is an invariant, the same in all reference frames, so it's a bona fide physical quantity. (An "inertial" worldline is just one for which A = 0.) For the specific worldlines I gave for Born rigid acceleration for each segment, you can compute this acceleration, and verify that it is equal to x_s for each segment (i.e., the original x-coordinate of the segment in the "initial" frame).

So aside from the description, the Born hypothesis is making a statement about the nature of the Lorentz contraction and is proposing that it will not take place through equal acceleration throughout a system , which will result in the opposite result . Catastrophic expansion.

You're correct that equal *proper* acceleration of all parts of a system will result in expansion, since equal *proper* acceleration of all parts is enough to specify the worldline of each part, and from there it's all kinematics, as I've said before. However, this doesn't mean that the Born hypothesis is just a statement about kinematics. I've already addressed this somewhat in my comments above, but let me restate them in more detail:

-- The original motivation for working out the specification of Born rigid acceleration was to determine what kind of accelerated motion would result in unchanging proper distances between the parts of a system.

-- The reason for wanting to know *that* was the *physical* assumption that internal stresses between the parts of a system depend on the proper distances between those parts. If that weren't the case, nobody would care about Born rigid acceleration; we'd be discussing some other specification (whatever one turned out to keep internal stresses constant).

So the original motivation for studying Born rigid acceleration was physical, not just kinematic. Also:

-- The proper acceleration of each part of a system depends on the *net* force on that part, *not* just the externally applied force. You must also take into account the internal forces between the parts, as we've seen already.

So, as we've seen before, equal proper acceleration does *not* necessarily mean equal rocket thrust. You need physical assumptions to determine *how* equal proper acceleration would be realized (or any other acceleration profile, for that matter).

It also seems to place a meaning to lines of simultaneity that I don't quite understand.

The meaning of each of the lines of simultaneity (lines of constant q in Greg Egan's diagram) is just what it seems: each line of constant q is the line of simultaneity for the MCIF of each worldline it crosses, at the point where it crosses that worldline. In other words, all events on a line of constant q will be seen to be simultaneous by observers traveling on any of the accelerated worldlines we've specified (the lines of constant s in Greg Egan's diagram).

What may be confusing you is the fact that all of those lines of simultaneity intersect at the origin of the "initial" frame, x = 0, t = 0. That may seem paradoxical, but it's not: it's just an unusual fact about accelerated worldlines. It does, however, mean that the "pivot point" (x = 0, t = 0 in this case), and the past and future light cone at that point (the lines t = x and t = -x; part of the t = x line appears as a 45-degree dotted line on Greg Egan's diagram) have a special meaning to observers traveling on those accelerated worldlines. Look up "Rindler horizon" for more information (Greg Egan's page has some discussion of this).

v_{1-before} &gt; v_{2-after} &gt; v_{3-after} &gt; v_{4-after} &gt; v_{5-after} &gt; v_{6-after} .... &gt; v_{N-after} = 0

...

Perhaps you could explain this . You say there are a finite number of objects among which the momentum can be shared but the momentum can never reach zero in that direction. Well that would seem to neccessitate an infinite number of objects or infinite division of the quantity of momentum .

I don't understand how you're coming up with these conclusions. Let me re-state once more what I've been saying:

-- If the total momentum of a system is p_x before some process, the total momentum of the same system (provided there are no external forces on the system) after the process will also be p_x (we're assuming here that all motion is in the x-direction only).

For example, in the case of N spheres, we have (when sphere #1 is the only one with any momentum before the process):

p_{1-before} = p_{1-after} + p_{2-after} + ... + p_{N-after}

We appear to have agreement on this; but you may not be realizing that when I say "the momentum can never reach zero in that direction", the above, conservation of momentum, is *all* that I'm saying. I'm *not* saying that the momentum of one particular sphere could not be zero; only that the *total* momentum, the sum of the momenta of all the spheres, must remain constant, so if it starts out non-zero, it must end up non-zero (meaning that at least one sphere must have a non-zero momentum after the process).

-- If the original momentum, which was carried by only one sphere before the process, is shared between two or more spheres *after* the process, we will have

p_{1-before} &gt; p_{i-after}

for any value of i from 1 to N (i being the number of any particular sphere). But this does *not* imply that the momenta are strictly decreasing as we go from left to right (i.e., as i increases from 1 to N). Nothing I've said implies that, and in general it won't be true. See below.

-- In the presence of dissipation, the rest mass of any given sphere after the process will be greater than it was before the process. Thus, the velocity of any given sphere after the process will be *less* than it would have been without dissipation (because some of the energy of the process goes into increasing the sphere's rest mass instead of its velocity). So even in a limiting case where virtually all of the original momentum ends up in a single sphere after the process, we will still have

v_{1-before} &gt; v_{i-after}

for any value of i from 1 to N. But this, again, does *not* imply that the velocities will be strictly decreasing as we go from left to right (i.e., as i increases from 1 to N). In fact, since the original momentum was to the right (increasing x-direction), it is easy to see that, since sphere i + 1 is to the right of sphere i, and since one sphere can't move through another, we must have

v_{i-after} &lt;= v_{i+1-after}

for any value of i from 1 to N - 1. If this were not true, we would have, for some i, sphere i moving to the right *faster* than sphere i + 1, which is to its right, *after* spheres i and i + 1 have collided. That's obviously impossible.

Does this help any?

Well if any of them were less than a later sphere then there were seem to be a definite violation of conservation of momentum. Energy out of nowhere

Once again, I don't know how you're coming up with this conclusion. Let's consider the simplest interesting case, with three spheres. Before the collisions, sphere #1 has velocity v_0 in the positive x-direction, and spheres #2 and #3 are at rest. After the collisions, sphere #1 has velocity v_1, sphere #2 has velocity v_2, and sphere #3 has velocity v_3 (all velocities are in the x-direction, but we don't yet know anything about their signs). Also, before the collisions, all spheres have the same initial rest mass M; after the collisions, dissipation has increased the rest masses of all the spheres by the same factor k > 1.

The equations for conservation of energy and momentum then look like this, using the fully relativistic formulas (I've eliminated the common factor M in both equations, since it appears in every term):

\gamma_0 + 2 = k \left( \gamma_1 + \gamma_2 + \gamma_3 \right)

\gamma_0 v_0 = k \left( \gamma_1 v_1 + \gamma_2 v_2 + \gamma_3 v_3 \right)

where we have written the relativistic \gamma factors with the same subscripts as their corresponding velocities. Unless I'm misunderstanding you, you're claming that it must be the case that p_{2-after} > p_{3-after}, or

\gamma_2 v_2 &gt; \gamma_3 v_3

or the above conservation equations will be violated. That's not correct; there are plenty of solutions of the above equations for which \gamma_2 v_2 &lt; \gamma_3 v_3. (In fact, if we consider the additional constraint I gave above, that the spheres can't move through each other, we see that we must have v_2 &lt;= v_3, meaning that

\gamma_2 v_2 &lt;= \gamma_3 v_3

must hold.)