Understand Special Relativity and Time paradox

[SJBauer]

Skipping the notion of velocity 0.5c (or half-light speed), in physics, the twin paradox is a fun thought experiment in special relativity involving identical twins, one of which makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more.

So while I have given the answer as provided on the internet, how does one arrive at this conclusion? The time differential is based on naive application of time dilation wherein the rate of time is influenced by the rate of gravitational acceleration. Time dilation postulates that the faster an object is moving, whether gravitationally accelerated or thrust propelled, then the slower time progresses for that object in relation to a more stationary observer. The proof for this concept has been demonstrated via twin atomic clocks appositely positioned, one off Earth in outer Space and another remaining on Earth.

And yet time dilation has more to do with impractical interpretations rather than applied science. This result appears puzzling because each twin sees the other twin as traveling, and so, according to a naive interpretation of time dilation, each should paradoxically find the other to have aged more slowly. However, this scenario can be resolved within the standard framework of special relativity because the twins are not within equal frames of reference; the space twin’s rate of time was gravitationally influenced by its acceleration. The naïve expectation is, of course, that the space traveling twin would maintain his time differential (younger self) once he joined his brother on Earth.

However this is not the conclusion of the analysis. Another experiment based its results on twin clocks placed on aircraft flying in opposing directions parallel with the spin of the Earth. The logic being the experiment was that gravitational acceleration follows the spin of the Earth. According to special relativity, the rate of a clock is greatest according to an observer who is at rest with respect to the clock. In a frame of reference in which the clock is not at rest, the clock runs more slowly. Considering this experiment in a frame of reference at rest with respect to the center of the Earth gravitational field, a clock aboard the plane moving eastward, in the direction of the Earth's gravitational rotation, had a greater velocity (resulting in a relative time loss) than one that remained on the ground, while a clock aboard the plane moving westward, against the Earth's gravitational rotation, had a lower velocity than one on the ground (resulting in a relative time gain). For some scientists, this was enough to expect that the space traveling twin would have to endure a like opposing gravitational resistance on his return to his brother’s Earth-bound inertial frame of reference. Ergo, both brothers would be the same age.

 

[PeterDonis]

So while I have given the answer as provided on the internet, how does one arrive at this conclusion?

The general principle that covers all of these kinds of cases is that the proper time experienced between two events depends on the path taken--i.e., different worldlines can have different lengths. That resolves the standard "twin paradox" as well as variants like the Hafele-Keating experiment (atomic clocks being flown around the world) and GPS (the natural "rate of time flow" aboard the GPS satellites is different than on the ground).

This result appears puzzling because each twin sees the other twin as traveling, and so, according to a naive interpretation of time dilation, each should paradoxically find the other to have aged more slowly. However, this scenario can be resolved within the standard framework of special relativity because the twins are not within equal frames of reference

This doesn't matter; you can do the entire analysis in a single frame, using the general principle I gave above. See, for example, here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

The naïve expectation is, of course, that the space traveling twin would maintain his time differential (younger self) once he joined his brother on Earth.

This is also the correct answer, derived using the analysis as above.

Another experiment based its results on twin clocks placed on aircraft flying in opposing directions parallel with the spin of the Earth. The logic being the experiment was that gravitational acceleration follows the spin of the Earth.

You're talking, I assume, about the Hafele-Keating experiment:

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

I'm not sure what you mean by "gravitational acceleration follows the spin of the Earth". The point of the experiment was that there are *two* effects involved: what the Wikipedia calls "kinematic time dilation", due to relative motion, and gravitational time dilation, due to altitude. The experiment tested the prediction resulting from combining these two effects; and the prediction was confirmed.

Considering this experiment in a frame of reference at rest with respect to the center of the Earth gravitational field, a clock aboard the plane moving eastward, in the direction of the Earth's gravitational rotation, had a greater velocity (resulting in a relative time loss) than one that remained on the ground, while a clock aboard the plane moving westward, against the Earth's gravitational rotation, had a lower velocity than one on the ground (resulting in a relative time gain).

Yes, but both aircraft were also at a higher altitude than the ground clock, which resulted in a gravitational time gain for both. For the eastward moving clock, the kinematic time loss was greater than the gravitational time gain, for a net time loss; for the westward moving clock, both the kinematic and the gravitational effects gave a time gain. But to get the right answer as shown by the experiment, you have to take *both* effects into account. Just the kinematic effect alone does not give the correct answer.

For some scientists, this was enough to expect that the space traveling twin would have to endure a like opposing gravitational resistance on his return to his brother’s Earth-bound inertial frame of reference. Ergo, both brothers would be the same age.

This is not correct. Why do you think this would happen?

 

[bobc2]

What is the "Andromeda paradox"?

I’ll try to illustrate the Andromeda Paradox with the illustration below (patterned after Roger Penrose’s book “The Emperor’s New Mind”).

Ruth and Bill walk past each other on the sidewalk. They move at the same speed in opposite directions with respect to the rest frame of the earth. Thus, each of them occupies a different simultaneous space (they are associated with two different inertial frames). Now, consider the worldline of the distant Andromeda galaxy (we will assume the Andromeda Galaxy is at rest in the rest frame of the earth). There is a point on the Andromeda worldline that is in the simultaneous space of Bill as Bill and Ruth pass each other on the street. But, as they pass, there is a different point on the Andromeda worldline that is in the simultaneous space of Ruth (a point later in Andromeda time for Ruth than for Bill).

In Bill’s simultaneous space a committee in the Andromeda Galaxy is meeting to decide whether to attack Earth. However, in Ruth’s simultaneous space, the decision has already been made and the attack fleet has already been launched.

 

[zonde]

This should be expected. When anyone performs a measurement the outcome of that measurement is frame invariant. Otherwise different frames would predict that the same experiment would generate different numbers. So there must be some mechanism for converting frame variant components into frame invariant scalars.

Yes, I was thinking about the same.
So what could be this mechanism? One simple way would be that frame variant represents a group of frame invariants. So the mechanism is just picking one certain invariant out of the group.

However, I think that it is important to note that the invariant quantity you labeled E is NOT the energy, except in the rest frame of the observer. So "energy" is frame variant, but a particular measurement of energy produces an invariant number. Other frames will disagree that the number produced by that measurement represents the energy.

But we can draw parallels with mass. And we do speak about rest mass that is mass in one particular frame.

 

[SJBauer]

PeterDonis - Per your question of why I would believe it would be possible for the twins to be of the same age when put back into the same frame of inertial reference. The concept of acceleration toward the speed of light puts the space traveling twin in a differing frame of reference. In order to be rejoined with his brother, he must decelerate from the speed of light. The process of acceleration causes a time dilation: i.e. whenever energy is added to a system, the system gains mass; whenever a system gains mass it further increases time dilation. On the other hand, the process of deceleration requires a resistance to acceleration momentum (kind of like putting on the brakes for the entire time of returning to the Earth-bound brothers frame of reference). The process of deceleration reverses time dilation:i.e. whenever energy is deleted from a system, the system loses mass; whenever a system loses mass it further decreases time dilation. As inducing decelerating momentum requires a stronger opposing force, it actually takes longer to decelerate back to the Earth-bound brother's frame of reference. This lag time accounts for the conservation of time that equalizes the space traveling twins age with that of his Earth-bound brother.

 

[Dale]

But we can draw parallels with mass. And we do speak about rest mass that is mass in one particular frame.

I prefer the term "invariant mass" over "rest mass" since a system may not be at rest in any inertial frame, but still has a well defined invariant quantity with units of mass. But your point is well taken. Similarly with proper time and proper length.

 

[Dale]

The process of acceleration causes a time dilation: ... The process of deceleration reverses time dilation:

These are both incorrect.

 

[Dale]

It was pretty clear to me. :)

Thanks! It felt awkward when I wrote it, but maybe it isn't as bad as it seemed.

 

[Alain2.7183]

In Bill’s simultaneous space a committee in the Andromeda Galaxy is meeting to decide whether to attack Earth. However, in Ruth’s simultaneous space, the decision has already been made and the attack fleet has already been launched.

Thanks for that description.

I see how the two different walkers can have different ideas of "now" at Andromeda. But I don't yet understand how you get from there to the "block universe" conclusion.

 

[bobc2]

Thanks for that description.

I see how the two different walkers can have different ideas of "now" at Andromeda. But I don't yet understand how you get from there to the "block universe" conclusion.

This is implied as soon as you acknowledge a couple of observations. First, Ruth and Bill exist together in the black rest frame as well as each of their respective rest frames at the event represented by the passing on the sidewalk. But, the only way that the event of the Andromeda committee meeting and the event of the space fleet launch can both exist is if all events are all there "at once."

That is, if the entire universe exists "all at once" as a 4-dimensional static structure ("block universe"), then there is no problem for two different observers (or any number of observers) to all exist in different cross-sections of the 4-dimensional universe "all at once."

The "all at once" phrase is used by Paul Davies in his book "About Time" when he describes the block universe. Some have trouble with that phrase because it is time related phrase. The question is raised, "To what kind of time does 'all at once' refer?" Is there some hypertime that flows as the static universe just sits there frozen? But, then the discussion seems to get too philosophical for a forum such as ours.

You should understand that most forum members here would insist that the Andromeda Paradox does not represent anything about physical reality. It could be taken as a pedagogical illustration to help graphically visualize aspects of the mathematics of special relativity. They consider that there are other equally valid interpretations of special relativity, such as the Lorentz Aether theory, that are in conflict with any notion that the Andromeda Paradox illustrates something about reality--thus, in that view, the block universe is not to be taken as a true characterization of external physical reality.

One of our members, Vandam, was very passionate about the true physical reality expressed in the block universe interpretation of special relativity. Someone posted a note to me the other day that Vandam has been banned from our forum. Evidently his continual dogmatic insistence about the reality of block universe was considered objectionable (and perhaps that along with some of the strong language used in exchanges with others who strongly disagreed with him--and in fairness to Vandam there was strong language directed at him as well). And as a note to Vandam, in case he still looks in on the forum: Sorry to see you go, colleague. We will miss your perspective on things.

I am not inclined to have any more to say about the block universe, except when feeling obligated to respond to direct questions.

[edit] By the way, some physicists get annoyed by Penrose ignoring the obvious changes to the picture if you were to take into account the tangential velocity of the Earth at the point of Bill and Ruth passing. Then, the orbiting of planets in the respective solar systems. And the relative motion between the Andromeda Galaxy and our Milky Way Galaxy along with the orbiting of our solar system as part of our rotating Galaxy, and the orbiting of the Andromeda solar system along with its rotating Galaxy where that meeting and launch takes place, etc., etc. Penrose would probably reply that it's not worth the distraction of taking all of those things into account since after all is done you would still have the basic Andromeda Paradox effect.

 

[ghwellsjr]

...I've prepared a set of sketches to help anyone interested visualize what is going on in the 4-dimensional universe with the twins in the context of simultaneity...

In an attempt to try to understand your sketches, I have redrawn my diagrams from post #23. I interchanged the colors of the twins to match what you drew and I eliminated the signals going between them for the same reason. In my diagrams, I mark yearly increments of Proper Time for both twins with dots so it was no problem adding in the black circles like you have in your Sketch I. Since simultaneity is marked by the Coordinate Time grid lines (or any other horizontal line), it was no problem adding the horizontal black lines and the blue circles placed on the traveling twin's world line. Here's what I've got so far:

But to add in the diagonal blue lines was not trivial because they are not coincident with the blue dots marking yearly increments of Proper Time for the traveling twin. Neither are they coincident with the blue circles. So what I did was transform from the IRF in which the black inertial twin is at rest to the IRF in which the blue traveling twin is at rest during the outbound portion of his trip. Then I added in horizontal blue lines going from the yearly increments of Proper Time for the black inertial twin and going over to the blue traveling twin's world line. These horizontal lines are simultaneous in this IRF. Here's that diagram:

Now, I look at the intersections of the horizontal blue lines in relation to the blue dots which enables me to know where to draw the blue diagonal lines on the first diagram. Here's the final diagram:

Now this leaves me with a question: what is the purpose of the blue circles?

[NOTE: I want to take this opportunity to point out an error I made in my first post (#7) of this thread. I incorrectly attempted to simplify the formula at the end of the post.]

 

[Austin0]

That is a philosophically based idea. Einstein cautioned against these kinds of ideas which trap you into solipsism.

You are saying that there is no reality to be associated with the hyperplanes of simultaneity for a given Lorentz frame. I don't think our monitor will want us to continue a discussion along those lines. .

Solipsism seems to be a recurring topic with you. I find this strange because no commentary from anyone in this forum has indicated to me any kind of solipsist bias whatsoever.
as I understand you you think that we can infer a picture of reality that goes beyond direct measurement and instantaneous information. I certainly agree on this point.
SO you ,based on the concept of absolutely invariant light propogation , have infered an instantaeous picture of an extended slice of the world.
Given points A and B an isotropic light burst at the midpiont M would arrive simultaneously at both A and B . Conversely signals from A and B received simultaneously at M would have been emitted simultaneously at A and B.
So then a system of clocks symnchronized on this basis would universally have simultaneously identical proper time readings. This is certainly logical as far as it goes and is of course simply the Einstein synchronization convention in action.
but your interpretation is totally dependent on the isotropic equality of light paths.
I.e., the distance from A to M is not only spatially equal to the distance from B to M but the light paths themselves must be actually equal. A-->M=B-->M
Assuming that this is the case , if we accelerate the system along the A-B vector to a new inertial velocity it is quite obvious that it is not possible that the paths could still be equal.
If the paths are unequal and light is absolutely invariant in speed then, equally obviously, the time between . A-->M could not equal the time bewteen B-->M. So the concept of actual simultaneity based on this premise could not be consistent with reality.

So it appears that you reject the reality infered by logic and physical conception of space and motion as we know it and replace it with a picture dependent on conventional clock readings which is a matter of practicallity and convention, based on an assumption of equal path lengths which we know is not actually true.
Do you have some kind of physical model where ALL light paths between equal distances in ALL frames could be isotropically equivalent?
Do you agree that the assumption of actual simultaneity based on equal light paths and the recognition that such paths cannot be equal in a system in motion are mutally exclusive?
You repeatedly assert that anyone who does not accept your interpretation of actual physical simultaneity of an instant defined by equal system clock readings,- is denying the actuality of a real 3-d slice of spacetime. But this is a gross misrepresentation. I for one have no problem whatsoever accepting such a slice. I assume that the physical structure and interrelationships within an inertial frame would insure some kind of fundamental simultaneity of existence or temporal connection. But this kind of "actual" simultaneity would not necessarily correspond with clocks haviong equivalent proper times which is what you are insisting without justification.
I can't speak for others but I would imagine that the majority simply feel that the surfaces of simultaneity do not represent such a slice of actual simultaneity not that such a slice does not exist.

So I wouldn't say you were a solopsist but you do seem to deny logical inferences regarding reality in favor of a metaphysical (non-logical) worldview.

 

[Austin0]

In the sketch below, I've added in a Red guy here in order to make a point about the interesting sequence of simultaneous worlds for the traveling twin as he does his turna-round. So, this additional red guy is at rest along with the black guy. As the Blue guy does his turn-around, the sequence of Black clock readings (as presented in the sequence of Blue's 3-D world cross-sections of the 4-D universe) progress into the future whereas the Red clock readings proceed into the past. As Blue enters the turn-around, the Black clock begins with event "black E" and at the end of the turn-around the last Black clock reading corresponds to event "black h." But, although Red's clocks begin with event "red E" (same simultaneous plane as "Black E") and end with event "red h", we see the sequence of Red clock readings going into the past along Red's X4 axis world line. The E, a, b, c, ... h designations identify the discrete hyperplanes (planes of simultaneity) in the movement through 4-D Space-Time of Blue as he progresses along his world line.

There seems to be confusion over the sequence of inertial frames associated with the traveling twin turnaround. I was simply pointing out what I thought was an interesting result for red clock readings as presented in the traveling twin’s sequence of inertial frames. So, here we will not be concerned about the period on the traveling twin's worldline during the turnaround. To avoid any further anguish over discretizing the turnaround, let’s just simplify the analysis by changing the focus away from the turnaround.

In the sketch below we show the same interesting feature by simply comparing readings on what I now show as the brown clock (red in the earlier sketch) as they are presented in the traveling twins two inertial frames, i.e., the purple frame (before turnaround) and the red frame (after turnaround). Notice that Event C presents a brown clock reading in the purple frame at the start of the outgoing trip of the traveling twin (the purple X1 axis represents the outgoing twin's simultaneous space at the start of his trip).

We show the beginning of the twin return trip as Event A (the traveling twin has just completed the turnaround and has started back home, i.e., the Red frame in the sketch). In this twin’s inertial frame the brown clock is presented on the worldline of the Brown frame as Event B.

When the traveling twin reaches Event D, the Event C is simultaneous with that same event (C simultaneous with D) in the twin’s inertial frame.

So, I am simply making the observation that as the traveling twin moves along his worldline, Event C is encountered at the start of the twin’s outgoing inertial frame before the Event B is encountered in the twin’s return inertial frame, even though as the Brown observer moves along his own X4 (time) axis, he (Brown) naturally experiences Event B before Event C. And Events D and C are simultaneous in the twin’s inertial frame. That’s all. No implications are drawn here—just an interesting observation for one to interpret however one pleases. Some may find nothing of interest here.

Well you have presented some intertesting observations here. I think there are some implications to be drawn.
Obviously there are some difficult complications regarding a non-inertial frame on several issues but certainly in principle we can assume a single co-accelerating frame for the traveler extending from the initial point to the turnaround.
Disregarding Born rigid acceleration for the turn around itself I for one have no real problem with taking your equivalence of MCIRF's for rough insights. Certainly such a frame would be at all times spatially aligned with the inertial CMRF's

So according to your interpretation at the outset (point O)the traveler at t'=0 at x=0 is simultaneous with event C with brown clock reading t=C at x=C

Then at event D on the travelers worldline at t'=D at x=D the traveler is simultaneous with event C

Later at event A on the traveler worldline at t'=A the traveler is simultaneous with event B at x=B , t=B (B<C)

So looking at this it means that event C is simultaneous with two different times at the same location I.e. at the traveler.
You may have no problem with this but I do.
SO at t'=0 there is a traveler clock at C that reads (C) t'=0 and at event A at t'=A there is also a traveler clock at C reading t'=A
So at a single location and a single instant you have two different observers from the traveler frame with different clock readings.
Quite clearly this is not merely counter intuitive it is simply impossible . Inconsistent with any fundamental conception of spacetime.

Then there is the intersection at event A with the earlier point on the brown worldline B
AT point O the traveler is simultaneous with C yet at later point A he is simultaneous with B which is earlier than C on the brown worldline.
Apparent time travel.
Even more , we can easily describe a situation where a traveler is momentarily co-located with a brown observer on the way out and then after turnaround is again co-located with the same observer at such an earlier time per your interpretation.
Clearly this physical interpretation is open for possible time paradoxes and not a viable possibility.
He receives a message from himself on the way out that he doesn't give the brown observer to forward until after turnaround , etc, etc etc,.

I think you would have to agree that every point in such a frame, even during acceleration,would not only be advancing forward in time but would actually observe all passing clocks also proceeding forward. That these apparent temporal shifts and impossibilities are artifacts of the conventions of Minkowski graphics and a misinterpretation of simultaneity.
They do not conform to any possible reality.
So as the man said"if you eliminate what is impossible , then what you have left ..."

What is quite possible and rational is this:

If we assume that the system clocks could instantaneously be conventionally synchronized during turnaround based on the traveler location as master ,then those outlying clocks would in fact be incremented forward.
So if the traveler clocks are being set ahead, this in a sense is equivalent to the brown clocks falling relatively behind. But the changes are all taking place in the traveler frame and are explicitly the result of human convention.
They are not the result of nor do they result in any change in the relationship between the world lines which both proceed blithely forward..

So it would appear that your illustration in fact provides a reductio ad absurdem argument countering the physical interpretation of planes of simultaneity you were promoting.

 

[SJBauer]

True, due to our understanding of special relativity, SpaceTime is relative to the observer. Ergo, the speed of light is same within your speed of light (SOL) reference frame as it would be for someone who was traveling much slower. But then, what of Space? Yet this whole frame of reference for mass at light speed is all theoretical since mass cannot exist at the speed of light. Since the concept for mitigating the differences between these two very different reference frames is the basis for this notion of Time slowing down or stopping, let’s explore this theory some more.

There are multiple perspectives to consider for mitagating Time between these two very different frames of reference: A) If as actual rate of Time changes with respect to his acceleration, his reality became part of this slower rate of time, then it would appear to him that his Earth-bound counterparts were milling about at incredible speed in relation to his own. OR B) Consider the SOL traveler is moving at the same rate as he would on Earth, but only the rate of Time changes with respect to his acceleration. In this respect the rate of Time is slowing down consistent with his acceleration, but his movements are still relative to his Earth-bound rate of Time; ergo, he is moving more in less time, or moving faster. Consistent with this notion, he is also able to process more information in less time, or he is thinking faster. Somewhat like a housefly that is able to react and move much faster relative to his frame of reference, the SOL traveler would perceive his Earth-bound counterparts to be moving more slowly or not at all.

The Theory of Relativity favors selection ‘A’ as the SOL traveler’s reality, allowing him to be younger when he returns to Earth. The logic behind this is that as the SOL is measured the same, relative to one’s frame of reference in the rate of time, so must the relative relationship among all of its participants. Subsequently the perspective relationship of man’s speed relative to the actual SOL should remain the same within any reference frame. Consequently if the rate of time slows with acceleration, then so must everything else in that frame of reference do likewise.

But one cannot consider Time without Space, so this brings up another aspect of relativity; i.e. mass increases with acceleration. This would suggest that the mass of the photon becomes greater upon acceleration, as well as everything else in this reference frame. Therefore to suggest that the SOL travelers would age less during the time of their acceleration and would be younger than their Earth-bound counterparts, one must also agree that the SOL travelers should have also retained their increased mass upon return to Earth. Since the most likely form of increased mass would be due to increased density, rather than increased volume, the SOL traveler would be a veritable superman upon his return.

 

[PeterDonis]

SJBauer, your post is a mixture of misstatements of SR and speculations which are not justified by SR.

SpaceTime is relative to the observer.

No, it isn't. Spacetime is a prime example of something that *isn't* relative to the observer: it's the same 4-dimensional object for everyone.

the speed of light is same within your speed of light (SOL) reference frame

There is no such thing as a speed of light reference frame. We have a FAQ on this:

https://www.physicsforums.com/showthread.php?t=511170

Since the concept for mitigating the differences between these two very different reference frames is the basis for this notion of Time slowing down or stopping

This might be a valid statement if you are referring to two ordinary inertial frames; but you're not.

as actual rate of Time changes with respect to his acceleration

"rate of Time" doesn't depend on acceleration. Time dilation in SR is purely a function of relative velocity. Acceleration only comes into it because it changes relative velocity. See here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

The Theory of Relativity favors selection ‘A’

The theory of relativity says neither of your "selections" are right. Time dilation is always something that an observer attributes to *other* observers moving relative to him. An observer always experiences his own time as "flowing" normally.

allowing him to be younger when he returns to Earth

You appear to be mixing up two different things here: time dilation due to relative motion, and different elapsed proper times along different worldlines that connect the same pair of events. These are different concepts, and you shouldn't confuse them. You can't account for the fact that the traveling twin in the standard twin paradox is younger when the twins meet up again just by using time dilation; there's more to it than that.

But one cannot consider Time without Space, so this brings up another aspect of relativity; i.e. mass increases with acceleration.

The "relativistic mass" of an object increases as it moves faster; but just as with time dilation, this depends only on relative velocity, not on acceleration. Acceleration comes into it only because it changes relative velocity.

This would suggest that the mass of the photon becomes greater upon acceleration

"Relativistic mass" is equivalent to "energy", and the energy of a photon does change from frame to frame; but again, this is purely due to the change in velocity from frame to frame; acceleration doesn't come into it. The photon's apparent speed does not change (it is always c); but its energy and momentum do.

to suggest that the SOL travelers would age less during the time of their acceleration and would be younger than their Earth-bound counterparts, one must also agree that the SOL travelers should have also retained their increased mass upon return to Earth. Since the most likely form of increased mass would be due to increased density, rather than increased volume, the SOL traveler would be a veritable superman upon his return.

This is unjustified speculation; it has nothing to do with SR or the solution of the twin paradox.

 

[BruceW]

@ SJBauer: I don't know what you mean by a SOL frame of reference. But it is interesting to think about what the change of the rest mass of the traveling twin would be. If the traveling twin was subject to a "pure four-force" (i.e. one where \vec{f} \cdot \vec{v} = 0) then the change in rest mass is zero. But if he were subjected to a non-pure four-force, then he would generally have some change in rest mass.

Luckily, the electromagnetic four-force is a pure four-force. And I guess we can assume that whatever spaceship he is using, that it is likely to be fundamentally using electromagnetic phenomena (for example, any rocket uses contact force, which is essentially electromagnetic). So unless the traveller has some advanced technology, his rest mass is going to remain unchanged along his journey.

 

[PeterDonis]

it is interesting to think about what the change of the rest mass of the traveling twin would be. If the traveling twin was subject to a "pure four-force" (i.e. one where \vec{f} \cdot \vec{v} = 0) then the change in rest mass is zero.

This, as you note, is the usual case.

But if he were subjected to a non-pure four-force, then he would generally have some change in rest mass.

Can you give a specific example of a non-pure four-force?

 

[BruceW]

Can you give a specific example of a non-pure four-force?

The only four-force I know of is the electromagnetic four-force. I guess the reason we usually assume the rest mass is constant is because we are usually interested in electromagnetic interactions only. If we consider weak and strong nuclear interaction, then it's a whole different kettle of fish. But I really don't know enough QM to be able to go into that. Anyway, I suppose the term 'four-force' is only used for classical (non-quantum) phenomena. Are there any other kinds of four-force apart from the EM four-force? I don't have a great deal of knowledge about this kind of stuff.

 

[Dale]

Then at event D on the travelers worldline at t'=D at x=D the traveler is simultaneous with event C

Later at event A on the traveler worldline at t'=A the traveler is simultaneous with event B at x=B , t=B (B<C)

When you say that two events are simultaneous, don't forget to specify the reference frame in which they are simultaneous. In this case, it is two different inertial frames.

 

[PeterDonis]

The only four-force I know of is the electromagnetic four-force.

Four-force is just the 4-D spacetime representation of force. Any force can be represented that way. The key is to look at the effects of the force. The general expression for 4-force is the rate of change of 4-momentum with respect to proper time:

f^{\mu} = \frac{d p^{\mu}}{d \tau}

4-momentum is a 4-vector with components (E, \vec{p}), and the rest mass is the Minkowski length of the 4-momentum: m = E^2 - p^2. If we look at the force in a frame in which the object starts out at rest before the force is applied, then the 4-momentum starts out as (E, 0), so the rest mass m is equal to the starting energy E; we will thus write the starting 4-momentum in this frame as (m, 0). (I'm using units in which c = 1 for simplicity.)

A pure 4-force has a zero dot product with 4-momentum. That means that, in the frame in which the object starts out at rest before the force is applied, the force cannot have a "time" component; it can only have a spatial component. In other words, the 4-force looks like (0, \vec{f}). The effect of the 4-force in this frame will be to change the 4-momentum of the object from (m, 0) to (\gamma m, \gamma m \vec{v}), where \gamma = 1 / \sqrt{1 - v^2} and \vec{v} = \vec{f} \tau / m, where \tau is the length of the proper time interval during which the force is applied. It should be evident that this will leave the Minkowski length of the 4-momentum unchanged; it will still be m.

An "impure" 4-force has a nonzero dot product with 4-momentum, so its 4-vector in the frame we're using looks like (f_0, \vec{f}). The effect of this 4-force will be to change the 4-momentum of the object from (m, 0) to (\epsilon + \gamma m, \gamma m \vec{v}), where \epsilon = f_0 \tau is the "internal energy" added to the object by the force. This "internal energy" is energy added in addition to the kinetic energy added by inducing motion in the object, and it should be evident that this will *increase* the rest mass of the object: the Minkowski length of the resulting 4-momentum is *greater* than m.

The term "internal energy" may make you think of things like heat, and indeed a good example of an "impure" 4-force is friction, which adds heat to the object in addition to changing its speed. Heat shows up as an increase in temperature, and in relativity temperature is part of "rest mass", because mass and energy are equivalent. You may also think of elastic vs. inelastic collisions, which is another good example: forces exerted in purely elastic collisions are pure, whereas forces exerted in inelastic collisions are impure.

I guess the reason we usually assume the rest mass is constant is because we are usually interested in electromagnetic interactions only. If we consider weak and strong nuclear interaction, then it's a whole different kettle of fish.

Not really, at least not at the level we're looking at here.

 

[BruceW]

The term "internal energy" may make you think of things like heat, and indeed a good example of an "impure" 4-force is friction, which adds heat to the object in addition to changing its speed. Heat shows up as an increase in temperature, and in relativity temperature is part of "rest mass", because mass and energy are equivalent. You may also think of elastic vs. inelastic collisions, which is another good example: forces exerted in purely elastic collisions are pure, whereas forces exerted in inelastic collisions are impure.

Right, I get it. So we can use a non-pure four-force, depending on the kind of situation we want to model. But we know that fundamentally, any four-force (not going into QM), is EM in nature, and therefore pure.
So an analogous example in classical, non-relativistic mechanics is when we model motion of projectiles near the surface of the earth, we say that there is a downward acceleration of g. So we are saying that vertical momentum is not conserved. But we really do believe that (in classical, non-relativistic mechanics), momentum is conserved, it is just that for the sake of our model, we don't bother to take the motion of the Earth into consideration, so we allow momentum to not be conserved.

 

[PeterDonis]

So we can use a non-pure four-force, depending on the kind of situation we want to model. But we know that fundamentally, any four-force (not going into QM), is EM in nature, and therefore pure.

Another way to put this is that you can always "break down" an impure 4-force into pure 4-forces at a more fundamental level. For example, friction is due to EM interactions between atoms, which are pure; the way the interactions fit together just happens to work out to an impure force at the level of macroscopic objects.

So an analogous example in classical, non-relativistic mechanics is when we model motion of projectiles near the surface of the earth, we say that there is a downward acceleration of g. So we are saying that vertical momentum is not conserved. But we really do believe that (in classical, non-relativistic mechanics), momentum is conserved, it is just that for the sake of our model, we don't bother to take the motion of the Earth into consideration, so we allow momentum to not be conserved.

Yes, this would be analogous to, for example, only looking at the effect of air resistance on a relativistic projectile, without looking at the effect on the air. When you include the air as well as the projectile, 4-momentum is conserved; but if you just look at the projectile, it isn't.

 

[Austin0]

Quote by Austin0

So according to your interpretation at the outset (point O)the traveler at t'=0 at x=0 is simultaneous with event C with brown clock reading t=C at x=C

Then at event D on the travelers worldline at t'=D at x=D the traveler is simultaneous with event C

Later at event A on the traveler worldline at t'=A the traveler is simultaneous with event B at x=B , t=B (B<C)

So looking at this it means that event C is simultaneous with two different times at the same location I.e. at the traveler.
You may have no problem with this but I do.
SO at t'=0 there is a traveler clock at C that reads (C) t'=0 and at [event A at t'=A] edit. This should be --event Dat t'=D there is also a traveler clock at C reading t'=D
So at a single location and a single instant you have two different observers from the traveler frame with different clock readings.
Quite clearly this is not merely counter intuitive it is simply impossible . Inconsistent with any fundamental conception of spacetime.

When you say that two events are simultaneous, don't forget to specify the reference frame in which they are simultaneous. In this case, it is two different inertial frames.

well I am just recovering from a long flu and made a serious mistake as I have corrected above.
As for the simultaneity---i was simply taking Bob2c's premise that the planes of simultaneity as indicated by CMRFs in a Minkowski chart are equivalent to 3-D slices of a single accelerated frame at face value and pointing out the impossible consequences of such an interpretation.
I should know from past arguments with Mike_Fontenot on this subject that those who cling to this interpretation are not to be dissuaded by mere details like clocks running backwards or multiple co-locations of disparate spatial points and times.
But there's always hope ;-)

 

[Dale]

I should know from past arguments with Mike_Fontenot on this subject that those who cling to this interpretation are not to be dissuaded by mere details like clocks running backwards or multiple co-locations of disparate spatial points and times.

That and they seem completely impervious to the fact that it violates one of the very few mathematical requirements that a coordinate chart must match. I don't understand why they "cling" so strongly, when there are so many obvious reasons against it.

 

[ghwellsjr]

For anyone needing a little more background on the space-time sketches and the 4-dimensional universe concept, you can go to this earlier post that outlines the concept, beginning with post #19 (be sure you understand relativity of simultaneity--if not ask questions):

https://www.physicsforums.com/showthread.php?p=4138802#post4138802

I followed your link and found another similar diagram to the one that I asked about in post #161. Here's a clip of the first one:

I asked why the blue circles were placed where they were.

Now the clip from the link:

Again, I wonder about the placement of the blue circles, why are they placed differently this time?

I really don't understand why you put the blue circles in either place, they have no significance either way to the blue traveler. And what further doesn't make any sense to me is why you draw the blue lines intersecting with the black circles. Now I realize that the slope of the blue lines are indicating simultaneity for the blue traveler no matter where they are placed but you must have some reason to pick these particular lines.

What would make sense to me is if you placed the blue circles at the same increments of Proper Time for the blue traveler as what you picked for the black twin like this:

Now you can easily see the symmetry of Time Dilation between the IRF's of the two twins. Or does that matter to what you are trying to convey?

 

[bobc2]

I followed your link and found another similar diagram to the one that I asked about in post #161.

I asked why the blue circles were placed where they wer

Again, I wonder about the placement of the blue circles, why are they placed differently this time?

I really don't understand why you put the blue circles in either place, they have no significance either way to the blue traveler. And what further doesn't make any sense to me is why you draw the blue lines intersecting with the black circles. Now I realize that the slope of the blue lines are indicating simultaneity for the blue traveler no matter where they are placed but you must have some reason to pick these particular lines.

Now you can easily see the symmetry of Time Dilation between the IRF's of the two twins. Or does that matter to what you are trying to convey?

Very good graphics, ghwellsjr. Your graphics make it very clear. Good job.

I was just trying to get across a concept of an observer's sequence of simultaneous spaces, and just happened to approach it this way. In both of the sketches I was simply indicating that the stay-at-home twin and the traveling twin each advances along his respective worldline (along the red and blue X4 axes). I wanted to illustrate the concept of the two different sequences of simultaneous spaces associated with each twin. In one case I put blue dots on the blue worldline to indicate where the black rest simultaneous spaces intersect with the blue worldline. In the other case I placed blue dots on the blue worldline simultaneous space sequence such that the blue simultaneous spaces would intersect with the black worldline. That’s all. It was just intended to help visualize the concept of simultaneous spaces. I wasn’t interested in numbers—just the concept.

The sketch below includes hyperbolic calibration curves to identify the ten year locations with respect to the rest frame origin. The sketch on the right shows the traveling twin’s two different inertial frames, one for the trip out and one for the return trip. I emphasize the these frames do not include the actual turnaround. The path length during turnaround is so short compared to the path length of the total trip out and back, that we would need a magnifying glass to see the curve. My interest was restricted to comparing the inertial trip out and the inertial trip back (after turnaround is complete).

I could overlay any inertial coordinates over the lower right sketch below to identify proper times for both rest system and any other inertial coordinates. I've given a short derivation of the hyperbolic curves. It begins with a sketch of red and blue guys moving at the same speed in opposite directions with respect to the black coordinates.

The sketch below gets messy, but it illustrates a couple of more details that one may or may not find interesting. Notice that the event A on the 2nd Red stay-at-home guy (displaced from the first red twin) is presented to the returning twin’s trip simultaneous space before it is presented to the outgoing twin’s simultaneous space. Notice that this does not in any way imply that the 2nd Red guy's time is flowing backwards for that Red guy sitting at rest in his own black inertail frame. It's just a feature of special relativity and is no more mysterious than the two twins having different ages after they reunite.

By the way, the blue dots on the traveling twin's worldline are placed with same proper time increments as the black worldline dots (one year intervals of proper time on both worldlines, in accordance with your preference). The hyperbolic calibration curves show the five year lapses.

 

[Austin0]

The sketch below gets messy, but it illustrates a couple of more details that one may or may not find interesting. Notice that the event A on the 2nd Red stay-at-home guy (displaced from the first red twin) is presented to the returning twin’s trip simultaneous space before it is presented to the outgoing twin’s simultaneous space. Notice that this does not in any way imply that the 2nd Red guy's time is flowing backwards for that Red guy sitting at rest in his own black inertail frame. It's just a feature of special relativity and is no more mysterious than the two twins having different ages after they reunite.

By the way, the blue dots on the traveling twin's worldline are placed with same proper time increments as the black worldline dots (one year intervals of proper time on both worldlines, in accordance with your preference). The hyperbolic calibration curves show the five year lapses.

I am now confused as to what you actually think.
i understood you were talking about an extended physical frame for the traveler.
SO the lines of simultaneity, the x₁s represent the actual traveling system of clocks and observers. A simultaneous 3-d slice of equal proper time readings.
So the post turnaround intersection of the x₁
with point A on the red observer's worldline meant an intersection with the actual traveling system at this time point on the red clock.Is this not what you are asserting?
If not what??
Your previous evaluation of the observed decreasing readings on the red clock at that point during turnaround was a completely accurate literal reading of the chart and what it implies about the observations of the passing travelers. The problem is that the literal interpretation and it's consequences is inconsistent with any possible real world accelerated system.
You make a casual assertion that " It's just a feature of special relativity and is no more mysterious than the two twins having different ages after they reunite." as if this was an obvious and accepted fact.That is simply a non sequitur. Assuming that which is to be demonstrated.

 

[bobc2]

I am now confused as to what you actually think.

Hi, Austin0. What I think (if I haven't messed up) is that I've used the Lorentz transformations to develop a Minkowski diagram for the twin's outgoing trip (up to the point of the turnaround) and then, picking up the traveling twin's trip after completion of turnaround, showing the diagram for the return trip.

The diagram shows the intersection of the simultaneous space of the outgoing X1 axis with the 2nd Red worldline right at the start of the outgoing trip. And then it shows the intersection of the outgoing X1 axis with the 2nd Red worldline right after the turnaround is complete. If one would like to nit-pick, they could point to the round off of the total round trip time given as 10 years for the traveling twin (the small turnaround time was not included, which would add some decimal value to the total 10 year number).

Now, I've added another intersection point, C, showing where the outgoing twin's simultaneous space intersects the 2nd Red worldline just before the outgoing twin goes into his turnaround. Folks can decide for themselves if they see anything interesting about this, I'm just presenting the Minkowski diagrams with the simultaneous spaces, that's all.

 

[Dale]

The diagram shows the intersection of the simultaneous space of the outgoing X1 axis with the 2nd Red worldline right at the start of the outgoing trip.

Where "simultaneous space" is defined by the momentarily co-moving inertial frame, which is only one of many possible definitions and which has some known problems.

 

[bobc2]

Where "simultaneous space" is defined by the momentarily co-moving inertial frame, which is only one of many possible definitions and which has some known problems.

I think that Minkowski diagrams are pretty well established and understood in the field of special relativity, as is the concept of simultaneous spaces.

 

[Dale]

I think that Minkowski diagrams are pretty well established and understood in the field of special relativity, as is the concept of simultaneous spaces.

Yes, but apparently not by everyone. Particularly since we get many novices and students, it is a point that bears mentioning and you didn't so I did.

 

[Austin0]

Hi, Austin0. What I think (if I haven't messed up) is that I've used the Lorentz transformations to develop a Minkowski diagram for the twin's outgoing trip (up to the point of the turnaround) and then, picking up the traveling twin's trip after completion of turnaround, showing the diagram for the return trip.

The diagram shows the intersection of the simultaneous space of the outgoing X1 axis with the 2nd Red worldline right at the start of the outgoing trip. And then it shows the intersection of the outgoing X1 axis with the 2nd Red worldline right after the turnaround is complete. If one would like to nit-pick, they could point to the round off of the total round trip time given as 10 years for the traveling twin (the small turnaround time was not included, which would add some decimal value to the total 10 year number).

`

Well I don't think you "messed up" as this is a perfectly standard chart for an instant turnaround. While I am sure all agree this is an accurate charting of two separate inertial frames the question at hand is how this relates to a single extended accelerated frame, yes?

The diagram shows the intersection of the simultaneous space of the outgoing X1 axis with the 2nd Red worldline right at the start of the outgoing trip. And then it shows the intersection of the outgoing X1 axis with the 2nd Red worldline right after the turnaround is complete.

Yes this is self evident but does not address my simple explicit question.

Given a single extended accelerated physical system of clocks and rulers do you think that right after turnaround this system would be congruent with x₁ intersecting event A?
Would there, or not, be a traveler at A with a clock reading of t'=5+(a hair)??

 

[bobc2]

Well I don't think you "messed up" as this is a perfectly standard chart for an instant turnaround. While I am sure all agree this is an accurate charting of two separate inertial frames the question at hand is how this relates to a single extended accelerated frame, yes?

You miss the point here. I explicitly indicated that this is not a chart for an instant turnaround. I emphasized that the curved path portion is so small on this scale that I couldn't represent it with the limited chart space. The Lorentz frames with the simultaneous spaces indicated actually occur before the turnaround and then after the turnaround.

If you don't get this sketch, I could easily select simultaneous spaces much farther away from the start of the trip and from the turnaround. I didn't think this would be such a problem. I'm not showing an accelerated frame at all, so that's not relevant here.

Yes this is self evident but does not address my simple explicit question.

Given a single extended accelerated physical system of clocks and rulers do you think that right after turnaround this system would be congruent with x₁ intersecting event A?
Would there, or not, be a traveler at A with a clock reading of t'=5+(a hair)??

I'm not talking about a single accelerated physical system here. In an earlier post, I analyzed the turnaround using a sequence of incremental inertial frames. We got tangled up with straw men, so now I've simplified the discussion to avoid arguing over single accelerated systems. And the fundamental point illustrated using two frames (outgoing and returning) which are clearly not accelerated is still the same as what I was trying to illustrate all along--just the fundamental concept of simultaneity and the interesting feature of events A, B, and C. Again, those features do not have to be interesting to you or anyone else. It was just a comment in the event anyone else might be interested.

 

[bobc2]

Yes, but apparently not by everyone. Particularly since we get many novices and students, it is a point that bears mentioning and you didn't so I did.

Do you have a problem with Minkowski space-time diagrams in general? Or, is it just when someone refers to the "simultaneous spaces" that show up in the space-time diagrams?

 

[Dale]

Do you have a problem with Minkowski space-time diagrams in general? Or, is it just when someone refers to the "simultaneous spaces" that show up in the space-time diagrams?

Neither. I have a problem when people make mathematically invalid statements and persist in doing so when their error is pointed out to them.

A "sequence of simultaneous spaces" is a simultaneity convention, in this case a non-inertial one. You may try to disguise it all you like, but that is what you are doing. Inventing new terms like "3D worlds" and "simultaneous spaces" doesn't change what it is.

 

[Dale]

And the fundamental point illustrated using two frames (outgoing and returning) which are clearly not accelerated is still the same as what I was trying to illustrate all along--just the fundamental concept of simultaneity and the interesting feature of events A, B, and C.

In neither of those frames does B come before A nor C before B.

 

[bobc2]

In neither of those frames does B come before A nor C before B.

We may not be talking about the same thing. I've made it clear before that in the 2nd Red guy's frame, as the Red guy moves along his worldline, he encounters event A first, then event B, then event C.

However, as the blue guy moves along his worldline, the event B is presented to his outgoing simlultaneous space first (right at the start of the outgoing trip). Then. event A is presented to blue's simultaneous space just after blue completes his turnaround.

Also, just before blue enters his turnaround path, event C is presented to blue's simlultaneous space. Then, event A is not presented to the blue simultaneous space until after the turnaround is complete.

 

[Dale]

We may not be talking about the same thing. I've made it clear before that in the 2nd Red guy's frame, as the Red guy moves along his worldline, he encounters event A first, then event B, then event C.

Not just the red guys frame, but all inertial frames.

However, as the blue guy moves along his worldline, the event B is presented to his outgoing simlultaneous space first (right at the start of the outgoing trip). Then. event A is presented to blue's simultaneous space just after blue completes his turnaround.

Also, just before blue enters his turnaround path, event C is presented to blue's simlultaneous space. Then, event A is not presented to the blue simultaneous space until after the turnaround is complete.

And here you go with your non inertial frame.

A sequence of simultaneous spaces is a simultaneity convention. And I have already told you that the naive simultaneity convention used here cannot cover the red worldline because it violates the few mathematical requirements of a coordinate system.

Your statement is mathematically invalid, as I pointed out well over 100 posts ago. I don't know why you persist in it.

 

[zonde]

I would like to bring up this earlier comment. I am trying to understand bobc2's position.

You should understand that most forum members here would insist that the Andromeda Paradox does not represent anything about physical reality. It could be taken as a pedagogical illustration to help graphically visualize aspects of the mathematics of special relativity. They consider that there are other equally valid interpretations of special relativity, such as the Lorentz Aether theory, that are in conflict with any notion that the Andromeda Paradox illustrates something about reality--thus, in that view, the block universe is not to be taken as a true characterization of external physical reality.

Paradoxes do not represent reality. Paradox reveals contradiction and we do not allow idea that reality could be self contradictory. Only models of reality can be self contradictory (and therefore flawed).

So I would ask in what model Andromeda Paradox is supposed to appear.

 

[PAllen]

However, as the blue guy moves along his worldline, the event B is presented to his outgoing simlultaneous space first (right at the start of the outgoing trip). Then. event A is presented to blue's simultaneous space just after blue completes his turnaround.

Also, just before blue enters his turnaround path, event C is presented to blue's simlultaneous space. Then, event A is not presented to the blue simultaneous space until after the turnaround is complete.

And this is the core of disagreement. You speak of blue's simultaneous space as if it has some physical meaning. Further, since blue, after turnaround, has a different past than the past of the post turnaround inertial frame, any physical procedure defining simultaneity will come out different for the blue observer than for an observer always at rest in the post turnaround inertial frame. Finally, even as a mathematical convention, talking about blue's simultaneous spaces does imply an overall simultaneity convention for the blue world line. For this, there are mathematical requirements - any region where a proposed simultaneity convention for blue has intersecting surfaces is outside the domain of that convention. If you want to talk about a blue simultaneity for such a region, you must adopt a different convention that does not have intersecting surfaces - of which there are many.

 

[Austin0]

Quote by Austin0

Well I don't think you "messed up" as this is a perfectly standard chart for an instant turnaround. While I am sure all agree this is an accurate charting of two separate inertial frames the question at hand is how this relates to a single extended accelerated frame, yes?

You miss the point here. I explicitly indicated that this is not a chart for an instant turnaround. I emphasized that the curved path portion is so small on this scale that I couldn't represent it with the limited chart space. The Lorentz frames with the simultaneous spaces indicated actually occur before the turnaround and then after the turnaround.

If you don't get this sketch, I could easily select simultaneous spaces much farther away from the start of the trip and from the turnaround. I didn't think this would be such a problem. I'm not showing an accelerated frame at all, so that's not relevant here.

yes of course ,instant turnaround is a convenient idealization,
it was understood that this depicted inertial phases before and after turnaround.
When you attribute the x ₁ axis to the traveler you are implicitly applying it to an accelerated frame. I.e. there is no turnaround without acceleration. So whether the traveler frame is accelerating at that time is not relevant , In the context of the overall trip it is non-inertial.
Quote by Austin0

Yes this is self evident but does not address my simple explicit question.

Given a single extended [edit out-accelerated] co-moving physical system of clocks and rulers do you think that right after turnaround this system would be congruent with x1 intersecting event A?
Would there, or not, be a traveler at A with a clock reading of t'=5+(a hair)??

I'm not talking about a single accelerated physical system here. In an earlier post, I analyzed the turnaround using a sequence of incremental inertial frames. We got tangled up with straw men, so now I've simplified the discussion to avoid arguing over single accelerated systems. And the fundamental point illustrated using two frames (outgoing and returning) which are clearly not accelerated is still the same as what I was trying to illustrate all along--just the fundamental concept of simultaneity and the interesting feature of events A, B, and C. Again, those features do not have to be interesting to you or anyone else. It was just a comment in the event anyone else might be interested.

Well you have completely avoided answering my question again.
Whether or not you are talking about a single accelerated system I am asking your thought regarding the x₁ axis as it would apply to such a system (with the edit above).

If there was such a co-moving system at that time after turnaround (inertial) would it correspond (be congruent) to the x₁ axis in your chart?
Would there, or not, be a traveler at A with a clock reading of t'=5+(a hair)??

 

[zonde]

When we speak about reality, do we mean only single moment of space or do we include all past and all future?
I believe that with reality we mean single slice of spacetime i.e. we do not include all past and all future.

 

[PAllen]

When we speak about reality, do we mean only single moment of space or do we include all past and all future?
I believe that with reality we mean single slice of spacetime i.e. we do not include all past and all future.

I value my past and future possibilities perhaps more than you More seriously what slice? Through a given event (e.g. me hitting submit for this post), there are uncountably infinite spacelike slices.

 

[bobc2]

You folks still seem to be troubled by the approximaty of the simultaneous spaces to the accelerated turnaround point in my sketches. So, here is a sketch where we consider the simultaneous spaces far far removed (years and millions of miles) from the turnaround neighborhood. We still have the same interesting feature about the order of events. Event A occurs before event B in the 2nd Black's rest frame. However, for the traveling twin moving along his worldline, event B is presented to his return trip simultaneous space before it is presented to his outgoing simultaneous space.

To emphasize the distinction between the outgoing frame and the return trip frame (not a single acceleration frame), I've colored the outgoing frame blue and the return frame red. The stay-at-home twin has the worldline along the black X4 axis.

 

[Dale]

However, for the traveling twin moving along his worldline, event B is presented to his return trip simultaneous space before it is presented to his outgoing simultaneous space.

Again, as we have already covered over and over again and again, your comments about simultaneous spaces are defining a simultaneity convention, and that convention is non-inertial. All of my previous comments hold. This approach violates the mathematical requirements in the region of the 2nd (now) black observer, so it is not a valid simultaneity convention for that region. If my many posts on this topic were not sufficiently clear, then please read PeterDonis' post 190, which is very well written.

You seem to think that I am having difficulty understanding your point. I understand your point quite clearly. Your point is not unclear, it is wrong.

To emphasize the distinction between the outgoing frame and the return trip frame (not a single acceleration frame), I've colored the outgoing frame blue and the return frame red. The stay-at-home twin has the worldline along the black X4 axis.

In neither the blue frame nor in the red frame does B come before A.

EDIT: oops, it is PAllen's post 190

 

[PeterDonis]

You folks still seem to be troubled by the approximaty of the simultaneous spaces to the accelerated turnaround point in my sketches...We still have the same interesting feature about the order of events.

Bobc2, a while back you did say that this "reversal of time" was just an interesting feature, nothing more. But now you appear to be saying that it *is* more; that there is some genuine physical meaning to the "interesting feature". And the pushback you are getting is because of the obvious paradoxical consequences of such a claim. So which is it?

 

[PAllen]

You folks still seem to be troubled by the approximaty of the simultaneous spaces to the accelerated turnaround point in my sketches. So, here is a sketch where we consider the simultaneous spaces far far removed (years and millions of miles) from the turnaround neighborhood. We still have the same interesting feature about the order of events. Event A occurs before event B in the 2nd Black's rest frame. However, for the traveling twin moving along his worldline, event B is presented to his return trip simultaneous space before it is presented to his outgoing simultaneous space.

To emphasize the distinction between the outgoing frame and the return trip frame (not a single acceleration frame), I've colored the outgoing frame blue and the return frame red. The stay-at-home twin has the worldline along the black X4 axis.

The question is the occurrence of this 'interesting feature' is part of the definition of the applicability of this simultaneity convention to the non-inertial observer. That is, the maximal spacetime region in which this simultaneity convention is applicable is the region in which there are no intersections of simultaneity surfaces. I could describe additional, physical plausibility criteria as well, but it is at least mathematically valid to use such a convention for the non-inertial observer as long as you don't try to cover a region of spacetime including such intersections.

Of course, I remain convinced that, even where applicable, making such statements as 'this is where the distant clock really runs faster than mine' are physically meaningless and conceptually grossly misleading.

 

[bobc2]

Again, as we have already covered over and over again and again, your comments about simultaneous spaces are defining a simultaneity convention, and that convention is non-inertial. All of my previous comments hold.

We will just have to agree to disagree. You concept of non-inertial simply does not apply to the two separate blue and red inertial frames in the above sketch.

This approach violates the mathematical requirements in the region of the 2nd (now) black observer, so it is not a valid simultaneity convention for that region

Neither of the separate individual frames violates the mathematical requirements. If I had been talking about a single non-inertial coordinates, I might have tried using Rindler coordinates or something, but then I would have to explain the Rindler horizen, etc. But, we are confronted with no such situation here.

If my many posts on this topic were not sufficiently clear, then please read PeterDonis' post 190, which is very well written.

There is no PeterDonis post no. 190.

You seem to think that I am having difficulty understanding your point. I understand your point quite clearly. Your point is not unclear, it is wrong.

In that case, your point is also clear--it is just wrong.

In neither the blue frame nor in the red frame does B come before A.

Just look at the space-time diagram. The intersections of the blue and red simultaneous spaces with the 2nd black worldline are there to see. There can be no mistaken about where the intersections are.

 

[bobc2]

The question is the occurrence of this 'interesting feature' is part of the definition of the applicability of this simultaneity convention to the non-inertial observer. That is, the maximal spacetime region in which this simultaneity convention is applicable is the region in which there are no intersections of simultaneity surfaces. I could describe additional, physical plausibility criteria as well, but it is at least mathematically valid to use such a convention for the non-inertial observer as long as you don't try to cover a region of spacetime including such intersections.

Of course, I remain convinced that, even where applicable, making such statements as 'this is where the distant clock really runs faster than mine' are physically meaningless and conceptually grossly misleading.

This makes no sense to me. By that logic you must dismiss the use of Minkowski space-time diagrams entirely, since by that definition there is probably no object in the universe that is inertial. All objects have accelerated at one time or another along the history of its worldline.

 

[bobc2]

Bobc2, a while back you did say that this "reversal of time" was just an interesting feature, nothing more. But now you appear to be saying that it *is* more; that there is some genuine physical meaning to the "interesting feature". And the pushback you are getting is because of the obvious paradoxical consequences of such a claim. So which is it?

Yes, an interesting feature. However, I did say that there is a sense in which one could consider his simultaneous space (as described by a Minkowski space-time diagram) as special. I described the experiences an observer could only have by experiencing these things in a Minkowski space-time frame. For example he measures the speed of light to be c, as do all other observers moving relative to him. And he experiences the laws of physics, as do all other observers living in Minkowski space-time frames. If he did not "live" in a Lorentz-Minkowski-Einstein frame he would not have those experiences, so that is the sense of which I referred.

However you misinterpreted my use of the word "experienced." Webster's dictionary gives two or three definitions for the use of that term. You chose the wrong definition where I thought the context made the definition I was applying very clear.

I'm not insisting anyone embrace that sense of specialness. Of course forum members can consider that observation or dismiss it--whatever their preference.