Velocity subtraction problem

[yuiop]

DO you have any specific objection or correction to the simultaneity analysis that I have totally taken account of?

In the frame of the tank, the arrival of the mark on the bottom track at the rear wheel (event 1 or E1) happens simultaneously with the arrival of the mark on the top track at the front wheel (event 2 or E2). The relativity of simultaneity tells us that if two events are simultaneous in one frame, then they will not be simutaneous in another frame.

... Taking Ich's derived velocity from the assumption of equal but opposite velocities in the tank frame ---- 0.74844 c for the measured velocity of the top track in the ground frame
and applying it within the ground frame, results in the marked point of the top track not completing the traversal from the top of the rear wheel to the top of the front wheel in the time that the bottom marked segment makes the complete translation.
Refering to the original drawing where the tank wheelbase moves its own length at 0.45 in the ground frame and the distance between the initial position of the top of the rear wheel and the final postion of the top of the front wheel is dx=17.860 as measured in the ground frame.
If we apply the derived velocity of 0.74844 to the marked top track segment and determine it's final position we get;... (dt19.844)*0.74844 = dx = 14.85204336 which is short of the opposite wheel center at x=17.860 .

During the same time interval the bottom marked sgment has traveled dx=17.860 reaching the opposite wheel center.

Clearly this is an asymmetric , physically untenable situation on every level.

As mentioned above, the relativity of simultaneity tells us that in another frame the two events E1 and E2 are not simultaneous. In the ground frame E1 happens before E2 which is what you have calculated, but you do not accept the result. It is not "untenable". It is what SR predicts. If you could show that the relativistic velocity addition formula contradicts the simultaneity equation then you might have an issue, but all you have done is shown is that they are consistent with each other. If you work out the additional time it takes the mark on the top track to arrive at the front wheel then you will find it is equal to the difference it simulataneity calculated by deltaT = Lv/c^2.

I welcome actual participation and critical analysis and if you find actual flaws in my analysis I will move onward

I did a lengthy analysis with detailed calculations earlier in this thread, showing how length contraction was consistent in this problem and you have chosen to ignore it and have not shown any flaws in it, so you are subject to the same criticism.

 

[yossell]

Austin0

I just can't get enough of that tank picture. I don't recognise the model though - is it one of those new top secret american ones, with the non-stretch tracks?

Refering to the original drawing where the tank wheelbase moves its own length at 0.45 in the ground frame and the distance between the initial position of the top of the rear wheel and the final postion of the top of the front wheel is dx=17.860 as measured in the ground frame.
If we apply the derived velocity of 0.74844 to the marked top track segment and determine it's final position we get;... (dt19.844)*0.74844 = dx = 14.85204336 which is short of the opposite wheel center at x=17.860 .

I've not checked the figures, but up to here I think I'm happy

During the same time interval the bottom marked sgment has traveled dx=17.860 reaching the opposite wheel center.

I wouldn't quite put it like that - we're analysing from the ground frame - so that part hasn't traveled at all. But it's ok - as your diagram shows, it's got the same x component.

Clearly this is an asymmetric , physically untenable situation on every level.

This is the point at which I get lost. Why is it untenable? I can think of two reasons (both have certainly worried me at some time).

(a) Because it's asymmetric? I don't expect a symmetry - the top part of the track is always moving wrt to the ground, the bottom part of the track is always stationary wrt to the ground. The set up is not symmetric, so I don't expect a symmetry in the result.

(b) Because, physically, as time passes, the top of the track will get more and more bunched up on itself while the bottom gets stretched out - eventually giving out?

But I don't think this will happen. Call the piece of track at T0, marked with a red arrow, A and the piece of track at B0, marked with a red arrow, B. Follow these round. It's true that, at the second moment you've drawn (the lower tank) A has `caught up' with B a little, as it chases it around the track. But if you continue the diagram for a full revolution around the track, and keep up with the analysis, drawing extra tanks where necessary, you'll find that, when B moves down to the lower track and A travels along the upper track, B `makes up the distance' so that, after a full revolution, A and B are both back to the beginning.

Or is it some other reason?

 

[yuiop]

Ken Natton
DO you understand that I am applying the velocity addition formula?

That absolutely everything that I have done is consistent with that formula?

That up until the last few posts I never argued that the top velocity must be 0.9 ??

Could you perhaps clarify the logic that the proposition that the top v is 0.9 , is inherently questioning the addition of velocities equation??
I would sincerely like to know.

The relativistic velocity addition formula is:

$$v= \frac{u+v'}{1+uv'/c^2}$$

where v' is the velocity of an object measured in frame S' and where u is the velocity of frame S' relative to frame S where v is measured.

Using the above formula the velocity of the top track in the ground frame (S) is (0.45+0.45)/(1+0.45*0.45/C^2)=0.74844C.

By declaring the velocity of the top track to be 0.45+0.45=0.9c you are choosing to ignore the validity of the relativistic addition formula.

As mentioned before, by declaring that the two marks should both arrive at their respective wheel tops simultaneously in both the tank frame and in the ground frame, you are choosing to ignore the validity of the relativity of simultaneity that declares that if two events are simultaneous in one frame, then the two events can NOT be simultaneous in a different frame that is not at rest wrt the first frame.

I just can't get enough of that tank picture. I don't recognise the model though - is it one of those new top secret american ones, with the non-stretch tracks?

Well it has to be streamlined to travel at nearly half the speed of light and the long-wheelbase-low-profile design is essential because at relativistic speeds the wheelbase gets shorter relative to its height and there is a danger of the tank toppling over.

I wouldn't quite put it like that - we're analysing from the ground frame - so that part hasn't traveled at all. But it's ok - as your diagram shows, it's got the same x component.

This is the point at which I get lost. Why is it untenable? I can think of two reasons (both have certainly worried me at some time).

(a) Because it's asymmetric? I don't expect a symmetry - the top part of the track is always moving wrt to the ground, the bottom part of the track is always stationary wrt to the ground. The set up is not symmetric, so I don't expect a symmetry in the result.

(b) Because, physically, as time passes, the top of the track will get more and more bunched up on itself while the bottom gets stretched out - eventually giving out?

But I don't think this will happen. Call the piece of track at T0, marked with a red arrow, A and the piece of track at B0, marked with a red arrow, B. Follow these round. It's true that, at the second moment you've drawn (the lower tank) A has `caught up' with B a little, as it chases it around the track. But if you continue the diagram for a full revolution around the track, and keep up with the analysis, drawing extra tanks where necessary, you'll find that, when B moves down to the lower track and A travels along the upper track, B `makes up the distance' so that, after a full revolution, A and B are both back to the beginning.

Your reasoning seems sound to me.

 

[Austin0]

Hi yossell
Likewise, in my original workup that derived a top track velocity of 0.756429 for the tank frame ;
If we instead insert the assumption of 0.45 velocity for the top marked segment in the tank frame , this results in (dt'=13.22)*0.45 =dx'= 5.94999 which is also only part of the way to the opposite wheel center while the bottom segment traverses the complete distance.

Additionally it appears that the percentage of the total distance traveled in each frame is significantly different suggesting a problem wrt frame agreement on local events although I haven't pursued this yet.

Austin0

I just can't get enough of that tank picture. I don't recognise the model though - is it one of those new top secret american ones, with the non-stretch tracks?

Glad to provide a smile. Yes you're right it's a yank tank,but I suspect they stole the design from the Israeli's. if you really liked it I have an even nicer design of a relativistic helocopter that doubles as a Schwarzschild observation platform.
Just send requests and donations to----- austin0 low Jester to the Court of 0reZ :tongue:
I hope you weren't implying I ever suggested any objection to the neccessity of non-rigid tracks

If we apply the derived velocity of 0.74844 to the marked top track segment and determine it's final position we get;... (dt19.844)*0.74844 = dx = 14.85204336 which is short of the opposite wheel center at x=17.860 .

I've not checked the figures, but up to here I think I'm happy

During the same time interval the bottom marked sgment has traveled dx=17.860 reaching the opposite wheel center.

I wouldn't quite put it like that - we're analysing from the ground frame - so that part hasn't traveled at all. But it's ok - as your diagram shows, it's got the same x component.

AH just a typo. Actually you're being generous here as this is not only clearly wrong but brainless.

Clearly this is an asymmetric , physically untenable situation on every level.

This is the point at which I get lost. Why is it untenable? I can think of two reasons (both have certainly worried me at some time).

(a) Because it's asymmetric? I don't expect a symmetry - the top part of the track is always moving wrt to the ground, the bottom part of the track is always stationary wrt to the ground. The set up is not symmetric, so I don't expect a symmetry in the result.

(b) Because, physically, as time passes, the top of the track will get more and more bunched up on itself while the bottom gets stretched out - eventually giving out?

c)But I don't think this will happen. Call the piece of track at T0, marked with a red arrow, A and the piece of track at B0, marked with a red arrow, B. Follow these round. It's true that, at the second moment you've drawn (the lower tank) A has `caught up' with B a little, as it chases it around the track. But if you continue the diagram for a full revolution around the track, and keep up with the analysis, drawing extra tanks where necessary, you'll find that, when B moves down to the lower track and A travels along the upper track, B `makes up the distance' so that, after a full revolution, A and B are both back to the beginning.

Or is it some other reason?

Hi yossell
a) There are all kinds of symmetry here. E.g. if we assume that prior to liftoff the two segments were marked symmetrically equidistant on the track then this implies that mechanically there are certain constraints on how much asymmetry is tolerable.
It is a given that at least the forward drive wheel must be a sprocket , yes?
In principle we could make both wheels geared as I have suggested . The actual areas of meshing are mostly transverse to motion so contraction should not be a problem and as per kev, the axles could be flexibly mounted.
In this instance the amount of asymmetry is limited and in any case it would seem that the point of orthoganal alignment between the top and bottom marks must occur at the midpoint of the wheel base. No?? In at least one frame.
If not perhaps an explanation.

b) No actually that was not my assumption. Particularly as JesseM had just demonstrated how this would not neccessarily be the case.BTW using a stretchable track which I had no problem with but which Ich took exception to.
Also we are all agreed that top to bottom and bottom to top feed must be equivalent , in fact , if not neccessarily as measured in both frames.
This a physical constraint.

c) This certainly a valid point of exploration although I am not quite sure of your explanation. The top mark has fallen behind the bottom as far as percentage of total circumferential travel.
I agree this is worth examining but just off the top, if we continue on just a bit to the point where the bottom mark is at the position of the top at the beginning, it is clear the top mark will not be at the comparable position of the bottom at the beginning.
SO in one half cycle the two points that were initially symmetrically equidistant, have gotten out of phase. I will work it out further
If you will look above I did the same analysis with the assumption of 0.45 measured in the tank frame. WIth a similar but quantitatively different result and significantly different percentage of circumference traveled
. ANY ideas how this would be reconciled?
Thanks for your criticism any additional input appreciated.

That's the spirit - rational, disinterested, unprejudiced inquiry.
.

WOrds to live by

 

[yossell]

I hope you weren't implying I ever suggested any objection to the neccessity of non-rigid tracks

Having seen how you've dealt with others, and having seen the kind of tanks you're equipped with, I wouldn't dare.

a) There are all kinds of symmetry here.

But I want an exact formulation of the problem or problems, just to understand and see if I can replicate and/or resolve it. Yes, there are some symmetries, but there are asymmetries too. If one (note: `one' - not you, not Austin0, not anybody, just a hypothetical *one*, not accusing you of anything, please believe me, no, NO! That rumbling??...not the tanks, not the tanks, noooo...) is worried by a certain asymmetry Asym, or one thinks it shouldn't be there, then I need to understand why, given that there are asymmetries in the set up, Asym is a problem.

E.g. if we assume that prior to liftoff the two segments were marked symmetrically equidistant on the track then this implies that mechanically there are certain constraints on how much asymmetry is tolerable.

`Prior to liftoff'? From your diagram, which is where I was starting, there was no prior to liftoff (bear with me). The tank was already in motion and we on the ground picked out a couple of points on the track at a time from our reference frame, and took things from there.

There are other interesting questions to be asked about the correct description of a tank that starts off stationary wrt the ground, and then accelerates to a constant velocity v. This involves a new component of acceleration - as well as the wheels spinning, the tank accelerates - so there is no straightforward analysis of the whole set up from the tank's point of view, as it is not an inertial object. For instance, if the people in the tank try and keep the front wheel and back wheel always in sync as they accelerate, the people from the ground will think they are not applying the forces at the same time, and the wheels will get out of sync. The analysis of the take off is very complex, and will depend upon the details of how the two wheels are accelerated.

At the very least, I see no obvious problem for relativity here. Not, of course, that you were suggesting that there was one.

It is a given that at least the forward drive wheel must be a sprocket , yes?

Uhh - my theoretical bent betrays me and I have no idea what a sprocket is. However, hopefully it doesn't matter for this problem and we can set the sprockets and brackets and thingies aside.

In this instance the amount of asymmetry is limited and in any case it would seem that the point of orthoganal alignment between the top and bottom marks must occur at the midpoint of the wheel base. No?? In at least one frame.

First, Bracket the `in at least one frame'. Then, off the top of my head, I don't see this. Do you have a calculation that shows it? Now, include the `in at least one frame'. I don't know what you mean - in other frames, the points that we're looking at from the point of view of the ground frame will not be simultaneous. So what are we looking at? By contrast to what was going on in your diagram, where I really felt I understood the situation we were analysing, I've now lost track. So: can you spell out the worries you have here more precisely.

SO in one half cycle the two points that were initially symmetrically equidistant, have gotten out of phase.

So it seems to me too - but I don't find this inherently problematic.

If you will look above I did the same analysis with the assumption of 0.45 measured in the tank frame. WIth a similar but quantitatively different result and significantly different percentage of circumference traveled
. ANY ideas how this would be reconciled?

Yes, I was aware that you had said this. Maybe it's already there in the picture. But to save me trawling back trying to find the argument, is it possible that, just as nicely as you did from the ground frame, you could give your analysis, highlighting the qualitative difference. Again, due to relativity of simultaneity and length contraction, none of which you are questioning or contesting, I expect qualitatively different answers from different frames. If you would show the analysis of this particular tank situation and the problematic discrepancy, then I could compare the two and see if it's resolvable.

 

[Austin0]

a) There are all kinds of symmetry here.

But I want an exact formulation of the problem or problems, just to understand and see if I can replicate and/or resolve it. Yes, there are some symmetries, but there are asymmetries too. If one is worried by a certain asymmetry Asym, or one thinks it shouldn't be there, then I need to understand why, given that there are asymmetries in the set up, Asym is a problem..

if we assume that prior to liftoff the two segments were marked symmetrically equidistant on the track then this implies that mechanically there are certain constraints on how much asymmetry is tolerable..

`Prior to liftoff'? From your diagram, which is where I was starting, there was no prior to liftoff (bear with me). The tank was already in motion and we on the ground picked out a couple of points on the track at a time from our reference frame, and took things from there.

Agreed on all counts. This complicated enough without worrying about acceleration.
On the other hand we can simply posit that the track was marked equidistantly beforehand and disregard how it got up to speed , with the simple assumption that there are an equal number of segments between the marks in both directions, acceptable?

It is a given that at least the forward drive wheel must be a sprocket , yes?
In principle we could make both wheels geared as I have suggested .

Uhh - my theoretical bent betrays me and I have no idea what a sprocket is. However, hopefully it doesn't matter for this problem and we can set the sprockets and brackets and thingies aside..

A sprocket is (as you're obviously anxious to know) simply a geared drive wheel for a chain or track.
It may be pertinent to the problem as, like stretchable track or springs between axles it matters to the mechanics of the contraption. Also as a boundary condition it might simplify
matters as gears on both wheels would mechanically guarantee equal feed in both directions top tobottom etdc. as well as equal numbers of segments on top and bottom.

In this instance the amount of asymmetry is limited and in any case it would seem that the point of orthoganal alignment between the top and bottom marks must occur at the midpoint of the wheel base. No?? [In at least one frame.].

First, Bracket the `in at least one frame'. Then, off the top of my head, I don't see this. Do you have a calculation that shows it? Now, include the `in at least one frame'. I don't know what you mean - in other frames, the points that we're looking at from the point of view of the ground frame will not be simultaneous. So what are we looking at? By contrast to what was going on in your diagram, where I really felt I understood the situation we were analysing, I've now lost track. So: can you spell out the worries you have here more precisely..

Yep bracketed! The basis for the assumption of the tank as a preferred frame is that mechanically it must be symmetrical in that frame. I.e. equal velocity top and bottom.
This implies that equidistant marks must be aligned mid point of wheel base as they make the transit full circle.. At this point contraction is balanced between front and back sections as delineated by the midpoint , yes?
This of course fine logic , it is hard to picture the mechanism working otherwise but this applies equally to the ground frame because there is only one track.
If it is accepted that this symmetry must actually be present we also know that it could only be measured as such in one frame. This of course the question at hand, which one?
My precise worry, which is daily growing is that I may be stuck in perpetuity jumping back and forth from the gound to an unlikely tank going nowhere at 0.45 c unable to find a clear reason to prefer a frame.

I agree this is worth examining but just off the top, if we continue on just a bit to the point where the bottom mark is at the position of the top at the beginning, it is clear the top mark will not be at the comparable position of the bottom at the beginning.
SO in one half cycle the two points that were initially symmetrically equidistant, have gotten out of phase. I will work it out further
. ANY ideas how this would be reconciled?.

So it seems to me too - but I don't find this inherently problematic..

If you will look above I did the same analysis with the assumption of 0.45 measured in the tank frame. WIth a similar but quantitatively different result and significantly different percentage of circumference traveled.

Yes, I was aware that you had said this. Maybe it's already there in the picture. But to save me trawling back trying to find the argument, is it possible that, just as nicely as you did from the ground frame, you could give your analysis, highlighting the qualitative difference. Again, due to relativity of simultaneity and length contraction, none of which you are questioning or contesting, I expect qualitatively different answers from different frames. If you would show the analysis of this particular tank situation and the problematic discrepancy, then I could compare the two and see if it's resolvable.

Point taken.
On simultaneity ; a thought.
If we posit sequential numbering on the track segments , in the ground frame the number of segments at rest is easily observed. At both ends there is a segment translating either up or down , these events must be simultaneous according to the ground frame's clocks and also simultaneous according to the physical mechanics of the situation. As the intervening track is actually at rest in this frame , it might suggest a preference for this simultaneity over the frame where it is all in motion . or maybe not??

Thanks and also for a good laugh

 

[Austin0]

Here is the problem analysed in terms of length contraction.

Let us say we have a simple tank with axle to axle proper length of 1.0 and a proper track length (considering only the horizontal portions of the track) of 2.0 when the tank engine is off.

In the tank frame (when the engine is on), the lower and upper parts of the track both have relative velocities of magnitude 0.45c so the half length of the track is 0.89302855 in the tank frame due to length contraction. One of the axles would have to have a tension control device to allow one of the axles to move inwards to prevent the track snapping under increased tension and so the axle to axle length in the tank frame would also have to be 0.89302855.

In the ground frame, the tank is moving at 0.45c and the axle to axle length is length contracted to 0.89302855*sqrt(1-0.45^2)= 0.7975.

The part of the track in contact with ground is stationary with respect to the ground frame so that part of the track has proper length 0.7975. The top part of the track also measures 0.7975 in the ground frame but because the top part of the track has relative velocity 0.7484407c in the ground frame (according to relativity), the proper length of the top part of the track is 0.7975/sqrt(1-0.7484407^2) = 1.2025. The total proper length of the track is therefore 0.7975+1.2025=2.0 in the ground frame.

Therefore a top track velocity of 0.7484407 is consistent with the relativistic addition laws and with relativistic length contraction if the total proper length of the track remains the same (2.0) in all frames. All this is totally consistent with relativity.

If the velocity of the top track was 0.9c, the total proper length of the track would not be 2.0 in both frames.

Therefore the velocity of the top track can not be 0.9c, if length contraction is a real physical effect and if proper length is invariant.

In the frame of the tank, the arrival of the mark on the bottom track at the rear wheel (event 1 or E1) happens simultaneously with the arrival of the mark on the top track at the front wheel (event 2 or E2). The relativity of simultaneity tells us that if two events are simultaneous in one frame, then they will not be simutaneous in another frame..

Of course you are saying this with the, already arrived at conclusion of the tank as preferred frame. Iti s equally applicable from either frame.

As mentioned above, the relativity of simultaneity tells us that in another frame the two events E1 and E2 are not simultaneous. In the ground frame E1 happens before E2 which is what you have calculated, but you do not accept the result. It is not "untenable". It is what SR predicts. If you could show that the relativistic velocity addition formula contradicts the simultaneity equation then you might have an issue, but all you have done is shown is that they are consistent with each other. If you work out the additional time it takes the mark on the top track to arrive at the front wheel then you will find it is equal to the difference it simulataneity calculated by deltaT = Lv/c^2. .

As I said, I did the analyses from both frames and am aware of this point . It is once again deciding which frame to be simultaneous.
To my understanding it would be an idiots understaking to try and prove the addition of velocities formula inconsistent with relative simultaneity as the latter is implicit in the additions formula. In fact the additions formula is just a convenience and this whole question could have been approached without its referernce or use. Yes??

I did a lengthy analysis with detailed calculations earlier in this thread, showing how length contraction was consistent in this problem and you have chosen to ignore it and have not shown any flaws in it, so you are subject to the same criticism.

You are right. I looked at it and thought it was interesting, but I was focused on simultaneity and thought contraction was a secondary issue so I didn't give it the proper thought I should have.
Having done so, I see how it is, an actually compelling argument, that perhaps could have saved me a lot of time and effort pursuing simultaneity.
I will give it some more thought but the only problem I see now is that the bottom section , at rest wrt the ground must be completely uncontracted. So in the ground frame there is 60% of the proper total length located in the top section while in the tank frame they are symmetrical.
This strange situation is of course equally a problem no matter what the assumed velocity is. If it is assumed that the segments are numbered it should be interesting to see how there could be frame agreement regarding observations and counts.
In any case you may have provided me with a ticket out of this tank as all other considerations were appearing completely reciprocal.

Thanks for your input

 

[Austin0]

The relativistic velocity addition formula is:

$$v= \frac{u+v'}{1+uv'/c^2}$$

where v' is the velocity of an object measured in frame S' and where u is the velocity of frame S' relative to frame S where v is measured.

Just a little calrification while I think about your contraction workup.

Using the above formula the velocity of the top track in the ground frame (S) is (0.45+0.45)/(1+0.45*0.45/C^2)=0.74844C.

This is totally inaccurate as applied in the ground frame. You must mean in the tank frame it is calculated that in the ground frame this measurement would apply.

By declaring the velocity of the top track to be 0.45+0.45=0.9c you are choosing to ignore the validity of the relativistic addition formula.

I certainly never declared any such thing. ANd definitely never stated or myself applied this ridiculous math, which is invalid in any frame.
I assumed an empirical measurement in the ground frame of 0.9 based on geometry and physics and applied the same formula that the OP originally used. 0.9 - 0.45 to derive a figure for the tank measurement.

You all seem to think this is somehow unusual or outside normal application of SR.

Given a problem clearly stated as being measured in the ground frame to determine the predicted measurement in the tank frame this is the natural order of the universe.
This application is just as consistent as the other approach because of course the additions equation is consistent.

Given the complexity of this unusual problem I don't question the expedient of jumping to another frame but it is retarded to claim that it that is more correct or more in compliance with the additions formula. It is just doing exactly the same thing. Using priciples of geometry and physics to assume a measured velocity for the top and then applying the equation.

It is a pure strawman to try to imply that disagreeing or questioning the appropriate frame to use, is questioning the validity of the formula or refusing to accept its conclusions

As mentioned before, by declaring that the two marks should both arrive at their respective wheel tops simultaneously in both the tank frame and in the ground frame, you are choosing to ignore the validity of the relativity of simultaneity that declares that if two events are simultaneous in one frame, then the two events can NOT be simultaneous in a different frame that is not at rest wrt the first frame.

Here again; I never declared that the two marks should be measured as arriving at the respective wheel tops in both frames. You are just ignoring what I calculated in black and white which was absolutely not simultaneous in one frame. SO you are just putting words in my mouth and declaring I am ignoring the validity of relative simultaneity.
Maybe we should just have a beer?

 

[phyti]

The track and tank do not move independently of each other, so it's not a case of a spaceship launching a space probe in the same direction requiring the composition formula.
The tank and track are parts of a composite object.
The tank requires constant acceleration to maintain a given speed, therefore it's not an inertial frame.
The tank and Earth are parts of a composite object.

 

[matheinste]

The track and tank do not move independently of each other, so it's not a case of a spaceship launching a space probe in the same direction requiring the composition formula.
The tank and track are parts of a composite object.
The tank requires constant acceleration to maintain a given speed, therefore it's not an inertial frame.
The tank and Earth are parts of a composite object.

Waht is your definition of a composite object. My intuitive and very very loose idea of such a body would be one whose "component" parts are constrained to remain at rest relative to each other when referred to an inertial frame in which one (any one) of the component parts is at rest. Is there an accepted definition?

Surely an object is moving inertially if the resultant forces acting upon it are zero. In this case the resultant of frictional forces and forward acceleration cancel out to give a constant forward speed for the body of the tanks. As in most cases of course we consider the earth, for our purposes, to be moving inertially

Matheinste

 

[Austin0]

Waht is your definition of a composite object. My intuitive and very very loose idea of such a body would be one whose "component" parts are constrained to remain at rest relative to each other when referred to an inertial frame in which one (any one) of the component parts is at rest. Is there an accepted definition?

Surely an object is moving inertially if the resultant forces acting upon it are zero. In this case the resultant of frictional forces and forward acceleration cancel out to give a constant forward speed for the body of the tanks. As in most cases of course we consider the earth, for our purposes, to be moving inertially

Matheinste

Are you saying that the tank is inertial because the constant acceleration is being evenly countered by friction and inertia to maintain a steady velocity??

Do you think that there is no difference between a tank suspended by Mike 's crane with no resistence on the bottom track and the same tank on the ground where the bottom track is under great force, between the friction and acceleration of the ground and the acceleration of the drive which is suffcient to overcome that and propel the mass forward??

DO you think the actual length contraction could be uneffected by the counter force of the Earth and the stress involved?
I.e. the air tank could have exactoly the same contraction of the bottom track as the ground tank where there is great force in effect, stretching or resisting contraction?

 

[matheinste]

Are you saying that the tank is inertial because the constant acceleration is being evenly countered by friction and inertia to maintain a steady velocity??

Do you think that there is no difference between a tank suspended by Mike 's crane with no resistence on the bottom track and the same tank on the ground where the bottom track is under great force, between the friction and acceleration of the ground and the acceleration of the drive which is suffcient to overcome that and propel the mass forward??

DO you think the actual length contraction could be uneffected by the counter force of the Earth and the stress involved?
I.e. the air tank could have exactoly the same contraction of the bottom track as the ground tank where there is great force in effect, stretching or resisting contraction?

Basic physics says that if the velocity of a body is constant the the resultant sum of the forces applied is to it zero and vice versa. Constant velocity of an object implies that the object is at rest in some inertial frame. The top and bottom tracks are not at rest in the same frame as each other or at rest in the same rest frame as the tank.

As for stressed contactions and stretchings, that's more of an engineering problem and too complicated.

Matheinste

 

[Austin0]

Basic physics says that if the velocity of a body is constant the the resultant sum of the forces applied is to it zero and vice versa. Constant velocity of an object implies that the object is at rest in some inertial frame. The top and bottom tracks are not at rest in the same frame as each other or at rest in the same rest frame as the tank.

As for stressed contactions and stretchings, that's more of an engineering problem and too complicated.

Matheinste

In an elevator rising at constant velocity wrt the Earth do you think you are in an inertial frame??

In the research I have done so far on the Ehrenfest question I found that they definitely took into consideration the centrifugal forces inherent in acceleration up to speed.
That the conclusion was that the force of expansion would override contraction until the cessation of acceleration. In the tank question acceleration never stops but is an ongoing application of force.
SO it would seem to be more than simply an engineering question and while certainly complicated is that reason enough to ignore it and assume that the effects are not relevant?

You did not answer the main point: DO you think the contraction that would take effect could be the same in a tank suspended without stress or resistence or acceleration from the ground effecting the track and the same tank on the ground with the bottom in contact with the ground and the whole track also being subjected to the stress between the acceleration of the drive wheel and the inertia of the tank body?

 

[yuiop]

In an elevator rising at constant velocity wrt the Earth do you think you are in an inertial frame??

In this case the elevator is not an inertial frame, but then again, even when the elevator is stationary with respect to the Earth, it is still not in an inertial frame. This is an effect due to the gravity and in SR problems we normally ignore gravity. In the elevator (moving with constant velocity or stationary with respect to the Earth) a vertical accelerometer would always show a non-zero reading. The tank in the OP is not moving vertically. A vertical accelerometer in the tank would also show a non-zero (but constant) reading, but a horizontal accelerometer in the tank body would always read zero and it is the horizontal motion we are mainly concerned with.

In the tank question acceleration never stops but is an ongoing application of force.
...
You did not answer the main point: DO you think the contraction that would take effect could be the same in a tank suspended without stress or resistence or acceleration from the ground effecting the track and the same tank on the ground with the bottom in contact with the ground and the whole track also being subjected to the stress between the acceleration of the drive wheel and the inertia of the tank body?

The "ongoing application of force" is mainly to overcome air resistance. In these idealised SR thought experiments we normally consider everything to be taking place in a vacuum (because the speed of light and the Lorentz transformations assume a vacuum). A car moving at 180 mph would not require any fuel to maintain that speed in a vacuum (if we ignore the rolling/bearing friction). Over 90% of the horsepower and fuel consumption of a supercar cruising at 180 mph is being used to overcome air resistance. However, air or no air, if the tank body has constant velocity relative to the Earth surface then the tank will have zero proper acceleration in the horizontal direction which is the motion we are mainly concerned with. An accelerometer attached to a tank track segment will show acceleration as it travels around the curved sections of the track, but when moving along the top section, or along the bottom section in contact with the ground, the track segment would show zero horizontal acceleration and we can consider the tank body and horizontal sections of the track to have inertial motion for our purposes here. If you are analysing a problem, with a view to elucidating some physical fact it is best to keep things as simple as possible.

If we want to obfuscate everything as much as possible, perhaps to cover a blunder, then we keep adding complications to cover our "tracks" to make the problem intractable so that no one can figure out what is going on. For example, we could say that at 0.45c the tank is exceeding the escape velocity of the Earth and as it travels along the curved surface of the Earth, it will probably take off and go into orbit. You have probably not noticed until now that we are assuming a flat Earth in this problem, which we know is not true. Anyway, if we remove the atmosphere, assume a flat Earth and negligible rolling resistance, then in this idealised thought experiment the answer to "DO you think the contraction that would take effect could be the same in a tank suspended without stress or resistance or acceleration from the ground effecting the track and the same tank on the ground with the bottom in contact with the ground and the whole track also being subjected to the stress between the acceleration of the drive wheel and the inertia of the tank body" is (to a good approximation) yes.

No amount of obfuscating, such as considering air resistance or curvature of the Earth's surface is going to make it a universal truth, that the top track moves at exactly twice the velocity of the tank body in the Earth frame and that is the question posed in the OP, no?

 

[matheinste]

In an elevator rising at constant velocity wrt the Earth do you think you are in an inertial frame??

In the research I have done so far on the Ehrenfest question I found that they definitely took into consideration the centrifugal forces inherent in acceleration up to speed.
That the conclusion was that the force of expansion would override contraction until the cessation of acceleration. In the tank question acceleration never stops but is an ongoing application of force.
SO it would seem to be more than simply an engineering question and while certainly complicated is that reason enough to ignore it and assume that the effects are not relevant?

You did not answer the main point: DO you think the contraction that would take effect could be the same in a tank suspended without stress or resistence or acceleration from the ground effecting the track and the same tank on the ground with the bottom in contact with the ground and the whole track also being subjected to the stress between the acceleration of the drive wheel and the inertia of the tank body?

As regards your first question, although it sounds simple I am not confident that I can give a correct answer as my knowledge as yet is limited to SR but when gravity is involved, as implied by the elevator, we are in the realms of GR.

As regards stresses and strains on the tank and tracks I still feel this is an engineering problem. You must of course be careful to distinguish between SR length contraction, which involves no stress, and mechanical contractions which are stressed but not caused by relativistic effects. I am no expert but I suspect that the relativistic effects involved look after themselves.

Matheinste.

EDIT After reading kev's response, I see that my response is probably redundant.

 

[Austin0]

In this case the elevator is not an inertial frame, but then again, even when the elevator is stationary with respect to the Earth, it is still not in an inertial frame. This is an effect due to the gravity and in SR problems we normally ignore gravity. In the elevator (moving with constant velocity or stationary with respect to the Earth) a vertical accelerometer would always show a non-zero reading. The tank in the OP is not moving vertically. A vertical accelerometer in the tank would also show a non-zero (but constant) reading, but a horizontal accelerometer in the tank body would always read zero and it is the horizontal motion we are mainly concerned with.

So you agree that constant motion in this context is not the only consideration for inertial motion.
I was thinking more of the track itself but then again the normal acceleration on the mass of the tank is a relevant factor.

The "ongoing application of force" is mainly to overcome air resistance. In these idealised SR thought experiments we normally consider everything to be taking place in a vacuum (because the speed of light and the Lorentz transformations assume a vacuum). A car moving at 180 mph would not require any fuel to maintain that speed in a vacuum (if we ignore the rolling/bearing friction). Over 90% of the horsepower and fuel consumption of a supercar cruising at 180 mph is being used to overcome air resistance. However, air or no air, if the tank body has constant velocity relative to the Earth surface then the tank will have zero proper acceleration in the horizontal direction which is the motion we are mainly concerned with. An accelerometer attached to a tank track segment will show acceleration as it travels around the curved sections of the track, but when moving along the top section, or along the bottom section in contact with the ground, the track segment would show zero horizontal acceleration and we can consider the tank body and horizontal sections of the track to have inertial motion for our purposes here. If you are analysing a problem, with a view to elucidating some physical fact it is best to keep things as simple as possible.

Can we not consider a spring a simple accelerometer?
DO you think that springs between the track segments would have equal extension or compression between the top and bottom track?
Do you think that they would have equal extension or compression between the front wheel and the back in the top track??
DO you disagree that there is rapid decceleration of the top segments when they reach the drive wheel and are redirected around to the bottom??
And likewise when moving from the bottom to the top??
DO you think this would continue purely on inertia without a constant input of energy?

If we want to obfuscate everything as much as possible, perhaps to cover a blunder, then we keep adding complications to cover our "tracks" to make the problem intractable so that no one can figure out what is going on. For example, we could say that at 0.45c the tank is exceeding the escape velocity of the Earth and as it travels along the curved surface of the Earth, it will probably take off and go into orbit. You have probably not noticed until now that we are assuming a flat Earth in this problem, which we know is not true. Anyway, if we remove the atmosphere, assume a flat Earth and negligible rolling resistance, then in this idealised thought experiment the answer to "DO you think the contraction that would take effect could be the same in a tank suspended without stress or resistance or acceleration from the ground effecting the track and the same tank on the ground with the bottom in contact with the ground and the whole track also being subjected to the stress between the acceleration of the drive wheel and the inertia of the tank body" is (to a good approximation) yes.

Very subtle kev. Almost creates the impression of a meaningful reductio ad absurdum argument against considering acceleration a valid consideration in this problem.
At the same time implying that I am making similar objections of a whimsical nature such as your suggestions and impugning my motive for offering any suggestions at all.
Good work.

No amount of obfuscating, such as considering air resistance or curvature of the Earth's surface is going to make it a universal truth, that the top track moves at exactly twice the velocity of the tank body in the Earth frame and that is the question posed in the OP, no?

I will say it once again; it was only very late in this thread that I ever suggested that the 0.9c velocity might be correct. After consideration of your contraction workup I put that back on hold and have been considering contraction and it's implications as well as studying the Ehrenfest problem which I was not really familiar with before.
Having gotten a little information on this it is clear that in that problem, which is actually much simpler than the tank question , it took many years of work by many brilliant people to resolve the question and reach concensus. It was not considered a simple question and as far as I know now, was not resolved simply through normal application of contraction but required advanced coordinate systems and perspectives.
And the system in question was completely independant, after spinup there was no acceleration involved. There was no part of the system that was at rest in another frame and there were only two frames involved in the first place.
Here there are a minimum of 4 frames, a complex physical interaction between various combinations and other unique features.
So if you think that taking this complexity into consideration is obfuscation , that questioning a quick simple evaluation derived through absolutely ignoring all these other factors is just contrariness or whatever you think is my motivation I don't know what to tell you.
All I can say is I did not dismiss your work on contraction but have given it a great deal of thought and am still. I really don't care what the actual answer to the question is at this point , in a real sense I am just stuck by my own inability to forget it until SOME really satisfactory , totally compelling resolution is found by somebody.

 

[yuiop]

Can we not consider a spring a simple accelerometer?

Depends on how you use it. If you fix one end to the tank and the other to a test mass, then if the spring bends, the tank is accelerating (which it is not in this thread). If you mean springs connecting track segments, then maybe not. The springs might be stressed due to length contraction at constant speed (as in the Ehrenfest paradox). If there is significant air resistance then the force of the drive required to overcome the air resistance will be measured in the springs, but that measured stress is not a reflection of the acceleration of the tank because it would be there when the tank has constant velocity.

DO you think that springs between the track segments would have equal extension or compression between the top and bottom track?

The extension is equal in the tank frame if we have negligible air and rolling resistance. If you insist of have non negligible resistance, then you would have to figure out how much of the extension is due to resistance and how much is due to length contraction. The question is complicated enough without that. The way to solve things is to make the situation as idealistic and as simple as is reasonable.

DO you disagree that there is rapid deceleration of the top segments when they reach the drive wheel and are redirected around to the bottom?? And likewise when moving from the bottom to the top??

I am not disputing the acceleration. See next comment.

DO you think this would continue purely on inertia without a constant input of energy?

Yes, if the system had perfect bearings. It is the same as a wheel. It will continue to spin forever if it had a perfect bearing despite parts on the rim constantly being accelerated in different directions. In the case of the wheel we can simulate a perfect bearing by spinning the wheel in space without an axle.

Very subtle kev. Almost creates the impression of a meaningful reductio ad absurdum argument against considering acceleration a valid consideration in this problem.
At the same time implying that I am making similar objections of a whimsical nature such as your suggestions and impugning my motive for offering any suggestions at all.

I though my comment was mischievous at the time, but I unreservedly apologise for any impugning implied. I wholeheartedly accept that your motives are good and you have a genuine desire to solve and understand this problem to your own satisfaction. My bad.

I will say it once again; it was only very late in this thread that I ever suggested that the 0.9c velocity might be correct. After consideration of your contraction workup I put that back on hold and have been considering contraction and it's implications as well as studying the Ehrenfest problem which I was not really familiar with before.
Having gotten a little information on this it is clear that in that problem, which is actually much simpler than the tank question , it took many years of work by many brilliant people to resolve the question and reach concensus. It was not considered a simple question and as far as I know now, was not resolved simply through normal application of contraction but required advanced coordinate systems and perspectives.

The Ehrenfest problem is really quite simple if you accept length contraction and the clock postulate. It only took so long to figure out the paradox, because people over complicated it and had ingrained misconceptions that were hard to shake off.

So if you think that taking this complexity into consideration is obfuscation , that questioning a quick simple evaluation derived through absolutely ignoring all these other factors is just contrariness or whatever you think is my motivation I don't know what to tell you.

Well as Einstein said, "physics should be made as simple as possible, but no simpler" or something like that. If the top track is moving at 0.45c in the tank frame then it is moving at (0.45+0.45)/(1+0.45*0.45) in the ground frame and that is the end of the story and what the track does as it rounds the wheel is irrelevant. Trying to analyse what happens at the wheel in the ground frame is horrendously complicated because the wheel is elliptical in the ground frame and translating and as well as rotating at the same time. To get an idea of the complexity, have a look at this post in an old thread https://www.physicsforums.com/showthread.php?t=233399&highlight=rolling+wheel. Why go to all that trouble when it is not even required in this problem? You would have to be some sort of masochist to figure it out and fortunately Dalespam is that sort of masochist and did figure it out for us, but it not helpful here. :tongue:

All I can say is I did not dismiss your work on contraction but have given it a great deal of thought and am still. I really don't care what the actual answer to the question is at this point , in a real sense I am just stuck by my own inability to forget it until SOME really satisfactory , totally compelling resolution is found by somebody.

I think what is confusing is that the Newtonian analysis is perfectly mathematically correct and consistent on its own terms and so is the relativistic analysis. You can not determine what is a correct theory by mathematics alone. You have to carry out some actual experiments and see if they agree with the theory and the experimental results suggest Newtonian theory is wrong and SR is correct.

The Newtonian solution suggests the top track moves at twice the speed of the bottom track in the ground frame and that is perfectly self consistent, but it means you have to reject:

1) The relativistic velocity addition law.
2) Relativistic length contraction.
3) The relativity of simultaneity.
4) The postulates of relativity.

and everything else about relativity. If you accept any of the above 4 items, then the top track does not move at twice the speed of the bottom track in the ground frame.