Is Velocity Addition in Special Relativity Contradictory?

[JesseM]

As you can see above, under my assumption the track would not be bunched up but would be counted as having more segments due to the desynchronization between front and back, as measured in the track frame.

What do you mean by "more segments" though? Suppose we paint two dots on the track, as in the example I posted in response to Ich--would you agree the number of segments between the two dots should stay constant as the pair of dots moves from the bottom to the top? If so, the segments would have to change length in the frame where both dots were at rest (the rest frame of that section of the material that makes up the tracks), depending on whether the dots were both on top or both on the bottom.

 

[Ich]

Not true

Oh no, JesseM, now we're 122 posts and a few lightyears from the original post.
We started with a tank, and a velocity addition problem. We had the assumption of vtop=0.9, and we actually had a paradox based on this assumption.
When you change the scenario now to the equivalent of two cannons shooting bullets back and forth at different speeds, there's not much left to discuss. This is trivially possible and therefore totally uninteresting, IMHO.

 

[JesseM]

Oh no, JesseM, now we're 122 posts and a few lightyears from the original post.
We started with a tank, and a velocity addition problem. We had the assumption of vtop=0.9, and we actually had a paradox based on this assumption.
When you change the scenario now to the equivalent of two cannons shooting bullets back and forth at different speeds, there's not much left to discuss. This is trivially possible and therefore totally uninteresting, IMHO.

OK, what was the original paradox if vtop was 0.9c? vtop is 0.9c in Austin0's example (in the ground frame), so if you say Austin0's example is "trivially possible and therefore totally uninteresting", what assumption is he making which is different from the original assumption? I've read the first few posts on this thread, it seems to me that calebhoilday was just mistaken in this paragraph:

The member of the design team states that if you treat the tank as the stationary observer, then what ever speed the top-side has the bottom-side needs to have, just in the opposite direction. If this is not the case then the tank tracks would rip apart, as either the top-side or bottom-side tracks would not feed enough track to the other.

As I showed in my example, as long as the material of the track is somewhat stretchable it's not actually necessary for the speed to be the same on top and bottom in the tank frame; if you have a series of dots drawn on the track which are 0.45 light-seconds apart when they're on the bottom, but 0.75630 light-seconds apart when they're on the top, then with the bottom side moving at 0.45c and the top side moving at 0.75630c, the top will feed the bottom a new dot once every second, and the bottom will feed the top a new dot once every second. So each side can feed the other the right amount of track and there's no need for the track to rip apart despite the difference in velocities in the tank frame.

In post #8 you responded to the part of calebhoilday's post above by saying:

From this (valid, btw) logic it is clear that the top side has v=0.45 and the bottom side has v=-.45.

But the logic isn't valid! Austin0's post, plus my addition of the dots on the track, shows that the amount of track on top and bottom will be perfectly stable even if the speed of the two sides is different in the tank rest frame.

 

[Austin0]

From the ground frame the clocks at the front wheelbase of the tank are running behind the clocks at the rear.
This would mean that measurement of track speed of the upper track would be faster than that measured for the bottom track.
As we know the bottom track velocity is 0.45c this would mean that the top track velocity must be greater than 0.45c
From the tank frame the reciprocal situation prevails.

The rear clocks are running behind so the measurement of the wheel base velocity must be lower. This is symmetrical with the clocks on the tank and both derive 0.45c as expected.

But in this case the top track is moving in the same direction as the wheel base so the desynchronization does not change the measurement as it is the exact same clocks measuring both.
So in this case the geometry and mechanics would seem to indicate twice the distance covered and hence twice the velocity i.e. 0.9c

It can be seen that in the tank frame the number of segments counted for the upper track would be increased due to simultaneity and therefore the conclusion would be they were contracted.
If there is a marked segment to start the velocity measurement and the rear observer starts counting segments from that point, he will count more segments passing in the time it takes the clock in the front to catch up as that first segment reaches his position.
But of course the opposite would seem to pertain wrt the bottom track where they would count fewer segments.

Likewise in the ground frame I can see no way that the count starting in the rear, where the clock is running behind could not count fewer segments in the interval between time there and the same time at the front where the clock is running ahead.

What do you mean by "more segments" though? Suppose we paint two dots on the track, as in the example I posted in response to Ich--would you agree the number of segments between the two dots should stay constant as the pair of dots moves from the bottom to the top? If so, the segments would have to change length in the frame where both dots were at rest (the rest frame of that section of the material that makes up the tracks), depending on whether the dots were both on top or both on the bottom.

No argument ; the number of segments between dots would be constant.
I think we may be taking two approaches to the same thing.
Having given more thought to my hasty "more segments" idea I have seen that I was simply wrong. In fact I got it reversed.
The count would be more on the bottom and less on the top.
Assume the rear tank observer on top ,starts counting segments from the marked segment at the beginning of the time trial. He just keeps a log of segment number and time.
When the marked segment reaches the midpoint of the front wheel the watch stops and from that elapsed time the rear observer can consult his log and see how many segtments had passed at that simultaneous time according to their clocks. This would mean that at the elapsed time at the front it would correlate to a past time at the rear i.e. less segments counted.
It seems that the reciprocal effect would pertain to the bottom , no??

How this would relate to contraction or stretching I am not yet sure but it seems like it should be a factor ?

Would you agree to my basic proposition that given the desynchronization between front and back , even if the actual velocities of the top and bottom were equal , the measured velocities could not be??

What conditions would have to exist for the two measured speeds to be the same??
It appears to me that it is clear; the top speed would have to be significantly lower than the bottom for the measurements to be the same. Yes?

This is pertinent to the tank frame where the two measurements are taken in different directions i.e. the top track is moving counter to the bottom so the desynchronization effect is in opposition.

Also counter to the ground frame velocity measurement.

In the ground frame the critical measurements are in the same direction;
the top track is moving in the same direction as the tank so the desynchronization effect is not in opposition but applies equivalently to both measurements.
But counter to the bottom track measurement so this is symmetrical with the tank frame
and both frames agree on the velocity of the bottom relative to the tank.

Does this all track??

[EDIT] scratch all the last part. I ran the numbers and the situation is completely symmetrical. Calculating simultaneity from the tank frame still leaves the problem intact. I.e. the ground frame calculates 0.74844074844074970834424639216174 for the top track. SO I am beginning to return to my original thought; that kinematics may not be able to resolve the problem and indicate a clear correct frame to apply the addition of v equation from.

But there is the clear preference for the ground frame on the basis of acceleration.
Not only has the tank undergone acceleration getting up to speed but it is under constant acceleration as it travels.
Normal force maintaining ground contact. Acceleration of the bottom track by the earth. Acceleration by the drive wheel that must propagate throughout the track smoothly, without slack or bunching, for the tank to move.
So to be consistent with the twins rationale the tank frame is not valid and the ground rules,,,maybe?

Additional thought; In the context of my workup above ,the velocity of the top track was not an argument. It was derived on the assumption of distance traveled . If you posit a velocity less than ..756430 like .45 this would mean that at the end point arrived at through the wheel base travel, the marked segment would be only approx .667 of the way there while the comparable bottom segment would have traversed the complete length. This is apparently unacceptable as far as viable mechanics. ANd throws out any ideas of equal distribution or equal transfer of segments between top and bottom. Or so it appears to me.
Of course the inverse is equally true. Given a lesser velocity of the top in the ground frame creates the same problem , no?

 

[Austin0]

Because it shows that the "2v-claim" - which seems so reasonable - would actually lead to breaking the tracks, while velocity addition preserves them. The proof is relatively simple, as neither the stretching of the tracks nor simultaneity are relevant.

That was not meant dismissively, it was meant to say that I posted all the formulas you need to follow the proof. You don't have to look elsewhere.

Then there's the issue of different track lengths (top vs bottom). Austin0 claimed this as evidence for undue stresses in the ground frame, without further thoughts or calculations. I dismissed that "evidence" claiming that he forgot the relativity of simultaneity.

Your example (which I understand now for the first time - if at all) is basically the same. Its resolution depends on physical stretching, lorentz contraction, and relativity of simultaneity, so it's necessarily complicated. I did not expect you to solve it.

First, the stretching of the track, in the tank frame:
I'm asking for the rest length of the track.
We have in the tank frame a length 2*l (top+bottom). But this is made up of Lorentz-contracted track.
If the contracted track has a length of 2*l, its rest length must be 2 l /0.893 = 2.24 l.
So either the track has a length of 2.24l from the beginning (a very loose fit), or it gets stretched to this rest length during spin-up.

Now in the ground frame:
The top track is .893l long in that frame, and is made up of contracted track with a factor of 0.663. So the rest length of the upper track is 0.893 l/0.6632 = 1.347 l
The bottom track is also .893l long in that frame, but it is not contracted at all. So its rest length is 0.893 l.

Consistency check: the sum of both rest lengths is still 2.24l. Of course, there is no difference if we look at the same situation in different frames.

Now, does the result mean that the top track is severely stretched, while the bottom track is compressed?
No. Here's the appearance of Relativity of Simultaneity:
In the tank frame, at the rear wheel, there's track going from bottom to top at a constant rate. The rate in terms of contracted track is v, in terms of track rest length it is therefore \gamma v = 1.12 v.
The same amount goes down at the front wheel.
Everything is balanced.

In the ground frame, however, with its simultaneity convention, we see the rear wheel at a later time than the front wheel. It is, so to speak, \Delta t =\gamma l_w v = l v =0.45 l older (tank time) than the front wheel.
In that tank time, a track rest length of dl=1.12 v \Delta t = 1.12 v^2 = 0.227 l went up.
That means, instead of the equilibrium amount of 1.12l we see 1.12+.227=1.347 track rest length at the top, and 1.12-.227 = 0.893 at the bottom.
Not because there's additional stress, it's because we use a different simultaneity in the ground frame, where we measure the track at the rear wheel at a later (tank) time, where some of it already went up that did not come down at the front wheel.

I think it's fair to say that this is complicated stuff. And if I say that a newcomer has exactly zero chance of getting such things right, that's neither arrogance nor ad hominem, it's how it is.

Could you explain how that works; I.e. How the rear wheel could be at a later time than the front wheel in the ground frame?
If you mean from the ground frame the clock on the tank in the rear would be running ahead it would make sense, but what does that have to do with measurements in the ground frame?
How in the ground frame the front clock could be running behind the rear clock??

 

[JesseM]

No argument ; the number of segments between dots would be constant.
I think we may be taking two approaches to the same thing.
Having given more thought to my hasty "more segments" idea I have seen that I was simply wrong. In fact I got it reversed.
The count would be more on the bottom and less on the top.

Yes. If we want to have an integer number of segments, we might say there are 200 segments on the bottom at any given moment in the tank frame, and 119 segments on the top. So each segment on the bottom would have a length of 10/200=0.05 in the tank frame, which means that moving at 0.45c, a new segment would reach the back wheel every 0.05/0.45 = 1/9 of a second. And each segment on top would have a length of 10/119 in the tank frame, which means that moving at 90c/119 = 0.75630c, a new segment would reach the front wheel every (10/119)/(90/119) = 10/90 = 1/9 of a second too. So the rate at which the back wheel was feeding new segments to the top would equal the rate at which the front wheel was feeding new segments to the bottom, and the number of segments on each side would remain constant.

Assume the rear tank observer on top ,starts counting segments from the marked segment at the beginning of the time trial. He just keeps a log of segment number and time.

OK, in my example above he counts 9 segments per second.

When the marked segment reaches the midpoint of the front wheel the watch stops and from that elapsed time the rear observer can consult his log and see how many segtments had passed at that simultaneous time according to their clocks. This would mean that at the elapsed time at the front it would correlate to a past time at the rear i.e. less segments counted.

As long as we're talking about observers at rest in the tank frame, and they both start and stop counting segments that pass them simultaneously in the tank frame, then each will count 9 segments/second and thus they should count the same total number of segments having passed them.

Would you agree to my basic proposition that given the desynchronization between front and back , even if the actual velocities of the top and bottom were equal , the measured velocities could not be??

What do you mean by "desynchronization"? If we have clocks at the front and back wheel, then what frame are they synchronized in? And what do you mean by "actual velocities of the top and bottom"--"actual" in what frame? And in what frame are you taking the "measured velocities"?

What conditions would have to exist for the two measured speeds to be the same??
It appears to me that it is clear; the top speed would have to be significantly lower than the bottom for the measurements to be the same. Yes?

Again I'm not clear on what frames you're using to talk about speeds and measurements. It would be possible to alter your example so that in the tank frame both the top and bottom were moving at 0.45c in opposite directions, if that helps.

This is pertinent to the tank frame where the two measurements are taken in different directions i.e. the top track is moving counter to the bottom so the desynchronization effect is in opposition.

Also counter to the ground frame velocity measurement.

In the ground frame the critical measurements are in the same direction;
the top track is moving in the same direction as the tank so the desynchronization effect is not in opposition but applies equivalently to both measurements.
But counter to the bottom track measurement so this is symmetrical with the tank frame
and both frames agree on the velocity of the bottom relative to the tank.

Does this all track??

[EDIT] scratch all the last part. I ran the numbers and the situation is completely symmetrical. Calculating simultaneity from the tank frame still leaves the problem intact. I.e. the ground frame calculates 0.74844074844074970834424639216174 for the top track.

I don't understand what you're calculating there.

SO I am beginning to return to my original thought; that kinematics may not be able to resolve the problem and indicate a clear correct frame to apply the addition of v equation from.

...and I don't understand what "problem" you think kinematics can't resolve. Keep in mind I haven't read the whole thread--can you summarize what you're trying to work out here, and what the problem in your mind is?

Additional thought; In the context of my workup above ,the velocity of the top track was not an argument. It was derived on the assumption of distance traveled . If you posit a velocity less than ..756430 like .45 this would mean that at the end point arrived at through the wheel base travel, the marked segment would be only approx .667 of the way there while the comparable bottom segment would have traversed the complete length.

Don't get this either. If both the top and bottom are moving at 0.45c in the tank frame, then in the tank frame a segment on the bottom will take the same time to travel from one wheel to another as a segment on top, right? If you agree with that, what are you talking about when you say "the marked segment would be only approx .667 of the way there while the comparable bottom segment would have traversed the complete length"?

 

[Ich]

Keep in mind I haven't read the whole thread

Yes, that's an issue here. We've very early said that the tank should behave generally like a track vehicle, not an extrusion machine. For example, Ken Natton https://www.physicsforums.com/showpost.php?p=2808361&postcount=22" it as two wheels with a track around them. Such specifications set certain restrictions to the velocities, which make the problem interesting. So I don't want to get things muddled up after the end of the initial discussion.

Of course, it's ok if you like to discuss your very general notion of a track vehicle now. in that case, I don't disagree with you.

 

[Ken Natton]

Perhaps I don’t need particularly to defend myself here, I realize you don’t particularly intend any criticism of me, Ich. But just for the record, I always understood that real tanks have a complex arrangement of multiple wheels. I just conceived it as a single front and back wheel because it seemed to me to distil the problem, as I saw it at least, down to its essentials. The original problem was specified by Cabelholiday, but the YouTube clip he referred us to was of a model that had just three wheels. The extra central wheel was actually just the device that enabled it to ascend and descend staircases, which was actually what the clip was meaning to demonstrate.

If I might presume to summarise for you JesseM, as the thread’s title suggests Cabelholiday’s original poser was meant to challenge the velocity addition / subtraction formula. A significant part of the thread has been a non too constructive argument about whether the top track runs at 2v or at the value arrived at by employing the velocity addition formula first derived by Einstein in his famous special relativity paper of 1905. I don’t wish to reopen that dispute but I did try to suggest that a questioning of that formula is nothing specific to the tank track problem in the hope of keeping attention on what I saw as the real tricky problem offered by this poser. Ich showed admirable patience in guiding me to the answer I sought. Some of the thread’s most prominent posters believe the problem remains unresolved. Several others have expressed their opinion that it has long since been addressed.

 

[Ich]

Perhaps I don’t need particularly to defend myself here, I realize you don’t particularly intend any criticism of me, Ich.

Not at all, what you stated is exactly how I (and, I'm sure, the other contributors who gave up earlier) understood the OP. This is what we've been discussing, not machines that forcefully stretch and shrink the track every turn. Just the essentials, they are complicated enough.

 

[JesseM]

Yes, that's an issue here. We've very early said that the tank should behave generally like a track vehicle, not an extrusion machine.

What's an "extrusion machine"? Are you saying it should be impossible for the track to be the slightest bit flexible, so two dots painted on the track should always be precisely the same distance apart in the mutual rest frame of the two dots? Perfectly rigid objects are impossible in relativity, so this doesn't seem realistic... For example, Ken Natton

https://www.physicsforums.com/showpost.php?p=2808361&postcount=22" it as two wheels with a track around them.

Yes, and if the track is just a bit rubbery, it's not at all unrealistic that the section of the track between the tops of the wheels could be slightly more stretched than the section of the track between the bottom of the wheels. You could demonstrate something like this with two spools and a rubber band, for example.

Of course, it's ok if you like to discuss your very general notion of a track vehicle now.

It's rather absurd to suggest that a slight amount of stretching/compression of the track requires a very general notion of a track vehicle. And in any case calebhoilday's logic here in the opening post, which you called "valid", was:

The member of the design team states that if you treat the tank as the stationary observer, then what ever speed the top-side has the bottom-side needs to have, just in the opposite direction. If this is not the case then the tank tracks would rip apart, as either the top-side or bottom-side tracks would not feed enough track to the other.

calebhoilday wasn't just saying that the tracks would rip apart if the material was more stretched on one side than the other, he was saying they would rip because "the top-side or bottom-side tracks would not feed enough track to the other". Do you agree that this is incorrect, since my analysis shows that if we have a series of regularly-spaced dots on the track, the front wheel is feeding dots to the bottom track at exactly the same rate that the back wheel is feeding dots to the top track?

 

[Ich]

What's an "extrusion machine"? Are you saying it should be impossible for the track to be the slightest bit flexible, so two dots painted on the track should always be precisely the same distance apart in the mutual rest frame of the two dots?

Extrusion machines are not impossible, and the track has to be stretched (or lengthened) anyway. It's just that the tank we considered does not act like an extrusion machine, i.e. stretching and lengthening segments every cycle.

Yes, and if the track is just a bit rubbery, it's not at all unrealistic that the section of the track between the tops of the wheels could be slightly more stretched than the section of the track between the bottom of the wheels. You could demonstrate something like this with two spools and a rubber band, for example.

Yes, no problem. But that's not a tank. Remember, both spools/wheels have contact to the ground, they run at the same peripheral speed, the speed of the tank. And this constraint makes the problem well-defined and interesting for a discussion of relativistic velocity addition.
You rubber band is an engineering problem, nothing to do with SR.

Do you agree that this is incorrect, since my analysis shows that if we have a series of regularly-spaced dots on the track, the front wheel is feeding dots to the bottom track at exactly the same rate that the back wheel is feeding dots to the top track?

What you are proposing is something different. You can build such machines in principle, and you can of course discuss them, but they are not the tanks we discussed. For tanks run on the ground and have front and bottom wheel synchronized.

 

[Austin0]

The count would be more on the bottom and less on the top.

Yes. If we want to have an integer number of segments, we might say there are 200 segments on the bottom at any given moment in the tank frame, and 119 segments on the top. So each segment on the bottom would have a length of 10/200=0.05 in the tank frame, which means that moving at 0.45c, a new segment would reach the back wheel every 0.05/0.45 = 1/9 of a second. And each segment on top would have a length of 10/119 in the tank frame, which means that moving at 90c/119 = 0.75630c, a new segment would reach the front wheel every (10/119)/(90/119) = 10/90 = 1/9 of a second too. So the rate at which the back wheel was feeding new segments to the top would equal the rate at which the front wheel was feeding new segments to the bottom, and the number of segments on each side would remain constant.

Agreed the number would remain constant

Assume the rear tank observer on top ,starts counting segments from the marked segment at the beginning of the time trial. He just keeps a log of segment number and time.
When the marked segment reaches the midpoint of the front wheel the watch stops and from that elapsed time the rear observer can consult his log and see how many segtments had passed at that simultaneous time according to their clocks. This would mean that at the elapsed time at the front it would correlate to a past time at the rear i.e. less segments counted.

OK, in my example above he counts 9 segments per second.

As long as we're talking about observers at rest in the tank frame, and they both start and stop counting segments that pass them simultaneously in the tank frame, then each will count 9 segments/second and thus they should count the same total number of segments having passed them.

My proposal is that the observer at the top rear wheel in the tank frame would count 9/s for dt' -(4.4s) while the observer at the front bottom would count bottom segments for dt'+ (4.5s) and would therefore have a significantly different count. Just as the derived velocity of the bottom was different from the top.

Would you agree to my basic proposition that given the desynchronization between front and back , even if the actual velocities of the top and bottom were equal , the measured velocities could not be??

What do you mean by "desynchronization"? If we have clocks at the front and back wheel, then what frame are they synchronized in? And what do you mean by "actual velocities of the top and bottom"--"actual" in what frame? And in what frame are you taking the "measured velocities"?

In this context of the simultaneity workup I did , the desynchronization is between the front and back clocks of the tank as pbserved from the ground frame. Of course they are synchronized within the tank frame.
Sorry about the "actual velocities" ,,within the context of this thread I meant the velocities not as actual quantitative values but actual in the sense of the constraints enforced by physical principles and logic. I.e. I don't think anyone really disagrees with the evident logic that, independent of measurement ,the top and bottom must actually travel the same distance per time in the tank frame to be functional. I certainly never questioned this logic. The problem of course arises from the fact that the exact same logic and physical constraint applies equally in the ground frame.

What conditions would have to exist for the two measured speeds to be the same??
It appears to me that it is clear; the top speed would have to be significantly lower than the bottom for the measurements to be the same. Yes?

Again I'm not clear on what frames you're using to talk about speeds and measurements. It would be possible to alter your example so that in the tank frame both the top and bottom were moving at 0.45c in opposite directions, if that helps.

Yes I am aware of that, in fact that is exactly what the numbers referred to below are about. I did the same simultaneity analysis from the tank frame with the assumption of equal distance traveled by top and bottom in that frame and got the same figure as applying the addtion equation from that frame. I.e. Once again kinematics did not prefer or eliminate either solution.

I ran the numbers and the situation is completely symmetrical. Calculating simultaneity from the tank frame still leaves the problem intact. I.e. the ground frame calculates 0.74844074844074970834424639216174 for the top track.

I don't understand what you're calculating there.

SO I am beginning to return to my original thought; that kinematics may not be able to resolve the problem and indicate a clear correct frame to apply the addition of v equation from.

.

..and I don't understand what "problem" you think kinematics can't resolve. Keep in mind I haven't read the whole thread--can you summarize what you're trying to work out here, and what the problem in your mind is?

My position is that the OP has presented a unique and problematic question and scenario.
That has two mutually exclusive but equally valid logical bases for the application of the Addition o' V formula from different frames.
My analysis is one veiwpoint. You have provided a counter to Ich's objection to this view but no basis for eliminating the other solution or a clear preference for either one.
AS you pointed out and I confirmed my analysis is also reciprocal.
To this point it seems clear that that applies to all analyses so far.
Ich's showed how that viewpoint could be workable in the tank frame but the physical implications in the ground were not so obviously workable.
So ,to this point kinematics being the A of V formula and simultaneity have not determined a clear answer or viewpoint. Neither has physical logic and mechanics.
The only somewhat unexplored factor is length contraction.

My "problem" is that, early on , a reasonable solution was presented and was accepted as "proven" on the basis that the numbers were consistent within the system.
Well of course the numbers were consistent. Having chosen a preferred perspective to apply the A of V equation they must be consistent.
But beyond that all , other considerations were dismissed and the case was determined to be closed.
But in fact the opposing perspective can be applied and the numbers come out just as consistent as has just been shown. SImply demonstrating consistency does not prove a proposition if there are alternative propositions that are equally consistent , true??
DO you think there is a clear cut answer among the alternatives so far presented?
Think of this. This question presents a scenario involving 3 frames. Tank and ground being somewhat inertial but the track being attached to both frames. The bottom is actually partially at rest in the ground frame while attached to the tank frame by drive wheels and in motion in that frame. The track is being accelerated by both frames through direct physical connection.
Can you think of a similar problem off hand?

Additional thought; In the context of my workup above ,the velocity of the top track was not an argument. It was derived on the assumption of distance traveled . If you posit a velocity less than ..756430 like .45 this would mean that at the end point arrived at through the wheel base travel, the marked segment would be only approx .667 of the way there while the comparable bottom segment would have traversed the complete length.

Don't get this either. If both the top and bottom are moving at 0.45c in the tank frame, then in the tank frame a segment on the bottom will take the same time to travel from one wheel to another as a segment on top, right? If you agree with that, what are you talking about when you say "the marked segment would be only approx .667 of the way there while the comparable bottom segment would have traversed the complete length"?

This was in the context of my analysis , as observed from the ground. It derived a top v of .7564 and a bottom v of .45 with both marked segments completely traversing from wheel to wheel top and bottom. If you then assume the bottom v stays .45 and the top v is reduced to .45 this would mean that for the same distance traveled by the wheel base the bottom would span the base but the top would only span some portion of the distance between the wheels.
Does this track?

Or comparably in the ground frame: If the wheel base travels its own length at .45 so v = dx/dt =.45 and the distance of the top of the rear wheel to the end position of the top of the front wheel is 2*dx as measured in the ground frame.
If you then use the .74844 velocity for the marked segment then .74844*dt must be significantly less than 2*dx...i.e somewhere in between the tops of the wheels. WOuld you agree??

 

[Austin0]

Perhaps I don’t need particularly to defend myself here, I realize you don’t particularly intend any criticism of me, Ich. But just for the record, I always understood that real tanks have a complex arrangement of multiple wheels. I just conceived it as a single front and back wheel because it seemed to me to distil the problem, as I saw it at least, down to its essentials. The original problem was specified by Cabelholiday, but the YouTube clip he referred us to was of a model that had just three wheels. The extra central wheel was actually just the device that enabled it to ascend and descend staircases, which was actually what the clip was meaning to demonstrate.

If I might presume to summarise for you JesseM, as the thread’s title suggests Cabelholiday’s original poser was meant to challenge the velocity addition / subtraction formula. A significant part of the thread has been a non too constructive argument about whether the top track runs at 2v or at the value arrived at by employing the velocity addition formula first derived by Einstein in his famous special relativity paper of 1905. I don’t wish to reopen that dispute but I did try to suggest that a questioning of that formula is nothing specific to the tank track problem in the hope of keeping attention on what I saw as the real tricky problem offered by this poser. Ich showed admirable patience in guiding me to the answer I sought. Some of the thread’s most prominent posters believe the problem remains unresolved. Several others have expressed their opinion that it has long since been addressed.

WHo do you think is questioning the Addition of Velocities formula ? It is a question of from what frame and on what assumptions of measured velocity in that frame you apply it.
It has NEVER been a question of the validity of the equation itself. If you think so then I think you have misunderstood most of what has transpired IMHO I certainly can't speak for CalebHolidays original intent but that quickly became irrelevant.

 

[yossell]

Hi Austin0

I've read these last couple of pages a number of times, and with all the changes and revisions to the example, the numbers, the frame, the problem, I'm unable to easily follow the discussion between you and Jesse. This could be my fault. Is that nice picture of the tank you posted with the figures still valid? Or have we moved on?

Anyway, it may be helpful for some of us - and certainly for me - if you could give a quick, clear summary of, in the light of the recent posts, what you now think the problem is, what the figures are, what the issue is.

 

[Austin0]

Hi Austin0

I've read these last couple of pages a number of times, and with all the changes and revisions to the example, the numbers, the frame, the problem, I'm unable to easily follow the discussion between you and Jesse. This could be my fault. Is that nice picture of the tank you posted with the figures still valid? Or have we moved on?

Anyway, it may be helpful for some of us - and certainly for me - if you could give a quick, clear summary of, in the light of the recent posts, what you now think the problem is, what the figures are, what the issue is.

Hi yossell Yes the "nice " pic and figures are still valid but sadly inconclusive.

I am unsure if you read my last post to JesseM as he asked the same question and I tried to briefly answer.
In short it appears to me we have a plethora of valid logics and viewpoints with no clearcut unambiguous means of deciding which is more valid or really falsifying either one.
I could certainly be satisfied with either resolution if sufficiently justified.

We seem to be running shy of possible approaches to resolve this. Length contraction may be the only contender on its feet .
I just think this question is unique and interesting enough to be worth seeking a real answer. If we don't do it now I suspect it will return later at some point.

PS as I remember Fredrik first mentioned stretching wrt contraction and kev brought in springs between axles
and if you don't assume some counter to a physical conception of contraction then the tank don't go in any frame

 

[Austin0]

Attached drawing.

Tank:wheelbase length 10 ls in rest frame.

Starting the velocity test from the marked track points at the top point of the rear wheel (T₀)

and the bottom point of the front wheel (B₀).

Ground frame: x=o, t=0 , x'=0,t'=0 at rear wheel center

x= 8.930, t=0, x'=10, t'= (-4.5) at front wheel center.

At the point where the top track marked point is coincident with the top of the front wheel (T₁)and the bottom marked point is coincident with the rear wheel center (B₁) the coordinates are:

x=17.860 ,t=19.844 , x'=10 , t'=13.22 at the front wheel ...and

x= 8.930 , t=19.844 , x'=0 ,t'= 17.720 at the rear wheel.
_____________________________________________________________________________

TOP TRACK:

dx' = (T₁)- (T₀)=10

dt' =t'(T₁)-t'(T₀)= 13.22

v=dx'/dt' =10/13.22= .756429

BOTTOM TRACK:

dx'=(B₁) - (B₀) = abs(-10) =10

dt'= t'(B₁) - t' (B₀) = 22.22

dx'/dt'=10/22.22 =.450045

The final velocity for the top track as measured in the tank frame is .756429 c

The velocity returned by the Addition of velocities function for (.9 +.45) is .75630 c


2) If there is no error , in your qualified opinion is this sufficient reason to consider the correct solution to be .9 c for the top track in the ground frame??

3) Can you think of any conditions that would be consistent with these events and still justify a measurement of .45 for the top in the tank frame?

In any case there are still interesting, unresolved questions regarding contraction and physics in this scenario.

Hi Austin0

I've read these last couple of pages a number of times, and with all the changes and revisions to the example, the numbers, the frame, the problem, I'm unable to easily follow the discussion between you and Jesse. This could be my fault. Is that nice picture of the tank you posted with the figures still valid? Or have we moved on?

Anyway, it may be helpful for some of us - and certainly for me - if you could give a quick, clear summary of, in the light of the recent posts, what you now think the problem is, what the figures are, what the issue is.

Hi yossell
I think I may possibly be able to offer a kinematic resolution to the basic question.
The correct application of the Addition of Velocities equation and physical assumptions are consistent with the original perspective I.e. As applied from the ground frame on the assumption of 0.9 c measured velocity of the top track.

The basis for this is as follows:
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.

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.

On the other hand , operating on the assumption of 0.9 for the top in the ground frame
results in equal distance traveled between the top and bottom in both frames, but with the measured velocity in the tank frame being greater than 0.45 as a purely kinematic consequence of the relativity of simultaneity.
Physical symmetry rather than measured symmetry in the tank frame.

Certainly I may be overly optimistic so I invite participation and critical analysis because I , for one ,would really like to get on to the questions still remaining regarding length contraction and how a workable tank might possibly arise from that conundrum.

 

[Ken Natton]

WHo do you think is questioning the Addition of Velocities formula ?

Austin0, I have had something of a debate with myself about whether there is any value in replying to you. I do not, for one moment suppose that I will persuade you of anything, and I don’t believe that there is any value in returning to the tenor of exchange that did occur at one point on this thread. The best I can hope for is a clarification of the difference between us that we can agree on.

Your contention is that it is consistent to argue that, in our example of v = 0.45c, the top track runs at 0.9c, and also to assert acceptance of the velocity addition formula. My contention is that contending that the top track runs at 0.9c is inherently a questioning of the velocity addition formula. Thus my answer to your question about who I believe is questioning it is anyone who contends that the top track runs at 0.9c. There are multiple contributors to this thread who have so contended.

 

[yuiop]

, 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.

As soon as you say "in the time" for events that do not happen at the same place, it immediately suggests that you have a simultaneity problem and you do not seem to have taken account of this.

, Certainly I may be overly optimistic so I invite participation and critical analysis because I , for one ,would really like to get on to the questions still remaining regarding length contraction and how a workable tank might possibly arise from that conundrum.

Designs for a workable tank have already been given. Either make the tracks elastic or mount one the axles on horizontal springs. Something has to give. A rigid tank with rigid track segments that cannot stretch can not operate at relativistic speeds. Relativity forbids the existence of rigid materials. The discovery of a material that can not compress or stretch would invalidate SR and GR.

 

[Austin0]

As soon as you say "in the time" for events that do not happen at the same place, it immediately suggests that you have a simultaneity problem and you do not seem to have taken account of this.

Designs for a workable tank have already been given. Either make the tracks elastic or mount one the axles on horizontal springs. Something has to give. A rigid tank with rigid track segments that cannot stretch can not operate at relativistic speeds. Relativity forbids the existence of rigid materials. The discovery of a material that can not compress or stretch would invalidate SR and GR.

You say "suggests" I do not seem to have taken account of simultaneity.

Have you actually read the foregoing posts where I presented a specific simultaneity workup from the ground frame and referenced a similar workup I did from the tank frame?

DO you have any specific objection or correction to the simultaneity analysis that I have totally taken account of?
Or is this just an a priori assumption based on taking a few words out of context and running with them??
I welcome actual participation and critical analysis and if you find actual flaws in my analysis I will move onward
but amorphous assertions based on your assumptions are not helpful.

Did you actually read enough to have any understanding of the basis of the conclusion?

As for the last part of your responce if you will look at the previous post to yossell you will find that I stated what you have said here and even mentioned you.
If you think that simply the assumption of some stretch or a couple of springs negates any questions wrt relative contraction you are entitled to your opinion.

 

[Austin0]

Austin0, I have had something of a debate with myself about whether there is any value in replying to you. I do not, for one moment suppose that I will persuade you of anything, and I don’t believe that there is any value in returning to the tenor of exchange that did occur at one point on this thread. The best I can hope for is a clarification of the difference between us that we can agree on.

Your contention is that it is consistent to argue that, in our example of v = 0.45c, the top track runs at 0.9c, and also to assert acceptance of the velocity addition formula. My contention is that contending that the top track runs at 0.9c is inherently a questioning of the velocity addition formula. Thus my answer to your question about who I believe is questioning it is anyone who contends that the top track runs at 0.9c. There are multiple contributors to this thread who have so contended.

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.

 

[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
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.

what 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