# Relativistic rolling tank

Discussion in 'Physics & Math' started by Pete, Jul 20, 2006.

1. ### PeteIt's not rocket surgeryRegistered Senior Member

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The premise of this thread is that the tank structure is much stronger than the track. This has been repeated many times.

We can discuss what happens if the tank structure is compressible and the track is very strong if you really want to, but perhaps we should get this one right first?

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3. ### DaleSpamTANSTAAFLRegistered Senior Member

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Ahh, I see your confusion now, this is your central mistake. If the distance between the axels is not strained (as has been consistently assumed in this thread) then the treads themselves must be strained. Yes, the distance between the axels is Lorentz contracted, but as Pete demonstrated above the Lorentz contraction of the axel spacing is insufficient to account for the Lorentz contraction of the tread spacing and thus, even from the ground frame of analysis, the tread must be strained. The only possible way for the tread to not be strained is for the axel spacing to strain. Something must give.

Yes, the strain in the rim of the wheel is caused by the fact that the spokes are unstrained. I explicitly mentioned that assumption. For the tank, the tread strain is caused by the explicit assumption that the axle spacing is unstrained. You certainly can run the numbers using any assumptions you desire. I have been explicit in my assumptions, if you prefer a different set of assumptions then simply be explicit and do the calculations.

I don't know why you included this quote, it supports my position as does the whole paper. That is why I referenced it above. The fact that Born rigid angular acceleration is not possible means that something must strain. If the spokes don't strain then the rim must (my assumption). If the rim doesn't strain, then the spokes must (Grünbaum and Janis' assumption). Similarly for the tank, if the axle spacing doesn't strain then the tread must and vice versa. Because the motion cannot be Born rigid then, by definition, something must inevitably strain.

-Dale

Last edited: Aug 6, 2006
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5. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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Pete, I think you are confused. The The contractions in the moving frame we are speaking of are due to relativistic effects, Lorentz contraction in specific. These effects, in themselves, have nothing to do with the strength of the tanks structure.The body of the tank contracts by a factor of two in the moving frame. Do you think the tank body is under strain, or is the contraction due to relativity of simultaneity, clock rates, etc.?

The wheel that is rotating in the moving frame has spokes that don't contract due to Lorentz contraction, but a rim that does. The spokes do not 'grow' in length, the radius of the wheel does not increase according to the ground observer. The 'strain' we are speaking of is due to a contracting rim circumference vs. non-contracting spokes or wheel diameter. Either (1)the rim has to 'stretch' to fit over the uncontracted diameter of the wheel, (2) the spokes and wheel diameter have to compress to fit within the contracted wheel diameter, or (3) the whole wheel assembly breaks due to the contraction paradox. The distance between the wheels holding the tread decreases in the ground frame, the tank tread itself is under no strain or stress due to Lorentz contraction of the wheel rim. The tank tread on the ground is not stretched or contracted, nor does any relativistic effect state it does. The top of the track appears to contract, again a relativity of simultaneity and clock effect. If not strain is placed on the tank body itself caused by inertial motion, then no strain will be placed on the top of the tank tread caused by inertial motion.

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In the rotating wheel example, the spokes did not contract causing strain on the rim, but the distance between the wheels does contract, so the track does not need to stretch to fit around the axels. The track around two wheels in a moving frame is not analogous to a rotating wheel, the relativistic predictions are different.

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7. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam,
Dale, you probably read my response to Pete, but I'll see if I can quickly expand on my line of thought.
I'm sure you and Pete both know I have always objected to the Lorentz contractions being anything other than a illusionary artifact cause by relativity of simultaneity, unsynchronized clocks, etc. used to 'measure' the distances. I firmly do not believe material things that go fast are physically contracted.

What you and Pete are envisioning is that the total length of the tread will be contracted, so it must 'stretch' to reach around the axels, because the total distance around the axels is contracted less than the tread length. Special Theory does predict this, as you say, but you claim this causes a physical strain on the tread, which must stretch or break. The contraction of the tread is on the top in between the axels, which is contracted by a factor of seven due to relative motion with respect to the ground frame, an inertial frame for both the top of the tread and the ground. As I stated earlier, we could make the top of the tread a light-year long if we wanted without effecting the basic gedankin premisis. I used a conveyor belt earlier for an example. My premisis is that if relative inertial motion causes a physical strain, then the atoms that make up the tread must be physically closer together, compressed, in an inertial frame just because different observers look at them. This physical strain must have infinite degrees of strain according to the infinate number of observers that view them from their rest frames. Do the treads stretch and beak according to one observer, but do not according to a different observer in a different state of relative motion? I think you get where I am headed.

8. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam,
You presented one interpretation only , the stretched rim, as if it were the only accepted interpretation that agrees with relativity theory. I presented another interpretation, the comperssed spokes, that also is in agreement with relativity theory. In fact, if the diameter of the rims decrease and the spokes compress, the tank tread we are discussing might fit around the wheels. Smaller wheels and closer axel spacing, the tread doesn't need to be as long.

9. ### DaleSpamTANSTAAFLRegistered Senior Member

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Sorry you misunderstood my presentation. I explicitly mentioned my assumptions and intended it to be understood, as usual, that other assumptions could have been made. As long as you are clear that something must strain then it is not conceptually important which part strains.

-Dale

10. ### DaleSpamTANSTAAFLRegistered Senior Member

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Yes, I see exactly where you are headed. Around in a circle back to the same stale arguments that I already refuted.

For at least the 4th time, Lorentz contraction is not the same as strain! You are once again demonstrating that you cannot get the two concepts correctly separated in your mind.

Let me try once again: relative inertial motion does not cause a physical strain, it causes length contraction. Length contraction is a frame-dependent disagreement about the length of an object or the distance between two points. Strain is a frame-independent measurement of the change in the proper length of a material object. All frames will agree on the amount of strain even though they disagree about the actual lengths. Because all frames agree on the strain they all will agree on things like failure characteristics.

In more mathematical terms: a length is a component of a four-vector, therefore lengths are clearly frame variant. A strain is a change in the proper distance between two material points, and the proper distance is a Minkowski norm (spacetime interval) which makes it frame invariant. The variant lengths do not determine failure characteristics, the invariant strains do. So stop making the silly and unsupported claim that SR predicts frame-variant failures. It is absurd, repetitive, and has already been thouroghly refuted.

The next time you claim that SR predicts some frame-dependent strain or failure back it up with some rigorous and quantitative math. Show that it is indeed a prediction of SR. Otherwise don't bother making the claim.

-Dale

11. ### PeteIt's not rocket surgeryRegistered Senior Member

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No, the tank is not strained. It is length contracted. But, it is under stress - the same stress that causes the tread to stretch. If the tank body was not a lot stronger than the track, then it would be strained by the applied stress, and be even shorter than its already length-contracted length in the ground frame.

Dale has said this many times. I've said it at least once:
Length contraction and strain are two separate effects. Don't confuse the two, and don't ignore either. You must consider both.

The length contraction of the tank is not sufficient to match the length contraction of the track. The distance between the wheels must decrease even more (ie strain the tank body), or the track must strain. Look at the numbers.

Last edited: Aug 7, 2006
12. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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Fellas, fellas, fellas, what causes your strain? Both of you 'claim', without proof, that your predicted Lorentz contraction of the rim of the wheel is offset because of the 'strain' exerted by the spokes resisting compression. The rim will Lorentz contract by different amounts according to different inertial observers.

Both of you 'claim', without proof, that the total length of the tread will be Lorentz contracted to a greater degree than a measurement around the wheels/axels indicates is necessary. You claim that leads to a 'stretching' of the tread, a strain. This Lorentz contraction of the tread is also dependent upon the relative velocities of all inertial observers. The amount of strain will vary according to how fast the inertial observer 'sees' the tank travel past his position. Two different observers in two different states of motion will disagree on the amount of strain the track of the tank is experiencing. Give it up, your 'stress and strain' arguement stinks. It cannot possibly be correct and it is in conflict with Special Theory. Ya'll have simply made a mistake, admit it.

Now, by using Grunbaum and Janis approach, the only sticking point may turn out to be the spoke compression or contraction. In order for Special Theory to be consistent, physicists must find some mechanism for causing rotating spokes to compress WITHOUT STRESS. I don't think it is possible as the theory is now structured, so Special Theory needs a revamp or replacement.

Strain arising because of an observer's relative motion? You've got to be kidding.

13. ### DaleSpamTANSTAAFLRegistered Senior Member

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No way, my derivation was careful and clear.

For at least the 5th time, the strain is frame invariant. Two different observers in two different states of motion will agree on the amount of strain the track is experiencing. This was the whole point of my work posted above. You already said that the work was correct, so it is logically inconsistent for you to say that the conclusions were wrong.

I have publicly posted the math supporting my claim that both frames agree on the strain, do the same with your claim that the strains are frame variant or stop claiming it.

The strain argument is fine, your comprehension stinks.

-Dale

Last edited: Aug 7, 2006
14. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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I didn't find fault with your maths, Dale, just the model you used. You assumed, without proof, that there will be a strain produced by incompressible spokes that causes the rim to stretch. If the rim actually does Lorentz contract, the spokes may compress and the wheel diameter decrease. I think for the gedankin to reflect what an inertial observer sees, the whole wheel should either contract without strain, or the physical size of the wheel simply does not change because some observers pass by it at different velocities. The direction the observer is travelling would also have an effect on the 'stretching' at different points along the rim. Suppose there were two inertial observers travelling in opposite directions at .866c relative to the same relativistically spinning wheel. The wheel would appear different to both because the velocity addition formula would have to be used for the part of the wheel that rotates toward each observer. Appearences are one thing, but physical deformations of the wheel are something else altogether.

15. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam,
OK, let me explain it to you this way. Suppose the track is 10 light-minutes long end-to-end. An observer on the ground could walk up to the track, remove a pin from the track holding the individual 'cleats' together, and there would be no increase in the gap while the track was still in contact with the ground. The track is at rest relative to the ground, no strain is evident. An observer riding on the top track watching the ground observer's actions will see the bottom tract contracted by a factor of seven. This inertial observer cannot see any strain on the track either, or the inertial frames are not equivelent.

Or do you believe when the ground observer removes the pin, the track gap will immediately widen? If the gap widens, the two sections of the track must slide along the ground in the rest frame. That would distinquish a true rest frame from a pseudo rest frame. Tank tracks are repaired all the time while they are at rest.

16. ### DaleSpamTANSTAAFLRegistered Senior Member

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Yes, and I specifically listed it as an assumption. If you want to assume something different, then be my guest. The assumption does not change the fact that the strains were shown to be frame invariant. If you make a different assumption then you will simply have frame invariant strains in a different part of the wheel or tank.

Do you not not realize that this is impossible? If you assume Born rigid (no strain) angular acceleration you get inconsistent boundary conditions. In other words, it is illogical and inconsistent to talk about this kind of motion without strain. Is it really so hard to believe that an object spinning up to relativistic speeds must inevitably undergo some strain?

Yes, and as I have stated 6 times now any observers would agree on all of the physical deformations (strains).

I disagree. State your assumptions, work it out rigorously, and demonstrate this.

You still don't get it. The fact that it is at rest has absolutely nothing to do with the fact that it is strained. You are saying that if I take a rubber band, stretch until it is strained to twice its length, and place it on the ground that there is no strain because it is at rest and that if the rubber band were cut it would not snap back. Come on! Get these concepts straight. Strain is not length contraction and it has nothing whatsoever to do with relative velocities.

2inquisitive, I must admit that I am rather frustrated here. Usually our conversations actually progress, but this one is just repeating itself. You are still stuck on your original misunderstanding. You claim to understand the difference between length contraction and strain and yet you continue to fill your posts with statements that clearly demonstrate that you do not. I am sorry that I cannot explain the difference clearly enough.

-Dale

17. ### PeteIt's not rocket surgeryRegistered Senior Member

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The track would fly apart like a breaking rubber band.

At rest does not mean no strain.Take a rubber band, stretch it to twice its length, and pin it to the ground in this stretched condition.

That's the track. It's at rest, and it is strained.

Cut the rubber band. That's the same as pulling a pin. Do you think the band will stay motionless?

No, it would distinguish a track under tension from an untensioned track.
Not while they're under tension!

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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I've just noticed this...
Terminology clarification:
In this context, strain means change in length. Strain is the result of stress, which is an applied force.

The rim applies compression stress to the spokes, the spokes apply tension stress to the rim.

The tank body applies tension stress to the track, the track applies compression stress to the tank body.

In practice, both the tank body and the track will be strained by these stresses. The end result depends on the relative strengths of tank and track. In this thread, we've been concentrating on the special case in which the strain on the tank can be neglected, to simplify the discussion. A general case would be interesting, but difficult, and not worth addressing if we can't sort out the special case.

19. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam,
You know we are speaking of strains arising from different Lorentz contraction factors in the system, not angular acceleration, torques, or any real physical strains.

DaleSpam,
Put it in caps next.

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You claim the strain on the rim is caused by incompressible spokes resisting a rim that has decreased in circumference. You claim all observers would agree on the amount of strain in the rim. The degree of strain is directly porportional to the rotational speed of the rim (the rims Lorentz contraction), so all observers must see the rim rotating at the same speed. Let me state at this junction that you are now agreeing with an earlier point of mine that a pulsar rotates at the same speed in all frames, regardless of the observer's relative velocity. The pulsar is a perfect clock, only the observer's clock rate can vary due to his own velocity. Thanks, I'll remember this!

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Does this mean the tank has an absolute velocity of .866c along the ground, since all observers agree the wheels are rotating at the same speed??? I'm having fun, how about you Dale?

DaleSpam,
No, you exerted a force to physically stretch the rubber band in the rest frame. Again, what is the source of this force in the inertial tank frame or inertial ground frame? Strain has nothing whatsoever to do with Lorentz contraction or relative velocities. Lorentz contraction cannot cause a strain on the spokes, the tank, or the tread. You are the one that keeps arguing that Lorentz contraction differences cause strain on the tread, 'stretching' the tread. The rigid disk cannot be spun up to relativistic velocities because the spokes must curve according to relativity theory, so rigid disks are incompatible with relativity theory, nothing to do with strains. Seems to me that relativistically rotating wheels are also incompatible with relativity theory, the Ehrenfest paradox.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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No. The degree of strain is the gamma factor of the rim speed in the axle frame. It is not directly proportional to the rotation speed, and it is not directly related to the rim speed in any other frame.

Lorentz contraction means that the track won't physically fit around the wheels. The strain arises because the wheels push against the track.

Consider the barn and pole paradox: because of length contraction, the pole will physically fit in the barn in the barn's frame, but if the barn doors close around the pole, the pole pushes against the doors and is strained by the resulting stress. The strain is not a direct result of length contraction. The strain is the result of the stress experienced because the track no longer fits the tank, just like the pole won't fit in the barn. Real physical strain.

No. The curvature of the spokes is frame dependent - the spokes are not curved in the spokes rest frame, or the axle frame, for example.

A disk or wheel can't be rigidly spun up to relativistic speeds because the rim and spokes must be strained.

Relativistically rotating unstrained wheels are certainly incompatible with relativity theory.

2inq, your problem appears to be that you refuse to consider the possibility that the track is strained simply as an article of faith. If you allow the possibility that the track is stretched, then there is no problem.

Last edited: Aug 7, 2006
21. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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Nevermind, the Ehrenfest paradox has been solved to my satisfaction.

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I just read the paper and found no contradictions with either the relativistic wheel or GPS. Great! The paper is by Robert D. Klauber and was first presented for publication on 6/30/2003. It comes highly recomended by ME!

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No, seriously, ya'll should read it if you haven't already.
http://freeweb.supereva.com/solciclos/klauber_d.pdf

22. ### DaleSpamTANSTAAFLRegistered Senior Member

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2inquisitive, yet again you are being illogical and not thinking this through clearly. If you wish to confine the discussion to a single state of rotation then there is no need to worry about strains (this is my preference). Rotation can be Born rigid, just not angular acceleration. Obviously the track motion cannot be rigid, but it can be approximately Born rigid as long as we are willing to ignore the ends where the track loops around the axles (like an ideal rope: rigid in tension, but not in bending).

However, if you choose this approach then you cannot infer anything whatsoever about the non-rotating geometry, which has been the primary point of contention so far. If you wish to relate the rotating geometry to the non-rotating geometry then there must have been real torques, and angular accelerations which caused real physical strains.

So what is your preference? Do you want to talk about a single state of rotation or multiple? If you want to talk about multiple states then there were obviously some angular accelerations and therefore necessarily some physical torques and strains. You logically cannot both talk about the rest geometry and also claim that there was no torque and no physical strain involved in getting to the moving state.

-Dale

23. ### DaleSpamTANSTAAFLRegistered Senior Member

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I finished the analysis starting from the ground frame. The most difficult part was figuring out a formula to use for the worldline of an arbitrary material point of the track in the ground frame. I finally found a reference on MathWorld that described the Fourier Series expansion of an asymmetric triangle wave. Using that I developed an analytical expression for the point's worldline in the ground frame:
s = {ct, vt + ∑(b sin[k(vt-x0)])}
where
b = 4 (-1)^n c^4 L sin[π n (c^2 + v^2)/(2c^2)]/(n^2 π^2 (c^4 - v^4))
k = n π/(L γ^2)
L is the length of the tank tread in the ground frame
v is the velocity of the tank in the ground frame
x0 is a term indicating the initial position and direction
and the summation is from n=1 to n=infinity

With this expression for s we can differentiate to get the four-velocity, u, and we can transform to the tank's rest frame to get s' and u'. We can select two different material points by choosing different values for x0, and we can calculate numerically the proper distance between these two points by following the same procedure as above.

For c=1, L=2, t=0, and v=.6 we get the following:

x0a=.5 and x0b=.6 are material points on the bottom of the tread, the plane of simultaneity from a intersects the worldline of b at t=0 in the ground frame, the two corresponding events are {0,.5} and {0,.6} and the proper distance is .1. Transforming to the tank's rest frame the plane of simultaneity from a intersects the worldline of b at t'=-.45, the two corresponding events are {-.375,.625} and {-.45,.75} and the proper distance is again .1.

x0c=2.5 and x0d=2.6 are material points on the top of the tread, the plane of simultaneity from c intersects the worldline of d at t=-.1875 in the ground frame, the two corresponding events are {0,.294} and {-.188,.082} and the proper distance is .1. Transforming to the tank's rest frame the plane of simultaneity from c intersects the worldline of d at t'=-.296, the two corresponding events are {-.221,.368} and {-.296,.243} and the proper distance is again .1.

In summary, both frames agree on the proper distance between two material points therefore both frames agree on the strain, both for material points on the top of the tread and for material points on the bottom. It doesn't matter which frame you begin in, the results are the same. Beginning in the ground frame complicates the math a lot, but does not change the results at all.

-Dale