Relativistic rolling tank

[DaleSpam]

When beginning in the ground frame, why would Hooke's law have anything to do with proper rest length spacing of the cleats? The cleats are not moving relative to the ground, thus they will have their proper length spacing. The total length of the track would be contracted by a factor of two at the bottom and seven at the top because it is moving wrt the ground, but the cleats, or distance markings are not.

First, it doesn't matter which frame you begin in. All of the kinematic laws are frame invariant. Your insistence on beginning in the ground frame complicates the math but doesn't change any results.

Why does Hooke's law come in? For simplicity consider a relativistic cartwheel. Let's say that the spokes are made of a much stronger material than the rim and actively controlled so that the radius of the wheel remains constant at all RPM regardless of any forces generated in the rim. Let's further attach a bunch of 1mm measuring rods around the rim in 1mm increments. The rods will only be attached to the rim on one end of each rod, not on both ends. At rest there is no gap between rods. As we spin the wheel up to relativistic speeds the rods length contract by γ and thus there is a gap between one rod and the next indicating that the rim has become physically strained by γ with accompanying stresses according to Hooke's law. In other words, there are two effects here, the geometric effect of length contraction as governed by the Lorentz transform and the stretching effect of strain as governed by Hooke's law.

Now, as the wheel rolls along the ground let's have it make marks every time the attached part of a rod strikes the ground. At low speeds the marks are 1mm apart, but at relativistic speeds the distance between marks is increased by γ. In other words, although the geometric effect does not occur at the ground the strain effect does. Therefore, even though there is no length contraction for the bottom tread that does not imply that the distance between cleats on the bottom tread is equal to the proper rest distance. Objects can certainly be strained at rest.

-Dale

 

[2inquisitive]

The track on the ground is not length contracted.
There is only 2m on the ground in the ground frame because the tank is length ontracted, but this on-the ground track is stationary in the ground frame - it is not length contracted from 4m.

You could consider this as an effect of relative simultaneity - in the tank frame, one cleat leaves the ground at the same time as the 4th cleat along touches the ground. But in the ground frame, that same cleat leaves the ground well before the 4th cleat along touches the ground.

The point is that whatever way you think about it, there are 2/S cleats on the ground at any time in the ground frame, right?

The cleats that are touching the ground are in the same frame of reference as the ground. Relativity of simultaneity does not apply to ground-touching cleats as a clock beats the same for all the cleats. SR's relativity of simultaneity would apply to any parts of the tank that are moving relative to the ground, such as the axels and the rounded end-portions of the track.

 

[2inquisitive]

DaleSpam,

Why does Hooke's law come in? For simplicity consider a relativistic cartwheel.

The relativistic cartwheel is moving (rotating) relative to the observer. The cleats on the bottom of the track aren't. Think of it this way. The cleats make a mark on the ground as they sink in. The marks on the ground are not moving relative to the ground observer. The marks on the ground will be at the same distance apart as their at-rest spacing unless they are length contracted along with the tank.

I'm sure you have seen animations of a car travelling at relativistic speeds on relativity websites. If the car is travelling at .866c, it is shown as being half the length in the ground frame as it is when at rest relative to the ground, and the wheels are also half the distance apart. The complicating factor in this tank gedankin is the track wrapped around the wheels, essentially a ruler. That ruler is at rest on the ground, moving at .866c from rounded end to rounded end (around the axel/wheels), and moving at 2 x .866c on the top of the track. The stationary ground observer sees three different states of motion from his rest frame, thus three different factors effecting distances he measures from his frame.

 

[DaleSpam]

The relativistic cartwheel is moving (rotating) relative to the observer. The cleats on the bottom of the track aren't. Think of it this way. The cleats make a mark on the ground as they sink in. The marks on the ground are not moving relative to the ground observer. The marks on the ground will be at the same distance apart as their at-rest spacing unless they are length contracted along with the tank.

The very bottom part of the cartwheel, the part in contact with the ground, is also not moving wrt the ground. Same with any marks made by cleats or the rods. The situation is exactly analogous in the two cases.

-Dale

 

[2inquisitive]

The very bottom part of the cartwheel, the part in contact with the ground, is also not moving wrt the ground. Same with any marks made by cleats or the rods. The situation is exactly analogous in the two cases.

-Dale

Yes, but there is no stretching at the bottom of the wheel in the ground frame. There is contraction in other parts of the wheel not in contact with the ground, correct?

 

[DaleSpam]

Yes, but there is no stretching at the bottom of the wheel in the ground frame. There is contraction in other parts of the wheel not in contact with the ground, correct?

That is not correct. There is strain (stretching) uniformly around the rim of the wheel in all frames. There is no length contraction at the bottom of the wheel in the ground frame. Strain and length contraction are separate things. Both occur in these examples.

You do understand that objects can be strained in their rest frame, right? So you can have strain without length contraction and length contraction without strain.

-Dale

 

[2inquisitive]

“ Originally Posted by 2inquisitive
Yes, but there is no stretching at the bottom of the wheel in the ground frame. There is contraction in other parts of the wheel not in contact with the ground, correct? ”

DaleSpam,
That is not correct. There is strain (stretching) uniformly around the rim of the wheel in all frames.

You seem to disagree with the standard explanation of the relativistic cartwheel:

Cartwheel


This is what a cartwheel looks like moving at 87% of the speed of light. The cartwheel appears Lorentz contracted by a factor of 2 along the direction of motion.

The bottom of the cartwheel, where it touches the road, is not moving, and is not Lorentz contracted. You might think that the top of the cartwheel would have to move faster than the speed of light to overtake the axle moving at 87% of the speed of light; but of course it can't.

The cartwheel offers another example of the impossibility of completely rigid bodies in special relativity. In the frame of reference of someone riding on the axle (but not rotating), the rim is whizzing around and is Lorentz contracted, while the spokes are moving transversely, and are not contracted. Something must give: the rim must stretch, or the spokes compress.

In the part I bolded, the cartwheel is moving at .87c relative to the stationary ground observer. The bottom of the wheel is not contracted or stretched, as I said.
In the part I placed in italics, the cartwheel is rotating relative to the observer on the axel, but not moving laterally. The observer on the axel sees the spokes as uncontracted. It is in this frame that the rim must stretch or the spokes compress. There are two frames of references discussed in the animation, the part of the wheel that touches the ground in the ground rest frame is not compressed nor stretched.
http://casa.colorado.edu/~ajsh/sr/contraction.html#cartwheel

 

[2inquisitive]

DaleSpam,

You do understand that objects can be strained in their rest frame, right?

Of, but an outside force is necessary. If an outside force is acting on the object, then it is not in an inertial frame, is it?

 

[Pete]

It is in this frame that the rim must stretch or the spokes compress.

The rim stretching and spokes compressing is not frame dependent, 2inq. If it happens in one frame, it happens in all frames.

The cleats that are touching the ground are in the same frame of reference as the ground. Relativity of simultaneity does not apply to ground-touching cleats as a clock beats the same for all the cleats. SR's relativity of simultaneity would apply to any parts of the tank that are moving relative to the ground, such as the axels and the rounded end-portions of the track.

2inq, relativity of simultaneity means that two events that happen in the same time in one frame don't happen at the same time in the other frame. Whether the cleats in question are moving in the ground frame or not is not relevant - what matters is that in the ground frame, one cleat leaves the ground simultaneously with the second cleat along reaching the ground.

 

[2inquisitive]

Pete,

The rim stretching and spokes compressing is not frame dependent, 2inq. If it happens in one frame, it happens in all frames.

Well then, it seems we have just found a way to distinguish between a true rest frame (one in which the rims do not stretch and the spokes do not compress) a moving frame with a pseudo rest frame ( a rest frame in which rims are stretched and spokes are compressed, under strain).
Pete,

what matters is that in the ground frame, one cleat leaves the ground simultaneously with the second cleat along reaching the ground.

Yes, that is what I agreed with, in the ground frame the two events are simultaneous.

 

[Pete]

Pete,

Well then, it seems we have just found a way to distinguish between a true rest frame (one in which the rims do not stretch and the spokes do not compress) a moving frame with a pseudo rest frame ( a rest frame in which rims are stretched and spokes are compressed, under strain).

The distinguishing feature is whether the wheel is rotating in any SR inertial reference frame (if it's rotating in one, then it's rotating in all). If it's rotating, then the rim is stretched and/or the spokes are compressed, regardless of the wheel's linear motion.

Yes, that is what I agreed with, in the ground frame the two events are simultaneous.

And in the tank frame they aren't, right?

 

[2inquisitive]

Pete,

The distinguishing feature is whether the wheel is rotating in any SR inertial reference frame (if it's rotating in one, then it's rotating in all). If it's rotating, then the rim is stretched and/or the spokes are compressed, regardless of the wheel's linear motion.

Pete, maybe you should read this page from wiki again. I thought we had agreed from our discussions of GPS that rotating frames are non-inertial frames? Here is a quote and a link, notice the parts I placed in bold:

A rotating frame of reference is a special case of a non-inertial reference frame in which the coordinate system is rotating relative to an inertial reference frame. An everyday example of a rotating reference frame is the surface of the Earth.

http://en.wikipedia.org/wiki/Rotating_reference_frame
However, the parts of the tank tread between the wheels (axels), top and bottom, are inertial frames. Just as on my conveyor belt, an observer attached to either surface feels no fictitious forces.

myself,

“ Yes, that is what I agreed with, in the ground frame the two events are simultaneous. ”

Pete,

And in the tank frame they aren't, right?

We aren't discussing what the tank observer sees, just the ground observer. The ground observer's frame of reference has to be correctly described before we can transform into the tank observer's frame.

 

[DaleSpam]

You seem to disagree with the standard explanation of the relativistic cartwheel:

In the part I bolded, the cartwheel is moving at .87c relative to the stationary ground observer. The bottom of the wheel is not contracted or stretched, as I said.
In the part I placed in italics, the cartwheel is rotating relative to the observer on the axel, but not moving laterally. The observer on the axel sees the spokes as uncontracted. It is in this frame that the rim must stretch or the spokes compress. There are two frames of references discussed in the animation, the part of the wheel that touches the ground in the ground rest frame is not compressed nor stretched.
http://casa.colorado.edu/~ajsh/sr/contraction.html#cartwheel

The only reason that I seem to disagree with the standard explanation is because you seem completely unable to distinguish between the concepts of strain and Lorentz contraction. The bottom part of the rim is not Lorentz contracted in the ground frame, it is strained in all frames. Lorentz contraction is a frame-dependent change in coordinate distances, strain is a frame-independent change in proper distances.

Do you know what a strain gauge is? Lorentz contraction does not cause any reading on a strain gauge because the gauge itself will contract by the exact same amount as the material it is measuring. Thus, strain gauges measure changes in proper distances, which are frame invariant.

Of, but an outside force is necessary. If an outside force is acting on the object, then it is not in an inertial frame, is it?

Not so. If an object is not accelerating then its rest frame is an inertial frame, that only implies that all forces and torques acting on it are balanced, not that there are no forces. The entire physics/engineering field of statics is concerned with exactly such objects: objects which deform under the application of external forces, but are not accelerating because all external forces are balanced. Consider a bar of metal in classical tension, there are equal and opposite forces on each end of the bar, so the sum of the forces is zero and the bar is not accelerating, but because of the strain the bar is longer than it was prior to the application of the tension forces.

-Dale

 

[Pete]

2inq, you're not making much sense... you're replies to my replies don't seem to have any bearing on the issue that my replies were addressing.

Yes, a rotating frame is non-inertial. I don't know why you think I implied otherwise.
Yes, the straight parts of the tread are inertial. No, an observer attached to the tread is not inertial - inertial for some of the time isn't enough.
Yes, the tank frame is still relevant, whether or not you want to think about it.

Yes, the ground frame has been fully described, here:

Let's say the track's proper length when not under tension is 8 metres, with 8 cleats, one every metre. Say the wheels are very small, and the tank is 4 metres long.
So when the tank is sitting still, there are 4 cleats on the bottom and 4 and top.

Now, let's look at the tank rolling at 0.866c in the ground frame. Let's call the stretch factor of the track S. If the track is not stretched, then S will be one. Note that this factor must be the same in every reference frame, because it determines obviously unambiguous things, like whether the track breaks or not.

The proper length of the stretched track will be 8S metres, with one cleat every S metres.

In the ground frame:
The tank will be 2 metres long.
There is 2 metres track on the ground, not length contracted. There will be 2/S cleats in this section.
There is 2 meters of track not on the ground, length contracted from 14 metres. There will be 14/S cleats in this section.

So, the total proper length of track (working from the ground frame) must be 16 metres, containing 2/S + 14/S cleats. This gives us to ways to work out S:

Total length of track = 8S metres = 16 metres, therefore S = 2.
Total number of cleats = 8 = 2/S + 14/S = 16/S, therefore S = 2.

So working purely in the ground frame, we find that the track must be under enough stress to stretch it by a factor of 2.​

 

[2inquisitive]

DaleSpam,

The only reason that I seem to disagree with the standard explanation is because you seem completely unable to distinguish between the concepts of strain and Lorentz contraction. The bottom part of the rim is not Lorentz contracted in the ground frame, it is strained in all frames. Lorentz contraction is a frame-dependent change in coordinate distances, strain is a frame-independent change in proper distances.

I know the difference between strain and Lorentz contraction, of course. What you do not seem to understand is that I don't agree that the bottom part of the 'rim' and the track between rims is strained in all frames. Hooke's law is a non-relativistic law, there is no accepted relativistic formulation of the law. You are trying to apply a non-relativistic law to a relativistic scenario. I remember there is a co-variant formulation by Goen (?) or something like that, but it is not accepted by everyone.
Back to GPS for a moment, for comparing different frames in rotating bodies. The rest frame , and ground frame, of the wheel is an inertial frame, inertial. The rotating frame of the same wheel is a non-inertial frame. There are no strains or torques in an inertial frame. In GPS, the inertial frame is the Earth-Centered Inertial (ECI) frame. It is a frame in which the surface of the Earth does not rotate, the frame in which clocks are synchronized. The frame in which the surface of the Earth does rotate is the Earth-Centered, Earth-Fixed frame, a non-inertial frame of reference. The non-inertial frame overlays the inertial frame in the backgound. Let me emphasize this statement from my previous post, along with a followup on the same page which you must not have read:

A rotating frame of reference is a special case of a non-inertial reference frame in which the coordinate system is rotating relative to an inertial reference frame.

All non-inertial reference frames exhibit fictitious forces.

To derive these fictitious forces, it's helpful to be able to convert between the coordinates of the rotating reference frame and the coordinates of an inertial reference frame with the same origin.

So, I believe you used the wrong approach by involking Hooke's law to try to solve a relativistic problem through 'stretching' of the rim, the track and everything else.

Edit: also remember our discussions of centrifugal force. There are no 'frame forces' in Special Theory. The ground observer in this scenario is in an inertial frame. He can describe no forces in the wheel, you have to be attached to the wheel to feel, and describe, any forces.

 

[2inquisitive]

Pete,

No, an observer attached to the tread is not inertial - inertial for some of the time isn't enough.

I will repost my statement:

However, the parts of the tank tread between the wheels (axels), top and bottom, are inertial frames. Just as on my conveyor belt, an observer attached to either surface feels no fictitious forces.

I disagree, a frame can be inertial, accelerate, then return to inertial again. The observer attached to the tread does just that, accelerating only when he 'rounds the bend' at the ends of the track.

Pete,

Yes, the tank frame is still relevant, whether or not you want to think about it.

I never said the tank frame wasn't relevant, I said we should try to model the ground frame correctly before transforming into the tank frame of reference.
The rest of your post about 'stretching' a track was addressed in my post to DaleSpam. In addition, there are two wheels in this scenario, with a tread encircling them. We could make the track a light-year long and a meter high if we wanted to.

 

[DaleSpam]

I know the difference between strain and Lorentz contraction, of course. What you do not seem to understand is that I don't agree that the bottom part of the 'rim' and the track between rims is strained in all frames. Hooke's law is a non-relativistic law, there is no accepted relativistic formulation of the law. You are trying to apply a non-relativistic law to a relativistic scenario. I remember there is a co-variant formulation by Goen (?) or something like that, but it is not accepted by everyone.

So, I believe you used the wrong approach by involking Hooke's law to try to solve a relativistic problem through 'stretching' of the rim, the track and everything else.

Care to back your claim (that the bottom part is not strained) up with something more than the claim itself?

I already demonstrated for the tank track that it is strained the same in both frames, and a simple symmetry argument suffices to show that all parts of the wheel rim are strained the same. If there is no relativistic formulation of Hooke's law that only means that we have no good way of relating the strains to stresses, it does not imply that the strains do not exist or that you can ignore them. The strains can be calculated kinematically.


-Dale

 

[2inquisitive]

DaleSpam,

Care to back your claim (that the bottom part is not strained) up with something more than the claim itself?

Yes, I can back it up. In Einstein's and Clark's explanation of Ehrenfest's paradox, which you used except for your addition of Hooke's law, the rods attached at one end to the rim are Lorentz-contracted, but the rim itself is not because the spokes of the rim cause a strain on the rim. There are no spokes along the part of the track at rest relative to the ground in between the rims. There is nothing to cause a strain on these cleats.
The distance between the axels is contracted according to the ground observer, so no strain is placed on the tread by the axel spacing either. Just the opposite, in fact, if you claim the tread is 'stretched' by a factor of two. The axel spacing was already less than the tread length by a factor before the tread is 'stretched' according to you. The explanation you are using, again, is applicable to a relativistic wheel, not two wheels with a tread encircling both wheels.

DaleSpam,

I already demonstrated for the tank track that it is strained the same in both frames, and a simple symmetry argument suffices to show that all parts of the wheel rim are strained the same.

Again, your strain is caused by the uncontracted spokes preventing rim from contracting on a spinning wheel. This has no effect on the parts of the tread not in contact with the wheel. Einstein's explanation itself was his intro into the non-Eculidean geometry of General Relativity, curved spacetime. The frame is non-inertial. The ground frame I speak of is inertial, with an inertial object (the tank and treads) moving relative to it, and in contact with the ground. Understand I said the tank and the flat sections of its tread were in a inertial frame, not the rotating wheels/axels.

DaleSpam,

The strains can be calculated kinematically.

By whom, H Nicolic? Gron, Clark and Einstein all used similar explanations for the relativistic wheel. I haven't seen any calculations of the relativistic kinematics addressed by any of them, except Gron. He stated a relativistic kinematical formulation of the wheel was impossible.

Whose version of the wheel do you support, Dale? Why not Grubaum and Janis?
Contracted rotating disk
Grünbaum and Janis [117] have considered a disk put into rotation in such a way that the
radius contracts and no tangential stresses appear. This means that the rest length of tangential
mass elements remains unchanged during the period of angular acceleration. At first moment
one might think that this is not possible. Due to the relativity of simultaneity the special
theory of relativity forbids, in the case of rotating motion with constant radius, to increase the
angular velocity of a rotating disk in a Born rigid way. Hence, tangential stresses will appear,
and the rest length of the periphery changes. There was a discussion of this in Foundations of
Physics [118, 119] and it became clear that the type of motion considered by Grünbaum and
Janis is indeed permitted by the theory of relativity.
http://freeweb.supereva.com/solciclos/gron_d.pdf

 

[Pete]

I disagree, a frame can be inertial, accelerate, then return to inertial again.

No. An object or observer can be inertial, accelerate, then return to inertial again. The rest frame of such an object is not inertial.
An observer attached on the track is sometimes inertial and sometimes accelerating. That observer's rest frame is noninertial.

But don't take my word for it - look up the meaning of an inertial frame.

The rest of your post about 'stretching' a track was addressed in my post to DaleSpam.

I don't think you even read it. You certainly didn't address it.

In addition, there are two wheels in this scenario, with a tread encircling them. We could make the track a light-year long and a meter high if we wanted to.

Yes... but I'm not sure why you feel the need to point it out to me?

 

[Pete]

Yes, I can back it up. In Einstein's and Clark's explanation of Ehrenfest's paradox, which you used except for your addition of Hooke's law, the rods attached at one end to the rim are Lorentz-contracted, but the rim itself is not because the spokes of the rim cause a strain on the rim. There are no spokes along the part of the track at rest relative to the ground in between the rims. There is nothing to cause a strain on these cleats.

The distance between the axels is contracted according to the ground observer, so no strain is placed on the tread by the axel spacing either. Just the opposite, in fact, if you claim the tread is 'stretched' by a factor of two. The axel spacing was already less than the tread length by a factor before the tread is 'stretched' according to you. The explanation you are using, again, is applicable to a relativistic wheel, not two wheels with a tread encircling both wheels.

You're not thinking this through, 2inq.
Firstly, in the classic rotating fixed disk, the rim is strained at all points - not just where the spokes attach. So whether spokes exist at a particular point isn't indicative of whether it is strained or not.

Secondly, you're forgetting the length contraction of the upper track. The upper track is contracted by seven, but the tank is only contracted by two - something has to give. Either the tank compresses, or the track stretches.

Thirdly, the track is continuous. A strain in one place will be distributed through the track unless something prevents it from happening.



However, it would be possible to mechanically control the track so that the lower track was not stretched. We can work through what SR would predict for the rest of the track in that situation, if you like?