Relativistic paradox involving torsion.

[PeterDonis]

I'm not sure about this. Stress and net force are pretty much independent of each other. For example, if the whole pipe were in compression then there would be stress between adjacent segments of the pipe despite the fact that the net force on a differential element remains in the y-z plane. I know this is the case in non-relativistic physics, but unfortunately I never took any relativistic statics courses!

I agree that the net force on a differential element is in the y-z plane. However, that force is not the "torsion" that Cleonis' "paradox" is referring to. That torsion is the stress in the pipe that the boosted observer would see as the pipe "trying to untwist itself"; the source of that stress is the force exerted on each other by pipe elements which are adjacent to each other in the x-direction. *That* stress therefore has a component in the x-direction in the pipe's CoM frame, which acquires a timelike component in the boosted frame.

I did phrase it somewhat poorly in my previous post; if the pipe is in a stationary state (i.e., it has a constant angular velocity in its CoM frame), the stresses in the x-t plane (the "torsion" ones) do *not* have any net force associated with them; the stresses in opposing directions balance out, unlike the ones in the y-z plane (the plane in which the net force due to the pipe's rotation acts). However, the stresses in the x-t plane are still nonzero, and those are what look like "torsion" in the boosted frame.

 

[PeterDonis]

Warning: long post!
(EDIT: I've added another post correcting some errors in this one--please read both before responding!)

After thinking this over some more, I decided to bite the bullet and actually look at the stress-energy tensor of the pipe in the CoM rest frame and in the boosted frame.

Once again I found that Greg Egan has done most of the hard work, at his page on http://www.gregegan.net/SCIENCE/Rings/Rings.html" . I'm not going to go to the level of detail he does, but I recommend looking at his work as a check on what I'm going to say below about which components of the stress-energy tensor are nonzero.

Egan treats the case of a thin disk or hoop, but that will work fine for dealing with a thin slice of our pipe that's at a particular longitudinal coordinate. I'm going to use z instead of x for the longitudinal direction (the direction in which we're going to boost the frame), since the coordinates we'll end up using will be cylindrical, with r being the radial coordinate and $$\theta$$ being the angular coordinate. All we'll have to do is add one term to Egan's stress-energy tensor, to deal with stresses in the z-direction (which he doesn't include).

Egan's web page actually expresses the stress-energy tensor in a frame which is "co-rotating" with the disk or hoop, but he gives enough information to make it easy to re-express it in the non-rotating frame which is at rest with respect to the pipe's CoM. When we do that, we find that the only nonzero components are (I'm not writing down the detailed expressions that Egan does; we're only interested here in which components are nonzero, not what their specific values are):

$$T_{tt} = E$$
-- energy density

$$T_{t \theta} = \Phi$$
-- angular momentum

$$T_{\theta \theta} = p_{\theta}$$
$$T_{rr} = p_r$$
$$T_{zz} = p_z$$
-- pressure in each of the coordinate directions.

The absence of any other nonzero components in the pipe's CoM frame corresponds to our assumption that there is no torsion in that frame. (I should note, here, that until I took this detailed look I hadn't really thought specifically about which stress-energy tensor components would indicate torsion; some of the discussion in this thread has indicated that torsion might have a timelike component--i.e., one of its indices would be t--but that is *not* correct, as we will see.)

Now, what happens when we boost this tensor in the z-direction? Boosting a tensor is kind of like boosting a vector squared; each component gets two coefficients from the boost matrix. Since we're boosting along the z-direction, our boost matrix, which I'll call L, has the following nonzero components:

$$L^{t}_{t'} = L^{z}_{z'} = \gamma$$

$$L^{t}_{z'} = L^{z}_{t'} = - \beta \gamma$$

$$L^{r}_{r'} = L^{\theta}_{\theta'} = 1$$

We then transform the tensor components using the boost matrix as follows: for a generic tensor component with indices a', b' in the boosted frame, transformation matrix L, and indices m, n in the original frame, we have:

$$T_{a' b'} = L^{m}_{a'} L^{n}_{b'} T_{mn}$$

Note that we are using the Einstein summation convention, so the right-hand side of the above is a sum over all m, n. So the components of our transformed tensor will be (writing only the non-zero components of each sum):

$$T_{t' t'} = L^{t}_{t'}^2 T_{tt} + L^{z}_{t'}^2 T_{zz}$$

$$T_{t' z'} = L^{t}_{t'} L^{t}_{z'} T_{tt} + L^{z}_{t'} L^{z}_{z'} T_{zz}$$

$$T_{z' z'} = L^{t}_{z'}^2 T_{tt} + L^{z}_{z'}^2 T_{zz}$$

$$T_{t' \theta'} = L^{t}_{t'} L^{\theta}_{\theta'} T_{t \theta}$$

$$T_{r' r'} = L^{r}_{r'}^2 T_{rr}$$

$$T_{\theta' \theta'} = L^{\theta}_{\theta'}^2 T_{\theta \theta}$$

Substituting, we obtain:

$$T_{t' t'} = \gamma^2 \left( E + \beta^2 p_z \right)$$

$$T_{t' z'} = - \beta \gamma^2 \left( E + p_z \right)$$

$$T_{z' z'} = \gamma^2 \left( \beta^2 E + p_z \right)$$

$$T_{t' \theta'} = \gamma \Phi$$

$$T_{\theta' \theta'} = p_{\theta}$$

$$T_{r' r'} = p_r$$

Notice that the only other nonzero term that appears is the t-z term, which corresponds to the pipe's momentum in the z-direction in the boosted frame. It is negative because in this frame the pipe is moving backwards (i.e., in the negative z' direction). As I noted above, this term does *not* represent torsion; it's just linear momentum.

Notice again that there are *no* other nonzero terms; i.e., *there is no torsion*! Torsion (i.e., shear stress) would look like a term with two different spatial indices, such as r-z, r-$$\theta$$, or z-$$\theta$$. There are no such terms in the boosted tensor above.

What does this mean? It means that, although the pipe "looks twisted" in the boosted frame (in the sense that a line marked on the pipe, which is parallel to the z-axis in the CoM frame, appears twisted around the pipe in the boosted frame), there is *no* torsional stress in the pipe corresponding to the apparent "twist". So the twist is, in that sense, an "illusion".
(EDIT: No it isn't--I missed a nonzero term in the boosted tensor above. See my next post.)

 

[PeterDonis]

OOPS!

I missed a term! (Serves me right for not checking thoroughly enough.) Here's what I left out:

$$T_{z' \theta'} = L^{t}_{z'} L^{\theta}_{\theta'} T_{t \theta}$$

or, substituting,

$$T_{z' \theta'} = - \beta \gamma \Phi$$

So I was wrong in the above post: there *is* a torsional stress in the z-$$\theta$$ plane. The negative sign means that the stress is opposing the rotation of the pipe.

So what keeps this stress from slowing the pipe's rotation in the boosted frame? I think the answer is that the angular momentum, $$\Phi$$, is *larger* in the boosted frame (by the factor $$\gamma$$) than in the CoM frame. I think this extra angular momentum is what compensates for the torsional stress and keeps it from untwisting the pipe.

 

[Dale]

Hi Peter, thanks for the very nice derivation!

So what keeps this stress from slowing the pipe's rotation in the boosted frame?

Stress wouldn't slow the rotation anyway since stress is a pair of forces that balance out to produce no net force. AFAIK the only way to get a net force is for the stress to be non-constant, which is not the case here, it is constant in both z and theta.

 

[PeterDonis]

DaleSpam: Yes, you're right; there would have to be a gradient in the stress in the negative $$\theta$$ direction to slow down the spin, and there isn't; the only variation in any of the tensor components is radial.

 

[Dale]

To me it seems like it all fits.

 

[Cleonis]

A paradox is a logical self contradiction.
In this case there never was a paradox established anyway. "Not twisted in frame A" and "twisted in frame B" are not mutually exclusive propositions. [...] Since it is simply Minkowski geometry it is completely self consistent.

Overall I share your opinion about the word 'paradox'.
For example, I prefer the expression 'the twin scenario' for what is most often called 'the twin paradox', because the twin scenario doesn't involve a self-contradiction.

That said, very often the word 'paradox' is used in a wider meaning, referring to a counter-intuitive situation, regardless of whether there's actual self-contradiction. In the physics literature it is common to refer to 'the twin paradox'.

Finally, I think the fact that Minkowski spacetime geometry is a self-consistent geometry does not in itself guarantee that applying special relativity will give self-consistent results. Electromagnetism is Lorentz-invariant, but what if some other fundamental law is ultimately not Lorentz-invariant?

[...] that Cleonis' "paradox" is referring to [...]

If this long spinning pipe scenario is to have a name I think 'Cleon's paradox' is neat. My first name is Cleon, the nick 'Cleonis' is a contraction of my first name and part of my last name. So if in the future you happen to discuss this scenario, then please use the name 'Cleon's paradox'. (I don't really think this paradox will make it to posterity, but hey, nothing ventured, nothing gained.)

Cleonis

I almost forgot: thank you both for your contributions. I'm not sufficiently familiar with this stuff to fully understand the stress-tensor material, but I find it very exciting and stimulating

 

[Dale]

That said, very often the word 'paradox' is used in a wider meaning, referring to a counter-intuitive situation, regardless of whether there's actual self-contradiction.

Yes, many words used in science and math have other non-technical meanings. For example, police "force", political "power", "work" of art. These alternate common usages don't change their specific technical meanings.

Finally, I think the fact that Minkowski spacetime geometry is a self-consistent geometry does not in itself guarantee that applying special relativity will give self-consistent results.

Yes it does.

Electromagnetism is Lorentz-invariant, but what if some other fundamental law is ultimately not Lorentz-invariant?

Then SR would still be self-consistent; it would just be inconsistent with experiment. In other words, it would not have any paradoxes, it just wouldn't correctly predict the result of experiments.