# If stress vanishes in one frame, it vanishes in all frames, but

1. Mar 16, 2014

### bcrowell

Staff Emeritus
This paper discusses the Bell spaceship paradox:

Petkov, "Accelerating spaceships paradox and physical meaning of length contraction," http://arxiv.org/abs/0903.5128

PF has a FAQ on this paradox: https://www.physicsforums.com/showthread.php?t=742729 [Broken] My purpose in creating this thread is *not* to open yet another general discussion of the paradox. My purpose is to discuss one specific passage in the Petkov paper.

The following occurs on p. 4:

Petkov's reasoning seems clearly wrong to me on at least one count, since nobody ever claimed that length contraction was the *only* way that stress could occur. In one frame there is stress because of Lorentz contraction. In another frame there is stress because the ships have gotten farther apart.

Petkov also seems to impute to Bell the claim that the thread is stressed in one frame but not another. This also seems clearly wrong to me. Bell claimed that the thread would be stressed in all frames -- hence all observers agree that it breaks.

However, the reason I'm posting here is yet a third logical point, which is Petkov's statement that if stress occurs in one frame, it must occur in all frames. He argues that this is because the "stress tensor," by which I assume he means the stress-energy tensor, must vanish in all frames if it vanishes in one frame. This argument doesn't make a lot of sense to me. It's certainly true that if a tensor vanishes in one frame it must vanish in every other. But the stress-energy tensor of the thread is not going to vanish in any frame, since the thread has mass. It's perfectly possible, for example, that we have $T^{xx}=0$ in one frame but $T^{x'x'}\ne0$ in another. I suppose the correct test to see whether the thread is under stress is to check the eigenvalues of the stress-energy tensor. If they look like $(\rho,0,0,0)$, then I guess there would be a frame, interpreted as the rest frame of the thread, in which it experienced no stress; and it would then follow that the thread would not stretch, break, etc. Both the test (eigenvalues) and the interpretation (whether it stretches, breaks, etc.) are frame-independent.

Is my interpretation correct? If so, then how are we to interpret the case where some matter has $T^{xx}=0$ in one frame but $T^{x'x'}\ne0$ in another? Should we interpret $T^{x'x'}$ as some kind of pressure, or not? If so, then why is it a pressure that doesn't produce effects on the matter (e.g., breaking)? After all, if you take dust in its rest frame and transform to another frame, you have $T^{xx}=0$ in the rest frame but $T^{x'x'}\ne0$ in any other. This $T^{x'x'}$ is a transport of x-momentum in the x direction, but I probably wouldn't interpret it as pressure.

One often sees the elements of the stress-energy tensor in a given frame described in a certain way, with, e.g., the space-space components being interpreted as pressure. E.g.: http://en.wikipedia.org/wiki/File:StressEnergyTensor_contravariant.svg This doesn't seem quite right in the example of dust in a frame other than its rest frame.

Last edited by a moderator: May 6, 2017
2. Mar 16, 2014

### Staff: Mentor

I think there's a key distinction that needs to be made here, between "tensor components", numbers describing a tensor in a particular coordinate chart, and "observed stresses", something someone actually measures. The observed stresses are contractions of the stress-energy tensor with some pair of vectors: for example, the energy density measured by a particular observer is $\rho = T_{ab} u^a u^b$, where $u^a$ is the observer's 4-velocity. Similarly, if that observer carries with him a spatial vector $\hat{e}_x$, with components $e^a$, the pressure he will measure in his $x$ direction is $T_{ab} e^a e^b$.

Now consider how that equation looks in two different frames. In the observer's instantaneous rest frame, $e^a = \partial / \partial x$, so the pressure he measures in the $x$ direction is simply $p_x = T_{\hat{x} \hat{x}}$, where I have put the hats on the indexes to make it clear that this is the tensor component in the coordinate chart adapted to the observer's instantaneous rest frame. But in some other coordinate chart, the observer's spatial vector $\hat{e}_x$ may have multiple components, so the equation for $p_x$ in that chart will be a more complicated contraction of the SET with the spatial $x$ vector as expressed in that chart. The more complicated equation will yield the same *number* for $p_x$, but it will be by a different route, so to speak.

Now suppose the stuff the observer is trying to measure the pressure of is dust, and suppose the observer is comoving with the dust particles in his vicinity. Then the above procedure should give $p_x = 0$, even in a coordinate chart where $T_{x x} \neq 0$. (In the observer's instantaneous rest frame, of course, $T_{\hat{x} \hat{x}} = 0$, so $p_x = 0$ is obvious.) That expresses the physical fact that the comoving observer measures zero pressure in the $x$ direction for the dust.

But if there is a second observer who is *not* comoving with the dust, his $\hat{e}_{x'}$ vector will be different from $\hat{e}_x$ (the $x$ vector for the comoving observer), so his $p_{x'}$ might not be zero. Why would that be? Because he is moving through the dust, therefore the dust is moving relative to him, therefore it can exert force on him. If, for example, the second observer was exposing a large flat surface area in his $y' - z'$ plane to the oncoming dust, a sort of "ram" effect would, I expect, cause him to measure $p_x' > 0$. And this would be true even if we evaluated the measurement in the "comoving" coordinate chart, where $T_{xx} = 0$. (I put "comoving" in quotes here because the second observer's large flat surface would change the motion of the dust as well; but I don't think we need to delve into the details of that here.)

The procedure I've described may be equivalent to checking the eigenvalues of the SET. If so, consider the above a long-winded agreement with that procedure for determining what the SET means, physically. Which also implicitly means that just looking at individual tensor components, in an arbitrary coordinate chart, is not a good route to physical interpretation; you have to look at what is actually going on, physically, with the observer as well as the substance itself.

3. Mar 16, 2014

### bcrowell

Staff Emeritus

The remaining question for me is whether there is any interpretation under which this statement by Petkov makes sense and is applicable to the topic he's discussing:

Clearly the thread has mass-energy, so even if it's not under stress (as measured by evidence like stretching or breaking) its stress-energy tensor won't be zero. Therefore I don't see how any sense can be made out of this statement of Petkov's.

4. Mar 16, 2014

### WannabeNewton

The author might be referring instead to the expansion tensor $\theta_{\mu\nu}$ which codifies radial and shear stresses and which in the case of the Bell spaceship paradox is non-vanishing.

5. Mar 16, 2014

### pervect

Staff Emeritus
My first thought as an interpretation is that Petkov may be considering the stress-energy tensor of a string which is assumed to be massless (and stressless). Or perhaps he is just making an error, it's hard to say.

I'm not sure this makes much physical sense. If we solve for the path of a string via $\nabla_a T^{ab}$ when we have a nonzero SET we have non-trivial equations which define a specific path.

But there is no physical content to the expression if the stress-energy tensor vanishes, any path through space-time will satisfy the equation. So I don't think that a massless, stressless "string" makes much physical sense. Any arbitrary curve could be considered a string under no tension, no matter how oddly shaped. It's not until you put the string under some tension that the notion of a string logically defines a path that must minimize or extremize some sort of distance functional.

Intuitively, if we have a tangled mess of a coil of wire of negligible mass, you can, I suppose, call it a string, but until you pull on the ends to straighten it out, the tangled heap of wire doesn't have anything to do with the notion of distance. When you pull on the ends (and unknot it) you start to have an entity which gives a unique solution that defines a specific path through space-time that extremizes some sort of functional over the path, which we can interpret as minimizing the "distance" of the path. But then the SET isn't zero.

Last edited: Mar 17, 2014
6. Mar 17, 2014

### bcrowell

Staff Emeritus
I guess that's conceivable, although I would consider the expansion tensor to be a measure of strain, not stress (in the sense that strain quantifies deformation while stress measures the forces that caused the deformation). Although it seems unlikely that that's what he intended, it would patch up that part of the argument and make it work.

Another possibility that occurred to me was that maybe he meant the 3-dimensional stress tensor from continuum mechanics. That tensor, unlike the 4-dimensional stress-energy tensor, could vanish for a material object. But then his argument would still be invalid because the 3-dimensional stress tensor isn't a relativistic tensor.

7. Mar 17, 2014

### Staff: Mentor

You could, I suppose, cobble together a "relativistic stress tensor" by taking the full SET in the momentarily comoving frame and subtracting the rest energy density in that frame; that would leave a tensor with only space-space components in that frame, and those components would be zero for an unstressed object (i.e., one with only rest energy density). You could then transform this object as a 4-tensor to an arbitrary frame. But I don't see any indication that Petkov was thinking along these lines.

8. Mar 17, 2014

### Staff: Mentor

I see another questionable statement at the end of Petkov's paper. He gives what actually appears to be a good discussion of Minkowski's view of objects as "world tubes" in 4-D spacetime, and how length contraction is really a consequence of observers in different states of motion measuring different 3-D projections of the world tubes. The world tubes are the invariants, and as he says on p. 8 that, because of the acceleration of the spaceships,

If he were to put this into proper mathematical language, he would basically be saying that the expansion scalar is positive (which is the invariant way of saying that "the proper distance between the ships increases"), and that's what causes the string to break.

But then, in the very next paragraph (the last one of the paper), he says:

But earlier in the paper he had said that the increasing proper distance between B and C because of their acceleration had no classical analogue, which means that in classical theory, the acceleration of B and C would *not* break the wire. (I don't know for sure that that's true, btw; but it would have to be true if what he says here is correct.) But if there were no length contraction, classical theory would have to be true, not relativity; so his statement that I just quoted is false: if there were no length contraction, the acceleration of B and C would *not* break the wire, on his view of classical theory.

In other words, he appears not to realize, even after his discussion of the spacetime "world tube" viewpoint, that length contraction is a necessary part of relativistic kinematics, just as the positive expansion scalar for the congruence described by the world tubes of B and C is. They have to go together.