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Confusion about Ohanian's one-dimensional gas

  1. Jan 23, 2014 #1

    bcrowell

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    This is about Hans Ohanian, "Einstein's E = mc2 mistakes," http://arxiv.org/abs/0805.1400 , a historical paper that discusses whether Einstein's original proofs of E=mc2 were really valid.

    My general understanding of the topic is as follows. I'm just discussing SR, not GR. What's really foundational is that the stress-energy tensor has zero divergence. Using this fact, along with Gauss's theorem, you can prove that the energy-momentum vector [itex]p^\mu=\int_S T^{\mu0} d^3(\text{vol})[/itex] is (a) conserved and (b) a four-vector, in the case of an isolated system. Rindler, Introduction to Special Relativity (1982 edition), has proofs in section 50. Basically you extend the three-surface S in the time dimension to make a four-volume R bounded by timelike surfaces A (earlier) and B (later). In the case where A and B are both surfaces of simultaneity for the same frame of reference, you get a proof of conservation. By allowing B to be a surface of simultaneity for some *other* observer, you get a proof that p transforms like a four-vector. The proof depends on the fact that it's an isolated system, because the three-volume S is assumed to be big enough so that the stress-energy tensor vanishes outside S. Only by making this assumption can we use Gauss's theorem to equate the flux of energy-momentum through A to minus the flux through B.

    Ohanian's example (p. 4) is the following. You let the matter contained in S be a set of identical particles, half of which move in the +x direction with speed u and half of which move with that speed in the -x direction. Basically it's a one-dimensional gas. Ohanian analyzes this without the stress-energy tensor (he's discussing a paper that predates the stress-energy tensor), just by using combination of velocities. He finds that when you start in the gas's rest frame, then apply a boost v, the low-velocity behavior of the kinetic energy is different depending on whether v is applied in the x direction or the y or z direction. In the x direction you get an excess mass of [itex]2mu^2[/itex], where m is the total mass of all the particles. In the y or z direction you get what you'd expect.

    I worked this out using the stress-energy tensor and got the same result. In the rest frame, the stress-energy tensor has the form [itex]\operatorname{diag}(\rho,P,0,0)[/itex], where P is the pressure. When you boost in the x direction, [itex]\rho[/itex] goes to [itex]\gamma^2(\rho+v^2P)[/itex]. The volume has been decreased by a factor of gamma, so integrating this over the volume kills off one of the factors of gamma. The quadratic term in the energy is then [itex](1/2)(\rho+2P)Vv^2[/itex] (expressed in terms of the volume V in the c.m. frame). To lowest order in u, [itex]P=mu^2/V[/itex], so this gives the same result as Ohanian's for the extra inertia in the x direction. When you instead boost the stress-energy tensor in the y or z direction, you don't pick up the extra term, since the yy and zz parts of the stress-energy are zero.

    The result of all this is that the kinetic energy is not isotropic, even to leading order in v, which means that the total energy-momentum p isn't transforming like a four-vector.

    Ohanian says that the resolution of the problem is that you have to include the container of the gas in the picture. The container is under tension (negative pressure), so when you include that tension in the stress-energy tensor, it just cancels out the effect of the gas's pressure, and you recover the usual isotropic behavior of kinetic energy. I've seen other examples in which it seems clear that this type of effect is necessary and sufficient to fix the problem, so that p transforms as a four-vector. (A nice one involving a capacitor is http://arxiv.org/abs/physics/0609144 .)

    Here's what's confusing me about this example. The question is whether, in Ohanian's example, the tension in the container is necessary and sufficient to fix the problem. Let's say the container simply doesn't exist. Then we have two interpenetrating clouds of particles, which will fly off in the +x and -x directions. Nevertheless it seems that by making S big enough, we could arrange things so that R would still have zero flux through its "sides" (everything except the surfaces A and B). The proof using Gauss's theorem would then hold, and the energy-momentum p should transform as a four-vector. The only difference this makes in my analysis using the stress-energy tensor is that if you're not in the c.m. frame, the volumes of the two interpenetrating clouds are no longer equal to the volume of the nonexistent container. The way I've attempted to account for this is as follows. Let V be the volume of either cloud in the c.m. frame. Then under a boost v in the x direction, they have volumes [itex]V_+'[/itex] and [itex]V_-'[/itex], where

    [tex]\frac{V_+'}{V}=\frac{\gamma_u}{\gamma_{u\parallel v}}=1-\frac{1}{2}v^2-uv+u^2v^2+\ldots[/tex]

    and [itex]V_-'[/itex] works the same way except for the sign of u. So after a boost v we have

    [tex]E' = \frac{1}{2}\gamma_v^2(\rho+v^2P)V_+'+\frac{1}{2}\gamma_v^2(\rho+v^2P)V_-'[/tex]

    Letting [itex]E=\rho V[/itex] be the total energy in the original c.m. frame, we have

    [tex]\frac{E'}{E} = 1+\frac{1}{2}\left(1+4u^2+\ldots\right)v^2+O(v^4),[/tex]

    which doesn't seem right. The two clouds have energy-momentum vectors in the c.m. frame that are (E/2,+...) and (E/2,-...), for a total of (E,0). This is just the sum of two four-vectors, so it should also transform as a four-vector, and therefore we should have

    [tex]\frac{E'}{E} = 1+\frac{1}{2}v^2+O(v^4),[/tex]

    regardless of the internal stresses. I stared at the [itex]4u^2[/itex] for a really long time, willing it to be a 2-2 rather than a 2+2...but it just wouldn't cooperate.

    Can anyone help me figure out what I've done wrong here? Thanks in advance!
     
  2. jcsd
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