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Using Einstein's Field Equations for a specific metric I am interested in, I have calculated the first diagonal term of the stress energy tensor $T$, which is the energy density. In my calculations, the covariant density ($T$ with lower indices 00) does not show a singularity, but the contravariant density ($T$ with upper indices 00) does show a singularity. I need to know whether this singularity is physical. Which form of the stress energy tensor represents actual physical density, the covariant or the contravariant?

Qmechanic
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Karl Pomeroy
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    $\uparrow$ Which specific metric? – Qmechanic Dec 17 '16 at 20:21
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    Neither one. You need to calculate invariant quantities from the Riemann tensor. Or the stress energy tensor. Most of the physical definitions refer to the contravariant components, see at http://en.wikipedia.org/wiki/Stress–energy_tensor, but the components are not invariant and a different coordinate system wil give you different values. The energy density, $\rho$, even though it has a physical name, is not invariant, it is part of that tensor. Even in special relativity, energy is part of a four vector. – Bob Bee Dec 17 '16 at 20:55
  • I'm confused because I would have expected $T^{\mu\nu} = g^{\mu\alpha} g^{\nu\beta} T_{\alpha\beta}$ and so a singularity in $T^{\mu\nu}$ should appear in either $T_{\mu\nu}$ or $g^{\mu\nu},$ no? Granted you only claimed $T^{00}$ was singular, and that could hide in the other components of $T_{\mu\nu},$ but if it appears in $g^{\mu\nu}$ then it sounds pretty physical... – CR Drost Dec 17 '16 at 21:42
  • @BobBee: that should be an answer – John Rennie Dec 18 '16 at 11:51
  • To the first reply "neither one": Energy density isn't invariant, so it seems the invariant or mixed form of T00 would not be the answer. You say energy density is "part of that tensor". Sorry, which tensor do you mean? To the second reply "I'm confused": Actually you're not. There IS a singularity in the metric, but it might only be a coordinate singularity. To be more precise, there are actually 2 singularities (r=0, r=R) in the contravariant form of T00 but only one (r=0) in the covariant. I need to know if the r=R singularity is a real physical divergence in energy density. – Karl Pomeroy Dec 18 '16 at 21:10
  • On the first one, the stress energy tensor of course. On the second, the singularity in the metric may be simply a coordinate singularity, as you seem to know. There is nothing inherent in the contravariant, covariant, or mixed components of the stress energy tensor that cannot be changed with a coordinate transformation, EXCEPT for the scalars. It does depend on the metric ALSO. See also Crowell's answer at http://physics.stackexchange.com/questions/63749/is-it-possible-to-have-a-singularity-with-zero-mass which deals with parts of this. Continued next comment – Bob Bee Dec 19 '16 at 01:25
  • Now,$\rho$ could be infinite also, but careful and look for invariants like the trace of the stress energy tensor. In cosmology FRW solutions the trace is 3p-$\rho$, and with p proportional to density that is then infinite at t=0 (except the obvious case, but it turns out also there so one invariant is not suffieceint). TOV black holes, have infinite density at r=0, and not at the horizon. The infinite density also has infinite curvature (invariants).Local energy is not a well defined nor conserved quantity in GR unless a time symmetry, global is for flat asymptotics. – Bob Bee Dec 19 '16 at 01:49
  • So, be specific on the metric, or there is no absolute answer. – Bob Bee Dec 19 '16 at 01:50

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Generally speaking the contravariant form $T^{\alpha\beta}$ is the one most naturally related to what we think of as physical observables. My favourite way of thinking about this is to start with the stress-energy tensor of a point particle, and we can understand the more complicated forms as made up from point particles.

But we should note Bob Bee's comments. When you write down $\mathbf T$ you have to choose a coordinate system to do so, and the individual components of the tensor will depend on the coordinates you choose. If your choice of coordinates results in coordinate singularity then you may well find your stress-energy tensor behaves strangely at that point.

It's hard to comment without knowing the details of the geometry you are studying, but if $T_{\alpha\beta}$ isn't singular this suggests very strongly that you have a coordinate singularity rather than a real one. Try calculating the trace of $\mathbf T$ to see if it diverges.

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John Rennie
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  • I think you can make an argument that the mixed index form $T^\alpha_\beta$ is the "most natural," since it is the thing you use to construct fluxes of a Noether current associated with some translation direction ($J^\alpha = \xi^\beta T_\beta^\alpha$, where $\xi^\beta$ is the vector defining the translation). This is mostly a side comment, but may be relevant to the question since you might be able to check that these fluxes are nonsingular as a way of getting something that is coordinate invariant. – asperanz Dec 19 '16 at 17:43
  • Thanks all for the discussion. To John Rennie, I think the covariant form of T is physical in this case. You may have the key, though, saying "you have to choose a coordinate system." I did, and since the dx's are contravariant ( as is customary), the invariance of the line element means the metric tensor g is covariant. Covariant g is therefore physical (think of 1-2m/r in the time slot of the Schwarzschild metric), and since covariant g is on the left side of Einstein's Field Equations, T on the right side is covariant and physical. (Maybe the opposite if you chose covariant coordinates.) – Karl Pomeroy Dec 19 '16 at 18:04
  • Correction to my previous comment. James Hartle's textbook "Gravity, An Introduction to Einstein's General Relativity" confirms more or less explicitly on pp474-477 that the physical energy density is the first diagonal element of the contravariant form of T, not the covariant as I had stated. Can I assume Hartle is correct? (He's been known to make an occasional mistake.) – Karl Pomeroy Dec 19 '16 at 19:38
  • Addendum to my previous correction: Stephen Weinberg, in "Gravitation and Cosmology" (1972) says on p. 152 that the COVARIANT form of T is the physical energy density---the opposite of what James Hartle said (see my last comment). So the plot thickens. Sorry for the confusion. – Karl Pomeroy Dec 19 '16 at 19:53