# Which form of Einstein's stress-energy tensor $T$ is physical, the covariant or the contravariant?

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?

• $\uparrow$ Which specific metric? Dec 17 '16 at 20:21
• 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 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. 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... Dec 17 '16 at 21:42
• @BobBee: that should be an answer 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. Dec 18 '16 at 21:10

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.
• 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. Dec 19 '16 at 17:43