In General Relativity, a perfect fluid has vanishing viscous shear and vanishing heat flux respectively, so its stress-energy tensor is given by the well known expression:

$$T_{\mu\nu}=(\;\rho(r)+p(r)\;)u^{\mu}u^{\nu}+p(r)g^{\mu\nu}.$$

This expression is tensorial, so it doesn't depend on the system of coordinates. When dealing with a static and spherically symmetric spacetime, like with the metric element:

$$ds^{2}=-e^{\nu(r)}dt^{2}+e^{\lambda(r)}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2},$$ in a co-moving frame, the above stress energy tensor becomes

$$T_{\mu\nu}={\rm diag}(\,\rho(r)\; e^{\nu(r)},p(r)\;e^{\lambda(r)}, p(r)\;r^{2},p(r)\;r^{2}\sin^{2}\theta\,).$$

What do they mean the components of $T_{\mu\nu}$?:

$T_{tt}=T_{00}=\rho(r)\; e^{\nu(r)}$ energy flux (mass density since the fluid is at rest).

$T_{rr}=T_{11}=p(r)\;e^{\lambda(r)}$ a radial pressure?

$T_{\theta\theta}=T_{22}=p(r)\;r^{2}$ an angular-azimuthal pressure?

$T_{\phi\phi}=T_{33}=p(r)\;r^{2}\sin^{2}\theta$ an angular-circular pressure?

Physically what do $T_{\theta\theta}$ and $T_{\phi\phi}$ mean?

Could one propose an anisotropic stress-energy tensor like

$$T_{\mu\nu}={\rm diag}(\,\rho(r)\; e^{\nu(r)},p_{r}(r)\;e^{\lambda(r)}, p_{\theta}(r)\;r^{2},p_{\theta}(r)\;r^{2}\sin^{2}\theta\,),$$

compatible with the static and spherical symmetry as long as $p_{r}(r)$ and $p_{\theta}(r)$ are connected through the compatibility relation: $\nabla_{\mu}T^{\mu\nu}=0$?

## Best Answer

Components of Stress-Energy Tensor, in any arbitary coordinates, are given by, $$ T_{{\mu}{\nu}} = T(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}) .$$ One can physically interpret them as follows: $ T_{{\mu}{\nu}}$, at a point P of space-time, tells the flow of $\mu ^{th} $ component of four momentum along the $\frac{\partial}{\partial x^{\nu}}$ direction. For example, $T_{00}$ denotes how much energy per unit volume is flowing in time direction, which is same as energy density. Similarly $T_{i0}$ denotes flow of momentum (not four momentum) per unit volume along time direction, that is momentum density.

Thus, $T_{ii}$ denotes flow of $i^{th}$ component of momentum along $\frac{\partial}{\partial x^i}$ direction. But that's the definition of pressure. Since pressure is a local phenomenon, even in curved space-time, it does not matter whether you work in curvilinear or rectilinear coordinates. Locally every transformation is linear enough to define pressure as we usually do.

In your example, $(\frac{\partial}{\partial r},\frac{\partial}{\partial \theta},\frac{\partial}{\partial \phi})$, at a point could be thought of constituting a Cartesian system. The radial direction could very well be defined as x direction, locally. Since pressure is also local, $T_{rr}$ denotes pressure along the radial direction. Same goes with other two directions. $T_{\theta \theta}$ and $T_{\phi \phi}$ denotes pressure along $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \phi} $ directions respectively.

For the second part of your question, No, you cannot do that. That would break spherical symmetry. A concise explanation is: metric and matter are coupled by Einstein Field Equation. If one is not symmetric under rotations then other cannot be symmetric.

A detailed explanation: Assume metric is spherically symmetric and matter is given by, $$T_{\mu\nu}=diag(\,\rho(r)\; e^{\nu(r)},p_{r}(r)\;e^{\lambda(r)}, p_{\theta}(r)\;r^{2},p_{\theta}(r)\;r^{2}\sin^{2}\theta\,).$$ If metric is invariant under rotation so is Einstein Tensor, $G_{\mu \nu}$, as Einstein Tensor is made from metric tensor. We have following equation, $$ G_{\mu \nu}=\kappa T_{\mu \nu} .$$ The left hand side is invariant under rotation but right hand side is not. A contradiction! Therefore our assumption is wrong. Matter cannot be given by our proposed form.