1
$\begingroup$

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: $\nabla_{\mu}T^{\mu\nu}=0$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In the second part of the answer you said "The left hand side is invariant under rotation but right hand side is not." In this point I'm not quite sure since the spherical symmetry is encoded in the fact that the density $\rho(r)$ and the pressures $p_{r}(r)$ and $p_{\theta}(r)$ depend only on the radial coordinate and not in the angular ones; otherwise, Einstein equations and the Continuity Equations are not a complete system because one has to also assume this "spherical symmetry" on the equations, which will be an additional constrain. $\endgroup$ – user115376 Dec 1 '17 at 12:34
  • $\begingroup$ @user115376: "$p_{\theta}(r)$ depend only on the radial coordinate and not in the angular ones". I might have misinterpreted your notation but I thought the subscript $\theta$ signifies angular dependence. If not then what does that signify? How will you make stress-energy tensor anisotropic without making it depend on angular coordinates? And yes spherical symmetry is a constraint. It reduces degrees of freedom in metric. $\endgroup$ – utagr Dec 1 '17 at 14:05
  • $\begingroup$ the way is defined the stress-energy tensor (the same as the metric tensor) with $\ldots p_{\theta}(r)(r^{2},r^{2}\sin^{2}\theta)$ the spherical symmetry is manifest. $\endgroup$ – user115376 Dec 4 '17 at 14:47
  • $\begingroup$ Oh, I see now. Yeah, your proposed form of stress tensor is spherically symmetric. I see no reason to rule out such a form of the stress tensor other than that if one wants a theory for perfect fluid. Maybe there is some argument against stresses of this form in theory of elasticity but I am not aware of that. $\endgroup$ – utagr Dec 4 '17 at 19:07
  • $\begingroup$ Thanks for you answer dear @utagr about the interpretation of the components, I was not totally convinced but now I've cleared up concepts. I've found a paper where they talk about this issue hindawi.com/archive/2012/965164 $\endgroup$ – user115376 Dec 5 '17 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.