Stress-energy tensor & mass term in metric

1. Dec 26, 2013

gabeeisenstei

I'm trying to clarify for myself the relation between the stress-energy tensor and the mass scalar term in metric solutions to Einstein's equations. Maybe I should also say I'm trying to understand the energy tensor better, or how it relates to boundary conditions on the solutions.

My question is, given a nonzero s-e tensor, can one solve for the mass term that will show up in the metric? (If the tensor were zero, one would have a "vacuum solution" in which the mass term is only specified as a boundary condition.) Or, given the energy density and pressures of the tensor, are there still degrees of freedom for mass-energy configurations producing that tensor?
(I'm assuming that the mass in question is coextensive with the tensor volume--so a simple case like the interior of a star or planet.)

Part of my confusion stems from the fact that the tensor is sometimes defined in terms of the metric (for a fluid in equilibrium, T=(ρ+p)vv + p*g, where g is the metric), so that the metric now appears on both sides of Einstein's equations. How does one eliminate that circularity?

2. Dec 26, 2013

WannabeNewton

Hey there! Let's stick to the simple case of a perfect fluid source. A perfect fluid is one which has no viscosity or heat conduction in a frame momentarily at rest with respect to the fluid 4-velocity $u^{\mu}$. The dynamical variables of interest then for describing the physics of fluid elements are the mass density $\rho$ and isotropic pressure $p$ in the momentary rest frame. From the definition of the stress-energy tensor, one can then show that $[T^{\hat{\mu}}{}{}_{\hat{\nu}}] = \text{diag}(\rho,p,p,p)$ where I have used hats on the Lorentz indices to indicate that we are in the momentary rest frame of the fluid and I have used brackets to indicate a matrix representation.

From this one can then easily write down the stress-energy tensor for a perfect fluid that is valid in any frame: $T^{\mu\nu} = \rho u^{\mu}u^{\nu} + p(g^{\mu\nu} + u^{\mu}u^{\nu})$.

The local conservation of energy, $\nabla_{\mu}T^{\mu\nu}$, then leads straightforwardly to the equations of motion for the perfect fluid: $\nabla_{\mu}(\rho u^{\mu}) = -p\nabla_{\mu}u^{\mu}$ and $(\rho + p)u^{\nu}\nabla_{\nu}u^{\mu} = -(g^{\mu\nu}+ u^{\mu}u^{\nu})\nabla_{\nu}p$.

Up till now we've basically derived equations governing the local kinematics of the fluid. Likewise $T^{\mu\nu}$ is a characterization of the pressure and 4-momentum density associated with the perfect fluid so it is also local. Notice how both $T^{\mu\nu}$ and the equations of motion for the fluid depend on $g_{\mu\nu}$ (the metric enters the equations of motion through $\nabla_{\mu}$) but at the same time $g_{\mu\nu}$ depends on $T^{\mu\nu}$ and the kinematics of the fluid. This interdependence of $T^{\mu\nu}$ and $g_{\mu\nu}$ is actually a general characteristic of Einstein's equations and is one of the reasons why Einstein's equations are so hard to solve.

Now say we have a static spherically symmetric perfect fluid (describing a fluid star of the same nature); the exterior space-time geometry will be the Schwarzschild solution and the interior solution can be determined from Einstein's equation in a very similar manner (see section 6.2 of Wald "General Relativity" and the notes below). The Schwarzschild mass of the fluid star can be determined entirely from $T^{\mu\nu}$ and the proper mass of the star can be determined from $T^{\mu\nu}$ and $g_{\mu\nu}$ where the latter enters the picture because the proper mass includes the gravitational binding energy of the self-gravitating system defined by the star.

See here for a detailed exposition of relativistic fluid stars: http://www.pma.caltech.edu/Courses/ph136/yr2004/0425.1.K.pdf

Last edited: Dec 26, 2013
3. Dec 27, 2013

gabeeisenstei

Thanks very much, I'm reading both the Wald and the Thorne now. It's a lot to digest, especially as regards the interdependence of T and g. But one thing I glean right away is that the time-time component of T is more or less the same as the M in the metric (with relativistic correction, or integrated over shells if nonuniform density).

4. Dec 28, 2013

WannabeNewton

In a sense yes that's true of the stress-energy tensor for a perfect fluid as evidenced by expressions 6.2.10 and 6.2.11 in Wald.

If you want a much more general and a much more detailed discussion of the mass-energy of space-times then read chapter 11 of Wald's text and feel free to ask any questions you may have regarding the chapter. Cheers.