# 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.