- #1
gabeeisenstei
- 37
- 0
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?
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?