# Deriving the Metric from the Energy-Momentum Tensor

by Airsteve0
Tags: deriving, energymomentum, metric, tensor
Thanks
P: 2,951
 Quote by atyy Doesn't the energy-momentum tensor contain the metric?
No. They are related by the Einstein Field Equation, which is a set of differential equations for the metric in terms of the stress-energy tensor, but the EFE alone does not contain enough information to completely specify the metric. Like any reasonable differential equation, the EFE gives you a family of solutions; the boundary conditions are required to know the specific member of that family that describes the physical situation at hand.
P: 8,005
 Quote by Nugatory No. They are related by the Einstein Field Equation, which is a set of differential equations for the metric in terms of the stress-energy tensor, but the EFE alone does not contain enough information to completely specify the metric. Like any reasonable differential equation, the EFE gives you a family of solutions; the boundary conditions are required to know the specific member of that family that describes the physical situation at hand.
How about Eq 8.15 of http://ned.ipac.caltech.edu/level5/M.../Carroll8.html? Isn't that the metric in the stress-energy tensor?
P: 2,470
 Quote by atyy How about Eq 8.15 of http://ned.ipac.caltech.edu/level5/M.../Carroll8.html? Isn't that the metric in the stress-energy tensor?
This has to do with choice of perfect fluid as the source. What this really tells you is that when you'll be solving for metric, you'll be solving it as a system. You'll be trying to find T and g that satisfy both the equation for fluid and the Einstein Field Equation.
P: 8,005
 Quote by K^2 This has to do with choice of perfect fluid as the source. What this really tells you is that when you'll be solving for metric, you'll be solving it as a system. You'll be trying to find T and g that satisfy both the equation for fluid and the Einstein Field Equation.
But isn't that the OP's question? I guess I don't understand what he means by "expression for the energy momentum tensor" since the expression seems to already contain the metric.

 Quote by Airsteve0 Say we were given an expression for the energy-momentum tensor (also assuming a perfect fluid), without getting into an expression with multiple derivatives of the metric, are there any cases where it would be possible to deduce the form of the metric?
PF Gold
P: 4,862
 Quote by atyy But isn't that the OP's question? I guess I don't understand what he means by "expression for the energy momentum tensor" since the expression seems to already contain the metric.
The original question was can you guess the metric from T, for some simple cases, without solving differential equations (and I consider guessing a method of solution; perhaps the most common method - guess and verify). I sidetracked that by suggesting it was an implausible expectation, because it couldn't be done for the trivial case of T=0.

As for g within T, the physical parts of it (e.g. pressure and density) do not contain the metric. This is what you might specify; then the EFE contain g on both sides as an undetermined variable to solve for. Alternatively, if you are given the full T as function in some arbitrary coordinates, you have no idea a priori how the the metric figures in T.

I guess one special case is you know the form of T in terms of g and physical quantities, you are given the physical quantities and the complete expression of T. Then, you could get g. Of course, if you pick these things arbitrarily, the chance of satisfying the EFE is zero.
P: 8,005
 Quote by PAllen The original question was can you guess the metric from T, for some simple cases, without solving differential equations (and I consider guessing a method of solution; perhaps the most common method - guess and verify). I sidetracked that by suggesting it was an implausible expectation, because it couldn't be done for the trivial case of T=0. As for g within T, the physical parts of it (e.g. pressure and density) do not contain the metric. This is what you might specify; then the EFE contain g on both sides as an undetermined variable to solve for. Alternatively, if you are given the full T as function in some arbitrary coordinates, you have no idea a priori how the the metric figures in T. I guess one special case is you know the form of T in terms of g and physical quantities, you are given the physical quantities and the complete expression of T. Then, you could get g. Of course, if you pick these things arbitrarily, the chance of satisfying the EFE is zero.
So is the last what the OP had in mind, since he did specify that it's the stress tensor of a perfect fluid?