# Deriving the Metric from the Energy-Momentum Tensor

by Airsteve0
Tags: deriving, energymomentum, metric, tensor
 P: 85 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: 5,027
 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?
I assume you mean stress energy tensor(?)

I don't see how. Even the case of stress-energy tensor = 0 => vacuum, produces non-trivial system of second order partial differential equations for the metric. The only case, even for vacuum, that is simple to derive, is if you add spherical symmetry. Axial symmetry is already much harder, even for vacuum.
Thanks
P: 4,160
 Axial symmetry is already much harder, even for vacuum.
All you have to do is solve Laplace's Equation.

P: 2,470
Deriving the Metric from the Energy-Momentum Tensor

 Quote by Bill_K All you have to do is solve Laplace's Equation.
Laplace equation has nothing to do with gravity. Einstein Field Equations are much more difficult to solve.

 Quote by Airsteve0 are there any cases where it would be possible to deduce the form of the metric?
You can find a numerical solution in some cases. Even then, you almost always have to rely on symmetry of some sort, simply because of dimensionality. Say you want 100 grid points per dimension. This would already require you to solve the equations numerically over 108 grid points. If your stress-energy tensor is time-independent, and you consider only steady-state solutions, then you can assume metric to be time-independent, and that reduces it to 106 points. Much more reasonable, but still extremely difficult to work with due to non-linearity. And how many points are sufficient to get a convergent and useful solution will depend on specific distribution of stress-energy density, and again, since it's non-linear, it might be hard to predict in advance.
PF Gold
P: 5,027
 Quote by Bill_K All you have to do is solve Laplace's Equation.
The simplest axial symmetric vacuum case I know of is the Kerr metric. The derivations of I've seen are far more complex than for spherical symmetry.
PF Gold
P: 5,027
 Quote by K^2 You can find a numerical solution in some cases. Even then, you almost always have to rely on symmetry of some sort, simply because of dimensionality. Say you want 100 grid points per dimension. This would already require you to solve the equations numerically over 108 grid points. If your stress-energy tensor is time-independent, and you consider only steady-state solutions, then you can assume metric to be time-independent, and that reduces it to 106 points. Much more reasonable, but still extremely difficult to work with due to non-linearity. And how many points are sufficient to get a convergent and useful solution will depend on specific distribution of stress-energy density, and again, since it's non-linear, it might be hard to predict in advance.
Right, and I was addressing: "without getting into an expression with multiple derivatives of the metric", for which I still can't think of any answer other than: No.
P: 2,470
 Quote by PAllen I was addressing: "without getting into an expression with multiple derivatives of the metric", for which I still can't think of any answer other than: No.
Have to agree on that one. Trying to solve differential equation without taking derivatives seems like a fool's quest under best of circumstances.
 P: 1,020 outside a given source,i.e. if energy momentum tensor is zero for a region with some specified symmetry(say spherical).one can guess the form of metric but with given energy momentum tensor one has to go with what others have said.
Thanks
P: 4,160
 All you have to do is solve Laplace's Equation.
Laplace equation has nothing to do with gravity. Einstein Field Equations are much more difficult to solve.
 The simplest axial symmetric vacuum case I know of is the Kerr metric. The derivations of I've seen are far more complex than for spherical symmetry.
The general nonrotating time-independent axially symmetric vacuum solution to Einstein's Equation is given very simply by a solution of Laplace's Equation. This is a standard topic covered in any introductory course in General Relativity. For example, see here.
PF Gold
P: 5,027
 Quote by Bill_K The general nonrotating time-independent axially symmetric vacuum solution to Einstein's Equation is given very simply by a solution of Laplace's Equation. This is a standard topic covered in any introductory course in General Relativity. For example, see here.
I was thinking only of the general case - nothing assumed but symmetry.. In the spherical symmetry case, you get static for free. In the axial case you don't. However, if you add staticity + (axial) symmetry, that obviously simplifies it a lot.
 Sci Advisor Thanks P: 4,160 The Weyl family of exact solutions were among the earliest found. Just as for Schwarzschild, in the derivation of the Weyl solutions there is nothing assumed but symmetry, namely the two Killing vectors that generate axial rotations and time translations. Even for the stationary case there are a great many exact solutions known. Kerr turns out to be just the simplest member of an infinite family of solutions.
PF Gold
P: 5,027
 Quote by Bill_K The Weyl family of exact solutions were among the earliest found. Just as for Schwarzschild, in the derivation of the Weyl solutions there is nothing assumed but symmetry, namely the two Killing vectors that generate axial rotations and time translations. Even for the stationary case there are a great many exact solutions known. Kerr turns out to be just the simplest member of an infinite family of solutions.
But time like killing vector is the extra symmetry. For spherical symmetry you don't need to assume it, you discover it without assuming it. For the axial case, that is not so. You have to assume it, and it greatly simplifies the solution.
Mentor
P: 6,232
 Quote by PAllen But time like killing vector is the extra symmetry. For spherical symmetry you don't need to assume it, you discover it without assuming it.
Is the interior of a Schwarzschild black hole spherically symmetric? Does this spacetime have a timelike Killing vector?
PF Gold
P: 5,027
 Quote by George Jones Is the interior of a Schwarzschild black hole spherically symmetric? Does this spacetime have a timelike Killing vector?
No, but it has an extra killing vector which you don't have to assume; it can be derived only assuming spherical symmetry. In the interior, the extra killing vector happens to be spacelike.
P: 424
 Quote by Bill_K The general nonrotating time-independent axially symmetric vacuum solution to Einstein's Equation is given very simply by a solution of Laplace's Equation. This is a standard topic covered in any introductory course in General Relativity. For example, see here.
I have a question about the form of the metric shown on page two of the link. It appears that the time dilation there is set equal to the tangent length contraction as inferred by a distant observer as a coordinate choice for the metric form. But if that is the case, then upon referring back to Schwarzschild coordinates, where the invariants should remain the same, with limits working toward a finite surface area, zero time dilation, and infinite local acceleration for a static observer very near the event horizon, with the Weyl solution, it would instead work toward limits of infinite surface area at the place with zero time dilation at any r>0 and with zero local acceleration at any r>0 also. The only remedy that I can see so far, at least there is one, is if with the Weyl solution, the event horizon can only be at r = 0 always, so never with any mapped interior coordinates. Only then can the invariants work toward a finite surface area and infinite acceleration there. Is that right?
P: 424
 Quote by PAllen No, but it has an extra killing vector which you don't have to assume; it can be derived only assuming spherical symmetry. In the interior, the extra killing vector happens to be spacelike.
When one assumes spherical symmetry, what are they really assuming? Is that inferred by a distant observer's coordinate system? If so, then it seems to me that we can always assume spherical symmetry in a relativistic universe just by taking all of the points around a mass (although the mass would have to be spherically symmetric initially and collapse uniformly) that have the identical invariants and map them out to some spherical r according to the distant observer. Otherwise, the universe would be "lopsided" in some way by some influence external to the mass that we do not account for with Relativity, much for the same reason that two inertial observers in a purely relativistic universe, both applying the same synchronization of their clocks, will measure the same speed 'v' of each other, correct?