Deriving the Metric from the Energy-Momentum Tensor

[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?

 

[PAllen]

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.

 

[Bill_K]

Axial symmetry is already much harder, even for vacuum.

All you have to do is solve Laplace's Equation.

 

[K^2]

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.

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.

 

[PAllen]

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.

 

[PAllen]

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.

 

[K^2]

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.

 

[andrien]

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.

 

[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.

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.

 

[PAllen]

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.

 

[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.

 

[PAllen]

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.

 

[George Jones]

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?

 

[PAllen]

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.

 

[grav-universe]

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?

 

[grav-universe]

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?

 

[PAllen]

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?

We are speaking here of pure vacuum solutions. There are coordinate independent ways to state spherical symmetry, but it does suffice to use coordinates. For a give 4-manifold with everywhere vanishing Ricci curvature but non-vanishing Weyl curvature, either it is possible to introduce coordinates such that the metric expressed in those coordinates takes a certain form or it is not. It turns out that there is exactly one maximally extended pseudo-Riemannian 4-manifold with vanishing Ricci curavature and non-vanishing Weyl curvature that admits a coordinate system manifesting spherical symmetry (up to a single parameter, that can be interpreted as mass). For axial symmetric vacuum, there are both static and non-static solutions, and even the static solutions cannot be characterized by a simple parameter choice.

 

[atyy]

Doesn't the energy-momentum tensor contain the metric?

 

[Nugatory]

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.

 

[atyy]

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/March01/Carroll3/Carroll8.html? Isn't that the metric in the stress-energy tensor?

 

[K^2]

How about Eq 8.15 of http://ned.ipac.caltech.edu/level5/March01/Carroll3/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.

 

[atyy]

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.

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?

 

[PAllen]

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.

 

[atyy]

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?

 

[PAllen]

So is the last what the OP had in mind, since he did specify that it's the stress tensor of a perfect fluid?

Well, we'll have to hope the OP answers that.

I assumed you had just T, not T and and density and pressure functions that you knew all constituted a valid solution. With just T, you've got no way to read off the metric. And the only way you have pressure, density, and T that satisfy the EFE is to first solve the EFE with metric undetermined. I believe my interpretation of the question is the most likely - but again, we'll never know without input from OP.