# How does one actually use the Einstein field equations?

I'm experimenting with the Einstein field equations (EFE) and I'm wondering how to actually use them. It seems to me that the simplest are those evolving a vacuum, which imply that the stress-energy-momentum tensor $T_{mn}$ equals zero, because there is no matter, nor energy to curve space-time. How does one solve for a metric for that given situation? If $T_{mn}$ the the entire Einstein tensor $G_{mn}$ equals zero, right? Does flat space-time imply that the Riemann curvature tensor $R^t_{mnk}$ also equal zero? This would imply that its contraction, the Ricci curvature tensor $R_{mn}$, would equal zero, which would also imply that its trace (or its contraction, the Ricci scalar $R$) would also equal zero. Or does the metric equal zero and $R^t_{mnk}$ has a value?

Also, given an a metric (which I invent using basis vectors of my choice), is it possible to find the the energy distribution $T_{mn}$ for that metric?

• With great difficulty? Commented Feb 7, 2016 at 18:09
• "But does flat space-time imply that the Riemann curvature tensor $R^t_{mnk}$ also equals zero?" yes, thats the definition of being flat. "Or does the metric equal zero and $R^t_{mnk}$ has a value?" this makes no sense: the metric can never be zero, and if $g$ were zero, then so would $R$, because $R\sim \partial^2 g$ (i.e., $R$ is constructed from derivatives of the metric). "given an a metric, is it possible to find the the energy distribution for that metric?" yes: just compute $G$; by the EFE, $T_{mn}=G_{mn}/\kappa$. Commented Feb 7, 2016 at 18:14
• Possible duplicate of: Finding the metric tensor from the Einstein field equation? Commented Feb 7, 2016 at 19:03
• You can do multiple subscripts like this $T_{mn}$. Commented Feb 8, 2016 at 0:56
• I ''invented'' a metric; a diagonal non-zero metric, and I discovered that the Riemann tensors are equal to zero which implies the Einstein tensor equals zero (right?). Hence, $T_{mn}$ also equals zero. But how do i know what kind of space(time) I'm looking at? Commented Mar 7, 2016 at 1:52

In practice, given a stress-energy tensor $T_{\mu\nu}$, we may attempt to find solutions to the Einstein field equations using perturbation theory. The basic idea is to expand around a known solution $g_{\mu\nu}$ by a perturbation $h_{\mu\nu}$. In the case of a flat background,

$$\delta G_{\mu\nu} = 8\pi G \delta T_{\mu\nu} = \partial_\mu \partial_\nu h - \partial_\mu \partial_\alpha h^\alpha_\nu -\partial_\nu \partial_\alpha h^\alpha_\mu + \partial_\alpha \partial^\alpha h_{\mu\nu} - \eta_{\mu\nu} \partial_\alpha \partial^\alpha h + \eta_{\mu\nu} \partial_\alpha \partial_\beta h^{\alpha \beta}.$$

In some cases, one may solve the equations exactly, or more typically, employ numerical methods. Another alternative approach when faced with a stress-energy tensor is to try to determine the symmetries the metric may have, and then plug in an ansatz for the metric to yield a set of differential equations which may be more tractable, analytically and numerically.

Yet another alternative is to generate new solutions from old ones through the introduction of pseudopotentials, a method due to Harrison, Eastbrook and Wahlquist, called the method of prolongation structures.

There are several other methods such as those that rely on Lie point symmetries of differential equations. There is also a Backlund transformation, which relies on identifying a simpler differential equation whose solution satisfies a condition involving the solution to the harder problem.

These methods are too involved to present here and require a significant background. They are explained in Exact Solutions to the Einstein Field Equations by H. Stephani et al.

Addressing your other question, if given a metric $g_{\mu\nu}$, of course one can compute $T_{\mu\nu}$ through the Einstein field equations, you just plug it in and tediously compute all the curvature tensors. There are strictly speaking probably some requirements on the functions in $g_{\mu\nu}$, but you can get away with most things, even distributions, such as a delta function, which may lead to a stress-energy tensor describing a brane.

• What's the delta in front of the G? Commented Feb 7, 2016 at 22:30
• @Investor $\delta G$ is the difference between the Einstein tensor of the fully perturbed system, and the Einstein tensor of the background, evaluated at linear order in the perturbation. Commented Feb 7, 2016 at 23:46

But, how does one solve for a metric for that given situation?

There are many possible metrics. For vacuum solutions you have Minkowski space as a solution. Another is a solution is a spacetime with a gravitational wave going in the $+\hat x$ direction as a plane wave filling all of spacetime. Another is like Minkowski space locally, but which is spatially like Pac-Man. Another is like Minkowski space locally, but which is temporally like Pac-Man. And note that the two Pac-Man spacetimes and the Minkowski spacetime all have metrics that are identically zero at every place and time. Yet they are different spacetimes.

One way to solve the equation is the guess and check method. Make up a manifold and a metric tensor on it and a stress-enwrgy tensor on it. Use the metric tensor to compute the Einstein tensor at every event. Check that it equals the Stress-Energy tensor at every event. To do even one better, you can verify that the Stress-Energy tensor obeys a corresponding evolution equation as well.

But does flat space-time imply that the Riemann curvature tensor $R^t_m$$_n$$_k$ also equals zero?

Yes. But don't think flat means the same thing you might naively think it means. For instance a flat spacetime could be a Pac-Man universe, it could have time repeat. It could be embedded into a larger manifold in a way that "looks" unflat. For instance a 2d cylinder is flat even though as a surface in 3d it might "look" curved.

Or does the metric equal zero and $R^t_m$$_n$$_k$ has a value?

If the metric tensor is zero at every event then the Riemann tensor and the Ricci tensor and the Ricci scalar are all zero and the Einstein tensor is zero as well.

Also, given an a metric (which i invent using basis vectors of my choice), is it possible to find the the energy distribution $T_m$$_n$ for that metric?

Given the metric you can find the Riemann tensor and the Ricci tensor and the Ricci scalar and thus find the Einstein tensor. However you won't then be able to find the Stress-Energy tensor unless you also know the cosmological constant and know that you have a solution to Einstein's Equation. And even so, that would only tell you the total Stress-Energy tensor. You wouldn't know the Stress-Energy of matter or the electromagnetic field or anything. Or know whether the Stress-Energy tensor is evolving correctly.

Another way to use the Einstein Equation is in numerical relativity. You have a Cauchy surface and on it you have some initial data and some Stress-Energy and some evolution equations for the Stress-Energy and then you use the Einstein Equation to figure out how to evolve the initial slice into a whole solution. This isn't always possible. And some times that solution is not the only one that agreed on that slice. That's the Cauchy problem.

If I set the Cosmological constant to zero, then do I know more about the stress-energy tensor I obtain?

If you know 1) the manifold and metric, and 2) the cosmological constant, and 3) that your manifold and metric is a solution to Einstein's Equation. Then yes, you can know the total Stress-Energy tensor.

• If I set the Cosmological constant to zero, then do I know more about the stress-energy tensor I obtain? Commented Feb 8, 2016 at 20:47
• If you know 1) the metric, and 2) the cosmological constant, and 3) that your manifold/metric is a solution to Einstein's Equation. Then yes, you can know the total Stress-Energy tensor. Commented Feb 8, 2016 at 22:27
• If the metric tensor is NOT a function of position, then there is no curvature, right? Therefore, the Riemann tensor equals zero, then so do the Ricci tensor and scalar. So with no cosmological constant (which is obtained how?) there is then no matter/energy distribution, right? Commented Feb 9, 2016 at 1:34
• @Investor If the metric doesn't depend on the event, then the Riemann tensor is zero and if lambda is zero too then the total Stress-Energy tensor is zero. That doesn't mean no matter, just that the total Stress-Energy is zero. And no curvature doesn't exclude space being like a cylinder. Commented Feb 9, 2016 at 6:13