3
$\begingroup$

This question already has an answer here:

I have watched and read a lot on the topic of General Relativity and the geometry behind it. I am confident that I can derive an approximation of the the stress-energy-momentum tensor with just the metric tensor, but if I was given the stress-energy-momentum tensor I'm not sure how I would go back to find the metric. How can I use Einstein's field equations to find the metric tensor? Would I have to expand all of the notation? Or is there some equation I can plug into my computer to get it? Or is there some other way?

$\endgroup$

marked as duplicate by Kyle Kanos, Qmechanic Mar 4 '15 at 1:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

As you described, if you have a metric, then you also have the manifold itself and all the points in it, and can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming no cosmological constant).

But the reverse direction is very different. For instance, if you started with the manifold (but no one gave you the metric, only the topological space) and then you had a stress-energy tensor defined at every point on your manifold, you already have a lot since you have the manifold. So this might be too much information, and be considered cheating. I'm also not sure it is enough information. The stress-energy tensor is the source term, it's like in electromagnetism if someone gave you the charge and current, that isn't actually enough to find the fields, you need boundary conditions too.

Let's look at a simple but troublesome example. Your manifold is $\mathbb{R}^4$, your stress-energy tensor is $T^{\mu \nu}=0$. There are many possible metrics. You could have a metric corresponding to a gravitational wave travelling left, one travelling right, one up, one down, one forwards, one backwards. Or just an empty flat space, the spacetime of special relativity. This is entirely like the situation with no charge and no current, you could have no fields, or an electromagnetic wave going in any direction. It's just not enough information.

So having the manifold might be "cheating" because it is "too much" information, but even having the manifold and the stress-energy tensor might not be enough information to get a unique solution.

If you question is about how people solve it. People look for solutions, and they find many (though it is quite hard). And they look for solutions that adequately describe experimental setups and observations.

And yes, numerical solutions are one way to go. There can still be many or no solutions depending on your setup, even when done numerically. And the calculations are very very hard to do well, or even to do them OK.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.