Finding the metric tensor from the Einstein field equation?

Question

I have have set my self a challenge to learn all the maths behind the Einstein field equation (EFE), and from reading it seems that the Metric tensor is the thing we are trying to find (from the 10 equations that forms the EFE). but since nearly all the terms in it are functions of this metric tensor, this seems very hard. What type of maths do we use to do this? (I have copied and pasted the EFE just for reference):

$$R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R+g_{\mu\nu}\Lambda=\frac{8 \pi G}{c^4}T_{\mu \nu}$$

Edit: Thanks for your comments. Just as I have no one else to ask, is this the equation that we try to solve:

$$\left(\frac{\partial}{\partial x^{\lambda}}\Gamma^{\lambda}_{\mu \nu}- \frac{\partial}{\partial x^{\nu}}\Gamma^{\lambda}_{\lambda \mu}+\Gamma^{\lambda}_{\lambda \rho}\Gamma^{\rho}_{\nu \mu}-\Gamma^{\lambda}_{\nu \rho}\Gamma^{\rho}_{\lambda \mu}\right)-\frac{1}{2}g_{\mu \nu}g^{ab}\left(\Gamma^c_{\enspace ab,c}-\Gamma^c_{\enspace ac,b}+\Gamma^d_{\enspace ab}\Gamma^c_{\enspace cd}-\Gamma^d_{\enspace ac}\Gamma^c_{\enspace bd} \right)+g_{\mu \nu}\Lambda=\frac{8 \pi G}{c^4}T_{\mu \nu}$$

with the $R_{μν}$ and $R$ terms expanded and where $g_{μν}$ and $T_{μν}$ are the μν th components from the related tensors? (a yes or no answer will be fine, thanks).

Answer 1

This is really a comment, but it got a bit long for the comment field.

I'd guess that, like me, your experience in physics is from an area where solving differential equations is a routine part of the job. We're used to analysing a problem, writing down a differential equation that encapsulates the physics and solving it, analytically if we're lucky or in the worst case throwing it at a computer.

What struck me very forcefully when I started reading up on GR is that this is hardly ever the approach used. The equations are so hard that in almost every case the metric is obtained either by ingenious use of symmetry or just guessing answers until one fits. If you read the derivation of the Schwartzschild metric, which is probably the simplest one most of us meet, Schwarzschild obtained the answer by guessing at a basic form for the metric then using the high symmetry to eliminate all the possibilities but one. Kerr seems to have arrived as his result by inspired guesswork (though inspired by vast amounts of effort!!).

This all feels somehow unsatisfactory for us hedge physicists. It feels as if general relativists just cheat all the time and never do things methodically like we do. That's an unfair impression of course, and born out of ignorance, but nevertheless I'm willing to bet that's how you will feel as you start reading around the subject.

If you want to start learning GR in anger then I strongly recommend A first course in general relativity by Bernard Schutz. This will get you from a starting point of knowing basic calculus to the point where you're comfortable doing basic GR calculations. Note however that even after 376 pages you will still not have seen the Einstein equation written out in full.

Answer 2

If you have a metric, then you also have the manifold itself and all the points in it, and you 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 you know the value of your 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. But it can also be too little 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 with no gravitational waves, the spacetime of special relativity. This is entirely like the corresponding situation in electromagnetism when there is no charge and no current, you could have no fields, or you could have an electromagnetic wave going in any direction. It's just not enough information.

Answer 3

The standard procedure to find the components of metric tensor has the following steps.

1. Guess a trial formula for dynamic evolution of the metric tensor components ($g_{\mu\nu}$), considering the symmetries your given system possesses. The trial formula should include enough unknown parameters.
2. Calculate the connection coefficients from your trial formula.
3. Calculate the components of Riemann curvature tensor from the connection coefficients.
4. Calculate the components of Einstein tensor from the Riemann tensor.
5. Substitute the Einstein tensor in the Einstein Field Equation.
6. By matching the boundary conditions (spacetime becomes flat at infinity), and find the unknown parameters that you had set in your metric tensor earlier.
7. If you find a solution to the metric tensor that satisfies the Einstein equation and as well as the boundary condition, then the solution you got is the correct solution. If not, change the trial and repeat the steps above.

You can see, the procedure is heavily dependent on (educated) guessing. Therefore, it can not be applied to a situation where the spacetime lacks symmetry. For this reason, we do not have many analytic solutions to Einstein's equation.