Minkowski metric derivation

Question

Can we derive the Minkowski metric from the Einstein field equations? I am still feeling unclear about the general relativity.

Answer 1

“Derive the metric” is maybe not the best description of what we do. You can take any metric you want. Then you can put it into the Einstein field equations to determine the stress energy tensor that produces that metric.

In the case of the Minkowski metric $R^\rho{}_{\sigma\mu\nu}=0$. So $G_{\mu\nu}=0$ and therefore $T_{\mu\nu}=0$. So the Minkowski metric is a vacuum solution of general relativity.

Answer 2

The Einstein field equations in vacuum (i.e. with $T_{\mu\nu}=0$ everywhere) are $$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0. \tag{1}$$

Because the tensors in (1) are symmetric $4\times 4$ tensors, these are $10$ equations. The $4$ contracted Bianchi identities $$\nabla_\rho R^\rho{}_\mu=\frac{1}{2}\nabla_\mu R$$ reduce the number of independent equations from $10$ to $6$. So we have only $6$ independent equations for the $10$ unknown $g_{\mu\nu}$. Thus the general solution $g_{\mu\nu}$ is not unique, but depends on $4$ arbitrary functions.

It is trivial to check that the flat Minkowski metric in Cartesian coordinates $(ct,x,y,z)$ $$g_{\mu\nu}=\text{diag}(1,-1,-1,-1) \tag{2}$$ is a special solution of (1). We can get more solutions by applying any coordinate transform $(ct,x,y,z) \to (x^0,x^1,x^2,x^3)$ to this. For example, by transforming from Cartesian coordinates to spherical coordinates $(ct,r,\theta,\phi)$ we get the metric $$g_{\mu\nu}=\text{diag}(1,-1,-r^2,-r^2\sin^2\theta)$$ which is also a solution of (1).

The coordinate transforms are $4$ arbitrary functions. From the reasoning above we know the most general solution $g_{\mu\nu}$ depends on $4$ arbitrary functions. So we can be sure, the flat Minkowski metric (2) and its coordinate transforms are all the possible solutions. And there are no other solutions besides these.