The Hole Problem of Classical General Relativity

A covariant equation cannot define a complete evolution equation for all the components of the metric tensor $$g_{\mu\nu}(x,t)$$. The reason is that a local transformation of the coordinates, which can leave the initial values as well as the boundary conditions untouched, allows to define another valid solution of the same covariant equations. But that means that the equations of motion do not fix the solution completely.

Le'ts see how it works. Assume we have a non-trivial coordinate transformation, $$x\to x'$$, which is nonetheless trivial at the initial values $$x^0< t_0$$ as well as outside a hole $$|x^i|> l_0$$. Let's consider a solution $$g_{\mu\nu}(x)$$ of the Einstein equations, and rewrite this same solution in the other coordinates: $g_{\mu\nu}(x) \to g'_{\mu'\nu'}(x') = \frac{\partical x^\mu}{\partical x^{\mu'}}\frac{\partical x^\nu}{\partical x^{\nu'}} g_{\mu\nu}(x(x'))$

This gives some metric $$g'_{\mu\nu}(x')$$. Given that outside the hole $$x^0< t_0,|x^i|> l_0$$ we have $$x'=x$$, no difference will be visible outside. But inside, where the coordinates differ, above expressions look different - as they should, once they are written in different coordinates.

But now let's, in a second operation, simply replace the denotation of the changed coordinates back to the old ones, but leaving the expression as it is: $g'_{\mu'\nu'}(x') \to g'_{\mu\nu}(x)$

In this second replacement, we have not followed the rules how to transform the representation in coordinates if one simply wants to describe the same solution in other coordinates. So, we have changed the metric. The metric $$g'_{\mu\nu}(x,t)$$ is different from $$g_{\mu\nu}(x,t)$$, even if now written in the same coordinate $$x$$. Nonetheless, the equations are covariant, thus, have the same form in all coordinates. So, the equation is the same, the solution looks the same too, thus, $$g'_{\mu\nu}(x)$$ is also a solution of the covariant Einstein equations.

Moreover, by the assumption that the coordinates agree outside the hole, we also have $$g'_{\mu\nu}(x) = g_{\mu\nu}(x)$$ outside the hole. And that means we have found two different solutions of the Einstein equations which have the same initial values as well as the same boundary values.