So I'm reading A relativist's toolkit (Page 21) it states:

$$ \Gamma^{\alpha}_{\gamma \beta} = \Gamma^{\alpha}_{ \beta \gamma}$$


$$ g_{\alpha \beta ; \gamma} = 0 $$

In general relativity, these properties come as a consequence of Einstien's equivalence principle.

where $\Gamma$ is the connection and $g$ is the metric. Using these properties he then derives the connection. It is not obvious to me how the equivalence principle implies this. How does the equivalence principle imply this?


For completeness's sake, let me recall what the equivalence principle states

In an arbitrary gravitational field, at any given spacetime point, we can choose a locally inertial reference frame such that, in a sufficiently small region surrounding that point, all physical laws take the same form they would take in absence of gravity, namely the form prescribed by Special Relativity.

So, the first identity can be derived in the following way: consider a general scalar field $\Phi$. It's first covariant derivative is a one form and coincides with the normal derivative. It's second covariant derivative is a $(0,2)$-tensor with components $\Phi_{;\mu;\nu} = \Phi_{,\mu; \nu} $. For the equivalence principle we can always find a locally minkowskian frame in which the components of the second covariant derivative are just $$\Phi_{,\mu;\nu} = \Phi_{,\mu,\nu} = \frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}\Phi$$ which is symmetric. But if a tensor is symmetric in one basis it's symmetric in any basis and so $$\Phi_{,\mu;\nu} = \Phi_{,\mu,\nu}-\Phi_{,\alpha}\Gamma^{\alpha}_{\mu\nu} = \Phi_{,\nu,\mu}-\Phi_{,\alpha}\Gamma^{\alpha}_{\nu\mu} = \Phi_{,\nu;\mu} $$ and since the equality holds for any scalar field, we get that $$\Gamma^{\alpha}_{\mu\nu} = \Gamma^{\alpha}_{\nu\mu} $$

The second identity follows from the same line of reasoning. We know from the equivalence principle that we can always find a locally minkowskian frame in which the metric tensor $g_{\mu\nu}$ reduces to $\eta_{\mu\nu}$ (the Minkowski metric). The coordinate basis associated with the minkovskian frame coordinates has constant basis vectors, so the affine connections vanish. In this frame $$g_{\mu\nu;\alpha} = \eta_{\mu\nu;\alpha} = \eta_{\mu\nu,\alpha}-\Gamma^{\beta}_{\mu\alpha}\eta_{\beta\nu}-\Gamma^\beta_{\nu\alpha}\eta_{\mu\beta} = 0$$ If all the components of a tensor are zero in a frame will be zero in any coordinate frame, and so $$g_{\mu\nu;\alpha} = 0$$

The general idea of the equivalence principle is just to have at hand, in any point in space-time, a locally minkowskian frame in which the laws of special relativity are true. Then, once you do your calculations in the local frame, you can get results in any other frames.


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