The Equivalence Principle

The Einstein Equivalence Principle requires that there must exist a transformation of coordinates in which the physics is that of special relativity. This means that the gravitational field strengths, which are expressible in terms of gradients of the metric tensor, must vanish. So what we want to do is show that under certain conditions, there always exists a coordinate transformation which reduces these gradients to zero at any chosen point. Let us designate the chosen point by "$P$". Then consider the coordinate transformation from $S$ to $S'$ given by:

$$x^{\mu}=x'^{\mu}-x'_P^{\mu}+{1 \over 2}(x'^{\alpha}-x'_P^{\alpha})(x'^{\beta}-x'_P^{\beta})A_{\alpha \beta}^{\mu}$$ (13)

Note that the coefficients $A_{\alpha \beta}^{\mu}$ must be symmetric with respect to interchange of the two lower indices:

$$A_{\alpha \beta}^{\mu}=A_{\beta \alpha}^{\mu}$$ (14)

because the product $(x'^{\alpha}-x'_P^{\alpha})(x'^{\beta}-x'_P^{\beta})$ is symmetric.

The tensor transformation law for the metric tensor is:

$$g'_{\alpha \beta}={\partial x^{\mu} \over\partial x'^{\alpha}}{\partial x^{\mu}\over \partial x'^{\beta}}g_{\mu \nu}.$$ (15)

Take the partial derivative of the above equation with respect to $x'^{\sigma}$ at the point $P$, use the chain rule, and remember that the transformation coefficients are functions of position:

$$g'_{\alpha \beta , \sigma}={\partial x^{\mu} \over \partial x'^{\... ...{ \partial^2 x^{\nu} \over \partial x'^{\sigma} \partial x'^{\beta}}g_{\mu \nu}$$ (16)

$$=g_{\alpha \beta , \sigma}+A_{\sigma \alpha}^{\mu}\delta_{\beta}^{\nu}g_{\mu \nu}+A_{\sigma \beta}^{\nu}\delta_{\alpha}^{\mu}g_{\mu \nu}$$ (17)

$$=\Gamma_{\alpha \sigma}^{\mu} g_{\mu \beta} + \Gamma_{\beta \sigm... ...}+A_{\alpha \sigma}^{\mu} g_{\mu \beta} + A_{\beta \sigma}^{\mu} g_{\alpha \mu}$$ (18)

i So if the Affine Connection $\Gamma_{\alpha \beta}^{\mu}$ is symmetric with respect to interchange of the two lower indices, and then if we choose

$$A_{\alpha \beta}^{\mu}=-\Gamma_{\alpha \beta}^{\mu}$$ (19)

Then we immediately see that all the gradients of the metric tensor will be zero:

$$g'_{\alpha \beta, \sigma}=0.$$ (20)

We shall show in the next section that this result implies that in the new coordinate system, all the affine connections also vanish.