PeroK said:
You mean you can generate the correct geodesic equation from the existence of local inertial reference frames and instantaneous inertial motion therein?
I haven't seen that done.
I'm sceptical because the LIF has no information on the spacetime curvature at that point.
Of course you can do this, but it's very tedious of course. I think Weinberg, Gravitation and Cosmology derives in this way the generally covariant formalism. What you get of course is the usual geodesic equation for the motion of a test particle in an electromagnetic field,
$$\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
The dot denotes derivatives wrt. proper time. That the affine parameter is proper time leads to the condition
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2.$$
If you consider a massless particle ("naive photon" for the derivation of light bending in gravitational fields), you have an arbitrary affine parameter ##\lambda## instead of the proper time (because there's no proper time for massless particles) and the additional constraint
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=0.$$
Formally the derivation of the general covariant tensor formalism given the equivalence principle, i.e., the existence of a local inertial frame in any space-time point goes as follows. Let the local coordinates in one specific point, where the particle is momentarily located, be ##\xi^{\mu}## and ##\tau## the proper time, such that ##\dot{\xi}^{\mu} \dot{\xi}^{\nu}\eta_{\mu \nu}=c^2##. The equation of motion for a free particle in this point (i.e., a particle being under no other influence than the graviational field) reads as in special relativity simply
$$\dot{u}^{\mu}=\ddot{\xi}^{\mu}=0.$$
Now how looks this same equation in arbitrary coordinates ##x^{\mu}##? Then you get
$$\dot{u}^{\mu}=\mathrm{d}_{\tau} \left ( \frac{\partial \xi^{\mu}}{\partial x^{\alpha}} \dot{x}^{\alpha} \right) = \frac{\partial \xi^{\mu}}{\partial x^{\alpha}} \ddot{x}^{\alpha} + \frac{\partial^2 \xi^{\mu}}{\partial x^{\alpha} \partial x^{\beta}} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
Just multiplying this by ##\partial x^{\gamma}/\partial \xi^{\mu}## (implying of course the summation over ##\mu##) leads to
$$\ddot{x}^{\gamma} + \frac{\partial x^{\gamma}}{\partial \xi^{\mu}} \frac{\partial^2 \xi^{\mu}}{\partial x^{\alpha} \partial x^{\beta}} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
In this way the Christoffel symbols are given by
$${\Gamma^{\gamma}}_{\alpha \beta}=\frac{\partial x^{\gamma}}{\partial \xi^{\mu}} \frac{\partial^2 \xi^{\mu}}{\partial x^{\alpha} \partial x^{\beta}}.$$
The usual relation between the pseudo-metric components and Christoffel symbols follow from
$$\mathrm{d} s^2=\eta_{\mu \nu} \mathrm{d} \xi^{\mu} \mathrm{d} \xi^{\nu} =g_{\alpha \beta} \mathrm{d} x^{\alpha} \mathrm{d} x^{\beta},$$
from which
$$g_{\alpha \beta}=\eta_{\mu \nu} \frac{\partial \xi^{\mu}}{\partial x^{\alpha}} \frac{\partial \xi^{\nu}}{\partial x^{\beta}}.$$
Taking the derivative of this wrt. to ##x^{\gamma}## and writing ##\partial_{\gamma} = \partial/\partial x^{\gamma}## from now on gives
$$\partial_{\gamma} g_{\alpha \beta} = \eta_{\mu \nu} [(\partial_{\gamma} \partial_{\alpha} \xi^{\mu}) \partial_{\beta} \xi^{\nu}+(\partial_{\alpha} \xi^{\mu}) \partial_{\gamma} \partial_{\beta} \xi^{\nu}] = \eta_{\mu \nu} [{\Gamma^{\delta}}_{\alpha \gamma} (\partial_{\delta} \xi_{\mu}) \partial_{\beta} \xi^{\nu} + {\Gamma^{\delta}}_{\beta \gamma} (\partial_{\alpha} \xi^{\mu}) \partial_{\delta} \xi^{\nu}] $$
or
$$\partial_{\gamma} g_{\alpha \beta}= {\Gamma^{\delta}}_{\alpha \gamma} g_{\beta \delta} + {\Gamma^{\delta}}_{\beta \gamma} g_{\alpha \delta}.$$
Now we just cyclically change the free indices of this equation to get
$$\partial_{\alpha} g_{\beta \gamma}={\Gamma^{\delta}}_{\beta \alpha} g_{\gamma \delta} + {\Gamma^{\delta}}_{\gamma \alpha} g_{\beta \delta}$$
$$\partial_{\beta} g_{\gamma \alpha}={\Gamma^{\delta}}_{\gamma \beta} g_{\alpha \delta} + {\Gamma^{\delta}}_{\alpha \beta} g_{\gamma \delta}.$$
Then adding the first two and subtracting the third of these equations leads to
$$\partial_{\gamma} g_{\alpha \beta} + \partial_{\alpha} g_{\beta \gamma} - \partial_{\beta} g_{\gamma \alpha}=2{\Gamma^{\delta}}_{\alpha \gamma} g_{\beta \delta} .$$
Eliminating the ##g_{\beta \delta}## factor and dividing by 2 finally leads to the well-known expression
$${\Gamma^{\delta}}_{\alpha \gamma}=\frac{1}{2} g^{\beta \delta} (\partial_{\gamma} g_{\alpha \beta} + \partial_{\alpha} g_{\beta \gamma} - \partial_{\beta} g_{\alpha \gamma}).$$
So we find the geodesic equation from the equivalence principle, i.e., the existence of a local inertial reference frame in terms of arbitrary coordinates.
