Well, in an inertial frame, we adjust the coordinates to make g = diag[1, 1, 1, -1] at the origin of our coordinate system. The first requirement of an inertail frame is that, (0, 0, 0, 0) = 0. This you can certainly do, if you like; it doesn't effect the ensuing calculation at all. The next requirement is more serious: that the partial derivatives of the gij vanish. This would force the geodesics to be uninteresting (straight) at the origin, since the Christoffel symbols vanish, and (I) becomes

Pi|4P4 = 0,

that is, since P4 m0 and Pi|4 = d/dx4(m0vi) = rate of change of momentum, that

rate of change of momentum = 0,

so that the particle is experiencing no force (even though it's in a gravitational field).

Question But what does this mean? What is going on here?

Answer All this is telling us is that an inertial frame in a gravitational field is one in which a particle experiences no force. That is, it is a "freely falling" frame. To experience one, try bungee jumping off the top of a tall building. As you fall, you experience no gravitational force -- as though you were in outer space with no gravity present. This is not, however, the situation we are studying here. We want to be in a frame where the metric is not locally constant. so it would defeat the purpose to choose an inertial frame.

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