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

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

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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