Weinberg writes in his book:
Consider a clock in an arbitrary gravitational field, moving with arbitrary velocity, not necessarily in free fall. The equivalence principle tells us that its rate is unaffected by the gravitational field if we observe the clock from a locally inertial coordinate system $\xi^\alpha$, the space-time interval $d\xi^\alpha$ between ticks is governed in this system by
$ \Delta t=(-\eta_{\alpha\beta}d\xi^\alpha d\xi^\beta)^{1/2} $,
where $\Delta t$ is the period between ticks when the clock is at rest in the absence of gravitation.
I really don't understand why $\Delta t$ is given by the formula above and not just by $\Delta t=\xi^0(\sigma)-\xi^0(\sigma+d\sigma)=\frac{d\xi^0}{d\sigma}d\sigma$, where $\sigma$ parameterizes the worldline of the clock in the $\xi$ coordinate system. And $\sigma$ is the parameter at which the worldline looks locally flat from the point of view of an observer in the $\xi$ coordinate system.
Best Answer
The reason is because the clock is not necessarily in free fall and its velocity is not necessarily zero. Therefore, when the clock passes by the observer, it could have a non-zero spatial velocity. This is the same situation as special relativity, where the proper time on the clock $\tau$ is related to the observer's time $t$ by $$\text{d}\tau^2 = -\eta_{\mu\nu} \text{d}x^\mu \text{d}x^\nu = \text{d}t^2 - \text{d}x^2 - \text{d}y^2 - \text{d}z^2$$ as stated in the book. You can't say that $\text{d}t$ and $\text{d}\xi_0$ (which correspond to $\text{d}\tau$ and $\text{d}t$ here) are the same, which is what you implicitly did when you wrote $\text{d}\tau = (\text{d}\tau/\text{d}\sigma)\text{d}\sigma =(\text{d}t/\text{d}\sigma)\text{d}\sigma$. The only time they are the same is when the observer and the clock are in the same frame (for example, if the observer is holding the clock). In this special case, $\text{d}\tau=\text{d}t$.