Acceleration from General Relativity

Solution 1:

In the video see the definitions of

• (at 1:39) four velocity $U\equiv\frac{d}{d\tau}$;

• (at 2:00) four acceleration $A\equiv\frac{dU}{d\tau}=\frac{d^2}{d\tau^2}$;

In this notation however it is a bit unclear of what that proper time $\tau$ is taken. It is standard to consider a curve (world line) $\gamma$ which then naturally has a proper time $\tau$ by which it can be parametrized. Thus,

• the velocity $\frac{d}{d\tau}\gamma(\tau)$ and the acceleration $\frac{d^2}{d\tau^2}\gamma(\tau)$ of that curve are well defined.

When instead you have a field $U$ of four velocities, i.e., a four velocity $U(x)$ at every point of space time its field of four accelerations is $$\tag{1} A=\nabla_UU=U^\mu\nabla_\mu U^\nu $$ (the directional derivative of $U$ in the direction of $U$).

Of course, the field of four velocities $U$ is a familiy of derivatives $\frac{d}{d\tau}\gamma(\tau)$ where $\gamma$ runs through a family of curves that can be imagined as the many world lines of particles filling the entire space time. Each of those has its own proper time.

I don't think the notation $U\equiv\frac{d}{d\tau}$ and $A\equiv\frac{dU}{d\tau}=\frac{d^2}{d\tau^2}$ is very helpful esp. when it is mixed up with (1).

Furthermore, in curved space time the correct way of defining four acceleration of a curve $\gamma$ is not $A^\mu=\frac{d^2\gamma^\mu}{d\tau^2}$ but rather $$\tag{2} A^\mu\equiv\frac{d^2\gamma^\mu}{d\tau^2}+\Gamma^\mu_{\rho\sigma}\frac{d\gamma^\rho}{d\tau}\frac{d\gamma^\sigma}{d\tau} $$ (see [1]).

[1] S. Carroll, Space Time and Geometry.