Acceleration from General Relativity

Question

I was watching a video on General Relativity that defines velocity as $v=\frac{d}{d \tau}$ where $\tau$ is the proper time.

Then acceleration would just be its covariant derivative $a=\nabla_{\frac{d}{d \tau}} \frac{d}{d \tau}$

However, the video denoted it as $a=\frac{d^2}{d \tau ^2}$

My question is if the two are equal: $\left(\nabla_{\frac{d}{d \tau}} \frac{d}{d \tau}\right)f= \frac{d^2 f}{d \tau ^2}$

Where $f$ is a scalar function.

The man claimed that the two are equal as the connection coefficients act as the second derivative. He said that the normal components of the acceleration disappear because we’re working in intrinsic geometry, but how would they disappear. If I were to compose the velocity vectors, how would it make sense?

The video is https://youtu.be/2pPvUx-EUqE at 2:00

Answer 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.