Fermi-Walker transportation

Question

Consider a timelike curve C in a curved spacetime with unit tangent vector $u^\alpha$. A vector $v^\alpha$ is said to be Fermi-Walker transported along C if: $$ u^\alpha\nabla_\alpha v^\beta = u^\beta A^\gamma v_\gamma - u^\gamma A^\beta v_\gamma $$ with $A^\gamma = u^\alpha \nabla _\alpha u^\gamma$ the acceleration of $u^\alpha$.

How can I show that the tangent vector $u^\alpha$ is always Fermi-transported (obeys the equation above) along $C$ (any timelike curve)?

Analytical solution:

First we substitute $u^\mu=v^\mu$ which gives the following terms:

LHS: $$ u^\alpha\nabla_\alpha u^\beta = A^\beta $$ RHS: $$ u^\beta A^\gamma u_\gamma-u^\gamma A^\beta u_\gamma=u^\beta u_\gamma A^\gamma+A^\beta $$ (using the fact that $u^\gamma u_\gamma = -1$ for tangent vetorrs on a time-like curve) $$ =u^\beta u^\alpha(\nabla_\alpha u^\gamma) u_\gamma +A^\beta $$ substituting the acceleration formula $$ =\frac{1}{2}u^\beta u^\alpha (\nabla_\alpha u^\gamma u_\gamma) +A^\beta $$ using the multiplication rule for differentiation $$ =\frac{1}{2}u^\beta u^\alpha (\nabla_\alpha (-1)) +A^\beta=A^\beta $$ Hence the tangent vector $u^\alpha$ is always Fermi-transported!

Answer 1

One can plug in $v=u$ in the condition for Fermi-Walker transport and get

$$u^\alpha\nabla_\alpha u^\beta = u^\beta A^\gamma u_\gamma - u^\gamma A^\beta u_\gamma$$

Notice that the left hand side is exactly $A^\beta$, then use the fact that $u^\alpha$ is a unit vector tangent to a timelike curve, so that $u^\alpha u_\alpha = -1$, to take care of the second term in the right hand side.

The only thing left to do is to show that $A^\gamma u_\gamma=0$. (Hint: start from $u^\alpha u_\alpha = -1$ and take a covariant derivative)