# Fermi-Walker transportation

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!

• I suppose you could see if they satisfy the teleparallism within general relativity? Dec 16, 2019 at 14:59
• Could you explain that in an answer please? Dec 16, 2019 at 16:01
• I would imagine it would be a fairly easy exercise to do, if you look into it. By the way my suggestion was just that, just a suggestion, I wouldn't know if there was a more standard way to approach your question. Dec 16, 2019 at 16:15
• Have you tried plugging in $v=u$? Dec 21, 2019 at 17:33

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)