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!

  • $\begingroup$ I suppose you could see if they satisfy the teleparallism within general relativity? $\endgroup$ Dec 16, 2019 at 14:59
  • $\begingroup$ Could you explain that in an answer please? $\endgroup$ Dec 16, 2019 at 16:01
  • $\begingroup$ 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. $\endgroup$ Dec 16, 2019 at 16:15
  • $\begingroup$ Have you tried plugging in $v=u$? $\endgroup$
    – John Donne
    Dec 21, 2019 at 17:33

1 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)

  • $\begingroup$ Thank you very much for your answer, I have updated my question with a full derivation. $\endgroup$ Dec 23, 2019 at 18:22

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