# How to reparametrize a geodesic such that $\nabla_V V = 0$

In Nakahara's section on geodesics he says that

$\nabla_V V = fV$

can be reparametrized to give

$\nabla_V V = 0$

if we use the reparametrize the tangent vector components $t \rightarrow t'$ such that the tangent vector components change as

$$\frac{dx^\mu}{dt} \rightarrow \frac{dt}{dt'} \frac{dx^\mu}{dt}$$

and if $t'$ satisfies

$$\frac{d^2t'}{dt^2} = f \frac{dt'}{dt}$$

I am unable to show this at all.

I tried to expand it as follows, setting $A=\frac{dt}{dt'}$:

$$\nabla_{(AV)} (AV) = A(V[A]V + A \nabla_V V) = A(V[A]V + AfV)$$

Here I am stuck: I need this to equal to 0 (i.e. a geodesic) but I don't know how to use the $t'$ condition above especially since it is defined as $\frac{dt'}{dt}$ and not $\frac{dt}{dt'}$ which was present in the reparametrizing of the tangent vector components. Even worse, there is no minus sign that appears anywhere so I can't cancel anything in the brackets regardless?

What am I doing wrong here?

• if $\gamma$ is a geodesic, $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$ whatever the ($C^1$) parametrization Commented Apr 27, 2016 at 22:08

It looks to me like there is a typo in Nakahara. Suppose we go the other way first. Start out with an affine parametrization $$\nabla_v v=0$$ and reparametrize $v$ so that we get something that is not affinely parametrized. I'm not going to use $t$ as the parameter because it is too easy to confuse $t$ with time, which in general it isn't, so let's call the two parameters along the curve $\lambda$ and $\lambda'$.

Expanding out the affinely parametrized geodesic equation in coordinates we have

$$\frac{d^2x^{\mu}}{d\lambda^2} +\Gamma^{\mu}_{\alpha\beta} \frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda} =0.$$

Now change $\frac{d}{d\lambda} \to (\frac{d\lambda'}{d\lambda})\frac{d}{d\lambda'}$

$$\left(\frac{d^2\lambda'}{d\lambda^2}\right)\frac{dx^{\mu}}{d\lambda'} + \left(\frac{d\lambda'}{d\lambda}\right)^2 \frac{d^2x^{\mu}}{d\lambda'^2} +\left(\frac{d\lambda'}{d\lambda}\right)^2 \Gamma^{\mu}_{\alpha\beta} \frac{dx^{\alpha}}{d\lambda'}\frac{dx^{\beta}}{d\lambda'} =0.$$

Rearranging, we have

$$\frac{d^2x^{\mu}}{d\lambda'^2} + \Gamma^{\mu}_{\alpha\beta} \frac{dx^{\alpha}}{d\lambda'}\frac{dx^{\beta}}{d\lambda'} = - \left(\frac{d\lambda'}{d\lambda}\right)^{-2} \left(\frac{d^2\lambda'}{d\lambda^2}\right)\frac{dx^{\mu}}{d\lambda'}.$$

Letting

$$f = - \left(\frac{d\lambda'}{d\lambda}\right)^{-2} \left(\frac{d^2\lambda'}{d\lambda^2}\right)$$

Gives you the parametrization you were looking for

$$\nabla_v v = f v.$$