Alternative form of geodesic equation

[rohanlol7]

Homework Statement

We are asked to show that:
$\frac{d^2x_\mu}{d\tau^2}= \frac{1}{2} \frac{dx^\nu}{d\tau} \frac{dx^{\rho}}{d\tau} \frac{\partial g_{\rho \nu}}{\partial x^{\mu}}$
( please ignore the image in this section i cannot remove it for some reason )

Homework Equations

The Attempt at a Solution

[/B]I have tried substituting $x_\mu = g_{a\mu} x^{a}$ in the geodesic equation, but all that i end up with is a complete mess.
Maybe I am understanding it all wrong, for instance i think that $\frac{dx_\mu}{d\tau} = \frac{ d g_{a\mu} x^{a}}{d\tau}$, Is this correct or is the $g_{a\mu}$ supposed to go outside of the derivative ?

 

[Orodruin]

You cannot put images in your posts like that. Please upload the relevant images.

 

[rohanlol7]

You cannot put images in your posts like that. Please upload the relevant images.

I have added the images properly now

 

[TSny]

I have tried substituting $x_\mu = g_{a\mu} x^{a}$ in the geodesic equation, but all that i end up with is a complete mess.
Maybe I am understanding it all wrong, for instance i think that $\frac{dx_\mu}{d\tau} = \frac{ d g_{a\mu} x^{a}}{d\tau}$, Is this correct or is the $g_{a\mu}$ supposed to go outside of the derivative ?

For general coordinates, $x^{\mu}$, it is not true that $x_\mu = g_{\mu \alpha } x^{\alpha}$. The $x^{\mu}$ are not components of a contravariant vector. So, it doesn't make sense to lower the index using the expression $x_\mu = g_{\mu \alpha} x^{\alpha}$.

However, the differentials $dx^{\mu}$ are the components of a contravariant vector. So, you can lower the index on these according to $dx_\mu = g_{\mu \alpha} dx^{\alpha}$ to make a covariant vector. Likewise, you can write $\frac{dx_\mu}{d \tau} = g_{\mu \alpha} \frac{dx^{\alpha}}{d \tau}$.

Apply this to $\frac{d^2 x_{\mu}}{d \tau^2}$ by writing $\frac{d^2 x_{\mu}}{d \tau^2} = \frac{d}{d \tau} \left( \frac{dx_\mu}{d \tau} \right)$.