# Geodesic Equation from Coordinate Transformation

Let $$\xi^a$$ be the usual coordinates and $$x^\mu$$ the new coordinates, both flat. Now we know that since the metric is flat, $$\frac{d^2\xi^a}{d\tau^2} = 0$$ $$\Rightarrow \ \frac{\partial}{\partial \tau} \left( \frac{\partial \xi^a}{\partial x^\mu} \frac{x^\mu }{\partial \tau} \right) = 0$$ Now the proof I have skips directly to $$\frac{\partial \xi^a}{\partial x^\mu} \left[ \frac{\partial^2 x^\mu}{\partial \tau^2} + \frac{\partial x^\mu }{\partial \xi^b} \frac{\partial^2 \xi^b}{\partial x^\lambda \partial x^\nu} \frac{\partial x^\nu}{\partial \tau } \frac{\partial x^\lambda}{\partial \tau} \right] = 0$$ And from by equating the brackets to zero one gets the geodesic equation, $$\ddot{x}^\mu + \Gamma^\mu_{\lambda \nu} \dot{x}^\nu\dot{x}^\lambda = 0$$ For all of the above, I am very comfy, but I was not able to cover the skipped line especially since we now have two indices for x and an extra index for $$\xi$$. So how do I do this?

• In short, if the metric is flat, i.e. $g_{ij}$ are constant functions in the neighborhood of the tangent vectors, then $\Gamma^{k}_{ij}=\frac{1}{2}\sum_{m}g^{km}\lbrace \frac{\partial g_{jm}}{\partial x^{i}} + \frac{\partial g_{im}}{\partial x^{j}} - \frac{\partial g_{ij}}{\partial x^{m}}\rbrace=0$. – Cinaed Simson Oct 5 '19 at 7:47

## 1 Answer

This is often a tricky (but easy) calculation in GR. Let's start with the fact that in your problem, as one can infer, the metric $$g_{\mu \nu} (\xi^\alpha)$$ is flat, not $$g_{\mu \nu} (x^\alpha)$$. In fact, the Simson's comment is wrong since the partial derivatives should be w.r.t. $$\xi^\alpha$$-coordinates.

Since $$g_{\mu \nu} (\xi^\alpha)$$ is flat then one has $$\frac{d^2\xi^a}{d\tau^2} = 0 = \frac{\partial}{\partial \tau} \left( \frac{\partial \xi^a}{\partial x^\mu} \frac{\partial x^\mu }{\partial \tau} \right) \,.$$ We can now calculate the r.h.s. of the above: \begin{align} \frac{\partial}{\partial \tau} \left( \frac{\partial \xi^a}{\partial x^\mu} \frac{x^\mu }{\partial \tau} \right) &= \frac{\partial x^\nu}{\partial\tau} \frac{\partial^2 \xi^a}{\partial x^\nu \partial x^\mu} \frac{\partial x^\mu}{\partial \tau} + \frac{\partial \xi^a}{\partial x^\mu} \frac{\partial^2 x^\mu }{\partial \tau^2} \\ &= \frac{\partial \xi^a}{\partial x^\mu} \frac{\partial^2 x^\mu }{\partial \tau^2} + \delta^a_b \frac{\partial^2 \xi^b}{\partial x^\lambda \partial x^\nu} \frac{\partial x^\nu}{\partial \tau } \frac{\partial x^\lambda}{\partial \tau} \,. \end{align} Using the fact that $$\delta^a_b = \frac{\partial \xi^a}{\partial x^\mu} \frac{\partial \xi^\mu}{\partial \xi^b}$$ and factoring out $$\frac{\partial \xi^a}{\partial x^\mu}$$ we arrive at: $$\frac{\partial}{\partial \tau} \left( \frac{\partial \xi^a}{\partial x^\mu} \frac{x^\mu }{\partial \tau} \right) = \frac{\partial \xi^a}{\partial x^\mu} \left[ \frac{\partial^2 x^\mu}{\partial \tau^2} + \frac{\partial x^\mu }{\partial \xi^b} \frac{\partial^2 \xi^b}{\partial x^\lambda \partial x^\nu} \frac{\partial x^\nu}{\partial \tau } \frac{\partial x^\lambda}{\partial \tau} \right] = 0 \,.$$ This in turn means that $$\frac{\partial^2 x^\mu}{\partial \tau^2} + \frac{\partial x^\mu }{\partial \xi^b} \frac{\partial^2 \xi^b}{\partial x^\lambda \partial x^\nu} \frac{\partial x^\nu}{\partial \tau } \frac{\partial x^\lambda}{\partial \tau} =0 \,.$$ Now, from the behavior of the Christoffel symbol under coordinate transformation (see the Wikipedia for instance) we know that $$\frac{\partial x^\mu }{\partial \xi^b} \frac{\partial^2 \xi^b}{\partial x^\lambda \partial x^\nu} = \Gamma^\mu_{\lambda \nu}(x^\alpha) \,,$$ since $$\Gamma^\mu_{\lambda \nu}(\xi^\alpha) = 0$$, as mentioned earlier. Now you have the geodesic equation.

• Can you please elaborate on why in your second line you had two indices of x and not just one. Why create mu and nu of x? – RetroCausalElectron Oct 6 '19 at 7:51
• Chain rule! To differentiate $\xi^a$, we differentiate it first w.r.t. to $x^\nu$ and then $x^\nu$ w.r.t. the affine parameter. – Astrolabe Oct 6 '19 at 8:46