Form of affinely parametrized geodesic equation under an arbitrary coordinate transformation?

Question

This question is based off of Chapter 3 of Hobson M.P., G. P. Efstathiou, A. N. Lasenby General Relativity An Introduction for Physicists (2006). The exact question is Q3.15. Not for homework though, but studying for exams.

For an affinely parameterised geodesic $x^a(u)$ that is parameterised by an affine parameter u.

The geodesic satisfies:

$$ \frac{d^2x^a}{du^2 }=\Gamma^{a}_{bc}\frac{d^2x^b}{du^2}\frac{d^2x^c}{du^2}$$

Under an arbitrary coordinate transformation $x^a \rightarrow x'^a$, the form of the equation should be unchanged. How do I show this? I am a little mixed up, does an arbitrary coordinate transformation just entail a change of parameter, e.g. $u \rightarrow u' $ or is the parameter changed to the new coordinate e.g. $x^a(u) \rightarrow x'^a(x^a) $ ? How do the coordinate vectors (e.g. $e^a $)change under a transformation?

Some thoughts I had on this:

• If a coordinate transformation is just a parameter change, could I just represent the derivatives as: $$ \frac{d^2x^a}{du^2 } \rightarrow \frac{d^2x^a}{du^2 }\frac{d^2u}{du'^2 }$$

  And the connection changes to: $$\Gamma^{a}_{bc} \rightarrow \Gamma'^{a}_{bc}=\frac{\partial x'^a }{\partial x^d }\frac{\partial x^f }{\partial x'^b }\frac{\partial x^g }{\partial x'^c } \Gamma'^{d}_{fg}+\frac{\partial x'^a }{\partial x^d }\frac{\partial^2 x'^a }{\partial x'^c \partial x'^b }$$ and then insert into the differential geodesic equation and evaluate?

• Or do I need to specify a general coordinate transformation equation? e.g. a linear transformation $x'^1= \alpha x^1 + \beta $ and then insert this into the original geodesic differential equation?

A quick explanation of the form of a geodesic $x^a(u)$ (i.e. is this similar to an equation of a line like $y=mx+c$) and what happens under arbitrary coordinate transformations in general, would be very useful in understanding this. Any help is greatly appreciated.

Answer 1

The geodesic equation is $$ \frac{ d^2 x^\lambda }{ d\tau^2} + \Gamma^\lambda_{\mu\nu}(x) \frac{ d x^\mu }{ d \tau} \frac{ d x^\nu }{ d \tau} = 0 ~. $$ Under $x^\mu \to x'^\mu(x)$, we note that the Christoffel symbol does not transform like a tensor. Rather, $$ \Gamma'^\lambda_{\mu\nu}(x') =\frac{ \partial x^\alpha}{ \partial x'^\mu} \frac{ \partial x^\beta}{ \partial x'^\nu} \left[ \frac{\partial x'^\lambda}{\partial x^\rho} \Gamma^\rho_{\alpha\beta}(x) - \frac{ \partial^2 x'^\lambda }{ \partial x^\alpha \partial x^\beta } \right] $$ We also have \begin{align} \frac{d x'^\lambda }{ d\tau} &= \frac{ \partial x'^\lambda }{ \partial x^\rho } \frac{ d x^\rho }{ d\tau}~, \\ \qquad \frac{d^2 x'^\lambda }{ d\tau^2} &= \frac{d}{d\tau} \left( \frac{ \partial x'^\lambda }{ \partial x^\rho } \frac{ d x^\rho }{ d\tau} \right) = \frac{ \partial x'^\lambda }{ \partial x^\rho } \frac{ d^2x^\rho }{ d\tau^2} + \frac{ \partial^2 x'^\lambda }{ \partial x^\alpha \partial x^\beta } \frac{ d x^\alpha }{ d\tau} \frac{ d x^\beta }{ d\tau} ~. \end{align} Thus, $$ \Gamma'^\lambda_{\mu\nu}(x') \frac{ d x'^\mu }{ d \tau} \frac{ d x'^\nu }{ d \tau} = \left[ \frac{\partial x'^\lambda}{\partial x^\rho} \Gamma^\rho_{\alpha\beta}(x) - \frac{ \partial^2 x'^\lambda }{ \partial x^\alpha \partial x^\beta } \right] \frac{ d x^\alpha }{ d \tau} \frac{ d x^\beta }{ d \tau} $$ Putting this altogether, we find \begin{align} &\frac{ d^2 x'^\lambda }{ d\tau^2} + \Gamma'^\lambda_{\mu\nu}(x') \frac{ d x'^\mu }{ d \tau} \frac{ d x'^\nu }{ d \tau} \\ &\qquad= \frac{ \partial x'^\lambda }{ \partial x^\rho } \frac{ d^2x^\rho }{ d\tau^2} + \frac{ \partial^2 x'^\lambda }{ \partial x^\alpha \partial x^\beta } \frac{ d x^\alpha }{ d\tau} \frac{ d x^\beta }{ d\tau} + \left[ \frac{\partial x'^\lambda}{\partial x^\rho} \Gamma^\rho_{\alpha\beta}(x) - \frac{ \partial^2 x'^\lambda }{ \partial x^\alpha \partial x^\beta } \right] \frac{ d x^\alpha }{ d \tau} \frac{ d x^\beta }{ d \tau} \\ &\qquad= \frac{ \partial x'^\lambda }{ \partial x^\rho } \left[ \frac{ d^2x^\rho }{ d\tau^2} + \Gamma^\rho_{\alpha\beta}(x) \frac{ d x^\alpha }{ d \tau} \frac{ d x^\beta }{ d \tau} \right] \\ \end{align}