1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Thank you for the explanation, that cleared up the confusion I was having. $\endgroup$ – gline Jul 6 '17 at 17:13
  • $\begingroup$ Your equation that describes the transformation property of the Christoffel symbol seems to differ from the one that I know (see this MathSE post). Cloud you maybe explain shorty where the change in sign comes from? $\endgroup$ – Sito Dec 28 '19 at 10:37
  • $\begingroup$ @Sito you can work out what the sign should be using the definition. $\endgroup$ – Prahar Dec 28 '19 at 13:04
  • $\begingroup$ This is how I would derive it. The expression at the end (eq. 16) is obviously different to the one that you stated. Is there a way to show that they are equivalent? $\endgroup$ – Sito Dec 28 '19 at 15:24
  • $\begingroup$ @Sito - The first line (and therefore all subsequent lines) of the equation under equation (3) in your link is wrong. You do not have the current transformation law for the metric. On the LHS, your primed indices are lowered whereas on the RHS they are raised. There might be other mistakes but I haven't looked at your derivation beyond this point. A quick glance tells me that you have similar index problems all over the place (for example - your equation (17) is MOST definitely wrong -- the indices just don't match up on both sides). $\endgroup$ – Prahar Dec 28 '19 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.