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I'm stuck on the following problem;

a) Calculate the Christoffel symbols for the metric

$dr^2 + G(r,\phi)d\phi^2$.

b) Write down the geodesic equations and verify the identity

$(r')^2 + G(\phi')^2 = constant$

is satisfied by virtue of these equations.

c) Show that the curves $\phi = constant$ are geodesics.

So for a) I have

$\Gamma^r_{rr} = 0, \Gamma^\phi_{rr} = 0, \Gamma^r_{r\phi} = 0, \Gamma^\phi_{r\phi} = G_r/2G, \Gamma^r_{\phi \phi} = -GG_r/2G = -G_r/2, \Gamma^\phi_{\phi\phi} = G_\phi/2G$

and for b), I get for the geodesic equations

$-\frac{G_r}{2}(\phi')^2 + r'' = 0$

$\frac{G_r}{G}r'\phi' + \frac{G_\phi}{2G}(\phi')^2 + \phi'' = 0$.

But after this Im stuck. The hint is that you differentiate the identity with respect to $t$ and sub in $r''$ and $\phi''$ from the geodesic equations. So I try that:

$\frac{d}{dt}\left(\frac{dr}{dt}\right)^2 + \frac{d}{dt}G\left(\frac{d\phi}{dt}\right)^2 = 0$

to get

$2r'r'' + 2G\phi'\phi'' = 0$

but when I substitute in $r''$ and $\phi''$ I get

$-G_rr'(\phi')^2 - G_\phi(\phi')^3$

and I don't see how this can equal zero? Have I calculated the Christoffel symbols correctly? Also Im not sure I differentiated the identity correctly - in that sense is $G$ a functions of $t$ through $r$ and $\phi$? If anyone can help me it would be greatly appreciated.

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You forgot to differentiate $G$ with respect to time. Let $I = r'^2 + G\phi'^2$ then

$$I' = 2r'r'' + G_r r' \phi'^2 + G_\phi \phi'^3 + 2G\phi'\phi''$$

The missing term in your equation is the term $G'\phi'^2 = [G_rr' + G_\phi\phi']\phi'^2$. Now by insering the geodesic equations $2G\phi''\phi' = -G_\phi\phi'^3 - 2G_rr'\phi'^2$ and $2r'r'' = G_rr'\phi'^2$ we obtain $I' = 0$.

For c) take $\phi = \text{constant} \implies \phi' = 0,~~\phi''=0$ in the geodesic equation to obtain $r'' = 0 \implies r = r_0 + at$. You can also use the equation derived in b) for this purpose.

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  • $\begingroup$ Thanks! Yeah, that makes sense $\endgroup$ – Mr Man Feb 8 '15 at 0:58

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