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Show for the null geodesic in 3D flat spacetime using polar coordinates so the line element is $ds^2=-dt^2+dr^2+r^2d\phi^2$. Do light rays move on straight lines?

My question is that I only learned how to find geodesic using Lagrangian, but ds^2=0 in this case, which I have no idea what to do now. Also, how many geodesic equations are there in this case?

Qmechanic
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lol
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1 Answers1

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The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line')

In the LC connection they are the integral curves of some vector field $V$ with $ \nabla_V V = 0$ (interpretation: the parallel transport of $V$ along itself is $0$): in flat Euclidean space this means that the integral curves of the vector field are just 'straight' lines (everywhere "parallel to itself").

As to finding the null geodesics: don't worry about them, just do it! The spacetime is flat so its quite trivial ($\Gamma^a{}_{bc} =0$). Indeed $ds^2=d\tau^2=0$ so they are not that handy as affine parameter of the curve; just use a general $\lambda$ to parametrize them. The EoM become: $$ \ddot x = 0 , \ddot y = 0, \ddot t = 0 \quad \quad \text{where } \dot q \equiv \frac{d q}{d\lambda} $$

123hoedjevan
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  • Does this mean that $\ddot{t}=0,\ddot{r}=0,\ddot{\phi}=0$? I don't know how to calculate $\Gamma^a_{bc}$ if ds^2=0. – lol Apr 29 '16 at 01:11
  • For null geodesic you can use any parameter as @123hoedjevan has already mentioned. Then you calculate affine parameter by comparing as you did before.(they will not be zero as affine parameter depend on coordinate and not on space). I think the geodesic will still be a straight line just written in polar coordinates, – AMIT SINGH Apr 29 '16 at 04:51
  • I agree with Amit Singh, you should just be able to take $ x=r \cos \phi $ and $ y = r \sin \phi $ and plug it in $ \ddot x $ and $ \ddot y $ to get the EoM in polar coordinates (they will be not nearly as nice as in xy coords). As to the fact that $ ds^2= 0 $, that is not a bit of information you use a priori; the geodesic equations give you the the EoM for ANY geodesic (time,null,or spacelike). The null property you really only use to determine for constraining the Lagrangian: $ g_{\mu\nu} u^\mu u^\nu = -1,0,1 $ (0 in this case) – 123hoedjevan Apr 29 '16 at 10:55