The metric below was considered in [Spacetime perspective of Schwarzschild lensing]

$ds^{2}=2f du^{2}-\frac{2}{l^{2}} du dl-\frac{1}{2l^{2}}(d\theta^{2}+\sin^{2}\theta\, d\phi^{2})$

The authors gave null geodesics for this metric as

enter image description here

where $A,B,C$ are three first integrals of the null geodesics.

I am not sure how the authors got the above equations because from what I understand the null geodesics are given by the equation $g_{\mu\nu}\frac{dx_{\mu}}{dt}\frac{dx_{\nu}}{dt}=0$ but I can't see how one can get the above equations from this.

Does anyone know how the authors got the above equations?

  • $\begingroup$ Please type out the equations instead of using pictures. It is clear that you already know how to use MathJax, so please do so consistently. $\endgroup$ – Danu Aug 15 '16 at 12:37
  • $\begingroup$ @Danu, I am just trying to show exactly how the authors presented the equations. $\endgroup$ – gbd Aug 15 '16 at 12:42
  • $\begingroup$ I know that; I'm just saying that you should type out all your equations ;) $\endgroup$ – Danu Aug 15 '16 at 12:42

The equation you list $$g_{\mu\nu}\partial_\tau x^\mu\partial_\tau x^\nu=0$$ is satisfied by all null-trajectories. However, a null geodesic must also satisfy the geodesic equation $$\partial^2_\tau x^\mu +\Gamma^\mu_{\nu\rho}\partial_\tau x^\nu\partial_\tau x^\rho=0$$ These two equations, taken together, should suffice to determine the null geodesics. The $\Gamma$ symbols are in principle easy to calculate, either by hand or using some computer algebra program.

Since the results in the paper are given by first order differential equations, while the geodesic equation is of second order, one has to integrate once (this explains the presence of the integration constants $A,B,C$).


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