Consider a photon coming from the infinity in a unbounded orbit to a Schwarzschild black hole (Schwarzschild radius rs) (see this for illustration). Its impact parameter is b and its distance of closest approach is r0 with b2=r30r0−rs.
Then its trajectory in polar coordinates is defined by :
dφdr=1r2√1b2−(1−rsr)1r2
Consequently : φ(r)=∫rr0dpp2√1b2−(1−rsp)1p2
and one can compute the total deviation using : Δφ=2×(lim
But my question is : how can I plot/draw the trajectory using the integral expression of \varphi\left(r\right) ?
Because if I compute : f\left(r\right) = 2\times\left(\int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}-\frac{\pi}{2}\right) I obtain f\left(r_{0}\right) = -\pi, and then f increases up to zero, crosses zero, and tends to its positive value at infinity \Delta\varphi. It does not make sense for me and I do not understand how to compute the trajectory from that.
If I compute : g\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}} it starts from 0, and increase up to \frac{\pi}{2}+\frac{\Delta\varphi}{2}.
I would like to compute the trajectory in the \left(x, y\right) plane, so how to use the values of f\left(r\right) or g\left(r\right) to compute the function y\left(x\right) ?
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