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I have to calculate the deflection of light by a star (using a given metric, not Schwarzschild) and have produced the equation: $$ \frac{d\phi}{dr} = \frac{1}{r^2} \left( \frac{1}{b^2} - \frac{1}{r^2} \right)^{-\frac{1}{2}} $$ where $b$ is the impact parameter.

I can't seem to find a way to solve this for $\phi$, for a photon approaching from infinity. I have my thoughts about what the result should be but I'm trying to do the maths without bias.

Any insights or tips on how to solve this would be appreciated!

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We can directly integrate the differential equation

$$ \frac{d\phi}{dr} = \frac{1}{r^{2} \sqrt{\frac{1}{b^2} - \frac{1}{r^2}}} $$

and seek the following antiderivative:

$$ \int \frac{dr}{r^{2} \sqrt{\frac{1}{b^2} - \frac{1}{r^2}}}$$

Substitute $u=\frac{b}{r}$ and obtain $du = - \frac{b}{r^{2}}dr$, so that $dr=-\frac{r^{2}}{b}du = -\frac{b}{u^{2}}du$.

Plug it into the integral:

$$ \int \frac{dr}{r^{2} \sqrt{\frac{1}{b^2} - \frac{1}{r^2}}} = \int \frac{-b}{u^{2}}\frac{u^{2}}{b^{2}} \frac{1}{\sqrt{ \frac{1}{b^{2}}(1 - u^{2})}}du = \int -\frac{1}{b} \frac{1}{\frac{1}{b}}\frac{1}{\sqrt{1 - u^{2}}}du= \\[3em] - \int \frac{1}{\sqrt{1-u^{2}}}du = -\arcsin(u) + C = -\arcsin(\frac{b}{r}) + C . $$

Now to extract the angle of deflection I think we can use the notion of distance of closest approach $r_{0}$. This is not the same as the impact parameter $b$, see answer by Rennie.

The closest-approach distance $r_{0}$ and its relation to $b$ can be perhaps determined if you provided the metric. I suppose that it has two Killing vector fields $\xi^{t}$ and $\xi^{\phi}$ that correspond to time-translation and rotation. This serve as two constants of motion for your null geodesic $\mathbf{k}$ and: $$-E = g(\xi^{t},\mathbf{k}) \\ L= g(\xi^{\phi},\mathbf{k}) $$ Pairing that with the $g(\mathbf{k}, \mathbf{k}) = 0 $ gives a set of equations that can be inspected for the closest approach $r_{0}$ at which $\frac{dr}{dt}=0$, as in this answer.

I am inclined to say that we need to assume axisymmetry and time-independence of the metric here - in the stronger form, i.e. that the metric is static, not only stationary. This I believe is reflected in the very fact that you could write your equation like this, without any time or free angular dependence on the RHS.

Then, with respect to this $r_{0}$, the photon's trajectory is symmetric and we can simply take twice the integral from infinity to $r_{0}$.

The total deflection angle is thus:

$$ \Delta \phi = -2 \cdot \arcsin(\frac{b}{r})\vert^{r=r_{0}}_{r=\infty}$$

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  • $\begingroup$ Incredibly helpful @K.T. The radius of closest approach in this metric does equal b, hence I knew what to expect but couldn't find the substitution needed for that integral. Much appreciated. $\endgroup$
    – stooks
    Commented Sep 30, 2021 at 23:14

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