# Deflection of light by star

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!

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:

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

• 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. Commented Sep 30, 2021 at 23:14