# Gravitational lensing on galaxy cluster with given potential

I am having a problem with gravitational lensing question where we are interested in deflection angle of light traveling in potential of galactic cluster, described with tensor

$h_{00}=\frac{a}{\sqrt{1+\left(\frac{r}{r_0}\right)^2}}$, which makes

$h_{\mu \nu}= \begin{bmatrix} h_{00} & 0 & 0 & 0 \\ 0 & h_{00} & 0 & 0 \\ 0 & 0 & h_{00} & 0 \\ 0 & 0 & 0 & h_{00} \end{bmatrix}$.

The idea is to use Lagrangian function and solve the differential equation derived from it to get the orbit of photons. Lagrangian function is

$L=\frac{1}{2}m\left[h_{\mu \nu} + \eta_{\mu \nu}\right]\dot{x}^{\mu}\dot{x}^{\nu}$,

so we get

$L = \frac{1}{2}m \left[ -(1-h_{00})\dot{t}^2 + (1+h_{00})(\dot{x}^2 + \dot{y}^2 +\dot{z}^2)\right]$.

Since we're working with photons, we can rewrite that as

$L' = \frac{L}{2m} = \left[ -(1-h_{00})\dot{t}^2 + (1+h_{00})(\dot{x}^2 + \dot{y}^2 +\dot{z}^2)\right]$. This is supposedly right, but I am having trouble converting it to spherical coordinates. I've used

$x = r\sin{\theta}\cos{\phi}$, $y = r\sin{\theta}\sin{\phi}$, $z = r\cos{\theta}$,

but I was told the result

$L' = \left[ -(1-h_{00})\dot{t}^2 + (1+h_{00})\left(\dot{r}^2 + r^2\left( \dot{\theta}^2 +\sin^2{\theta}\dot{\phi}^2 \right)\right)\right]$

is not correct.

What would be correct version of Lagrangian function in spherical coordinates? Could it be treated as 2D motion and reduced to polar coordinates? Is there maybe a better way to approach this problem than through Lagrangian? How would one calculate deflection angle for this case? By the way, I did look for help in literature. I think I understand basic derivation of gravitational lensing for point mass. I have a problem with this particular example of potential and answers in form of general definitions don't help me.

• Is $h$ zero except for $h_{00}$? In that case the $(1+h_{00})$ would be incorrect since the space components of $h+\eta$ reduce to those of $\eta$. Also, to find the path, you're optimizing the spacetime distance, for which a square root is needed in your Lagrangian. – brm Aug 5 '15 at 14:31
• I have edited the question with complete $h$. – Ivana Aug 5 '15 at 14:38
• Ok, but I think the Lagrangian still needs the square root. As you suggested, the space part can be limited to $\theta = \frac{\pi}{2}$ since everything happens in a plane. – brm Aug 5 '15 at 15:57
• How would Lagrangian with the square root look like? – Ivana Aug 5 '15 at 16:11
• By taking the square root of what you've got now. I'd lose the $\frac{1}{2}m$ prefactor though, where did that come from? Although multiplying by a constant doesn't change the equations that follow from the Lagrangian, the $m$ suggests the rest mass of a test particle (and you're talking about fotons, things without rest mass) – brm Aug 6 '15 at 14:55