Computing Null Geodesics in Schwarzschild Geometry

[stevendaryl]

Computing timelike geodesics in the Schwarzschild geometry is pretty straightforward using conserved quantities. You can treat the problem as a variational problem with an effective Lagrangian of

$\mathcal{L} = \frac{1}{2} (Q \frac{dt}{d\tau}^2 - \frac{1}{Q} \frac{dr}{d\tau}^2 - r^2 (\frac{d\theta}{d\tau}^2 + sin^2(\theta) \frac{d\phi}{d\tau}^2))$

where $Q = 1 - \frac{2GM}{r}$

This "lagrangian" leads to the following conserved quantities:

1. $K = Q \frac{dt}{d\tau}$
2. $L = r^2 \frac{d\phi}{dt}$ (You can choose $\theta$ and $\phi$ so that $\theta = \frac{\pi}{2}$, so all the radial motion is due to changing of $\phi$
3. $\mathcal{L}$ itself, which is always equal to 1/2.

In terms of these conserved quantities, the geodesics are completely determined by $\frac{dr}{d\tau}$, which satisfies the one-D equation:

$\frac{1}{Q} (K^2 - \frac{dr}{dt}^2 ) - \frac{L^2}{r^2} = 1$

My question is: How are things changed if we are computing a null geodesic, instead of a timelike geodesic? The biggest change is that you can't use proper time as the parameter (since it's identically zero for null geodesics, by definition). If you replace $\tau$ by a different parameter, $\lambda$, I'm assuming that it's still true that there is something like angular momentum that is conserved, but I'm not sure about the first equation, which is about the conservation of the $t$ component of momentum.

 

[Ibix]

The 1 becomes a zero, I think.

Carroll does it by observing that there is a Killing field parallel to $\partial_t$ and another parallel to $\partial_\phi$, hence writing expressions for $dt/d\lambda$ and $d\phi/d\lambda$ in terms of conserved quantities along a geodesic, restricting himself to the equatorial plane, noting that $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$ is either 0, -1, or +1, and solving for $dr/d\lambda$. I think that's your last equation, give or take the value on the RHS.

 

[Orodruin]

Indeed, the 1 becoming a zero is the only change. The conserved ”energy” and ”angular momentum” quantities take the same form as they are the inner product between an affinely parametrized geodesic’s tangent and a Killing field.

 

[stevendaryl]

Indeed, the 1 becoming a zero is the only change. The conserved ”energy” and ”angular momentum” quantities take the same form as they are the inner product between an affinely parametrized geodesic’s tangent and a Killing field.

So what is the parameter $\lambda$ for a null geodesic?

 

[Orodruin]

So what is the parameter $\lambda$ for a null geodesic?

It does not have direct physical meaning the same way proper time does as you can reparametrize the geodesic without changing that the tangent vector us null.

If you want you could choose it such that the tangent vector is the 4-frequency of a light signal following that geodesic. However, there can be several such signals following the same geodesic.