Null geodesics of a Kerr black hole

[Dick Taid]

Homework Statement

Hi,

From the Kerr metric, in geometrized units,

[tex]\left(1 - \frac{2M}{r}\right) \left(\frac{dt}{d\lambda}\right)^2
+ \frac{4Ma}{r} \frac{dt}{d\lambda}\frac{d\phi}{d\lambda}
- \frac{r^2}{\Delta} \left(\frac{dr}{d\lambda}\right)^2
- R_a^2 \left(\frac{d\phi}{d\lambda}\right)^2 = 0[/tex]

where [itex]R_a^2 = r^2 + a^2 + \frac{2Ma^2}{r}[/itex] is the reduced circumference, [itex]a \equiv \frac{J}{M}[/itex] is the spin parameter and [itex]\lambda[/itex] is some affine parameter. I need to calculate the equations of motion.

Homework Equations

I want to solve the Lagrange equations

[tex]-\frac{d}{d\sigma}\left(\frac{\partial L}{\partial\left(dx^\alpha/d\sigma\right)}\right)
+ \frac{\partial L}{\partial x^\alpha} = 0[/tex]

for the Lagrangian

[tex]\mathcal{L}\left(\frac{dx^{ \alpha}}{d\sigma},x^{\alpha}\right)
= \left(-g_{\alpha\beta}\frac{dx^{\alpha}}{d\sigma}\frac{dx^{\beta}}{d\sigma}\right)^{1/2}[/tex]

The Attempt at a Solution

The problem is, that the metric is null so the Lagrangian is as well (?). Is it possible to calculate the equations of motion using this approach, or am I "forced" to do it the hard way, using

[tex]\frac{d^2x^{\alpha}}{d\lambda^2}
= -\Gamma_{\beta\gamma}^{\alpha}\frac{dx^{\beta}}{d \lambda}\frac{dx^{\gamma}}{d\lambda}[/tex]

finding the Christoffel symbols, and so forth?

 

[Oxvillian]

Yes - you can use the Lagrangian approach.

Use the version of the Lagrangian without the square root - you already have it in your first equation.

Unless you really like differentiating square roots

 

[Dick Taid]

Hmm, thanks. :)
I'm not entirely convinced though. But that's probably more due to my inability to state what it is I want with it. Here's everything (well, most of it):

I'm doing a little project right now, where I'm trying to plot the timelike and lightlike orbits of some test particle. I've done the timelike part by using the Lagrangian

[tex]\mathcal{L}(r,\dot{r},\phi,\dot{\phi}) = \frac{d\tau}{dt} = \left[\left(1 - \frac{2M}{r}\right)
+ \frac{4Ma}{r} \dot{\phi}
- \frac{r^2}{\Delta} \dot{r}^2
- R_a^2 \dot{\phi}^2\right]^{1/2}[/tex]

Solving Lagrange's equations for this gives the equations of motion (as functions of the observer time [itex]t[/itex])

[tex]\begin{align*}\ddot{r} &= -\frac{1}{r^4\Delta}\left[r^2\dot{r}^2\left(a^2(r - 2M) - 3Mr^2 + 2Ma\left(a^2 +
3r^2\right)\dot{\phi}\right)\right. \nonumber\\
&\qquad\qquad \left. + \Delta^2\left(M - 2Ma\dot{\phi} + \left(Ma^2 - r^3\right)\dot{\phi}^2\right)\right],
\label{eq:EqMr}\\
\ddot{\phi} &= -\frac{2\dot{r}}{r^2\Delta}\left[\left(Ma + a^2(r - 2M)\right)\dot{\phi}
+ Ma\left(a^2 + 3r^2\right)\dot{\phi}^2\right].\end{align*}[/tex]

Those can easily be transformed into first order differential equations by setting [itex]v = \dot{r}[/itex] and [itex]\omega = \dot{\phi}[/itex] and using

[tex]\frac{dt}{d\tau} = \frac{1}{\Delta} \left[R_a^2e - \frac{2Ma}{r}\ell\right],[/tex]
[tex]\frac{d\phi}{d\tau} = \frac{1}{\Delta} \left[\frac{2Ma}{r}e + \left(1 - \frac{2M}{r}\right)\ell\right][/tex]
and
[tex]\mathcal{E} \equiv \frac{e^2 - 1}{2} = \frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 + V_\text{eff}(r,e,\ell),[/tex]
where
[tex]V_\text{eff}(r,e,\ell) = -\frac{M}{r} + \frac{\ell^2 - a^2(e^2 - 1)}{2r^2} - \frac{M(\ell - ae)^2}{r^3}[/tex]

to calculate the initial conditions of the first order equations (i.e. [itex]v_0 = \frac{dr/d\tau}{dt/d\tau} = \frac{dr}{dt}[/itex] and so on), and then using MATLAB to compute the orbits. I want to do the same with photon orbits. That is, compute Lagrange's equations for a Lagrangian for the lightlike orbits, as functions of time.

Another thing, I've seen many removing the square root in the Lagrangian. Why is that allowed? Not that it matter to my calculations, as those are done with Mathematica anyways -- I'm just wondering.

 

[Oxvillian]

Another thing, I've seen many removing the square root in the Lagrangian. Why is that allowed? Not that it matter to my calculations, as those are done with Mathematica anyways -- I'm just wondering.

I think the easy answer is to vary the action, turn the crank as usual, and discover that the Lagrangian without the square root does indeed yield the geodesic equation as its equations of motion.

If I remember correctly, you can also use some integral inequality (Cauchy maybe?) to show that if the action integral with the square root is extremized, then so too is the action integral without the square root. Perhaps someone wiser than me can clarify...

 

[paranormal]

Hello,

Thank you for your question. The null geodesics of a Kerr black hole are an important aspect of the black hole's spacetime geometry. In order to calculate the equations of motion for these geodesics, it is necessary to use the geodesic equation you have mentioned, which involves finding the Christoffel symbols and solving for the second derivatives of the coordinates. This is the standard approach for calculating geodesics in general relativity.

Regarding the Lagrangian approach, it is possible to use this method to calculate the equations of motion for geodesics in curved spacetimes. However, in the case of null geodesics, the Lagrangian is indeed zero, making it difficult to use this approach. Instead, one can use the Hamiltonian formalism to calculate the equations of motion, but this is also a more advanced technique.

In summary, while the Lagrangian approach may not be the most efficient method for calculating the equations of motion for null geodesics in a Kerr black hole, it is still possible to do so. However, the standard approach of using the geodesic equation is likely the most straightforward method for this particular problem. I hope this helps.

Best regards,