# Null geodesics of a Kerr black hole

• Dick Taid
In summary, the Lagrangian for the equations of motion for a particle in a gravitational field can be written as-\frac{d^2x^{\alpha}}{d\lambda^2} = -\Gamma_{\beta\gamma}^{\alpha}\frac{dx^{\beta}}{d \lambda}\frac{dx^{\gamma}}{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
Dick Taid

## Homework Statement

Hi,

From the Kerr metric, in geometrized units,

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

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

## Homework Equations

I want to solve the Lagrange equations

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

for the Lagrangian

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

## 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

$$\frac{d^2x^{\alpha}}{d\lambda^2} = -\Gamma_{\beta\gamma}^{\alpha}\frac{dx^{\beta}}{d \lambda}\frac{dx^{\gamma}}{d\lambda}$$

finding the Christoffel symbols, and so forth?

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

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

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

Solving Lagrange's equations for this gives the equations of motion (as functions of the observer time $t$)

\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*}

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

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

to calculate the initial conditions of the first order equations (i.e. $v_0 = \frac{dr/d\tau}{dt/d\tau} = \frac{dr}{dt}$ 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.

Dick Taid said:
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...

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,

## 1. What are null geodesics of a Kerr black hole?

Null geodesics are the paths followed by massless particles, such as photons, in the curved spacetime around a rotating black hole. In the case of a Kerr black hole, these geodesics follow a curved path due to the warping of spacetime caused by the black hole's rotation.

## 2. How do null geodesics differ from timelike and spacelike geodesics?

Null geodesics differ from timelike and spacelike geodesics in that they are not affected by the gravitational pull of the black hole. Timelike and spacelike geodesics can be influenced and even captured by the black hole's gravity, while null geodesics remain on a fixed path due to their lack of mass.

## 3. Can null geodesics escape from the event horizon of a Kerr black hole?

No, null geodesics cannot escape from the event horizon of a Kerr black hole. The event horizon is the point of no return for anything, including light, that enters it. Once a photon crosses the event horizon, it is pulled towards the singularity at the center of the black hole and cannot escape.

## 4. How do the properties of a Kerr black hole affect the behavior of null geodesics?

The properties of a Kerr black hole, such as its mass and spin, can greatly affect the behavior of null geodesics. The spin of the black hole can cause frame-dragging, which affects the path of the geodesics, and the mass of the black hole determines the strength of its gravitational pull on the geodesics.

## 5. What are some applications of studying null geodesics of Kerr black holes?

Studying null geodesics of Kerr black holes has many practical applications, including understanding how light is affected by the curvature of spacetime and the warping caused by massive objects. This can also help in the development of advanced technologies such as gravitational wave detectors and black hole imaging.

Replies
11
Views
2K
Replies
0
Views
446
Replies
10
Views
1K
Replies
1
Views
485
Replies
2
Views
912
Replies
19
Views
811
Replies
5
Views
2K
Replies
1
Views
924