How to Derive Radial Geodesics in Kerr Metric?

• JD_PM
In summary, the Kerr metric is a spacetime metric described by a set of equations, including the Delta and rho variables, that can be simplified when the theta angle is equal to pi/2. The metric coefficients are all independent of the coordinates t and phi, and thus any time or spatial translation will leave the metric unchanged. This leads to the existence of two Killing-vector fields, K^mu and R^mu, which are conserved along the geodesic. The geodesic equation and metric compatibility condition also imply an extra conserved quantity, epsilon, which is given by a specific equation. By rearranging this equation, the result can be simplified to show that the bracketed term is equal to (E-LW_+
JD_PM
Homework Statement
Consider the orbits of massless particles, with affine parameter ##\lambda##, in the equatorial plane (i.e. ##\theta = \pi /2##) of a Kerr black hole

Show that

\begin{equation*}

\left( \frac{dr}{d \lambda} \right)^2 = \dot r^2 = \frac{\Sigma^2}{\rho^4}\left( E- L \ W_+ (r)\right)\left(E-L \ W_- (r)\right)

\end{equation*}

To do so, you will have to explicitly find the expressions for ##W_+ (r)## and ##W_- (r)##

This is exercise 2, section a) Chapter 6 in Carroll's book
Relevant Equations
N/A
The Kerr metric is given by

\begin{align*}
(ds)^2 &= -\left(1-\frac{2GMr}{\rho^2} \right)(dt)^2 - \frac{2GMar \sin^2 \theta}{\rho^2}(dt d\phi + d\phi dt) \\ &+ \frac{\rho^2}{\Delta}(dr)^2 + \rho^2 (d \theta)^2 + \frac{\sin^2 \theta}{\rho^2} \left[ \underbrace{(r^2+a^2)^2-a^2 \Delta \sin^2 \theta}_{\Sigma^2} \right] (d \phi)^2
\end{align*}

Where

\begin{equation*} \Delta = r^2 -2GMr+a^2 \end{equation*}

\begin{equation*} \rho^2 = r^2+a^2 \cos^2 \theta \end{equation*}

\begin{equation*}
\Sigma^2 = (r^2 + a^2)^2 - a^2 \Delta \sin^2 \theta
\end{equation*}

The Kerr metric at ##\theta = \pi /2## gets simplified to

\begin{align*}
(ds)^2 &= -\left(1-\frac{2GM}{\rho} \right)(dt)^2 - \frac{2GMa}{\rho}(dt d\phi + d\phi dt) \\ &+ \frac{r^2}{\Delta}(dr)^2 + \frac{\Sigma^2}{\rho^2} (d \phi)^2
\end{align*}

Where we used

\begin{equation*} \rho^2 = r^2 \end{equation*}

\begin{equation*}
\Sigma^2 = (r^2 + a^2)^2 - a^2 \Delta
\end{equation*}

The metric coefficients in Kerr's metric are all independent of the coordinate ##t##, which means that any time translation will leave the metric invariant. Then the vector field ##\partial / \partial t## is a (time-like) Killing-vector field i.e.

\begin{equation*}
K^{\mu} = (1,0,0,0)
\end{equation*}\begin{equation*}
K_{\mu} = g_{\mu \nu}K^{\nu}=\left( -\left(1-\frac{2GM}{\rho} \right),0,0,0\right)
\end{equation*}

The metric coefficients are all also independent of the coordinate ##\phi##. Thus

\begin{equation*}
R^{\mu} = (0,0,0,1)
\end{equation*}

\begin{equation*}
R_{\mu} = g_{\mu \nu}R^{\nu}=\left( 0,0,0, \frac{\Sigma^2}{\rho^2} \right)
\end{equation*}

Parameterizing a geodesic by an affine parameter ##\lambda## and defining a four-velocity vector ##U^{\mu}=dx^{\mu}/d \lambda := \dot x^{\mu}##, given a Killing vector ##K_{\mu}##, the have that the following quantity (along the geodesic) is conserved (Carroll Chapter 5, EQ. 5.54)

\begin{equation*}
K_{\mu}U^{\mu} = \text{constant}
\end{equation*}

Thus we get the following conserved quantities

\begin{equation*}
E= -K_{\mu}U^{\mu} = \left(1-\frac{2GM}{\rho} \right) \dot t \Rightarrow \dot t = \frac{E}{1-\frac{2GM}{\rho}}
\end{equation*}\begin{equation*}
L=R_{\mu}U^{\mu}=\frac{\Sigma^2}{\rho^2} \dot \phi \Rightarrow \dot \phi = \frac{\rho^2}{\Sigma^2} L
\end{equation*}

Where ##E## stands for the conserved energy (where we introduced a negative sign to turn ##E##'s sign positive) and ##L## for the conserved ##z-##component of the angular momentum.

The geodesic equation and the metric compatibility condition imply an extra conserved quantity (Which is given in equation ##(5.55)## in Carroll's book. Besides, the parameter ##\lambda## is chosen such that ##\epsilon=1## for massive particles and ##\epsilon=0## for massless particles. The case ##\epsilon=-1## simply corresponds to spacelike geodesics).

\begin{equation*}
\epsilon=-g_{\mu \nu}\frac{d x^{\mu}}{d \lambda}\frac{d x^{\nu}}{d \lambda}
\end{equation*}

Expanding it yields

\begin{align*}
\epsilon &= -g_{tt} \dot t^2 -g_{rr} \dot r^2 -2g_{t \phi} \dot t \dot \phi -g_{\phi \phi} \dot \phi^2 \\
&= \left(1-\frac{2GM}{\rho} \right) \dot t^2 -\frac{\rho^2}{\Delta} \dot r^2 + \frac{4GMa}{\rho}\dot t \dot \phi -\frac{\Sigma^2}{\rho^2} \dot \phi^2
\end{align*}

Plugging ##\dot t## and ##\dot \phi## into above's expression and rearranging yields

\begin{align*}
&\large \epsilon = \frac{E^2}{1-\frac{2GM}{\rho}} - \frac{\rho^2}{\Delta} \dot r^2 + \frac{4GM a \rho}{\Sigma^2 \left(1-\frac{2GM}{\rho}\right)} EL - \frac{\rho^2 L^2}{\Sigma^2} \Rightarrow\\
&\Rightarrow \dot r^2 = \frac{\Delta}{\rho^2}\left[ \frac{E^2}{1-\frac{2GM}{\rho}} + \frac{4GM a \rho}{\Sigma^2 \left(1-\frac{2GM}{\rho}\right)} EL - \frac{\rho^2 L^2}{\Sigma^2} -\epsilon \right] \Rightarrow\\
&\Rightarrow \dot r^2 = \frac{\Sigma^2}{\rho^4}\left[ \frac{\rho^2 E^2 \Delta}{\Sigma^2 \left(1-\frac{2GM}{\rho}\right)} + \frac{4GM a \rho^3 \Delta}{\Sigma^4 \left(1-\frac{2GM}{\rho}\right)} EL - \Delta \frac{\rho^4 L^2}{\Sigma^4} -\frac{\rho^2 \Delta \epsilon}{\Sigma^2} \right]
\end{align*}

So my result is (recalling that we deal with massless particles i.e. ##\epsilon =0##)

\begin{equation*}
\boxed{\dot r^2 = \frac{\Sigma^2}{\rho^4}\left[ \frac{\rho^2 E^2 \Delta}{\Sigma^2 \left(1-\frac{2GM}{\rho}\right)} + \frac{4GM a \rho^3 \Delta}{\Sigma^4 \left(1-\frac{2GM}{\rho}\right)} EL - \Delta \frac{\rho^4 L^2}{\Sigma^4} \right]}
\end{equation*}

Yikes! note my naïve attempt to match the provided solution but I see no straightforward way to get ##W_{\pm} (r)##...

So my question is: how to show that the bracketed term is equal to

\begin{equation*}
\left( E- LW_+ (r)\right)\left(E-LW_- (r)\right)
\end{equation*}

?

I wish you a good 2021

Thank you!

PS: There is another method to get the radial geodesic of the Kerr metric and that is explicitly solving the geodesic equation for ##\mu = r##. That method is much more tedious since it implies computing (some) Christoffel symbols for the Kerr metric... which is not really enjoyable!

docnet and etotheipi
$$\text{E}= -K_{\mu}U^{\mu}= -g_{\mu \nu}K^{\nu}U^{\mu}=-g_{t \mu}U^{\mu}= -g_{tt}U^t - g_{t\phi}U^{\phi}= -g_{tt}\dot t - g_{t \phi} \dot \phi$$Here only ##t## component contributes in 2nd step.

Similar mistake in computation of ##L##

EDIT:
Corrected Typos

Last edited:
JD_PM
Hello @Abhishek11235!

Abhishek11235 said:
$$\text{E}= -K_{\mu}U^{\mu}= -g_{\mu \nu}K^{\nu}U^{\mu}=-g_{t \mu}U^{\mu}= -g_{tt}U^t - g_{\phi \phi}U^{\phi}= -g_{tt}\dot t - g_{\phi \phi} \dot \phi$$
Here only ##t## component contributes in 2nd step.
Similar mistake in computation of ##L##

You are absolutely right! I think you made a typo in one of the indices though (?) i.e.

\begin{equation*}
\text{E}= -K_{\mu}U^{\mu}= -g_{\mu \nu}K^{\nu}U^{\mu}=-g_{t \mu}U^{\mu}= -g_{tt}U^t - g_{t \phi}U^{\phi}= -g_{tt}\dot t - g_{t \phi} \dot \phi
\end{equation*}

Abhishek11235
OK so let's start over

The conserved quantities are

E = -g_{tt}\dot t - g_{t \phi} \dot \phi = \left( 1 - \frac{2GM}{\rho} \right) \dot t + \left( \frac{2GMa}{\rho} \right) \dot \phi \tag{1}

L = g_{\phi \phi}\dot \phi + g_{t \phi} \dot t = \frac{\Sigma^2}{\rho^2} \dot \phi + \left( \frac{2GMa}{\rho} \right) \dot t \tag{2}

We need to solve for ##\dot t## and ##\dot \phi## to plug their values into ##\epsilon## (which should be OK as posted in #1).

Hence we solve the system of equations. Performing ##\large \frac{\Sigma^2}{\rho^2} (1) - \frac{2GMa}{\rho} (2)## leads to

\begin{equation*}
\large \dot t = \frac{\frac{\Sigma^2}{\rho^2}E - \frac{2GMa}{\rho}L}{\frac{\Sigma^2}{\rho^2}\left( 1 - \frac{2GM}{\rho} \right)-\left(\frac{2GMa}{\rho}\right)^2} = \frac{\frac{\Sigma^2 E - 2GMa \rho L}{\rho^2} }{\frac{\Sigma^2 \rho - 2GM \Sigma^2 - 4G^2 M^2 a^2 \rho}{\rho^3}} = \frac{\Sigma^2}{\rho^2} \left[ \frac{\rho^3 \left( E - 2MGaL/ \Sigma^2 \right)}{\Sigma^2 \rho - 2GM \Sigma^2 - 4G^2 M^2 a^2 \rho} \right]
\end{equation*}

Plugging this into ##(2)## and solving for ##\dot \phi##

\begin{equation*}
\dot \phi = \frac{\rho^2}{\Sigma^2} \left[ L - 2 GM a \Sigma^2 \left( \frac{E - 2MGaL/ \Sigma^2}{\Sigma^2 \rho - 2GM \Sigma^2 - 4G^2 M^2 a^2 \rho}\right) \right]
\end{equation*}

I am afraid that plugging these into ##\epsilon## is going to look even uglier...

If anyone sees a more convenient way to right them down please let us know.

Meanwhile I'll be trying to get the radial geodesic.

docnet
I do think I am missing a piece of the puzzle.

The radial geodesic is given by

\begin{equation*}
\dot r^2 = \frac{\Delta}{\rho^2} \left[ \left( 1- \frac{2GM}{\rho} \right)\dot t^2 + \frac{4GMa}{\rho}\dot t \dot \phi - \frac{\Sigma^2}{\rho^2} \dot \phi^2 \right]
\end{equation*}

Plugging and chugging with the expressions we have just got for ##\dot t## and ##\dot \phi## looks like a messy idea.

What do you all think?

Last edited:
Abhishek11235
Hello @TSny and @PeterDonis

Sorry for bothering you, but I was wondering if you happened to have time to have a look at this problem.

Thank you

PhDeezNutz
JD_PM said:
OK so let's start over

The conserved quantities are

E = -g_{tt}\dot t - g_{t \phi} \dot \phi = \left( 1 - \frac{2GM}{\rho} \right) \dot t + \left( \frac{2GMa}{\rho} \right) \dot \phi \tag{1}

L = g_{\phi \phi}\dot \phi + g_{t \phi} \dot t = \frac{\Sigma^2}{\rho^2} \dot \phi + \left( \frac{2GMa}{\rho} \right) \dot t \tag{2}
Looks like the sign of the last term in equation (2) is wrong. ##g_{t \phi}## has a negative sign.

I don't see any shortcuts to what you are doing. It seems like a very tedious calculation!

JD_PM
JD_PM said:
This paper works it all out. Getting to equation (4.24) of the paper is straightforward. The cleverness comes in the steps that lead from (4.24) to (4.25). After that, it seems fairly straightforward again. Since Carroll is asking about null geodesics, you can set ##\kappa## in the paper equal to zero.

JD_PM
Thank you @TSny , your help is priceless.

JD_PM said:
I do think I am missing a piece of the puzzle.

What I should have done is defining variables before solving for ##\dot t## and ##\dot \phi## i.e.

\begin{equation*}
E = -g_{tt}\dot t - g_{t \phi} \dot \phi = \underbrace{\left( 1 - \frac{2GM}{\rho} \right)}_{:=A} \dot t + \underbrace{\left( \frac{2GMa}{\rho} \right)}_{:= B} \dot \phi \tag{1}
\end{equation*}\begin{equation*}
L = g_{\phi \phi}\dot \phi + g_{t \phi} \dot t = \underbrace{\frac{\Sigma^2}{\rho^2}}_{:= \ C} \dot \phi - \underbrace{\left( \frac{2GMa}{\rho} \right)}_{= B} \dot t \tag{2}
\end{equation*}

Where

\begin{align*}
\Sigma^2 &= \rho^4 + 2\rho^2 a^2 + a^4 - a^2 \rho^2 + 2GM \rho a^2 - a^4 \\
&= \rho^4 + \rho^2 a^2 + 2GM \rho a^2 \Rightarrow \\
&\Rightarrow \frac{\Sigma^2}{\rho^2} = \rho^2 + a^2 + \frac{2GMa^2}{\rho} = C
\end{align*}

We are now ready to solve the system

\begin{equation*}
C(1) - B(2) \Rightarrow CE - BL = \left( AC + B^2 \right) \dot t
\end{equation*}

\begin{equation*}
B(1) - A(2) \Rightarrow BE - AL = \left( AC + B^2 \right) \dot \phi
\end{equation*}

Where (as shown in the paper) ##AC + B^2 = \Delta##

Now we get much better looking equations for ##\dot t## and ##\dot \phi##

\begin{align*}
\dot t &= \frac{1}{\Delta} \left[ CE-BL \right] \\
&= \frac{1}{\Delta} \left[ \left( \rho^2 + a^2 + \frac{2GMa^2}{\rho} \right)E - \frac{2GMa}{\rho} L \right]
\end{align*}

\begin{align*}
\dot \phi &= \frac{1}{\Delta} \left[ BE-AL \right] \\
&= \frac{1}{\Delta} \left[ \frac{2GMa}{\rho} E + \left( 1 - \frac{2GM}{\rho} \right)L \right]
\end{align*}

TSny said:
The cleverness comes in the steps that lead from (4.24) to (4.25).

I see, so the key idea was rewriting ##\epsilon## in function of the parameters ##A, B, C##. This is something I would have not come up with... as I was not even defining them in the first place! I will certainly learn from the experience.

We get

\begin{align*}
\epsilon &= -g_{tt} \dot t^2 -g_{rr} \dot r^2 -2g_{t \phi} \dot t \dot \phi -g_{\phi \phi} \dot \phi^2 \\
&= \left(1-\frac{2GM}{\rho} \right) \dot t^2 -\frac{\rho^2}{\Delta} \dot r^2 + \frac{4GMa}{\rho}\dot t \dot \phi -\frac{\Sigma^2}{\rho^2} \dot \phi^2 \\
&= A \dot t^2 + 2B \dot t \dot \phi - C\dot \phi^2 - \frac{\rho^2}{\Delta} \dot r^2 \\
&= \underbrace{\left( A \dot t + B \dot \phi\right)}_{E} \dot t + \underbrace{\left( B \dot t - C \dot \phi\right)}_{-L} \dot \phi - \frac{\rho^2}{\Delta} \dot r^2
\end{align*}

Solving for ##\dot r^2## for massless particles (i.e ##\epsilon = 0##)

\begin{align*}
\dot r^2 &= \frac{\Delta}{\rho^2} \left( E \dot t - L \dot \phi \right) \\
&= \frac{1}{\rho^2}\left[ E\left( CE - BL \right) - L\left( BE + AL\right) \right] \\
&= \frac{1}{\rho^2}\left[ CE^2 - 2BLE - AL^2 \right] \\
&= \frac{C}{\rho^2}\left[ \left(E-LW_+ \right)\left(E-LW_- \right) \right] \\
&= \frac{\Sigma^2}{\rho^4}\left[ \left(E-LW_+ \right)\left(E-LW_- \right) \right]
\end{align*}

Where ##W_{\pm}(r)## are solutions to the following equation

\begin{equation*}
CE^2 - 2BE - A = 0
\end{equation*}

\begin{equation*}
W_{\pm}(r) = \frac{B \pm \sqrt{B^2 + AC}}{C} = \frac{1}{C} \left(B \pm \sqrt{\Delta} \right)
\end{equation*}

Nice! To check our answer (as the paper indeed does) it is good practice to explicitly show that our potential reduces to the Schwarzschild potential under ##a \to 0##

The Schwarzschild potential for massless particles is given by (Carroll's equation (5.66)

\begin{equation*}
V(r)= \frac{L^2}{2 r^2} - \frac{GML^2}{r^3}
\end{equation*}

The energy equation for the Kerr radial geodesic is given by

\begin{equation*}
\frac 1 2 \dot r^2 + U(r) = \mathcal{E}
\end{equation*}

Where

\begin{equation*}
U(r) = \frac 1 2 \frac{A}{\rho^2} L^2
\end{equation*}

\begin{equation*}
\mathcal{E} = \frac{E}{2 \rho^2} \left( CE - 2BL\right)
\end{equation*}

Taking ##a \to 0## yields

\begin{equation*}
\frac 1 2 \dot r^2 + \frac 1 2 \frac{A}{\rho^2} L^2 = \frac{E^2}{2}
\end{equation*}

Thus we see that

\begin{align*}
\frac 1 2 \dot r^2 + \frac{1}{2\rho^2}\left( 1 - \frac{2MG}{\rho} \right) L^2 = \frac{E^2}{2} \Rightarrow \\
\Rightarrow \frac 1 2 \dot r^2 + \underbrace{\frac{L^2}{2\rho^2} -\frac{GM}{\rho^3} L^2}_{V(r)} = \frac{E^2}{2}
\end{align*}

So indeed Kerr's potential reduces to Schwarzschild's.

There's a section b) as well

'Using the previous result and assuming ##\Sigma^2 >0## show that the orbit of a photon in the equatorial plane cannot have a turning point inside the outer event horizon ##r_+##'

Where

\begin{equation*}
r_+ = GM + \sqrt{G^2 M^2 - a^2}
\end{equation*}

I better study further (paper, section 4.1.1 and Carroll's section 6.6) before replying.

JD_PM said:
\begin{align*}
\epsilon &= -g_{tt} \dot t^2 -g_{rr} \dot r^2 -2g_{t \phi} \dot t \dot \phi -g_{\phi \phi} \dot \phi^2 \\
&= \left(1-\frac{2GM}{\rho} \right) \dot t^2 -\frac{\rho^2}{\Delta} \dot r^2 + \frac{4GMa}{\rho}\dot t \dot \phi -\frac{\Sigma^2}{\rho^2} \dot \phi^2 \\
&= A \dot t^2 + 2B \dot t \dot \phi - C\dot \phi^2 - \frac{\rho^2}{\Delta} \dot r^2 \\
&= \underbrace{\left( A \dot t + B \dot \phi\right)}_{E} \dot t + \underbrace{\left( B \dot t - C \dot \phi\right)}_{-L} \dot \phi - \frac{\rho^2}{\Delta} \dot r^2
\end{align*}
If we were to stop at the next to last line above, and then try to solve for ##\dot r## we would be calculating ##\dot t^2##, ##\dot t \dot \phi##, and ##\dot \phi^2##. That would be a royal mess.
But the way he factored the expression in the last line saves the day.Yes, it's nice to see that you get back Schwarzschild for ##a## going to zero.

JD_PM
JD_PM said:
I better study further (paper, section 4.1.1 and Carroll's section 6.6) before replying.

OK, there we go.

Actually I did not use paper's approach.

When studying orbits I am really used to first derive the effective potential ##U(\rho=r)## and then plot it to see what kind of orbits it represents.

We have

\begin{equation*}
U(r) = \frac 1 2 \frac{A}{r^2} L^2 = \frac{L^2}{2 r^2} - \frac{GML^2}{r^3}
\end{equation*}

This potential is precisely the same that massless particles experience around Schwarzschild black hole i.e. (page 211 in Carroll's book)

OK we have just checked that the potential ##U(r)## explains the behavior of the photon outside ##r_+## but we are rather interested in the potential inside ##r_+##.

My guess is that we will have to modify the energy equation, defining a new effective potential that describes the behavior of the particle in the region between ##r_-## and ##r_+##...

Do you think this is indeed the way to go? If yes, how could we redefine the potential?

1. What is a radial geodesic in Kerr's metric?

A radial geodesic in Kerr's metric is a path or trajectory that a particle would follow in the spacetime surrounding a rotating black hole, according to the equations of general relativity. It describes the motion of a particle that is moving only in the radial direction, without any angular motion.

2. How is a radial geodesic different from a geodesic in flat spacetime?

A geodesic in flat spacetime is a straight line, while a radial geodesic in Kerr's metric is curved due to the presence of a rotating black hole. In flat spacetime, the curvature of spacetime is zero, while in Kerr's metric, the curvature is determined by the mass and spin of the black hole.

3. What is the significance of the Kerr metric in general relativity?

The Kerr metric is a solution to Einstein's field equations in general relativity that describes the spacetime around a rotating black hole. It is an important solution as it provides a better understanding of the effects of gravity on objects in the presence of a rotating mass, and has been used to make predictions about observed phenomena such as gravitational lensing and the motion of stars around black holes.

4. How does the spin of a black hole affect the radial geodesic in Kerr's metric?

The spin of a black hole affects the shape and curvature of the radial geodesic in Kerr's metric. As the spin of the black hole increases, the curvature of spacetime also increases, causing the geodesic to become more tightly curved. This can result in the geodesic spiraling towards the black hole or being ejected from the black hole's vicinity.

5. Can a particle on a radial geodesic escape from a rotating black hole?

Yes, a particle on a radial geodesic can escape from a rotating black hole if it has enough energy to overcome the gravitational pull of the black hole. This is known as the Penrose process, where a particle can gain energy by falling into the black hole's ergosphere and then escaping with a portion of the black hole's rotational energy. However, the particle must follow a specific trajectory to achieve this, known as a "critical geodesic".

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