Radial Geodesic in Kerr's metric

[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 appreciate your help.

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!

 

[Abhishek11235]

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

Similar mistake in computation of $L$

EDIT:
Corrected Typos

 

[JD_PM]

Hello @Abhishek11235!

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

 

[JD_PM]

OK so let's start over

The conserved quantities are

\begin{equation}
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}
\end{equation}

\begin{equation}
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}
\end{equation}

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.

 

[JD_PM]

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?

PS: here's a PDF about https://www.roma1.infn.it/teongrav/leonardo/bh/bhcap4.pdf that may be helpful.

 

[JD_PM]

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

 

[TSny]

OK so let's start over

The conserved quantities are

\begin{equation}
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}
\end{equation}

\begin{equation}
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}
\end{equation}

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!

 

[TSny]

PS: here's a PDF about https://www.roma1.infn.it/teongrav/leonardo/bh/bhcap4.pdf that may be helpful.

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.

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*}

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.

 

[JD_PM]

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.

 

[TSny]

\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]

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?