- #1
JD_PM
- 1,131
- 158
- 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!
\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!