- #1
latentcorpse
- 1,444
- 0
Hi.
I'm trying problem 1 on p138 of this
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
Now when I try and get the Euler Lagrange equation for [itex]\phi[/itex] I get
(the Kerr metric in BL coordinates can be found at the bottom of p77)
[itex]\frac{\partial L}{\partial \phi} = \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{\tau}} \right)[/itex]
[itex]0= \frac{d}{d \tau} \left( 2 \frac{(r^2+a^2)^2- \Delta a^2 \sin^2{\theta}}{\Sigma} \sin^2{\theta} \frac{d \phi}{d \tau} - \frac{2a \sin^2{\theta} ( r^2 + a^2 - \Delta ) }{\Sigma} \frac{dt}{d \tau} \right)[/itex]
But the question says we are in the equatorial plane and so when I impose [itex]\theta=\frac{\pi}{2}[/itex] all the sine terms cancel and I'm left with [itex]0=\frac{d}{d \tau} (0)[/itex] which is useless - what am I doing wrong here?
I'm trying problem 1 on p138 of this
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
Now when I try and get the Euler Lagrange equation for [itex]\phi[/itex] I get
(the Kerr metric in BL coordinates can be found at the bottom of p77)
[itex]\frac{\partial L}{\partial \phi} = \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{\tau}} \right)[/itex]
[itex]0= \frac{d}{d \tau} \left( 2 \frac{(r^2+a^2)^2- \Delta a^2 \sin^2{\theta}}{\Sigma} \sin^2{\theta} \frac{d \phi}{d \tau} - \frac{2a \sin^2{\theta} ( r^2 + a^2 - \Delta ) }{\Sigma} \frac{dt}{d \tau} \right)[/itex]
But the question says we are in the equatorial plane and so when I impose [itex]\theta=\frac{\pi}{2}[/itex] all the sine terms cancel and I'm left with [itex]0=\frac{d}{d \tau} (0)[/itex] which is useless - what am I doing wrong here?