latentcorpse
Feb27-11, 04:23 PM
Show that the extreme Reissner Nordstrom metric in isotropic coordinates is
ds^2=-(1+\frac{M}{\rho})^{-2}dt^2 + ( 1+ \frac{M}{\rho})^2 ( d \rho^2 + \rho^2 d \Omega^2)
I have done this using the substitution
r=\rho + M + \frac{M^2-Q^2}{4 \rho}=\rho + M since M= | Q | for the extreme case
The next part of the question asks me to verify that \rho=0 is an infinite proper distance from any finite \rho along any curve of constant t.
Now I had a bit of a problem here. I had to also assume that \theta, \phi are constant as well. This then gave
ds=( 1 + \frac{M}{\rho} ) d \rho
s=\int_0^\rho (1 + \frac{M}{\rho} ) d \rho = \rho + M \ln{\rho} -M \ln{0} = \infty since \ln{0}=-\infty
Does anyone know how to do this without having to make the extra assumption that d \Omega=0?
Then I am asked to verify that |t| \rightarrow \infty as \rho \rightarrow 0 along any timelike or null curve but that a timelike or null ingoing radial geodesic reaches \rho=0 in finite affine parameter.
I have no idea how to do this. What do I take as the form of my curve?
Thanks.
ds^2=-(1+\frac{M}{\rho})^{-2}dt^2 + ( 1+ \frac{M}{\rho})^2 ( d \rho^2 + \rho^2 d \Omega^2)
I have done this using the substitution
r=\rho + M + \frac{M^2-Q^2}{4 \rho}=\rho + M since M= | Q | for the extreme case
The next part of the question asks me to verify that \rho=0 is an infinite proper distance from any finite \rho along any curve of constant t.
Now I had a bit of a problem here. I had to also assume that \theta, \phi are constant as well. This then gave
ds=( 1 + \frac{M}{\rho} ) d \rho
s=\int_0^\rho (1 + \frac{M}{\rho} ) d \rho = \rho + M \ln{\rho} -M \ln{0} = \infty since \ln{0}=-\infty
Does anyone know how to do this without having to make the extra assumption that d \Omega=0?
Then I am asked to verify that |t| \rightarrow \infty as \rho \rightarrow 0 along any timelike or null curve but that a timelike or null ingoing radial geodesic reaches \rho=0 in finite affine parameter.
I have no idea how to do this. What do I take as the form of my curve?
Thanks.