- #1
Jimmy Snyder
- 1,127
- 21
Here is a question from problem 26 part a on page 246 of "A First Course in GR" by Schutz. I doubt that it can be answered by someone without a copy of the book, but I have thought that before and been wrong. If someone without the book wants to help and needs more information then of course I would promptly provide it.
Eq. (9.58) in the vacuum region outside the source - i.e., where [itex]S_{\mu \nu} = 0[/itex] - can be solved by separation of variables.
Eq (9.58) (edited) follows:
[tex]
(\nabla^2 + \Omega^2)(\bar{h}_{\mu \nu}e^{i\Omega t}) = 0
[/tex]
Assume a solution for [itex]\bar{h}_{\mu \nu}[/itex] has the form
[tex]
\Sigma_{km}A^{km}_{\mu \nu}f_k(r)Y_{km}(\theta, \phi)/\sqrt{r}
[/tex]
where [itex]Y_{km}[/itex] is the spherical harmonic. (The book uses l as does everyone else on the planet, but I changed l to k so that this post would read more easily).
(a) Show that [itex]f_k(r)[/itex] satisfies the equation:
[tex]
\ddot{f}_k + \frac{1}{r}\dot{f}_k + [\Omega^2 - \frac{(k+\frac{1}{2})^2}{r^2}]f_k = 0
[/tex]
where dot means differentiation with respect to r. Without even trying to solve this problem, my question is simply this: how can k show up in the differential equation? Neither A nor Y are functions of r, and k is just a subscript on f.
Eq. (9.58) in the vacuum region outside the source - i.e., where [itex]S_{\mu \nu} = 0[/itex] - can be solved by separation of variables.
Eq (9.58) (edited) follows:
[tex]
(\nabla^2 + \Omega^2)(\bar{h}_{\mu \nu}e^{i\Omega t}) = 0
[/tex]
Assume a solution for [itex]\bar{h}_{\mu \nu}[/itex] has the form
[tex]
\Sigma_{km}A^{km}_{\mu \nu}f_k(r)Y_{km}(\theta, \phi)/\sqrt{r}
[/tex]
where [itex]Y_{km}[/itex] is the spherical harmonic. (The book uses l as does everyone else on the planet, but I changed l to k so that this post would read more easily).
(a) Show that [itex]f_k(r)[/itex] satisfies the equation:
[tex]
\ddot{f}_k + \frac{1}{r}\dot{f}_k + [\Omega^2 - \frac{(k+\frac{1}{2})^2}{r^2}]f_k = 0
[/tex]
where dot means differentiation with respect to r. Without even trying to solve this problem, my question is simply this: how can k show up in the differential equation? Neither A nor Y are functions of r, and k is just a subscript on f.