- #1
Bertin
- 11
- 6
- Homework Statement
- Given the affine parameter [itex]\lambda[/itex] of a null geodesic on the equator ([itex]\theta = \pi/2[/itex]), prove that that the [itex]r[/itex] coordinate satisfies the following equation:
$$\left(\frac{dr}{d\lambda}\right)^2 = \frac{\Sigma^2}{\rho^4}(E - L W_-(r))(E - L W_+(r))$$
for some [itex]W(r)[/itex] that might depend on [itex]E,L[/itex] and [itex]r[/itex], and for [itex]E, L[/itex] constants of motion.
- Relevant Equations
- The Kerr metric, in the Boyer-Lindquist coordinates and on the equator, reads
$$ds^2 = -(1 - \frac{R}{r})dt^2 - \frac{R}{r}a (dtd\phi + d\phi dt) + \frac{r^2}{r^2 + a^2 - R r} dr^2 + \frac{\Sigma^2}{r^2} d\phi^2$$
for [itex]\Sigma^2 = r^4 + a^2 r^2 + R r a[/itex].
By the symmetries of the metric, [itex]k = \partial_t[/itex] and [itex]l = \partial_\phi[/itex] are Killing vectors. Since they are Killing vectors, they satisfy [itex]k_\mu \dot{x}^\mu = E[/itex] and [itex]l_\mu \dot{x}^\mu = L[/itex], for the same constants appearing in the expression we must prove, and where the dot means the derivative w.r.t. to the affine parameter. Hence it follows that
$$E = -(1 - \frac{R}{r})\dot{t} - \frac{R}{r}a\dot{\phi}$$
$$L = -\frac{R}{r}a\dot{t} + \frac{\Sigma^2}{r^2}\dot{\phi}$$.
Moreover, since [itex]x(\lambda)[/itex] is a null geodesic, we have that [itex]\dot{x}_\mu\dot{x}^\mu = 0[/itex], whence
$$ 0 = \frac{r^2}{r^2 + a^2 - R r}\dot{r}^2 - (1 - \frac{R}{r})\dot{t}^2 - 2\frac{R}{r}a\dot{t}\dot{\phi} + \frac{\Sigma^2}{r^2}\dot{\phi}^2$$
We can then solve the equations of [itex]E[/itex] and [itex]L[/itex] for [itex]\dot{t}[/itex] and [itex]\dot{\phi}[/itex] to later replace those values inside last equation. Nevertheless, this leads to a very messy expression for [itex]\dot{r}^2[/itex] that does not look that the one we must prove, first and foremost because the resulting expression doesn't seem to include any [itex]\frac{\Sigma^2}{r^4}E^2[/itex] (unless both Mathematica and I are missing a possible simplification, which could be the case), so I probably have done some mistake (not calculatory, though, because my results agree with Mathematica) or I am missing something.
I would appreciate if someone could show me how do we derive above expression. Thank you in advance.
$$E = -(1 - \frac{R}{r})\dot{t} - \frac{R}{r}a\dot{\phi}$$
$$L = -\frac{R}{r}a\dot{t} + \frac{\Sigma^2}{r^2}\dot{\phi}$$.
Moreover, since [itex]x(\lambda)[/itex] is a null geodesic, we have that [itex]\dot{x}_\mu\dot{x}^\mu = 0[/itex], whence
$$ 0 = \frac{r^2}{r^2 + a^2 - R r}\dot{r}^2 - (1 - \frac{R}{r})\dot{t}^2 - 2\frac{R}{r}a\dot{t}\dot{\phi} + \frac{\Sigma^2}{r^2}\dot{\phi}^2$$
We can then solve the equations of [itex]E[/itex] and [itex]L[/itex] for [itex]\dot{t}[/itex] and [itex]\dot{\phi}[/itex] to later replace those values inside last equation. Nevertheless, this leads to a very messy expression for [itex]\dot{r}^2[/itex] that does not look that the one we must prove, first and foremost because the resulting expression doesn't seem to include any [itex]\frac{\Sigma^2}{r^4}E^2[/itex] (unless both Mathematica and I are missing a possible simplification, which could be the case), so I probably have done some mistake (not calculatory, though, because my results agree with Mathematica) or I am missing something.
I would appreciate if someone could show me how do we derive above expression. Thank you in advance.
Last edited: