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I am trying to derive the angular velocity of a circular orbit in Kerr geometry, eqn.(2.16) in Bardeen et al (1972) which reads $$\Omega=\dfrac{1}{r^{3/2}+a}$$ (Note that I am using the units in which $M=1$ and only direct orbits are considered i.e. only the upper sign is used)

MY APPROACH TO THE SOLUTION:

Now using the angular momentum per unit mass $(\mathscr{L})$ and energy per unit mass $(\mathscr{E})$, the conserved specific angular momentum can be defined as $$l=\dfrac{\mathscr{L}}{\mathscr{E}}$$

Using this definition, I had obtained that $$\Omega=\dfrac{d\phi}{dt}=\dfrac{\dot{\phi}}{\dot{t}}=-\dfrac{g_{t\phi}+lg_{tt}}{g_{\phi\phi}+lg_{t\phi}}$$

which for $l=0$ is simply $\Omega=-\dfrac{g_{t\phi}}{g_{\phi\phi}}=\omega$, i.e. the angular velocity that corresponds to frame dragging.

But, I am getting trouble how to find the result as given in the above paper. Any suggestions regarding the approach to derive the equation would be helpful.

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  • $\begingroup$ You need to use that you have a circular orbit, i.e. $\dot{r}=\ddot{r}=0$ $\endgroup$ – mmeent Sep 14 at 15:42
  • $\begingroup$ Did you try this ? $\dfrac{d\varphi }{dt}=\dfrac{d\varphi }{d\lambda}\times \dfrac{d\lambda }{dt}$ $\endgroup$ – Eli Sep 14 at 21:11
  • $\begingroup$ @Eli In the question $\dot{\phi}=\dfrac{d\phi}{d\lambda}$ and $\dot{t}=\dfrac{dt}{d\lambda}$, exactly as you mentioned. $\endgroup$ – Richard Sep 18 at 10:51
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Another solution for you:

consider a circular (r=const) timelike geodetic in the equatorial plane ($\theta=\pi/2,\dot{\theta}=0)$

if you calculate the geodesic equation for the r coordinate you get:

$$\frac{d}{d\lambda}\frac{\partial \mathcal{L}}{\partial \dot{r}}=\frac{\partial \mathcal{L}}{\partial r}=0\tag 1$$

where $\mathcal{L}=\frac{1}{2}\,g_{\mu\nu}\,\dot{x}^\mu\dot{x}^\nu$.

for circular geodesic is $\dot{r}=\ddot{r}=0$

so equation (1) :

$$\frac{\partial g_{tt}}{\partial r}\dot{t}^2+2\,\frac{\partial g_{t\varphi}}{\partial r}\dot{t}\dot{\varphi}+\frac{\partial g_{\varphi\varphi}}{\partial r}\dot{\varphi}^2=0$$

with

$\omega=\frac{\dot{\varphi}}{\dot{t}}$

and

$g_{tt}=-\left(1-\frac{2}{r}\right)$

$g_{t\varphi}=-\frac{2 a}{r}$

$g_{\varphi\varphi}=r^2+a^2+\frac{2 a^2}{r}$

$$\frac{\partial g_{tt}}{\partial r}+2\,\frac{\partial g_{t\varphi}}{\partial r}\omega+\frac{\partial g_{\varphi\varphi}}{\partial r}\omega^2=0\tag 2$$

the solution for $\omega$ give you:

$$\omega=\frac{1}{r^{3/2}+\,a}$$

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    $\begingroup$ Thanks for this answer. This is exactly that for which I asked the question. $\endgroup$ – Richard Sep 20 at 7:41
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Let $u^\mu= (\dot{t},\dot{r},0,\dot{\phi})$ be the four-velocity. Using the conservation of $\mathcal{E}$ and $\mathcal{L}$, you can find expressions for $\dot{t}$ and $\dot{\phi}$. As noted in the question the ratio of these will give the sought after frequency $\Omega$. However, this still depends on $\mathcal{E}$ and $\mathcal{L}$. To find $\mathcal{E}$ and $\mathcal{L}$ for a circular orbit with radius $r_o$ we need to impose two conditions. The first is that $\dot{r}=0$ at the radius $r_o$, the second that $\ddot{r}=0$ at the radius $r_o$. (The first condition by itself, only tells us that $r_o$ is a turning point for the orbit.)

You can find an expression for $\dot{r}^2$ by using that the four-velocity is normalized, $$ g_{\mu\nu}u^\mu u^\nu=-1.$$ You can find an expression for $\ddot{r}$ by taking a (proper) time derivative.

Setting both conditions to zero and solving for $\mathcal{E}$ and $\mathcal{L}$ yields:

$$ \mathcal{E}= \frac{a+(r_o-2) \sqrt{r_o}}{\sqrt{2 a r_o^{3/2}+(r_o-3) r_o^2}}$$ and $$\mathcal{L}= \frac{a^2-2 a \sqrt{r_o}+r_0^2}{\sqrt{2 a r_o^{3/2}+(r_o-3) r_o^2}}.$$ Plugging these back in the expression for $\Omega$ yields

$$\Omega = \frac{\dot{\phi}}{\dot{t}}=\frac{1}{r_o^{3/2}+a}$$

(as derived by Bardeen et al.)

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