# Kepler orbits

1. Nov 14, 2009

### stevebd1

The following is the equation for a Keplerian stable orbit at the equator around a Kerr black hole-

$$\tag{1}v_s=\frac{\pm\sqrt{M}(r^2\mp2a\sqrt{Mr}+a^2 )}{\sqrt{\Delta}(r^{3/2}\pm a\sqrt{M})}$$

where $M=Gm/c^2,\ a=J/mc$ and $\Delta=r^2-2Mr+a^2$

which for a static black hole would reduce to -

$$v_s=\frac{\pm\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}$$

and I would have thought that it was save to assume that for a static black hole-

$$\frac{v_s^2}{r}=\frac{M}{r^2}\sqrt{g_{rr}}$$

which is basically $a_c=a_g$ and where $g_{rr}=(1-2M/r)^{-1}$

rearranging the equation, the following should apply-

$$\frac{v_s^2r}{M}=\sqrt{g_{rr}}$$

but for some reason, the answer I get is simply $g_{rr}$ instead of $\sqrt{g_{rr}}$. Does anyone see why this might be the case?

http://www.iop.org/EJ/article/0067-...uest-id=99398a5d-17c3-4c96-a4c7-516cf1ef178b" for equations (1)

Last edited by a moderator: Apr 24, 2017
2. Nov 14, 2009

### mathman

You should give a step by step derivation. This would help in isolating the problem.

3. Nov 15, 2009

### stevebd1

Hi mathman, thanks for the reply.

Starting from where the stable orbit vs was established in the equatorial plane for a black hole-

$$v_s=\frac{\pm\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}$$

For a stable orbit, centripetal acceleration (ac) has to equal gravity (ag)-

$$a_c=a_g$$

where $a_c=v^2 m_2/r$ and $a_g=G m_1 m_2/r^2$ where v is the tangential velocity of the orbiting object (m2), m1 is the object being orbited and r is the radius of orbit.

rewriting $a_c=a_g$

$$\frac{v^2 m_2}{r}=\frac{G m_1 m_2}{r^2}$$

in the case of an extreme gravitational field, coordinate acceleration needs to be taken into account which for a static BH is-

$$dr'=dr\left(1-\frac{2M}{r}\right)^{-1/2}=dr\sqrt{g_{rr}}[/itex] so in an extreme field, [tex]\frac{v^2 m_2}{r}=\frac{G m_1 m_2}{r^2}\sqrt{g_{rr}}$$

re-arrange relative to $\sqrt{g_{rr}}$

$$\frac{v^2 r}{G m_1 }=\sqrt{g_{rr}}$$

substitute v for the equation for vs and replace Gm1 with M (geometric units) where M is the gravitational radius (as derived in post #1)-

$$\left(\frac{\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}\right)^2\frac{r}{M}=x$$

expand and cancel out-

$$\frac{r^2}{(r^2-2Mr)}=x$$

x should equal $\sqrt{g_{rr}}$ but instead equals $g_{rr}$. I'm beginning to suspect that the equation for centripetal acceleration needs to take into account relativity in some respect.

______________________________________________

There seems to be the suggestion that centripetal acceleration goes into reverse around BH's (and in SR in general), this may be equivalent to dividing ac by $\sqrt{g_{rr}}$ for a stable orbit which would explain the answer to the above equation.

one source- http://arxiv.org/abs/0903.1113v1

There is also the suggestion that ac becomes negative beyond the event horizon. Considering the change over from positive to negative ac would be smooth, it might be said that ac reduces relative to $1/\sqrt{g_{rr}}$ (or simply $\sqrt{g_{tt}}$) becoming zero at the event horizon and then negative at r<2M. It would also mean that ac was reasonably unaffected at r>20M.

Last edited: Nov 15, 2009
4. Nov 15, 2009

### mathman

I have to confess that I am lost when it comes to the details of gen. rel. However, I am confused about what you are doing in the middle of the derivation.

You have v2m2/r=one expression at one point and something different a few lines later. I don't understand what you are doing.

5. Nov 16, 2009

### stevebd1

I mathman, I think I may have answered my own question. It appears that centripetal acceleration also needs to be modified in GR. The reason why I change the derivation in post #3 is that I had initially assumed that $v^2 r/G m_1=\sqrt{g_{rr}}$, where v is the tangential velocity of the rotating object, would be correct but when introducing the result for v from the Kepler equation, this wasn't the case and I changed $\sqrt{g_{rr}}$ to x, x turning out to be simply $g_{rr}$. It appears a correct solution for centripetal acceleration that works with the Kepler equation for a static black hole is-

$$a_c=\sqrt{g_{tt}}\,\frac{v^2 m_2}{r}$$

where $g_{tt}=(1-2M/r)$

which means that centripetal acceleration begins to reduce significantly at approx. 20M, becoming zero at the event horizon and negative within the EH.

If we introduce vs from the Kepler stable orbit equation, we can say-

$$\sqrt{g_{tt}}\,\frac{v_s^2 m_2}{r}=\frac{G m_1 m_2}{r^2}\sqrt{g_{rr}}$$

where $g_{rr}=(1-2M/r)^{-1}$

The idea of ac becoming negative beyond the EH is also discussed to some degree in https://www.physicsforums.com/showthread.php?t=10369".

Last edited by a moderator: Apr 24, 2017