Kerr Black Hole & Keplerian Stable Orbit

[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/0067-0049/112/2/423/35032.pdf?request-id=99398a5d-17c3-4c96-a4c7-516cf1ef178b" for equations (1)

 

[mathman]

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

 

[stevebd1]

Hi mathman, thanks for the reply.

Starting from where the stable orbit v_s 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 (a_c) has to equal gravity (a_g)-

$$a_c=a_g$$

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-

[tex]dr'=dr\left(1-\frac{2M}{r}\right)^{-1/2}=dr\sqrt{g_{rr}}[/itex]<br /> <br /> so in an extreme field,<br /> <br /> $$\frac{v^2 m_2}{r}=\frac{G m_1 m_2}{r^2}\sqrt{g_{rr}}$$<br /> <br /> re-arrange relative to $\sqrt{g_{rr}}$<br /> <br /> $$\frac{v^2 r}{G m_1 }=\sqrt{g_{rr}}$$<br /> <br /> substitute v for the equation for v_s and replace Gm₁ with M (geometric units) where M is the gravitational radius (as derived in post #1)-<br /> <br /> $$\left(\frac{\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}\right)^2\frac{r}{M}=x$$<br /> <br /> expand and cancel out-<br /> <br /> $$\frac{r^2}{(r^2-2Mr)}=x$$<br /> <br /> 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.<br /> <br /> ______________________________________________<br /> <br /> 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 a_c by $\sqrt{g_{rr}}$ for a stable orbit which would explain the answer to the above equation.<br /> <br /> one source- <a href="http://arxiv.org/abs/0903.1113v1" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://arxiv.org/abs/0903.1113v1</a>There is also the suggestion that a_c becomes negative beyond the event horizon. Considering the change over from positive to negative a_c would be smooth, it might be said that a_c 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.[/tex]

 

[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.

 

[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 v_s 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 a_c becoming negative beyond the EH is also discussed to some degree in https://www.physicsforums.com/showthread.php?t=10369".