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The "Orbital motion" section of the Wikipedia entry corresponding to Schwarzschild metric reads:

A particle orbiting in the Schwarzschild metric can have a stable circular orbit with $r > 3r_s$. Circular orbits with $r$ between $1.5r_s$ and $3r_s$ are unstable...

  1. What does justify this instability?

Some follow-up questions:

  1. What would change if one consider a big object, say, a soccer ball, instead of a particle?

  2. Does the same result hold for rotating (Kerr) black holes?

Qmechanic
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  • Did you read https://en.wikipedia.org/wiki/Schwarzschild_geodesics ? – Alex Trounev May 04 '19 at 10:25
  • @AlexTrounev: That page somehow answers the first question. Though, I couldn't find anything related to the rest of my questions. –  May 04 '19 at 10:33
  • @AlexTrounev What about equatorial circular orbits? "Kerr black holes: II. Precession, circular orbits, and stability" (http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec27.pdf), or chapter 13 in Hobson et al's General Relativity: An Introduction for Physicists (2006) – Chiral Anomaly May 04 '19 at 20:45
  • @ChiralAnomaly Yes you are right. It is necessary to add: no circular orbits, excluding the equatorial plane. And then we immediately answer the third question: circular orbits in the Kerr metric exist only in the equatorial plane. The "stable" orbits of the particle depend on the metric parameter and can reach $r_{max}=M$, whereas in the Schwarzschild metric, $r_{max} = 3M$ (note $r_s = 2M$). See Figure 7 and section "Gravitational radiation" on https://authors.library.caltech.edu/14972/1/Ruffini2009p1645Phys_Today.pdf and eq (4.49) on http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap4.pdf – Alex Trounev May 04 '19 at 22:40
  • Consider to only ask 1 subquestion per post. – Qmechanic May 12 '19 at 13:41

1 Answers1

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Regarding your first question:

In general relativity the energy of a light object moving around a spherically symmetric mass can be written as:

$E=mc^2\left(\dfrac{\sqrt{1-\dfrac{2GM}{rc^2}}}{\sqrt{1-\dfrac{v^2}{c^2\left({\left(1-\dfrac{2GM}{rc^2}\right)}^2 {\left(\hat{r}\cdot\hat{v}\right)}^2+\left(1-\dfrac{2GM}{rc^2}\right) {|\hat{r}\times\hat{v}|}^2\right)}}}\right) $

The strange part above is because in the Schwarzschild solution the local velocity of light is different in the radial direction and transverse to the radial direction. For pure circular motion this expression reduces to:

$E=mc^2\left(\dfrac{\sqrt{1-\dfrac{2GM}{rc^2}}}{\sqrt{1-\dfrac{v^2}{c^2 \left(1-\dfrac{2GM}{rc^2}\right)}}}\right)$

This can be rewritten as:

$E=mc^2\left(\dfrac{{1-\dfrac{2GM}{rc^2}}}{\sqrt{1-\dfrac{2GM}{rc^2}-\dfrac{v^2}{c^2}}}\right)$

For pure stable circular motion we also have, just as classically, $v=\sqrt{GM/r}$ so we can write:

$E=mc^2\left(\dfrac{{1-\dfrac{2GM}{rc^2}}}{\sqrt{1-\dfrac{2GM}{rc^2}-\dfrac{GM}{rc^2}}}\right)$

From the last expression it can be seen that infinite energy is needed to stay in a circular orbit at the photon radius $r=\frac{3GM}{c^2}=1.5r_s$. By taking the derivative of the last expression with respect to r we see that it has a minimum for $r=\frac{6GM}{c^2}$. This means that as long as you are "higher up" than $r=\frac{6GM}{c^2}=3r_s$ circular orbits require less energy to sustain as you go closer to the black hole. However, closer to the black hole than that radial distance, circular orbits require more and more energy to sustain the closer you go to the black hole, and that is why they are unstable.

PointedEars
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Agerhell
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    Agerhell wrote: "the local velocity of light is different in the radial direction and transverse to the radial direction" - the local velocity of light is c in every direction, it's the shapirodelayed coordinate velocity measured by the far away observer that is different in different directions. – Yukterez Jul 15 '23 at 02:39