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How close to a black hole can an object orbit elliptically? I know circular orbits are no longer stable at distance less than 3 times the Schwarzschild radius. But what about elliptical orbits? Can an object have a semi-major axis or perihelion at distance of less than 3 times the Schwarzschild radius?

blademan9999
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2 Answers2

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A bound elliptical orbit around a Schwarzschild black hole must have $r > 2 r_s$ at all times (where $r_s = 2 M$ is the Schwarzschild radius). Deriving this result is a good exercise for students learning about the Schwarzschild geometry, so I won't go through all the details, but the basic sketch of the proof is as follows:

  • Recall that a massive particle moving in a Schwarzschild geometry is equivalent to a particle moving in a classical "effective potential" given by $$ V_\text{eff}(r) = - \frac{M}{r} + \frac{\ell^2}{2 r^2} - \frac{M\ell^2}{r^3}, $$ where $M$ is the mass of the black hole and $\ell$ is the specific angular momentum of the particle.
  • Note that for a bound orbit, we must have $V_\text{eff}(r) < 0$ at all times.
  • Find the points at which $V_\text{eff}(r) = 0$ for a given value of $\ell$. This will be the closest possible value of perihelion for a bound orbit for a particular value of $\ell$.
  • Find the value of $\ell$ that allows for the closest perihelion. It turns out to be $\ell = 4M$, and for that value of the angular momentum you must have $r > 2 r_s$ to satisfy $V_\text{eff}(r) < 0$.
Michael Seifert
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  • @safesphere : You have misparsed the text of the Answer. In "can get arbitrarily close to $r=2r_s$ (where $r_s=2M$ is the Schwarzschild radius) but no closer", the phrase "but no closer" is never interpreted as modifying the parenthetical; it modifies "can get arbitrarily close to $r=2r_s$", which is $4M$. – Eric Towers Nov 17 '21 at 07:24
  • Tangentially related: A mass in close orbit will radiate gravitational waves; is that energy loss relevant close to the minimal radius and does it therefore lead to a decay of such an orbit? – Peter - Reinstate Monica Nov 17 '21 at 12:26
  • @safesphere: I rephrased the first sentence; hopefully it's clearer now. – Michael Seifert Nov 17 '21 at 12:32
  • @Peter-ReinstateMonica: You're right that this would cause a gradual decay. But the rate of decay also depends on the orbiting mass (I believe it's proportional to $m^2$). I suspect that so long as $m \ll M$, the decay would still be rather slow even for an orbit that gets to $r \approx 2r_s$. But I will admit I don't have a definitive proof to back that up. It might be a good question to ask separately! – Michael Seifert Nov 17 '21 at 16:05
  • How smalle can the semi-major axis be? – blademan9999 Nov 19 '21 at 09:49
  • @blademan9999: I don't know an obvious way to define semi-major axis for a black hole spacetime. The easiest way I can think of is the average of the $r$ coordinates at perihelion and aphelion. But the $r$ coordinate doesn't directly measure "distance from the center" in a black hole spacetime due to the curvature of space, so it's not clear to me whether this quantity would be physically meaningful. That said, the orbit I described that gets to $r = 2 r_s$ is just barely bound, meaning its "aphelion" is very far away; any meaningful definition of the semi-major axis for it would be infinity. – Michael Seifert Nov 19 '21 at 12:51
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Yes, highly eccentric orbits can go deeper. (Kostic 2012) derives analytic solutions in terms of elliptic functions, noting (in footnote 4) that the closest approach would be 2 Schwarzschild radii out for a $l=2$ orbit. Such orbits may not be very elliptic-looking since there can be multiple turns per perimelasma approach (OK, periapse is the more common term).

Anders Sandberg
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    Note that $l$ here is the "reduced angular momentum", which Kostic defines as the specific angular momentum divided by the Schwarzschild radius. (Just in case the reader is confused about how your answer compares with mine.) – Michael Seifert Nov 16 '21 at 14:39

  • perimelasma

    what a lovely word!

     – neph Nov 16 '21 at 23:24
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    But this ignores radiating away energy through gravitational waves(?). And circularise the orbit(?). – Peter Mortensen Nov 16 '21 at 23:47
  • "Such orbits may not be very elliptic-looking since there can be multiple turns per perimelasma approach (OK, periapse is the more common term)" So, it'd look like a spirograph drawing, then? – nick012000 Nov 17 '21 at 10:58
  • @PeterMortensen - See https://physics.stackexchange.com/questions/458444/on-what-timescale-does-gravitational-wave-emission-circularise-an-orbit for the circularisation issue. Basically, this is a pretty slow process for non-heavy bodies. – Anders Sandberg Nov 17 '21 at 17:26
  • @nick012000 - Yes, there are spirograph-like precessing orbits, but also more flower-like ones: https://physics.stackexchange.com/questions/46332/what-happens-to-orbits-at-small-radii-in-general-relativity – Anders Sandberg Nov 17 '21 at 17:29