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I never heard of Schwarzschild diameters, only of Schwarzschild radii. Why is it like this? Wouldn't it be better to portray the (possible) size of a black hole in a diameter?

Either way, can we use the terms radius or diameter term precisely? Is an event horizon always perfectly spherical or can it have something like an equatorial bulge even though it's a black hole's immaterial event horizon.

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    $\begingroup$ A sphere is most simply described by means of radius, not by diameter: volume = 4/3 pi r^3. Also radius is a good choice in polar coordinates which are suitable to describe the non-spherical yet rotational symmetric deviations from a perfect sphere for rotating black holes $\endgroup$ Nov 10, 2020 at 7:25
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    $\begingroup$ Distance (& coordinates) in general relativity is a complicated topic. Especially near black holes! It's better to think of the Schwarzschild radius (of a non-rotating black hole) as a way to measure the circumference or area of the event horizon (using the usual formulae), rather than thinking of it as the distance to the centre of the black hole. There are numerous questions on this topic on the Physics stack, but they tend to get fairly technical very quickly. $\endgroup$
    – PM 2Ring
    Nov 10, 2020 at 10:01
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    $\begingroup$ Changing the conventions or units would have no practical effects. Why is it important to you? $\endgroup$
    – D. Halsey
    Nov 10, 2020 at 15:30
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    $\begingroup$ Our usual notions of geometry are practically useless near a black hole. It's bad enough near a SMBH (super-massive black hole) but near a tiny BH the spacetime curvature changes very rapidly. A rigid body can't get near a small BH without being spaghettified (but you can cross the horizon of a SMBH before getting spaghettified). According to vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator the tidal force from one atom to its neighbour atom in normal matter near a 1" radius BH is ~14 trillion g (Earth gravities). No hand or tape measure can cope with that. $\endgroup$
    – PM 2Ring
    Nov 11, 2020 at 0:44

2 Answers 2

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We use the Schwarzschild radius $r_s$ (rather than a diameter) because it's convenient. We're want to describe what happens in the vicinity of a black hole, so it's natural to talk about the distance from the black hole.

For example, there's the photon sphere at $3r_s/2$ (for a Schwarzschild BH), and the innermost stable circular orbit or ISCO, the smallest circular orbit in which a test particle can stably orbit a massive object, at $3r_s$

The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk. [...] Circular orbits are still possible between the ISCO and the photon sphere, but they are unstable. For a massless test particle like a photon, the only possible circular orbit is exactly at the photon sphere, and is unstable. Inside the photon sphere, no circular orbits exist. [...] The case for rotating black holes is somewhat more complicated. 

Another phenomenon which involves the distance from a BH is gravitational lensing. The amount of deflection of a light beam passing near a BH (or indeed any source of gravity) depends on the distance $r$ of the beam's path from the centre of mass. The deflection angle is given by $\theta=2r_s/r$. I guess you could use diameter in that formula, but it's more natural to think in terms of radius. It's a bit like shooting at a target (with light beams). When you shoot at a circular target, you're concerned with how far your shots are from the centre of the target.


As I said in the comments, our usual notions of geometry are not very useful in the highly curved spacetime near a BH. And if you want to get close to a BH, you need to be in an orbit travelling near the speed of light, and / or your spacecraft needs to be capable of the extremely high acceleration required to hover near a BH (and you & your ship need impossible strength to cope with the insanely huge g force resulting from such acceleration). Travelling at high speed adds additional relativistic effects on top of those GR effects due to the BH's gravity.

To describe the locations and times near the black hole, you need to choose a coordinate system. Schwarzschild coordinates are often used to describe a non-spinning black hole, but they have a coordinate singularity at the EH (event horizon), so they're annoying to work with when you want to talk about objects crossing the EH. It's a lot like how the latitude & longitude on Earth break down near the poles. You can't travel further south than the South Pole at 90°S, and the pole itself doesn't have a well-defined longitude. If I tell you that I'm 1 km north of the South Pole, I could be anywhere on a circle of 1 km radius.

Fortunately (as I mentioned in the comments), GR is very flexible regarding coordinates, and there are quite a few standard coordinate systems used in various circumstances, you can see a list of some of them on Wikipedia's page Category:Coordinate charts in general relativity. However, that flexibility makes things complicated, and difficult to describe properly without using advanced mathematics. So popular treatments of GR gloss over those difficulties (or avoid them completely), and that has led to a popular understanding of GR elements that is somewhat distorted.

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  • $\begingroup$ I believe photons don't exist. Light is an immaterial wave, consisting of itself, not of particles. The photon theory doesn't make any sense to me, and noone ever imaged photons in a microscope. $\endgroup$
    – John
    Nov 11, 2020 at 6:37
  • $\begingroup$ @John Modern science doesn't try to say what's really going on in the world, it leaves that job to the philosophers. Instead, it takes an indirect approach. It develops theories that attempt to explain some aspects of the world, and within that theoretical framework it develops models that are consistent with existing empirical measurements / observations and are hopefully consistent with future observations. So we aren't claiming that photons (or any other particles) truly exist, they're simply an element of our currently most successful theory of electromagnetic phenomena. $\endgroup$
    – PM 2Ring
    Nov 12, 2020 at 21:55
  • $\begingroup$ (cont) Photons are not like classical particles. They do have some properties that are reminiscent of such particles, and other properties that are quite unclassical. But all of that isn't really relevant to what I said in my answer. Light can orbit a black hole at the photon sphere, but such orbits are unstable. This "photon sphere theory" is consistent with our observations of actual black holes. $\endgroup$
    – PM 2Ring
    Nov 12, 2020 at 21:56
  • $\begingroup$ Perhaps because the spacetime gets curved light behaves like it does around compact stars. I think Einsteinian physics can explain it without photons, in contrast to Newtonian gravity. $\endgroup$
    – John
    Nov 13, 2020 at 7:41
  • $\begingroup$ @John Maybe I don't get what you're trying to say. Light gets gravitationally deflected in the vicinity of stars, but it can't orbit a normal star: their radius is too large compared to their Schwarzschild radius. (As Wikipedia says, it's possible that some neutron stars have a photon sphere. I expect that's most likely for the more massive neutron stars when they are new, and still have a high spin). $\endgroup$
    – PM 2Ring
    Nov 13, 2020 at 10:57
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The diameter of a black hole (the distance "through" it) or the radius (distance to the center) aren't really meaningful concepts. The singularity is in the future, not in any spatial direction. The $r$ Schwarzschild coordinate is actually a so-called "reduced circumference" – the metric length of a circle around the hole, divided by $2π$ – so if you wanted to replace $r$ by a value with more physical meaning, it would make more sense to multiply it by $2π$ than by $2$. The problem with doing either of these things is that they make the Schwarzschild metric uglier. You end up having to divide by $2$ or $2π$ again every time the coordinate appears, so it's just $r$ with extra steps.

The event horizon of an ideal eternal black hole is exactly spherical, even if the hole is rotating. In fact you can't detect the rotation locally near the event horizon (perfect frame dragging). The event horizon of more realistic holes isn't exactly spherical, but it's very close to spherical most of the time.

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    $\begingroup$ I have accepted the other answer but yours is good and interesting too, and it answers my 2nd question. $\endgroup$
    – John
    Nov 11, 2020 at 6:35

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