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September 13

The speed of electricity

How fast does electricity move? I've heard it said that electricity moves at the speed of light. If this is the case then surely there is something else, other than light, that moves at the speed of light. Does this not contravene the rules of physics? Thank you — Preceding unsigned comment added by 2A00:23C6:6884:6200:75D7:FCD:5331:8FC7 (talk) 09:02, 13 September 2020 (UTC)[reply]

Have you looked at Speed of electricity? When people say that electric current moves at the speed of light (even though that is not entirely accurate), what they mean is that the wave/signal travels at that speed; the electrons or ions that transport the electric charges move at a much slower rate. Anything with a non-zero mass that moves at the speed of light would violate the laws of physics, but electric current has no inherent mass, so it does not violate any physical law, even if it would move at the speed of light. - Lindert (talk) 11:21, 13 September 2020 (UTC)[reply]

It's the speed of light, in the conductor (if they weren't opaque). The linked article Velocity factors gives typical speeds. For copper it's around 2/3rds of the speed of light in a vacuum, or 200,000 km/sec. LongHairedFop (talk) 13:14, 13 September 2020 (UTC)[reply]

The "speed of light" is not inherently about light at all, which makes the name somewhat poor. (Like many names of things in science, it caught on by happenstance, and now we just have to put up with it.) Special relativity asserts that no two points in spacetime can exchange information faster than c. It just so happens that c is also the speed at which electromagnetic waves (including light) travel in a vacuum. Since "electricity" is an electromagnetic wave (or at least can be modeled as one), it will travel at the "speed of light" in whatever thing it's moving through; this is always less than c if it's not a vacuum. Your confusion is common and completely understandable, resulting from the somewhat misleading name. I suggest this video for more detail. --47.146.63.87 (talk) 07:02, 14 September 2020 (UTC)[reply]

black hole collision

[1] If two black holes collide and merge into a bigger black hole, during the approach does that at least temporarily create a region from which no light can escape, but whose shape is something like a figure 8 or dumbbell? And is that in conflict with the no-hair theorem that says black holes are always spherical? Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 15:27, 13 September 2020 (UTC)[reply]

Yes, it creates a toroid [2]. As for whether that violates the no hair theorem, I don't know. Handschuh-^{talk to me} 00:53, 14 September 2020 (UTC)[reply]

Only non-spinning black holes have a spherical event horizon. The no-hair theorem doesn't assert "black holes are spherical"; it just asserts "stable" black holes can be fully described by eleven parameters. Also, again, it only applies to "stable" black holes. --47.146.63.87 (talk) 06:48, 14 September 2020 (UTC)[reply]

Does this mean that the Einstein–Maxwell equations of gravitation and electromagnetism in general relativity only apply to stationary situations? If so, the text of the articles concerned should make that clear. --Lambiam 07:53, 14 September 2020 (UTC)[reply]

The equations apply to any situation, of course, stationary or not. But the idea was that the no-hair "theorem" (it's a conjecture, strictly speaking) only applies to stationary solutions of the equations. I thought so until five minutes ago, but apparently that's not correct. This paper says that the frequencies and decay rates of the overtone modes of the ringdowns signal are uniquely determined by the final hole’s mass and spin magnitude. Hence, while you can talk about non-spherical shapes during ringdown that does not give you any new information beyond mass and spin. The situation already applies to the stable Kerr black hole: the event horizon is non-spherical, so you can measure an ellipticity of the black hole. But again, this tells you nothing new as the ellipticity is uniquely determined by the spin. In that sense the no-hair theorem applies to the ringdown phase as well. --Wrongfilter (talk) 08:51, 14 September 2020 (UTC)[reply]

Continuity considerations suggest compellingly that there has to be a transitional phase between the event horizons of the merging black holes being a pair of perfect spheroids before the merger, and the event horizon being a single spheroid after the merger. This is, however, excluded by the formulation given by the No-hair theorem article. So which one is it: continuity but additional conditions needed for the theorem to apply as stated, or a physical discontinuity? --Lambiam 15:31, 14 September 2020 (UTC)[reply]

The transitional phase is called ringdown, and from the cited paper I gather that the way ringdown proceeds is uniquely determined by total mass and total spin (I wonder whether the mass ratio shouldn't have an influence, though). But nobody said anything about discontinuity. --Wrongfilter (talk) 16:56, 14 September 2020 (UTC)[reply]

Presumably nobody said anything about discontinuity because there is none. But if during the ringdown period, however brief, "any distortion in the shape is dissipated as more gravitational waves",^[3] the no-hair theorem is not unconditionally true in its stated form. (Some black holes do ring, others with exactly the same mass, electric charge, and angular momentum do not.) --Lambiam 17:35, 14 September 2020 (UTC)[reply]

Well, I only cited a peer-reviewed paper, so... --Wrongfilter (talk) 17:52, 14 September 2020 (UTC)[reply]

And yet another way to look at the ring down is that it is just an echo of the gravitational wave lensed by the black hole. You can expect infinite echos, getting weaker and weaker as light or gravitation waves passing very close in get bent more and more, and exceed 360°, 720° and even more extreme spiralling around the final black hole. I may have written the bit Lambiam quoted, but it may not be the right way to look at ringdown. Note that you will never observe an event horizon or its formation, and that the black hole (s) have not yet actually been formed, as they only do so in our infinite future. Timescales are seriously distorted. Graeme Bartlett (talk) 01:07, 18 September 2020 (UTC)[reply]

(Yes, you did). We lack the terminology to properly discuss black holes in English, so any such exposition going into detail is in some way wrong or misleading. I maintain that the idea that a pair of perfect spheroids "instantaneously" merges into one perfect spheroid is more misleading than others. This idea stems from a popular formulation of the no-hair theorem, which therefore is apparently also misleading. But all will become clear in our infinite future. --Lambiam 07:19, 18 September 2020 (UTC)[reply]