Plasma physicist here, I haven't tried 5.4 yet, but in general I am very impressed with the recent upgrades that started arriving in the fall of 2025: for tasks like manipulating analytic systems of equations, quickly developing new features for simulation codes, and interpreting and designing experiments (with pictures) they have become much stronger. I've been asking questions and probing them for several years now out of curiosity, and they suddenly have developed deep understanding (Gemini 2.5 <<< Gemini 3.1) and become very useful. I totally get the current SV vibes, and am becoming a lot more ambitious in my future plans.
If we don't need plasma physicists anymore then we probably have fusion reactors or something, which seems like a fine trade. (In reality we're going to want humans in the loop for for the forseeable future)
In a hypothetical pure GR universe what you're saying is correct, but our universe also includes QM and that makes BH physics much more subtle, e.g.: https://en.wikipedia.org/wiki/Firewall_(physics) and we can't state things with such certainty...
In pure GR an infalling observer will sail past the EH and not notice anything unusual since spacetime is locally Minkowski (ignoring tidal forces, which is valid e.g. for humans falling into supermassive BHs). If the (GR+QM) firewall hypothesis is correct (a big if), an infalling observer will instead be promptly incinerated within a Planck's length of the EH. The intuition one builds from a pure GR understanding of BHs may be dramatically wrong, not just at the singularity, but all the way out at the EH.
> If the (GR+QM) firewall hypothesis is correct (a big if)
A big if indeed, but if that hypothesis is correct, then the GR solution that applies is no longer the standard black hole solution. The "firewall" is not vacuum--more precisely, it does not have a vanishing stress-energy tensor. Which means "the intuition one builds from a pure GR understanding" for the "firewall" case will need to be a pure GR understanding of a different solution from the standard BH, and of course such an understanding can be perfectly correct.
In other words, if you're going to talk about a "firewall" solution, then saying "well, GR doesn't model that correctly because it's not a standard GR black hole" is simply wrong. GR can model lots of other things besides standard (vacuum solution) black holes. You just have to use the correct GR model for the actual stress-energy tensor that is present. Of course statements about a standard vacuum black hole will not be correct for a different non-vacuum solution; but that is not contradicting anything I said, because the post I was responding to was assuming a standard vacuum black hole, and my statement was correct for that case.
The real question is whether such a "firewall" model, with a nonzero stress-energy tensor, would even have an event horizon. As far as I know nobody has actually answered that question; the treatments I have seen have simply assumed that there is one without taking into account the fact that the "firewall" stress-energy tensor is non-vanishing. If there is an event horizon in such a model, then my statement would still be correct for that model, since my statement was based on general properties of event horizons.
Hi greysphere, you are definitely correct that one primary thing preventing velocity of the electron from exceeding than the speed of light is the presence of gamma in the relativistic force law, aka \partial_t (m_e \gamma v ) = q_e(E + v \times B), although the LHS doesn't quite equal \gamma m_e a, since \gamma also depends on v...
In general I think it's fine to use Coulomb's law as an approximation in this case because the proton is much heavier than the electron and so we can just stay in the proton's reference frame and let the electron fall in from infinity (and we're ignoring QM and just doing relativistic EM here). We could also switch to a tritium nucleus and make it a bit better of an approximation, or indeed add a whole bunch more neutrons and get lucky that they don't beta decay to make it an arbitrarily good one. It is true that if the proton starts moving that you will no longer have a pure Coulomb field with respect to the original reference frame, as after a Lorentz boost the E field gets squished into the transverse direction somewhat, and you'll gain a B field swirling around the proton...
Staying with the frozen proton approx, if we plug numbers in we get quite a bit of energy: set the proton radius r_p to 1E-15, and we get U = q_e^2 / ( 4 \pi \eps_0 r_p ) ~ 1.4 MeV, or a gamma of about 4, so yeah, it would be moving faster than c if we stayed with Newtonian mechanics. But there's another wrinkle: the 1.4 MeV of liberated potential energy won't all go into the electron's relativistic kinetic energy, because it is accelerating like crazy, especially in the final femtometers, and that acceleration (essentially Bremsstrahlung, although its not braking here) will generate an intense pulse of EM radiation as well - a decent fraction of the 1.4 MeV will go into that instead. You could perhaps estimate how much using the Larmor formula (in general calculating this radiation reaction force precisely becomes very complex, because the excitation of the EM wave modifies the acceleration, which modifies the excitation of the EM wave etc... And, now looking on Wikipedia, I'm not surprised to see that the first QM version of the calculation was done by Sommerfeld).
So yeah, the electron will zip through the proton, with much of the potential energy converted to an EM pulse that zips off to infinity, and so the electron is now bound to the proton, and will continue to zig zag back and forth, emitting more radiation until it comes to a rest inside the proton. So yeah, we do need QM after all.
Thanks for the explanation of some of the interactions I was missing! It's amazing the complexity of what's basically the simplest setup one could think of.
This is extremely unlikely unless the relative motion is very slow, or, to put it another way, the total center of mass energy is very close to the rest energy of electron + proton, so there is a significant probability amplitude for capture into a bound hydrogen atom.
The post I originally responded to was obviously not considering such a case since it claimed the electron could exceed the speed of light.
If the electron starts with zero energy at infinity (e.g. a parabolic orbit, a natural default assumption), and some of the potential energy is converted into free EM radiation due to acceleration of the electron as it is falling down the potential well, then it will become bound to the proton. My reading of phkahler's original statement is that the electron will wind up going faster than the speed of light (which is incorrect, due to gamma) due to falling down the potential well, and not due to having non-zero kinetic energy at infinity...
> If the electron starts with zero energy at infinity...it will become bound to the proton
Even if that's true (I'm not sure it always is--see below), that case is extremely rare. A much more common case is Bremsstrahlung, which you mentioned upthread--and as the Wikipedia article you referenced notes, the electron in this process starts out free and remains free after the radiation is emitted; it does not become bound to the proton.
> My reading of phkahler's original statement is that the electron will wind up going faster than the speed of light (which is incorrect, due to gamma) due to falling down the potential well, and not due to having non-zero kinetic energy at infinity...
That may have been the original intent, yes (and, as you note, it's wrong because it neglects the gamma factor). However, even in that case, what matters is not the electron's energy at infinity in the proton's rest frame, but its energy in the center of mass frame. If the electron is really falling in from far enough away that the relativistic gamma factor is relevant, which is what was implied by pkahler's original statement, then its energy in the center of mass frame (or more precisely the center of momentum frame, since in relativity you have to take momentum and energy into account) will be relativistic, i.e., large enough that it's by no means guaranteed that it will emit enough energy in radiation to become bound to the proton.
