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Comment by phkahler | Hacker News Reader

An electron falling (electrostatically) toward a proton will reach the speed of light at some point. This is of course the same distance where inside it would need an escape velocity greater than c. So that's an event horizon due to a different force.

Some claim matter falling into a black hole never really does from the point of view of an outside observer. I've seen weird sounding descriptions like it "spreads out over the surface". What if electron orbitals are some kind of equivalent to that?

When I ask these (admittedly naive) questions, physicists will usually say something like "oh you have to treat that with quantum mechanics". But why? Isn't trying to resolve it using more conventional means (including concepts from relativity) a good idea? I feel like it's not right to reject one approach simply because nobody has figured out how to make it work while another does. That's different from showing that it can't work. Or have such approaches somehow been categorically proven inviable?

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> An electron falling (electrostatically) toward a proton will reach the speed of light at some point.

No, it won't. A correct relativistic analysis of the relative motion of the electron and proton will show their relative speed never reaching c, let alone exceeding it. You can't just plug numbers into Coulomb's Law for this case, because Coulomb's Law by itself is not relativistically correct. You need to use the full Maxwell's Equations and the relativistic Lorentz force law.

> So that's an event horizon due to a different force.

No, it isn't. No force in the relativistic sense produces an event horizon. In relativity, gravity is not a force, it's spacetime geometry, and so is an event horizon in spacetimes where one is present.

> physicists will usually say something like "oh you have to treat that with quantum mechanics".

They are correct in the sense that once the electron and proton get close enough together, classical relativity and Maxwell's Equations are no longer a good model. But as above, you don't need to do that to realize that your claim about reaching the speed of light is wrong.

> You can't just plug numbers into Coulomb's Law for this case, because Coulomb's Law by itself is not relativistically correct.

Sorry if this is a bit pedantic, but as someone trying to study this at the moment, I don't see this the same way and I'd like to validate my interpretation: You can just plug numbers into Coulomb's law, that part is correct. But then the problem of infinite velocities comes from interpreting the 'F' side of the equation, assuming Newton's law (F=ma), rather than using its relativistic counterpart.

Coulomb's law: F = qq'/r^2

Lorentz force law: F = q(E + mu x B)

For the 2 particle case, both of these say the same thing (substitute into the Lorentz eq E = q'/r^2, B = 0 and you get the same thing).

The promotion from non-relativistic to relativistic mechanics is a change of what 'F' means.

nonrelativistically: F = p' = m v' = m x'' = m a

relativistically: F = p' = \gamma m v' = \gamma m x'' = \gamma m a

where \gamma is the Lorentz factor.

Interpreted this way, infinite velocities are avoided.

But, as r->0 we still have an infinity problem - namely infinite energy! This necessitates a quantum mechanical correction to both the Coloumb and Lorentz laws.

TLDR: relativity is necessary when things start to move 'very fast', qm is necessary when things are 'very small'

You can plug arbitrary values in, but you can not expect to gain any valid predictions or reasonable physical insight from Coulomb's law as soon as you are no longer dealing with static point charges. That's because B and E are not independent quantities but actually closely intertwined components of the electromagnetic field strength Tensor F. As soon as you start dealing with motion, these components will mix, preserving only certain quantities like the tensor contraction E^2-B^2. So even if you construct a case where B=0 at time t=0, that will no longer be true once you had any acceleration of your charge carriers.

In the fundamental quantum field theory picture you don't even have forces and particles in the original sense anymore. The dynamics are then described by interaction between the em field and charged fermionic fields. Stuff like Coulomb's law (or any other force potential) only emerges as a macroscopic low energy approximation for specific field configurations.

In a classical view of 2 particles accelerating towards each other v x r will always be 0 so B will always be 0 even if the particles are accelerating towards each other. I believe all this holds under QFT [1].

Looking further a redefinition E is necessary when including the \beta factor [2]. So that was a mistake on my part - relativity does change the rhs of Coulomb's law.

Admittedly the problem as stated (two particles falling towards each other) constrains things in such a way that there is no off-axis contribution. Or to put it another way, 1d electromagnetism doesn't have magnetism.

[1] https://physics.stackexchange.com/questions/142159/deriving-...

[2] https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity

> In a classical view of 2 particles accelerating towards each other v x r will always be 0

Only if the impact parameter is exactly zero. But that is an extremely unlikely edge case.

>I believe all this holds under QFT

It doesn't even hold in ordinary QM (because there are no point charges anymore, only charge densities) and it fails completely in QFT for interacting fields. What you linked is an intro example from many textbooks that shows how the tree level diagram in the nonrelativistic limit can indeed yield the Coulomb potential. At the tree level you will often see such "classical" behaviour. But if you consider higher order corrections, the picture changes rapidly. See the Uhling potential for a practical example, but for higher loops this gets analytically intractable very quickly. The world starts to work very differently once you reach these length scales.

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.

To follow on a bit, the wikipedia article: https://en.wikipedia.org/wiki/Bremsstrahlung links to a paper by Weinberg: https://arxiv.org/abs/1903.11168 and a quick skimming shows that he's perfectly happy to use the Coulomb field as an approximation...

> so the electron is now bound to the proton

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.

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.

> You can just plug numbers into Coulomb's law

Not if you want to get correct answers in problems that involve moving charges.

> relativity is necessary when things start to move 'very fast'

In electromagnetism, relativity is necessary whenever charges are in motion. Whether or not they are moving "ver fast" is irrelevant. Coulomb's Law is only valid when you have a charge at rest.

That said, the post I originally responded to obviously was talking about charges moving "very fast" since it claimed an electron falling towards a proton could exceed the speed of light.

You're trying to solve the two body problem of an electron and a proton classically including relativistic effects. But we know this is not describing reality, because an electron orbiting a proton should radiate energy in form of electromagnetic waves and quickly collapse into the proton. The orbit of an electron in the ground state is well outside the Schwarzschild radius of the proton.

Quantum mechanics successfully explains why the electron does not collapse: because its time evolution is given by the Schrödinger equation. Unlike your idea, it even correctly produces the energy level of the ground state and everything to an astonishing degree.

Quantum mechanics is arguably the most correct theory we ever had, so ignoring it and trying to find an alternative approach is extremely unlikely to work. People may start listening if you can also produce the correct energy level of the ground state.

An electron-proton pair approaching each other will not necessarily form a hydrogen atom by emitting radiation. They could just scatter off each other, and if the impact parameter is large enough, this process could be modeled reasonably well by an analysis using classical relativity. Or, at high enough energy, other particles could be produced, which would require quantum field theory to model.

You can't just reject evidence like this though: electrons orbiting protons don't emit synchrotron radiation. So whatever else you want to the prize, you have to be able to reproduce this result.

> electrons orbiting protons don't emit synchrotron radiation

Not in atoms, no. But I mentioned scattering, which is a different scenario.

And I have absolutely no idea what you think this explains. Which is the point of physics you know: to convey actionable concepts about the behavior of experiments. Not be cryptic and vague.

> I have absolutely no idea what you think this explains.

It explains the case where an electron and a proton scatter off each other and do not form a bound hydrogen atom.

It explains what happens if the electron goes too fast and instead of getting "trapped" it just changes it direction a little.

A problem with trying to use concepts like this and asking "what if?" is that it's reasonning and trying to extrapolate from an analogy

It's one thing to use analogies to guide your intuition, but physical theories are written in the language of math, and not the language of analogies!

You don't have to use QM to describe protons and electrons at a fine level, but it is very hard to do otherwise, because whatever new theory you want to invent would also have to agree with QM on all the experiments where we have observed quantum effects. You can make an even bigger theory, but you can't throw away the existing approach without reinventing most of its results.

You're welcome to try, of course. But be aware you'll need cold hard math, not just high-level ideas

Well said. I think Einstein himself used the same kind of analogous thinking to guide his intuition, and wouldn’t have been nearly as successful as he was without allowing himself this sort of unbounded thinking. In the end, however, he proved these intuitions true or false with the light of math.

Einstein followed where the results lead - his primary breakthrough was assuming no wave-propagation medium was necessary, and accepting an absurd result on it's face by exploring the mathematical consequence (namely: speed of light is constant, as a result the passage of time can vary, there is no medium through which light propagates unlike all other known wave phenomena).

You have grown up with a set of different analogies based on more modern theories: which is to say, you're letting the tail wag the dog: we have those analogies to explain physics to laymen, but the analogies don't inform the physics.

> Some claim matter falling into a black hole never really does from the point of view of an outside observer.

Such claims are wrong. The correct statement is that the outside observer never sees the matter reaching or falling inside the event horizon. But that's not because it never happens; it's because the spacetime geometry prevents light emitted at or beneath the horizon from getting back out to the outside observer.

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...

> our universe also includes QM and that makes BH physics much more subtle

In some ways, yes, but none of that changes the particular thing I said about BH physics.

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.

> An electron falling (electrostatically) toward a proton will reach the speed of light at some point.

Only in Newton's mechanics. With special relativity, it'll approach the lightspeed.

> Some claim matter falling into a black hole never really does from the point of view of an outside observer. I've seen weird sounding descriptions like it "spreads out over the surface".

It doesn't. To an outside observer, the object falling towards the black hole just becomes progressively dimmer and more red, until it disappears.

Even in Newton's mechanic, electron cannot accelerate further after reaching speed of light propagation in the medium. When electron exceeds speed of light it a medium (because of change in medium, for example), it produces Cherenkov radiation.

You can try to make them viable if you want: it's Physics, not Math. You can't really "categorically prove something inviable". But you'll also have to reproduce the results of Quantum Mechanics that predict experimental results to the 11th decimal place.

I think something to keep in mind is humility. In the Bayesian sense it's quite unlikely for you to have picked up something that physicists missed or didn't try to work out before accepting Quantum Mechanics.

If you make a game that is 4D, you can visualize moving 4-dimensionally via a 3 dimensional shadow. See the book Flatland for more concepts like this if this sounds interesting.

Now you can get quite good at predicting where you will end up after a while, and even be able to remember how to get places. But does that mean you are thinking 4 dimensionally? No, you’re still thinking in 3 dimensional shadows.

I get the feeling that is analogous to what happens when you try to do what you’re describing.

I thought it's impossible for any object with non-zero mass to reach the speed of light?

Another layman observation, based on your last paragraph: In terms of electron orbitals, the definition of what quantum mechanics means varies. For example: Are you using quantum mechanics when describing an electron in hydrogen's orbital?

I have heard both answers. It's spread out over space and is not like the classical pre-Bohr models, but it's described by a classical wave equation, and can be viewed at as a differential equation solution; a function over 3D space (For a time snapshot; or 4D spacetime with rotating phase). In this definition, you are not doing quantum mechanics until dealing with things like anti-symmetry, spin statistics, exchange interactions etc.

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