[greysphere]

> 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'

[sigmoid10]

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.

[greysphere]

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

[sigmoid10]

>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.

[pdonis]

> 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.

[pdonis]

> 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.

[tgarrett]

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.

[tgarrett]

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...

[greysphere]

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.

[pdonis]

> 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.

[tgarrett]

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...

[pdonis]

> 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.