5 ms·

> 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 rel

by pdonis 2y ago

> 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.

greysphere 2y ago
> 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 2y ago
  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 2y ago
    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-the-coulomb-force-equation-from-the-idea-of-virtual-photon-exchange https://physics.stackexchange.com/questions/142159/deriving-... [2] https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity
    - sigmoid10 2y ago
      >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.

by pdonis 2y ago

greysphere 2y ago

> 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 2y ago
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 2y ago
  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-the-coulomb-force-equation-from-the-idea-of-virtual-photon-exchange https://physics.stackexchange.com/questions/142159/deriving-... [2] https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity
  - sigmoid10 2y ago
    >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.