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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 n

• cyberax 2y ago
  > 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.
  • oneshtein 2y ago
    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.
• the__alchemist 2y ago
  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.
• mr_mitm 2y ago
  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.
  • pdonis 2y ago
    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.
    • XorNot 2y ago
      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.
      • pdonis 2y ago
        > electrons orbiting protons don't emit synchrotron radiation Not in atoms, no. But I mentioned scattering, which is a different scenario.
        • XorNot 2y ago
          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.
• 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'
    • 2y ago
• cheschire 2y ago
  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.
• pdonis 2y ago
  > 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.
  • tgarrett 2y ago
    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) https://en.wikipedia.org/wiki/Firewall_(physics) and we can't state things with such certainty...
    • pdonis 2y ago
      > 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.
      • tgarrett 2y ago
        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.
        • pdonis 2y ago
          > 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.
• tux3 2y ago
  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
  • IAmGraydon 2y ago
    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.
    • XorNot 2y ago
      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.
• csomar 2y ago
  I thought it's impossible for any object with non-zero mass to reach the speed of light?
• deleted 2y ago
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• drowsspa 2y ago
  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.