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I have been reading a lot of QED books lately, and understand (as well as possible anyway) the interaction between electrons and photons. But I can't seem to get a clear indication of the interaction between photons and protons. It seems that normal light (not talking about high-energy levels or anything exciting, just the stuff that comes out of a light bulb) would be insufficient to really produce a reflection, but, so far, it depends upon who I ask.

That said, to boil down what I am really trying to determine: Would otherwise-normal atoms (or matter, really) with no electrons be visible? Would the protons take up the role normally provided by the electrons and cause a similar scattering of light, or would it really just mess things up?

Jeff Davis
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    What are "otherwise-normal atoms without their electrons" like? Are they like controlled nuclear fusion, without the control? – Emilio Pisanty Nov 12 '12 at 19:28
  • @EmilioPisanty I suppose the question is only valid if we somehow also removed the charge that causes protons to repulse each other. But that would be the same charge which results in the light interaction, so it would again invalidate my original question! So it seems that stable protons would require electrons to also be present, which means that once again, no matter how you slice it, the protons would not be able to maintain any sort of visibility. – Jeff Davis Nov 12 '12 at 19:52

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Ordinary light has far too little energy to significantly affect protons. But gamma rays are the result of interactions between protons, neutrons and photons in an unstable nucleus (i.e., a radioactive atom).

Normal atoms without their electrons are positively charged and would not form ordinary matter but an exploding gas. Most ordinary experience would become invalid.

Arnold Neumaier
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  • Ok. Thank you. I am mostly interested in the phenomena of the visibility of matter. I am digging into the implications of the Higgs field. There are MANY things that would happen if it weren't there, and it seems me that since electrons would simply fly off (since they would be massless), one implication would be that everything would then be invisible (not to mention, like you said, exploding, but that's a separate implication). – Jeff Davis Nov 12 '12 at 19:21
  • @JeffDavis: Massless electrons would still be bound to nuclei by their charge, not fly away. - Visibility requires a working eye, which probably needs standard QED with massive electrons. – Arnold Neumaier Nov 12 '12 at 19:25
  • Ok. I was basing my understanding on what is said in http://www.amazon.com/Higgs-Discovery-Kindle-Single-ebook/dp/B008LUYZFM/ref=sr_1_3, in particular, this quote: "The Higgs mechanism is responsible for elementary particle masses, such as the mass of the electron. Mass is what provides resistance when a force is applied. If particles have no mass, they travel at the speed of light. A particle's mass tells us how it responds to forces and how it travels through space" Am I misunderstanding what she is saying? – Jeff Davis Nov 12 '12 at 19:32
  • although the working eye example is another way of coming at this problem, that is helpful as well. – Jeff Davis Nov 12 '12 at 19:39
  • I want to leave my question here in case you come back to answer it at some point, but I realized that both can be true. The massless electrons could travel at the speed of light and still be "captured" by any atom they came close enough to. But this begs the question: Would a massless electron interact with light at all? – Jeff Davis Nov 12 '12 at 20:29
  • Also found this: http://cms.web.cern.ch/news/why-would-i-care-about-higgs-boson "In a simple hydrogen atom for example, the radius of the electron's orbit is inversely proportional to the electron’s mass. A massless electron would therefore be at infinity from the proton, not allowing atoms to form at all." – Jeff Davis Nov 12 '12 at 20:34
  • @JeffDavis: Of course, electrons in massless QED also interact with photons, in just the same way. But I never checked whether atoms would be stable in massless QED,. – Arnold Neumaier Nov 13 '12 at 07:56
  • Note that for massless electrons one can no longer use the nonrelativistic treatment of hydrogen, so the extrapolation from the masive electron case is not necessarily correct. – Arnold Neumaier Nov 13 '12 at 10:12
  • You mean we haven't unified everything already? (Joking, of course) Thanks! – Jeff Davis Nov 13 '12 at 14:10
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I will take the example of a proton beam in an accelerator. Your question becomes:

will light scatter off a proton beam

The answer is yes, though one would need very sensitive equipment to catch the scattered photons since the density of the beam is nowhere near the density of outer electrons on a surface. It is calculable, but people are not interested for the very low energy photons of the visible spectrum.

anna v
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  • So, it seems that you could see "something," maybe something very faint, but it wouldn't be as clear or as well defined as what you get with electrons. Does that sound about right? – Jeff Davis Nov 12 '12 at 19:16
  • Yes, in my opinion. But it would be similar to the reflection from an electron beam of the same density as far as definition goes. – anna v Nov 12 '12 at 19:21
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The main thing that makes electrons, rather than nuclei, be the primary interlopers with electromagnetic radiation in an atom is their much smaller mass. This is because for a given electric field, the acceleration on a nucleus will be far smaller than on the electron, so that to a very good approximation we can picture the nucleus as static.

Thus one alternative question you can ask (since we're doing counterfactual physics) is, what would atom-light interactions be like if the electron and proton masses were comparable? This is very tricky to answer on one hand because if the ratio $m_e/m_p$ were not small then the Born-Oppenheimer approximation breaks down completely and we would have to completely re-write all of atomic and molecular physics.

On the other hand, at least in the lighter atoms (so the nucleus isn't 300 times heavier than electrons by dint of having so many nucleons), it is quite clear that the nucleus would also play an important role in the interaction dynamics. The way this would come about is in the calculation of atomic dipole moments, which would have to include an expectation value over the nuclear coordinates as well as the electronic ones (corresponding, physically, to the nucleus vibrating together with the electrons on atomic transitions).

Emilio Pisanty
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  • Thank you for illuminating the discussion. I now remember reading about how the nucleus does play a part, albeit a very small one. Like you said, we can pretty much discard it for most purposes. But that does hint at the crux of the issue. – Jeff Davis Nov 12 '12 at 20:23