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Will electrons, and other particles, also loose energy as they travel through the cosmos? They have wavelengths. Do they get "stretched"? My guess is that the EM force, somehow, counteracts this effect. What about neutrinos?

Reference found: http://books.google.com/books?id=AmlEt6TJ6jAC&pg=PA96&lpg=PA96&dq=redshift+of+de+broglie+wavelength&source=bl&ots=oAW0q9bmaj&sig=nAwE-ohARZ3YTCZrHDBBjZ_IYfQ&hl=en&sa=X&ei=0lQDVJjtLMKEjAL80IHQCw&ved=0CB4Q6AEwAA#v=onepage&q=redshift%20of%20de%20broglie%20wavelength&f=false

yalis
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    Are you asking if the de Broglie wavelength of matter will red-shift and effectively make the matter lose mass or momentum? – Jim Aug 30 '14 at 19:28
  • So I never considered this. I don't see why it wouldn't stretch the de Broglie wavelength, which would decrease the object's peculiar velocity. I looked around but I didn't see any mention of that sort of effect. Perhaps someone knows off-hand if that is an effect. It could be an interesting idea to explore. I doubt there would be much in the way of observational evidence. But it could be good for a fun weekend or something. – Jim Aug 30 '14 at 20:29
  • You don't have to appeal to quantum mechanics for this. This is a purely classical fact about general relativity. Cosmologists generally talk about cosmological models as having a component of "dust," i.e., nonrelativistic matter, where a dust particle is some object such as a galaxy or cluster of galaxies. During cosmological expansion, the velocities of dust particles relative to the Hubble flow have systematically decreased. This is a good thing for us -- otherwise we'd live among galaxies colliding at relativistic speeds. You can calculate the de Broglie wavelength of a galaxy if you like. –  Aug 31 '14 at 23:55

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The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse of the scale factor.

Neutrinos are an example of a particle with a non-zero mass (maybe of order 0.1 eV - see http://adsabs.harvard.edu/abs/2014PhRvL.112e1303B ). They decouple from the rest of the Universe at about 1 second and freely propagate. The expansion then reduces their momenta to the extent that they should have a temperatures $<2$K in the present-day Universe, typical kinetic energies of 0.2 meV (e.g. see http://adsabs.harvard.edu/abs/2010PhRvD..82f2001K ) and may have speeds of only (depending on their actual masses) perhaps $\sim 10^3-10^4$ km/s and so are non-relativistic.

Electrons would behave in the same way, if they could be considered not to be strongly (or rather electromagnetically!) interacting with other particles and photons. I don't think this can be satisfied except in the very early universe and a typical free, intergalactic electron in the present day universe has an energy of $\sim 0.1$ keV due to heating by radiation from stars and galaxies.

ProfRob
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  • Something seems off about this answer. I reread it and I can't put my finger on it but each time I go "OK, yeah, hmmmm still feels weird". You got some references for all those numbers? Perhaps reading those will ease this feeling. (not the 1s number, I'm a cosmologist, so I'm very familiar with numbers about decoupling) – Jim Aug 30 '14 at 21:20
  • The 2K is a well-known result, but assumes massless neutrinos. If the neutrino mass is >1meV then they will become non-relativistic. I think current estimates are higher ( http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.112.051303 ). If the neutrino mass is indeed of order 0.1eV, at $T \sim 2K$ they have speeds $<2\times 10^7$ m/s. – ProfRob Aug 30 '14 at 23:00
  • And if the effect did not occur, wouldn't the neutrinos still have energies of $>1$ MeV and be easily detectable? – ProfRob Aug 30 '14 at 23:27
  • @Jim: "Something seems off about this answer. I reread it and I can't put my finger on it [...]" -- I believe that I can put my finger on something specificly "off" with Rob Jeffries' answer:
    Rob Jeffries: "Neutrinos are an example of a particle with a non-zero mass. They decouple from the rest of the Universe at about 1 second and freely propagate. The expansion then reduces their momenta [...]" -- Why should a neutrino whose momentum (magnitude) and/or speed decreases in the course of a trial be called "freely propagating"??     – user12262 Aug 31 '14 at 11:15
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    @user12262 Freely propagating in exactly the same sense as a CMB photon is freely propagating and has a decreasing momentum; subject only to the expansion of the universe and not other interactions. – ProfRob Aug 31 '14 at 11:21
  • @RobJeffries and user12262 no, that wasn't what felt off. The answer seems right, I can't find anything wrong with it (not that I've looked), I think it's that no initial velocities are given and so giving the final velocities gives no information about the loss of momentum – Jim Aug 31 '14 at 13:18
  • Oh, I figured out what it was. That we would expect to observe a nearly isotropic value for the order of magnitude of the speed of a massive particle created 1s after inflation ended. The CMB makes sense because photons are massless, but neutrino aren't so an isotropic temperature and velocity profile doesn't make sense to me. Not saying it's wrong, just that it feels weird and I'd feel much better after reading something about it – Jim Aug 31 '14 at 13:31
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    Found a good explanation also here: http://books.google.com/books?id=AmlEt6TJ6jAC&pg=PA96&lpg=PA96&dq=redshift+of+de+broglie+wavelength&source=bl&ots=oAW0q9bmaj&sig=nAwE-ohARZ3YTCZrHDBBjZ_IYfQ&hl=en&sa=X&ei=0lQDVJjtLMKEjAL80IHQCw&ved=0CB4Q6AEwAA#v=onepage&q=redshift%20of%20de%20broglie%20wavelength&f=false – yalis Aug 31 '14 at 18:32
  • Rob Jeffries: "Freely propagating in exactly the same sense as a CMB photon is freely propagating and has a decreasing momentum; subject only to the expansion of the universe and not other interactions." -- Well, yes, that's what I meant by "being off": trying to answer a question by appealing to a notion (here "free-ness") whose description seems even more complicated than the question itself. Of course: certain attempts at "laying foundations of RT", such as those by EPS, are at least as "off". – user12262 Aug 31 '14 at 21:38
  • @yalis: "Found a good explanation also here: ["Principles of Physical Cosmology", Phillip James, Edwin Peebles; §5: "The Expanding Universe"; p. 96]" -- Of particular importance there seems the real number "ratio between observed to emitted wavelengths". How should the value of such a ratio be determined, in any one particlar trial, at least in principle ?? – user12262 Aug 31 '14 at 21:52
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    @user12262 The idea of a non-interacting particle seems quite straightforward to me. It has a simple wavefunction, and a wavelength and this wavelength is increased by the expansion of the universe. Perhaps I just don't see the subtleties. Jim Decoupling takes place at about $kT=2$ MeV so the lost momentum is about 2 MeV/c. Why would the average speed by anisotropic? – ProfRob Aug 31 '14 at 22:42
  • Another way of saying this is that their peculiar momenta with respect to a local volume decrease as the inverse of the scale factor. Particles don't have velocity relative to a region of space. Do you mean their velocity relative to the Hubble flow? –  Aug 31 '14 at 23:34
  • The expansion then reduces their momenta to the extent that they should have a temperatures <2K in the present-day Universe and may have speeds of only (depending on their actual masses) perhaps ∼103−104 km/s and so are non-relativistic. No, if this were true then, e.g., we could determine the masses of neutrinos by detecting photons from neutrino-antineutrino annihilation. In reality, cosmic neutrinos are all ultrarelativistic. This tells us that they are not in thermal equilibrium with the rest of the universe at 2 K. –  Aug 31 '14 at 23:42
  • Rob Jeffries: "The idea of a non-interacting particle seems quite straightforward to me." -- Apparently it should be possible to distinguish "non-interacting particle" from "interacting particle"; this requires some further definition (in, hopefully, even more straightforward terms). Generally: notions of dynamics would be defined in terms of notions of geometry/kinematics. "It has a simple wavefunction, and a wavelength and this wavelength is increased by the expansion of the universe." -- Alright; so what is (how to measure and compare) "wavelength" ?? (Should be asked first.) – user12262 Sep 01 '14 at 05:37
  • @Ben Crowell "No, if this were true then, e.g., we could determine the masses of neutrinos by detecting photons from neutrino-antineutrino annihilation. In reality, cosmic neutrinos are all ultrarelativistic.". Please provide a reference. This contradicts everything I have read. Relic neutrinos have a Fermi-Dirac distribution characteristed by a 1.95K temperature. They have typical energies of 0.2 milli-eV, so if their masses exceed this they are non-relativistic. http://adsabs.harvard.edu/abs/2010PhRvD..82f2001K – ProfRob Sep 01 '14 at 07:15
  • @user12262 "requires some further definition (in, hopefully, even more straightforward terms). ". How about - has not undergone any strong, weak or electromagnetic interaction that significantly changed its momentum between decoupling and the present day. – ProfRob Sep 01 '14 at 08:10
  • @RobJeffries Didn't say it would be anisotropic, just don't see any reason for it to be isotropic. I'm not saying I think this is wrong, it just feels off to me. But you seem to have your hands full with a couple people who do think it's wrong. So deal with that and maybe in the meantime that weird feeling of mine will go away. – Jim Sep 01 '14 at 15:38
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    @Jim I'm not cosmologically informed to argue much about isotropy. I see no reason why initially the C$\nu$B shouldn't be as isotropic as the CMB. As the $\nu$ become non-rel. they will become anisotropic as they fall into potential wells. This affects structure formation and offers a route to determining neutrino masses. Two more excellent presentations and a paper I have been looking at are http://www-physics.lbl.gov/seminars/old/Petr_Vogel.pdf and http://darkuniverse.uni-hd.de/pub/Main/WinterSchool08Slides/CosmologicalNeutrinos.pdf and http://adsabs.harvard.edu/abs/2014PhRvL.112e1303B – ProfRob Sep 01 '14 at 16:14
  • @user12262 The book referenced above by yalis also uses the term "freely moving particles" (p.95). If it's good enough for Peebles, it will do for me. – ProfRob Sep 01 '14 at 16:18
  • The notion of "free propagation" here is the same as it's meaning in ordinary mechanics or quantum mechanics: it means that we can neglect the forces acting on the particle. Nothing mysterious at all. The changing of the scale factor is not a force in the conventional sense and does not interfere with understanding the term in the usual way. – dmckee --- ex-moderator kitten Sep 01 '14 at 16:29
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    @RobJeffries: Oops, sorry, I was wrong about the cosmic neutrino background. Undid the -1. –  Sep 01 '14 at 16:48
  • @dmckee: "The notion of "free propagation" here [...] Nothing mysterious at all." -- Well, as far as I understand there are profound questions (if not mysteries) being addressed in investigations such as "Characterizability of Free Motion in Special Relativity", U. Schelb, FP 30(6):867-892 (2000); and they seem even more difficult outside SR. "The changing of the scale factor is not a force in the conventional sense [...]" -- In order to prove this assertion you'd have to consider in detail the definitions of (how to measure) both these quantities. – user12262 Sep 05 '14 at 04:59