Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does a particle's kinetic energy vary with a(t)?

  1. May 1, 2015 #1
    I am curious about how a particle's kinetic energy changes with the explansion of the unverse. I calculated that for a matter dominated universe it would vary as 1/a2, while the photon temperature varies as 1/a. Is this correct? Is this also true in general? If so, the implication would be that assuming equlibrium at the time of recombination, the temperature of the atomic hydrogen gas that dominates bryonic matter would about 0.003 oK. Would this be detectible by finding very narrow fraunhoffer lines of light passing through this gas?
     
  2. jcsd
  3. May 1, 2015 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Kinetic energy relative to what?

    I can't tell because I don't know what this kinetic energy is supposed to be relative to. What kind of particles are you assuming, what state of motion are they in, and what is the kinetic energy being measured relative to?

    This is certainly not correct. The temperature of the universe at the time of recombination was about 3,000 K. Since up to that point matter and radiation were in thermal equilibrium, that was the temperature of both. But this temperature is not calculated based on calculation of kinetic energy of a particle; it is simply the observed temperature of the CMBR today, 3 K, times the redshift factor of the CMBR, which is about 1000.
     
  4. May 1, 2015 #3
    My bad. I should have said that the kinetic energy is relative to the comoving coordinates of the explansion. I am assuming the particles are about 75% hydrogen atoms, 23% He atoms, ans 2% all other baryonic matter paricles formed by nuclear fusion during the early expansion, but the same relationship between kinetic energy and a(t) should also apply to dark matter particles.
     
    Last edited: May 1, 2015
  5. May 1, 2015 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    A comoving particle is at rest in the comoving coordinates of the expansion, so its kinetic energy in those coordinates is zero.

    In our cosmological models, we don't model the matter or radiation or dark matter or dark energy as individual particles; we model it as a fluid, with an equation of state relating pressure and energy density; "matter", "radiation", and "dark energy" simply correspond to different equations of state. The key relationship for each kind of fluid is how its energy density ##\rho## changes with the scale factor ##a##: for matter (ordinary or dark), we have ##\rho \propto 1 / a^3##; for radiation, we have ##\rho \propto 1 / a^4##; and for dark energy, we have ##\rho = constant##. But this energy density is not "kinetic", because the fluid is at rest in comoving coordinates; it is rest energy density, i.e., the energy density in the fluid rest frame.
     
  6. May 1, 2015 #5

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    I believe momentum falls off as 1/a. The kinetic energy of primoridal neutrinos for example would be falling off about the same way as the CMB temperature.

    I could be wrong of course. You might look up the Cosmic Neutrino Background
    http://en.wikipedia.org/wiki/Cosmic_neutrino_background
    analogous to the CMB except that "recombination" happened earlier and the temp is just a bit lower, around 2K.

    I'll try to check a little bit. I understand that Weinberg treats this in his cosmology textbook.
    Expansion drains particles of their momentum.
    Always relative to "comoving observers" of course. Momentum and kinetic energy as seen by observers at CMB rest.
     
    Last edited: May 1, 2015
  7. May 1, 2015 #6

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    https://www.physicsforums.com/threads/expansion-drains-momentum-from-massive-objects.249589/
    An alternative to looking it up in Steven Weinberg's textbook:

    http://arxiv.org/abs/0808.1552
    Note on the thermal history of decoupled massive particles
    Hongbao Zhang
    (Submitted on 11 Aug 2008)
    "This note provides an alternative approach to the momentum decay and thermal evolution of decoupled massive particles. Although the ingredients in our results have been addressed in [Weinberg's new Cosmology text], the strategies employed here are simpler, and the results obtained here are more general."

    Yes this confirms that momentum falls off as 1/a

    So it would seem that for slower moving particles (non-relativistic) the kinetic energy would fall off as 1/a2 as you conjectured!
     
    Last edited: May 1, 2015
  8. May 1, 2015 #7

    Chalnoth

    User Avatar
    Science Advisor

    The average kinetic energy of the particles in the co-moving frame is given by the temperature. When self-gravitation becomes important, there's no simple formula for how this temperature varies over time. When self-gravitation isn't important (i.e., before any bound objects form), the change in temperature should be an adiabatic process.
     
  9. May 1, 2015 #8
    I was not assuming the particles moved with the comoving coordinates. I assumed that a particle's velocity was relative to these coordinates.

    This is how I calculated the change in velocity.

    Assume that at time t1 a particle is at some point P1 in space and, P1 moves with the comoving coordinates, and at time t2 it is at another such point, P2. If v is the average velocity between P1 and P1, the distance the particle has moved is (t2-t1) v.

    During the period of time a has increased from a(t1) to a(t1).

    For relatively small t2-t1, h(t2) ~= (a(t2)-a(t1)) / (a (t2-t1)).

    The Hubble redshift velocity between these two points, corresponding to the distance (t2-t1) v, is H(t2) v (t2-t1).

    For a matter dominated universe, a = const t2/3, t = const a3/2,
    and H(t) = 2/(3t).

    Therefore the Hubble redshift velocity is 2/(3 t2) v (t2-t1). This represents the reduction in the particle's velocity v(t1)-v (t2) .

    Let t = t2, dt = t2-t1. Also let dv = v(t2)-v(t1).

    (1/v) dv = -(2/3) 1/t dt

    Integrating:
    log(v) = log(const) - 2/3 log(t) = log (const t-2/3) .

    Exponetiating:
    v = const t-2/3 = const/a .

    So, if the velocity (or momentum) varies inversely with a, the kinetic energy would vary inversely with a2.
     
    Last edited: May 1, 2015
  10. May 1, 2015 #9
    Hi marcus:

    Thank you for your post agreeing with my conclusion. I have downloaded the Zhang paper and plan to study it.

    Do you know if astronomers could detect this phenomenon by finding extremely small widths for Fraunhoffer lines in light passing through a relatively small regions of primordial gas?
     
  11. May 1, 2015 #10
    Hi again marcus:

    If a neutrino has rest mass, then its toral energy should be reduced to an extremely small amount more than its rest mass if its kinetic energy varies with a2.

    Do you know what assumptions are made (for example: a nutrino has zero rest mass an travels at light speed) when the current density for neutrinos is given as about the same value as for photons? (Please see https://en.wikipedia.org/wiki/Cosmic_neutrino_background .)
     
  12. May 1, 2015 #11
    Hi Chalnoth:
    As I have come to understand (with much uncertainty) an adiabatic process assumes equilibrium, and the universe had not been in a state of equilibrium since at least the time of recombination, and possibly not since nuclear synthesis. Am i wrong about this?
     
  13. May 1, 2015 #12

    Chalnoth

    User Avatar
    Science Advisor

    To a very good approximation, the universe prior to the emission of the CMB was in thermal equilibrium at each point in time. This early expansion was approximately adiabatic. I'm not completely sure how to extend this to non-interacting particles (e.g. neutrinos, dark matter), but I doubt it changes too much.

    This approximation breaks down pretty dramatically at later times, when gravitationally-bound systems start to form.
     
  14. May 1, 2015 #13

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    But this is because neutrinos are relativistic--they're not quite massless, but they're close to it, so their equation of state is very similar to that of radiation. That means their energy density falls off like that of radiation, i.e., as ##1 / a^4##. But the energy density of non-relativistic matter falls off as ##1 / a^3##.

    I understand the argument for momentum falling off as ##1 / a##, but I'm still trying to reconcile that, and the corresponding conjecture that kinetic energy should fall off as ##1 / a^2##, with the behavior of the energy density of matter and radiation cosmological fluids, which I described above.
     
  15. May 1, 2015 #14
    Hi Chalnoth:

    I am currently reading a very well written book: An Introduction to Cosmology by Derek Raine and Ted Thimas (2001). This source says that a very large majority of the universe's baryonic matter remains primoridal and not influenced by the gravity of galaxies, clusters of galaxies and super clusters. Also, the expansion of the universe does not effect these structures, For structures larger than super cluseters, the expnasnsion pushes these structures apart, and the greatest mass of primordial stuff is among these 4th order structures.
     
  16. May 1, 2015 #15

    Chalnoth

    User Avatar
    Science Advisor

    The kinetic energy for non-relativistic particles is a negligible component of their total energy. So if you're interested in how the kinetic energy of non-relativistic particles changes over time, it's not useful to examine their total energy.
     
  17. May 1, 2015 #16

    Chalnoth

    User Avatar
    Science Advisor

    I'm not sure what the ratios are, but I believe it.

    One issue is that within clusters, most of the baryonic matter consists of the diffuse cluster gas whose temperature is essentially set by the depth of the gravitational well. That temperature is so vastly higher than the temperature of the gas between clusters (it emits in the X-ray range) that the impact of expansion on temperature for such gas is negligible.

    It might be reasonable to use this sort of analysis to get an idea of the temperature of the matter that remains between galaxies and galaxy clusters, but for most purposes that temperature is going to be essentially zero. You can probably get a rough idea by assuming adiabatic expansion, though.
     
  18. May 1, 2015 #17

    Chalnoth

    User Avatar
    Science Advisor

    Just bear in mind that the intercluster medium is now ionized (and has been since enough stars turned on at about z=20 or so), so it should interact with the CMB. That complicates things a bit.
     
  19. May 1, 2015 #18

    Chalnoth

    User Avatar
    Science Advisor

    Actually, now that I think of it, the ionization of the intercluster medium indicates that it absorbs quite a bit of energy from stars, meaning its temperature is likely much higher than that of the CMB, or than you would naively expect from looking at expansion alone. I wasn't able to find any decent sources on this with a quick search, but my bet is that getting at the details of how temperature of baryonic matter has changed over time is going to be really, really difficult. For dark matter, the difficulty there is that we don't know the temperature at early times, so it's hard to say how that changes (in the simplest model, which seems to fit the data pretty well, the temperature is identically zero).
     
  20. May 1, 2015 #19

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Ah, of course; the ##1/a^3## is just the number density being diluted by the expansion.
     
  21. May 2, 2015 #20
    I hope someone can help me understand this. I read trhe Zhang article recommended by marcus. Here is a quote:

    Whence we know that for a freely traveling massive particle in
    an expanding FLRW universe, it is its momentum rather than energy that goes like
    p ∝ 1 / a (2.7).


    p here is momentum. Later the article says:

    So the number density of massive particles at the time t with momentum between p and p + dp would be

    Zhang1.PNG
    Zhang then discusses the use ot the Fermi-Dirac and Bose-Einstein distributions in the "second step", and observes:

    Therefore the form of the Fermi-Dirac and Bose-Einstein distributions are preserved for the thermal evolution of decoupled
    massive particle,
    ...

    The d subscripts represent variable values at the latest time of equilibriium. The e subscripts a later time when the particles are no longer in equilibrium. The article also shows the derived temperature realtion below.

    Zhang2.PNG

    This seems to say that the temperature of a collection of particles not in equilibrium varies inversly with a. Near the end of the article, the following is said:

    ... although the spectrum has still kept the form of the Fermi-Dirac and Bose-Einstein distributions
    since decoupling, it is not the thermal spectrum with the effective temperature and
    chemical potential since the effective mass is not equal to the static mass.


    What I think all this comes to, although it is not said explicitly (and I am not at all clear in my mind that I have it right) is the Fermi-Dirac and Bose-Einstein distributions form are preserved assuming an artificial temperature, while the real, Maxwell-Boltzmann distibution of the energy (in terms of the velocity-squared of the collection of particles) would correspond to a temperature that varied inversely with a2.

    Also, as I understand it, the Fermi-Dirac and Bose-Einstein distributions relate to leptons and bosons rather than to atoms, but the conclusions should apply to atoms also with respect to the Maxwell-Boltzmann distibution.

    I hope someone will comfirm my understanding if it is right, and correct it if I have it wrong.
     
    Last edited: May 2, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How does a particle's kinetic energy vary with a(t)?
  1. Does energy gravitate (Replies: 3)

Loading...