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Questions arisen while reading Weinberg's Cosmology

  1. Feb 14, 2013 #1
    I was reading Weinberg's book Cosmology. When I went to chapter 2, under the subsection 2.1 "Expectations and discovery of the microwave background" I could not understand one statement.

    The work done by pressure in an expanding fluid uses heat energy drawn from the fluid. The universe is expanding, so we expect that in the past matter was hotter as well as denser than at present. If we look far enough backward in time we come to an era when it was too hat for electrons to be bound into atoms. At sufficiently early times the rapid collisions of photons with free electrons would have kept radiation in the thermal equilibrium with the hot dense matter. The number density of photons in equilibrium with matter at temperature [itex]T[/itex] at photon frequency between [itex]\nu[/itex] and [itex]\nu+d\nu[/itex] is given by the black-body spectrum:

    [itex]\displaystyle n_T(\nu)d\nu=\frac{8\pi\nu^2d\nu}{\mathrm{exp}(h \nu/k_BT)-1}[/itex]​

    The followings are what I do not understand:
    When the temperature is sufficiently high, the bound electrons escape from the atoms and become free electrons in the universe.

    Yet, how come there are photons in the universe?
    Do they come from the radiation of the hot matter?
    Even thought there are photons and they collide with the free electrons, what does "kept radiation in the thermal equilibrium with the hot dense matter" mean?

    What I have learnt previously about black-body radiation is that a "hot" object in thermal equilibrium with its surrounding emits radiation.
    Last edited: Feb 14, 2013
  2. jcsd
  3. Feb 14, 2013 #2


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    hi kiwakwok! :smile:
    photons were created at the same time as all the other elementary particles

    everything was in thermal equilibrium with everything else, until matter started forming into atoms

    a bit like soup … if you stir hot soup, it'll all be at the same temperature ("thermal equilibrium") … but if it cools enough for lumps to form, the liquid soup and the lumps won't necessarily be at the same temperature! :wink:
    photons aren't conserved: the total number of photons can change

    so yes, some photons come from radiation of hot matter

    in black-body radiation, a "hot" object always (whether in thermal equilibrium or not) both emits radiation (photons) and absorbs radiation …

    when it emits and absorbs at the same rate (at all frequencies), that's thermal equilibrium :smile:
  4. Feb 14, 2013 #3
    I would like to ask one more question:

    As time pass, the matter became cooler and less dense, and eventually the radiation began a free expansion, but its spectrum has kept the same form. We can see this most easily under an extreme assumption, that there was a time [itex]t_L[/itex] when radiation suddenly went from begin in thermal equilibrium with matter to a free expression. Under this assumption, a photon that has frequency [itex]\nu[/itex] at some later time [itex]t[/itex] when photons are traveling freely would have had frequency [itex]\nu a(t)/a(t_L)[/itex] at the time radiation went out of equilibrium with matter and so the number density at time [itex]t[/itex] of photons with frequency between [itex]\nu[/itex] and [itex]\nu + d\nu[/itex] would be

    [itex]\displaystyle n(\nu,t)d\nu=\left(a(t_L)/a(t)\right)^3n_{T(t_L)}\left(\nu a(t)/a(t_L)\right)d(\nu a(t)/a(t_L))[/itex],​

    with the factor [itex]\left(a(t_L)/a(t)\right)^3[/itex] arising from the dilution of photons due to the cosmic expansion. Using the equation of black-body spectrum and the equation above, we see that beat the redshift factor [itex]a(t)/a(t_L)[/itex] all cancel except in he exponential, so that the number density at time [itex]t[/itex] is given by

    [itex]\displaystyle n(\nu,t)d\nu=\frac{8\pi\nu^2d\nu}{\mathrm{exp}(h \nu/k_BT(t)-1}=n_{T(t)}(\nu)d\nu[/itex],​



    The radiation became a free expansion, is it because there is vacuum in the universe?
    Also, How can I know "photons are traveling freely would have had frequency [itex]\nu a(t)/a(t_L)[/itex]"?
    Can you give me some hint, so that I can make [itex]\nu a(t)/a(t_L)[/itex] into [itex]\displaystyle n(\nu,t)d\nu=\left(a(t_L)/a(t)\right)^3n_{T(t_L)}\left(\nu a(t)/a(t_L)\right)d(\nu a(t)/a(t_L))[/itex]?
  5. Feb 14, 2013 #4


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    no, it's because matter became less dense

    photons produced at the centre of the sun take thousands of years to get out of the sun and come towards us …

    that's because they keep bouncing backwards and forwards off the matter in the sun, the matter is so dense that they hardly get anywhere (which is ideal for thermal equilibrium, since the energy produced stays more or less where it is)

    only when they get where the sun is less dense can they move in more or less a straight line (and the sun at last loses energy to the outside :wink:)

    the free expansion was for the same reason … matter was less dense, and the photons could actually keep going
    a(t)/a(tL) is the red-shift factor, which applies equally to all wavelengths

    so you take the present wavelength, divide it by the red-shift, and you get the wavelength at time L (the time radiation went out of equilibrium with matter) :smile:
  6. Feb 14, 2013 #5
    Do you mean?
    [itex]\frac{\lambda(t)}{a(t)/a(t_L)}=\lambda (t_L)[/itex]

    [itex]d\nu=-\frac{1}{\lambda^2}d\lambda=-\left(\frac{a(t)}{a(t_L)}\right)^2\cdot\frac{1}{ \lambda (t)^2}d \lambda [/itex]​
    Last edited: Feb 14, 2013
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