Wednesday, April 10, 2013

thermal equilibrium and the CMB

why does the CMB have a Planckian spectrum? after all, it has been generated over a range in redshift ($\Delta z\simeq 100$) over which the temperature changed, and there has been energy exchange with the electron plasma.

bonus question: can you show that the asymptote of the integral sine, $\int_0^x\mathrm{d}\ln(t)\:\sin(t)$, is $\pi/2$ for $x\rightarrow\infty$?

2 comments:

  1. Cryptic version of my reply: Because we live in a Friedmann universe. More extended one to follow in a few days if no one else replies.
    Uli Bastian

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  2. okay, then - if nobody else wants to raise his/her voice - my real reply is the following:
    As long as the universe is homogeneous and opaque, the spectrum is Planckian just because there is no reason for a deviation from local thermal equilibrium. The temperature of that spectrum decreases inversely proportional to the scale factor, just due to the expansion.

    During recombination the temperature continues to drop, so that in the end we (at the present time, long after recombination) receive radiation that was emitted at different temperatures originally. Thus the *emitted* radiation is a mixture of different Plancks, indeed. Yet, the *received* radiation is a pure Planck with a specific, uniqe temperature because anything emitted at a slightly higher temperature (i.e. at smaller scale factor) is redshifted by an exactly matching larger amount.

    The reason for this match is that the temperature drops exactly with the inverse scale factor in a Friedmann universe (as long as there are no significant sinks or sources of radiative energy).

    Uli Bastian

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