Wednesday, June 26, 2013

CMB as an ideal black body

why is the cosmic microwave background such a perfect black body with a Planck-spectrum and why don't we see the recombination lines of the atomic levels?

maths bonus question: what's the reason why Christoffel symbols $\Gamma^{a}_{bc}$ are required to be symmetric in the lower indices?

2 comments:

  1. From my point of view, the question about the lines is a bit of a "trap", since the key is the thermodynamic equilibrium, and not the many, many interactions that bring it about.

    But there is a puzzle: why the light emitted at different redshifts yields the picture of a CMB with well defined temperature. We already discussed it in spring on this blog. I'll try to link to the question, in order not to steal the praise from someone else ;)

    Now, about the Christoffels, they are symmetric for a torsion free connection. I don't know a reason for choosing such - besides simplicity. I have heard about theories with torsion, but not gone into it. If there's more to it, I'd really like to learn.

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  2. We can't see recombination lines in the CMB because the life-time of atomic excitations is shortened by induced emission due to the high photon density which results in a broadening of the lines, such that they merge into a continuum.

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