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On 2002-11-03 11:41, AgoraBasta wrote:
Then the BB universe must have both things - the somewhat averaged starlight and the CMB.
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That's right. Only the component from the averaged starlight is negligible. That is to say there aren't enough photons produced by starts to come close to the numbers of photons that we see in the CMB. They are of such a small number that we just cannot see them. They don't exist (well below the one part in 10^4 threshhold for the anisotropy). Models have been done for the Big Bang universe asking how many microwave photons should come from stars and the number is about 10 or 15 orders of magnitude lower than the number we see in the CMB. So, try again. No interaction because not enough photons: it's as simple as that.
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And they somehow avoid interacting. Hmm...
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Hmm... is right. You should get out a bit more and actually become FAMILIAR with the paradigm you are railing against. That might allow you a bit more integrity.
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Thus their IGM-interaction cross-section is a function of IGM and their wavelength only.
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If you mean by that the DENSITY of the IGM and the wavelength.
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If IGM is static - the IGM particles occupy the volume next to zero; if those particles move around, then they are being spread over the volume of universe and they are spread evenly starting from distances R = cV/(l<sup>2</sup>vN), where N/V - concentration of "protons", v - average speed of "protons", l - wavelength of interacting light.
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Not quite. Of course the IGM isn't static. No one said it is (it has to be at a finite temperature above zero). Your treatment of the mean free path is fine, but the conclusions you come to are utterly bogus. You are assuming that the only two things that are interacting are the background radiation and the IGM in this limitting case. This is one of the worst approximations you could make. The IGM may be diffuse, but its self-interactions are far more important than any equilbirium processes with the CMB.
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Assuming a terminal case of v = c gets R = 10 <sup>4</sup> metres; assuming v = 1cm/s gets R = 3x10 <sup>14</sup> metres (both figures for 1 proton/m^3, l = 1cm).
For effective mixing, it's better to multiply these results by a factor of (mc<sup>2</sup>/hf) ~ 10 <sup>13</sup>, where m is the mass of a "proton".
Thus, for some reasonable value of average "proton" speed, we get to total thermodynamical equilibrium of the "CMB" and the IGM well withing the visible universe limits.
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Except the IGM's temperature need not be reasonant with the CMB since there are other thermal processes. The IGM is NOT in thermal equilbrium with the CMB.
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If we use electrons instead of "protons", the same result comes at smaller distances (Good luck, Mr. Kierein)...
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Which is to say that if the universe is less massive, we get thermal equilibrium sooner in an expanding universe. This is a legitimate answer, but doesn't make sense in the context you put it in.
<font size=-1>[ This Message was edited by: AgoraBasta on 2002-11-03 12:02 ]</font>
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