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Quick question on CMB anisotropy in Earth frame

  1. Apr 19, 2014 #1

    WannabeNewton

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    Hi guys. Consider the mean occupation number and specific intensity of the CMB photons in the CMB frame as given by the blackbody formulas: ##\eta = \frac{1}{e^{h\nu/k_{B}T_0} - 1}## and ##I_{\nu} =\frac{2h\nu^3}{e^{h\nu/k_{B}T_0} - 1}## with ##T_0## the thermal bath temperature in the CMB frame and ##c = 1##.

    Now we consider the Earth as a Lorentz frame moving relative to the CMB frame with some velocity ##v## relative to the ##x##-axis of the CMB frame and a telescope in the Earth frame oriented at some angle ##\theta##. Note that the mean occupation number can be put in the frame-independent form ##\eta = \frac{1}{e^{-p_{\mu}u^{\mu}/k_{B}T_0} - 1}## where ##u^{\mu}## is the 4-velocity of the CMB frame and ##p^{\mu}## the 4-momentum of the photons.

    In the Earth frame, ##u^{\mu} = \gamma(1,-v)## and ##p^{\mu} = h(\nu', -\nu' \cos\theta, -\nu' \sin\theta)## so ##\eta = \frac{1}{e^{\gamma h\nu'(1 - v\cos\theta )/k_{B}T_0} - 1} = \frac{1}{e^{h\nu'/k_{B}T} - 1} ## where ##T = T_0 \frac{\sqrt{1-v^2}}{1 - v\cos\theta}##. Then the specific intensity in this frame would be ##I_{\nu'} = \frac{2h\nu'^3}{e^{h\nu'/k_{B}T} - 1}##. Here ##\nu = \gamma \nu' (1 - v\cos\theta)## so ##\nu' = \frac{\sqrt{1 - v^2}}{1 - v\cos\theta}\nu## is the doppler shifted frequency in the Earth frame.

    However, in p.16 of http://www.astro.princeton.edu/~jeremy/heap.pdf, one is given the result ##\eta = \frac{1}{e^{h\nu/k_{B}T} - 1}## instead (the paper's ##\hat{T}_{\text{CMB}}## is the ##T## above) so where did I go wrong?

    On the other hand in p.20 of http://www.staff.science.uu.nl/~proko101/JildouBaarsmaCMB.pdf, one is given ##\eta = \frac{1}{e^{|p'|/k_{B}T} - 1} = \frac{1}{e^{h\nu'/k_{B}T} - 1}## (the paper's ##T'## is the ##T## above) which seems to agree with what I have so I'm confused.

    Thanks in advance.
     
  2. jcsd
  3. Apr 19, 2014 #2

    George Jones

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    At a quick glance, it looks the same to me. Note carefully the notation for the frames given in the paragraph immediately after 1.29.
     
  4. Apr 19, 2014 #3
    you might want to also check out this handy article

    "Physics of the intergalactic medium." I'm still studying it myself as its fairly intense lol. However you may find it useful for your research

    http://arxiv.org/abs/0711.3358

    by the way nice articles I'll be adding them to my collection to study myself
     
    Last edited: Apr 19, 2014
  5. Apr 19, 2014 #4

    WannabeNewton

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    Oh haha I totally missed the difference in notation for the frames. Well that's embarrassing. Thanks George!

    Awesome, thanks Mordred!
     
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