- #1
- 5,848
- 552
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.
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.