Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power Spectrum of CMB angle-l

  1. Feb 1, 2015 #1


    User Avatar
    Gold Member


    Here one reads in sec. 26.2.4 that:

    However it states that a single [itex]Y_{lm}[/itex] corresponds to angular variations of [itex] \theta \sim \pi /l[/itex].

    I am not getting these statements. Also I find it difficult to understand, since most of the times, the power spectrum (eg. Fig. 26.1 in the above reference ) shows a figure of the amplitude versus on the lower axis the multipoles [itex]l [/itex] and on the upper axis the angles [itex] \theta [/itex]. If there is no one-to-one correspondence between [itex]\theta \text{-} l [/itex] how does these figures make sense?
  2. jcsd
  3. Feb 1, 2015 #2


    User Avatar
    Science Advisor

    It's an approximate correspondence. There can't be a one-to-one correspondence because the fact that it's on the surface of a sphere prevents that.
  4. Feb 2, 2015 #3


    User Avatar
    Science Advisor

    The [itex]C_\ell[/itex]'s receive contributions from perturbations across a range of scales. On large angular scales where the Sachs-Wolfe effect dominates, each [itex]C_\ell[/itex] is dominated by contributions from perturbations subtending an angle of [itex]\theta = \pi/\ell[/itex]. The approximate correspondence follows.
  5. Feb 2, 2015 #4


    User Avatar
    Science Advisor

    One way of looking at it is to examine the spherical harmonics where [itex]\ell = |m|[/itex]. For these spherical harmonics, all of the variation of the signal is in the [itex]\phi[/itex] direction (angle of longitude): [itex]e^{im\phi}[/itex]. The waves of the same [itex]\ell[/itex] but different [itex]m[/itex] represent different "directions" of an oscillation of approximately the same wavelength. The peak-to-trough distance of a wave [itex]e^{i\ell\phi}[/itex] is [itex]\pi/\ell[/itex]. I believe this is considered a full wavelength in context because it is the amplitude squared that represents the real field: [itex]Y_\ell^mY_\ell^{m*}[/itex], and squaring the amplitude halves the wavelength.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Power Spectrum angle Date
A Angular power spectrum, bias from N weighted events Mar 18, 2017
I Primordial Power Spectrum Nov 22, 2016
I Primordial power spectrum Oct 1, 2016
I Scale invariance in the power spectrum Sep 19, 2016
A Plotting the CS power spectrum Feb 17, 2016