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

Mathematics of superhorizon

  1. Mar 20, 2014 #1
    Hi, Im reading Tasi lectures on cosmology and in a lot of places states that certain magnitudes (example, comoving curvature R) work in someway (example "does not evolve") outside the horizon k<aH, where k is some fourier frequency of the magnitude we may be talking about. I understand the general idea (outside the horizon things cannot affect what happens inside the horizon -or something like that-. k is a frequency so 1/k > (aH)^-1 -that is wavelength > hubble radius should behave in a special way-) but:

    1 i dont see how that intuitive condition transform in "for k<aH" in mathematical language. I get what it tries to say but I cant see the full symmetry between the words and the math
    2 what "does not evolve" mean in this context (or whatever other condition is defined on any other magnitude)
    3 why does it have to mean that? I insist, i get the idea that things otuaide my horizon cant affect me, my question is how is that idea translated to mathematical language.

    I dont know if the question is clear, i hope you can help me! Thanks in advance for your usual useful help!
     
  2. jcsd
  3. Mar 21, 2014 #2

    bapowell

    User Avatar
    Science Advisor

    1) The wavenumber [itex]k[/itex] is comoving, meaning that it does not change as the universe evolves. This is helpful for placing a label on each Fourier mode that "stays with" the mode even though it is stretching. The physical wavenumber, [itex]k/a[/itex], does evolve with the expansion. Now, if [itex]k/a[/itex] is the physical wavenumber, then [itex]\lambda_{\rm phys} = a/\lambda[/itex] is the physical wavelength of the mode. Note that it increases [itex]\propto a[/itex]. Meanwhile, the inverse Hubble parameter is the horizon size, [itex]H^{-1} = d_H[/itex]. So, rewrite the condition as [tex]d_H < \lambda_{\rm phys}[/tex] and it should be clear that the condition k<aH is the same as saying that the physical wavelength of the mode is larger than the horizon.

    2) The amplitude of the mode is constant.

    3) It's not so much saying that things outside the horizon cannot affect things within. It's saying that this thing -- the Fourier mode -- does not evolve (at least not in a Hubble time). You've got a single fluctuations with a wavelength larger than the causal horizon, or equivalently, a thing with a frequency smaller than the age of the universe. Such a fluctuation is by all accounts frozen.
     
  4. Mar 22, 2014 #3
    Ok, thank you very much. I kept on reading and I have a lot of more doubts so I will keep on posting!

    Thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mathematics of superhorizon
  1. Mathematical Matter (Replies: 17)

Loading...