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

Comoving Hubble radius

  1. Aug 23, 2010 #1

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I know this is dumb, but I'm just not getting any sort of intuition for what the "comoving Hubble radius" is. I have the definition in front of me in a book which says that it is equal to (in c = 1 units):

    (aH)-1

    With a being the scale factor and H the Hubble parameter. So basically, it must be equal to dt/da. Later on, there is a statement that it is the "distance over which particles can travel in the course of one expansion time i.e. roughly the time in which the scale factor doubles." I'm not seeing how that follows from the definition. Later still, the book states: "...if [particles] are separated by distances larger than the Hubble radius, then they cannot currently communicate." I'm not seeing how this statement follows from the previous one. I'm not sure if I even understand what "currently communicate" means since communication can't happen instantaneously anyway.

    The book is careful to make a distinction between this and the comoving horizon scale, which I understand perfectly well. If particles are separated by a comoving distance greater than the comoving horizon scale, then they could never have communicated in the history of the universe, since it represents the largest distance over which information can have propagated at any time.
     
  2. jcsd
  3. Aug 24, 2010 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The book in front of you (which is also in front of me) is not very pedagogical. For a nice discussion of cosmological horizons (with different notation), see section 15.12 from General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lasenby. You might be able to read this section from Google Books.
    "one expansion time" suggests exponential growth, so take [itex]H[/itex] constant and

    [tex] a \left(t \right) = a_0 e^{Ht}.[/tex]

    The metric restricted to constant angular part (i.e., [itex]d \Omega^2 = 0[/itex]) is given by

    [tex]ds^2 = dt^2 - a\left(t\right)^2 d \chi^{2},[/tex]

    where [itex]\chi[/itex] is comoving distance.

    On a photon's worldline, [itex]ds^2 = 0[/itex], and

    [tex]d \chi = \frac{dt}{a\left(t\right)}[/tex]

    Suppose a photon is emitted at time [itex]t_e[/itex] and received at time [itex]t[/itex]. The comoving distance traveled by the photon is

    [tex]\chi = \int^t_{t_e} \frac{dt'}{a\left(t'\right)} = \int^a_{a_e} \frac{da'}{a' \dot{a}'}
    [/tex]

    This is a general expression. What to you get for exponential growth with with the scale factor at reception bigger by a factor of [itex]e[/itex] than the scale factor at emission?
    I don't like the way this is written. Proper distance is given by [itex]d = a \chi[/itex], and for a galaxy with no peculiar velocity, proper recession velocity is

    [tex]\dot{d} = \dot{a} \chi.[/tex]

    Thus, proper recession velocity is the speed of light at the comoving Hubble radius.
     
    Last edited: Aug 24, 2010
  4. Aug 24, 2010 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I appreciate the helpful response, and I want to look into this, I'm just bogged down with some other work right now. I'll post a proper response once I've had a chance to look at that other reference (it'll be a few days).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook