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

Conserved charge in FRW expansion

  1. Apr 29, 2014 #1
    Hi all,

    I'm reading Kinney's lectures on inflation: http://arxiv.org/abs/0902.1529

    and got stuck trying to show that for some comoving length scale [itex]\lambda[/itex], the quantity
    [tex]\left(\frac{\lambda}{d_h}\right)^2 |\Omega -1| [/tex]
    is conserved, if [itex]w[/itex] is constant in the equation of state. Here [itex]d_h[/itex] is the horizon size; it's not clear if he means the comoving or physical scale. I've been assuming he means the comoving scale; then, differentiating, it looks to me as if this is constant if
    [tex]\ddot{a}d_h +H=0[/tex]
    where a dot denotes a derivative with respect to time and H is the hubble constant, but I don't see why this should be true, or how to get rid of [itex]d_h[/itex] so that I could use the Einstein equations for the scale factor.

    Thanks in advance.
     
  2. jcsd
  3. Apr 29, 2014 #2
    There are a few odd statements in this article with regards to redshift, such as the "frequency of light increasing as the universe expands" then calling that redshift. Doesn't make sense.. its completely backwards. However that's a subject for another topic lol. However it does make me concerned on the validity of the equation

    [tex]\left(\frac{\lambda}{d_h}\right)^2 |\Omega -1|=constant [/tex]

    I can't follow how you arrived at the second equation you have, [itex]\lambda[/itex] he has a commoving length scale, might provide a clue to whether Dh is commoving or proper. It will probably help if you show how you arrived at equation 2.
     
  4. Apr 29, 2014 #3

    bapowell

    User Avatar
    Science Advisor

    [itex]d_H[/itex] is comoving: look at the definition Eq. 62. Where is the quote about the frequency increasing? I can't seem to find it.
     
  5. Apr 29, 2014 #4
    An important kinematic effect of cosmological expansion is the phenomenon of
    cosmological redshift: we will see later that solutions to the wave equation in an FRW space have constant wavelength in comoving coordinates, so that the proper wavelength of (for example) a photon increases in time in proportion to the scale factor section 16

    then section 18

    the wavelength of a photon traveling through the spacetime increases because the underlying spacetime is expanding. Another way to look at this is that a photon traveling through an FRW spacetime loses momentum with time.

    edit: might just be his descriptive comparison between commoving and proper wavelength. as his formulas appear correct.
     
    Last edited: Apr 29, 2014
  6. Apr 29, 2014 #5

    bapowell

    User Avatar
    Science Advisor

    Yes. "Wavelength" not "frequency" is mentioned. What's wrong with these statements? They sound perfectly fine to me.
     
  7. Apr 29, 2014 #6
    I think its just the way he worded it confused me lol
     
  8. Apr 29, 2014 #7
    Thanks for your replies.

    My confusion about co-moving or not came from equation 63, where he instates a superscript on *both* comoving and proper horizon distances, each of which has the apellation d_H; but it does seem more likely he's using the comoving definition.

    As for the equation I wrote down above:
    [tex]|\Omega-1|=\frac{1}{(aH)^2}=\frac{1}{\dot{a}^2} [/tex]
    so all the time-dependence looks to be in a factor
    [tex]\frac{1}{(\dot{a} d_H)^2}[/tex]
    and hence the derivative with respect to time is proportional to
    [tex]\frac{d}{dt}(\dot{a}d_H)=\ddot{a}d_H+\frac{\dot{a}}{a}[/tex]
    so that the quantity is constant if this vanishes.
     
  9. Apr 29, 2014 #8

    bapowell

    User Avatar
    Science Advisor

    Well, [itex]\lambda \sim a(t)[/itex], and the Hubble scale [itex]d_H \sim H^{-1}[/itex]. That gives [itex]\lambda^2/d_H^2 = a^2H^2[/itex]. But that would require [itex]d_H[/itex] to be the proper Hubble scale...not comoving ;) Probably just a notation snafu.
     
    Last edited: Apr 29, 2014
  10. Apr 29, 2014 #9

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think I managed to brute force the result. I hope to get a chance later to type in the steps here, and to see if if what i did works.
     
  11. Apr 29, 2014 #10
    bapowell: That'd do it- thanks!

    Is there a way of seeing in general that [itex]d_H \sim H^{-1}[/itex]? For a flat universe with vacuum energy it follows from his calculation of the conformal time [itex]\tau=\frac{1}{aH}[/itex], but I'm under the impression that this result holds more generally?
     
    Last edited by a moderator: Apr 29, 2014
  12. Apr 29, 2014 #11

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I interpreted the intent of the exercise a little differently, i.e., I took ##\lambda## to be a comoving distance (thus, a constant), and ##d_h## to be the horizon size, not the size of the Hubble sphere.
     
  13. Apr 29, 2014 #12
    George: on p.5 Kinney claims that the Hubble length "sets the scale of the observable universe"- is he not talking about the horizon there?

    In the concrete case of a flat universe with vacuum energy, it seems to be an exact equality: proper horizon size =1/H :confused:
     
  14. Apr 29, 2014 #13

    bapowell

    User Avatar
    Science Advisor

    No, not generally. The particle horizon gives the size of the observable universe; the Hubble sphere merely marks the distance at which objects recede from Earth with superluminal speed. During inflation, though, the Hubble sphere is generally taken to coincide with the cosmological event horizon (it only exactly coincides for de Sitter inflation) in which case it sets the scale of causal physics.

    EDIT: Actually, to be fair -- it does set the scale, as the particle horizon and the event horizon both are [itex]\mathcal{O}(1)H^{-1}[/itex]
     
  15. Apr 29, 2014 #14

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Drat! Looks like I don't have it. Now I have to catch my bus home, and I don't think I will have a chance to look at this tonight.
     
  16. Apr 29, 2014 #15

    bapowell

    User Avatar
    Science Advisor

    Yes, there is for a constant equation of state. Let's let [itex]d_{H,{\rm com}} = \int dt/a(t)[/itex] be the comoving particle horizon, and [itex]r_H = H^{-1}[/itex] be the (proper) Hubble radius. I will leave it to you to show that [itex]\dot{r}_{H,{\rm com}} = q/a[/itex], where [itex]q[/itex] is the deceleration parameter, which is constant for constant equation of state. Then, consider the comoving particle horizon:
    [tex]d_{H,{\rm com}} = \int \frac{dt}{a} = \frac{1}{q} \int d r_{H,{\rm com}}[/tex]
    which shows that [itex]d_H \sim H^{-1}[/itex].
     
  17. Apr 29, 2014 #16
    I think both λ and dH should be comoving. If the equation of state w is constant then you can in principle solve H and a explicitly from the Friedmann equations:

    [tex]H = \frac{H_*}{1+\frac{3}{2}H_*(1+w)(t-t_*)} \qquad a = a_* \Big[1+\frac{3}{2}H_*(1+w)(t-t_*)\Big]^{\frac{2}{3(1+w)}}[/tex]

    for some reference time t_* (which you can fix by requireing a=0 at t=0). From this you can also derive the particle horizon. Just by looking at it, it seems that the relation should hold though I haven't checked. There's probably a more elegant way of showing it by I couldn't think of one.
     
  18. Apr 29, 2014 #17

    bapowell

    User Avatar
    Science Advisor

    Yes, they should either be both comoving or both proper. The ratio shouldn't care.
     
  19. Apr 29, 2014 #18
    True. All I'm saying is that the lectures specify that they are supposed to be comoving.
     
  20. Apr 30, 2014 #19
    Actually, forget what I wrote. This only works in a flat universe.
     
  21. Apr 30, 2014 #20
    Thanks for all the replies!
    It looks to me as if the deceleration parameter
    [tex]q=-\frac{\ddot{a}a}{\dot{a}^2}=-\frac{\ddot{a}}{a H^2}=-\frac{1}{2}(1+3w)\left(1+\frac{k}{\dot{a}^2}\right)[/tex]
    is only constant in a flat universe?
     
    Last edited: Apr 30, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook