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Derivation of non decaying mode in cosmology

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Derive the following result : [itex] \phi_k=-C_1(k)\frac{\dot{H}(t)}{H(t)} [/itex].

    2. Relevant equations

    [itex]\phi_k=C_1(k)\Bigg(1-\frac{H}{a}\int\limits^{t}a(t)dt\Bigg)[/itex]

    [itex]a(t)= a(t_i)\exp\Bigg(H_i(t-t_i)+\dot{H}_i\frac{(t^2-t_i^2)}{2}\Bigg)[/itex]

    3. The attempt at a solution
    So I stuck a(t) into [itex]\phi_k[/itex] and then expanded the exponential and integrated. Just ended up with a mess really.
     
  2. jcsd
  3. Jul 23, 2015 #2

    fzero

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    I assume that ##\phi## is the metric perturbation and that you mean
    $$ \phi_k = - C_1(k) \frac{\dot{H}}{H^{2}}.$$
    Note the ##H^2## in the denominator.

    If so, there is kind of a bizarre trick to use here. First, the fact that ##H = \dot{a}/a##, implies that
    $$ a dt =H^{-1} da .~~~(*)$$
    The idea then is to integrate by parts to show that
    $$ \int a dt = \int H^{-1} da = H^{-1} a- \int a~ d \left( H^{-1} \right).$$
    It seems weird, but this is formally ok to do. Now, we want to write
    $$ d \left( H^{-1} \right) = \frac{ d \left( H^{-1} \right)}{dt} dt $$
    and use (*) again to write the ##dt## integral as one over ##da##. Then it's one more integration by parts to get the term you want, plus another integral that you can see is small for ## \dot{H} \ll H^2##.

    I had to give a lot of the problem away to explain the trick, but it's probably confusing enough that you still have a lot of work to do for it. I'll try to help with followup questions as needed.
     
  4. Jul 23, 2015 #3

    Hey,

    Thank you so much for the help but to solve the problem I have to put a(t) into [itex]\phi_k[/itex] or have I missed something? I know that at some point I need to do an expansion in [itex]\frac{\dot{H_i}}{H^2_i}[/itex]


    So following what you have said I have [itex]d(H^{-1})=\frac{d(H^{-1})}{dt}[/itex]

    and [itex]dt=\frac{H^{-1}}{a}da\longrightarrow d(H^{-1})=\frac{d(H^{-1})}{dt}H^{-1}\frac{da}{a}[/itex] which looks hideous. Do I then stick this into the second term on the RHS of the integral? and then integrate by parts?
     
    Last edited: Jul 23, 2015
  5. Jul 23, 2015 #4

    fzero

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    With this method of doing the problem, we don't need to know anything special about ##a(t)## other than ## H = \dot{a}{a}##. So you don't need to use the expression for ##a## that you quoted.

    Yes, integrate by parts
    $$\int u dv = uv - \int v du$$
    with ##u=(H^{-1}/a)( d(H^{-1})/dt)## and ##v=a##. The integrated part gives the expression that you were asked to show. The remaining integral is a mess of time derivatives that you should be able to argue is much smaller than the leading term.
     
  6. Jul 23, 2015 #5
    Ahh thank you! Will have a work through it and see what happens.
     
  7. Jul 23, 2015 #6
    Ok, so I've looked at it, and something doesn't seem quite right, I am integrating over da, and so if I set v=a, then dv=da but then what happens to my da and what happens to the second a in the integral?
     
  8. Jul 23, 2015 #7

    fzero

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    You've got a stray ##1/a##, so let me back up a step to where we had
    $$ \int a~d\left(H^{-1}\right) = \int a~\frac{d\left(H^{-1}\right)}{dt} dt = \int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da. $$
    We have to integrate this by parts, which gives
    $$ \int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da = \frac{d\left(H^{-1}\right)}{dt} H^{-1} a- \int a ~d\left( \frac{d\left(H^{-1}\right)}{dt} H^{-1} \right) .$$
    The integrand on the RHS should be thought of as ## d(\cdots) = [ d(\cdots)/dt] dt##.
     
  9. Jul 23, 2015 #8
    So I have now

    [itex]\int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da = \frac{d\left(H^{-1}\right)}{dt} H^{-1} a- \int a ~\frac{d}{dt}\left( \frac{d\left(H^{-1}\right)}{dt} H^{-1} \right)dt [/itex]

    And I want to integrate the integral on the RHS by parts. To do this, I (assume)
    [itex]dv=a[/itex] and

    [itex]u=\frac{d}{dt}\Bigg(\frac{d(H^{-1})}{dt}H^{-1}\Bigg)[/itex]
     
  10. Jul 23, 2015 #9

    fzero

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    Whenever you have ##a dt## you want to rewrite it as ##H^{-1} da##. If you keep integrating by parts you get a series with terms like
    $$\left( H^{-1} \frac{d}{dt} \right)^n H^{-1}.$$
    This series is what you want to truncate to the first term.
     
  11. Jul 25, 2015 #10
    Thank you so much, after doing the integrals about 4 times to see the series appear, I was able to see what I wanted. Thank you so much for your help and patience.
     
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