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General Relativity ODE

  1. Dec 15, 2009 #1
    im hoping someone has access to the book "General Relativity" by Wald as it'll be easier to see what's going on but notheless i'll try and put all the details in

    on p97 we have the equations

    [itex]3 \frac{\dot{a}^2}{a^2}=8 \pi \rho - \frac{3}{a^2}[/itex]
    [itex]3 \frac{\ddot{a}}{a}=-4 \pi \rho[/itex]

    where [itex]\dot{a}=\frac{da}{d \tau}[/itex]

    and i need to show the solutions are as shown in the table on p98:

    [itex]a=\frac{1}{2}C(1-\cos{\eta})[/itex]
    [itex]\tau=\frac{1}{2}C(\eta-\sin{\eta})[/itex]

    i've spent ages on thsi and keep getting nowhere
     
  2. jcsd
  3. Dec 15, 2009 #2

    Nabeshin

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    Science Advisor

    Well you obviously need to know something about the density to say anything about the solutions. If I'm not mistaken, those solutions you mentioned are those for a matter-dominated closed universe. Does that help?
     
  4. Dec 16, 2009 #3
    yes. that's correct, they are for a matter dominated universe. but i already took that into account and set P=0 where P is pressure in the equations i gave.
     
  5. Dec 16, 2009 #4
    Combine the Friedmann equations into one by eliminating the energy density then write this equation in terms of conformal time. After that you can just plug your expression for a into the equation and see that it is a solution. As for t, you can get that from the definition of conformal time.
     
  6. Dec 16, 2009 #5
    yeah so that gives

    [itex]\dot{a}^2-\frac{C}{a}+k=0[/itex]
    but k=+1 for 3-sphere (which is the one im trying to do)
    so i need to solve
    [itex]\dot{a}^2-\frac{C}{a}+1[/itex]

    i still don't know how to solve that
    especially seeing as they get an answer in terms of [itex]\eta[/itex] not [itex]\tau[/itex]
    surely there's a way of solving ti rather than just checking the expression for a is in fact a solution?
    also you mentioned getting [itex]\tau[/itex] from the definition of conformal time. is that in this same book?
     
  7. Dec 16, 2009 #6
    [tex]\dot{a}^2-\frac{C}{a}+k=0[/tex]

    is a separable diff. equation:

    [tex]\left( \frac{da}{d \tau} \right)^2 = \frac{C}{a} -k \Leftrightarrow \frac{da}{\sqrt{\frac{C}{a} -k}} = d\tau \Leftrightarrow d\tau = C^{-1/2} \frac{a^{1/2} da}{\sqrt{1-\frac{ka}{C}}}[/tex]

    which can be solved with a substitution

    [tex]\frac{ka}{C} = \sin^2 \phi[/tex].

    EDIT: Never mind, I see you've assumed that rho goes like a^-3 (matter domination). I was thinking of a more general case where rho is unknown. So just solve the equation using the method described above. Or simpler yet, rewrite that equation in terms of conformal time eta and solve directly for that using a similar technique. Conformal time:

    [tex]d\eta = \frac{1}{a}dt[/tex]
     
    Last edited: Dec 16, 2009
  8. Dec 16, 2009 #7
    so i was going to do it just for the 3-sphere geometry in which k=1 but if i leave it as k as you have above, will i get a more general answer that will save me ahving to do the integral multiple times.

    anyway that [itex]a^{1/2}[/itex] is annoying
    i get [itex]da=\sin{2 \phi} d \phi[/itex]

    and then [itex]d \tau = C^{-1/2} \frac{a^{1/2} \sin{2 \phi}}{\cos^2{\phi}}[/itex]
     
  9. Dec 16, 2009 #8
    [tex]
    a = C \sin^2 \phi \Rightarrow a^{1/2} = C^{1/2} \sin \phi
    [/tex]

    Where I have set k to 1. Note also the square root in the denominator, so there is only one power of cosine there which is cancelled by the cosine inside sin(2phi). However this will give you a as a function of t. If you want a as a function of eta then rewrite the differential equation in terms of eta (chain rule), which incidentally leads to a simpler integral:

    [tex]
    d\eta = C^{-1/2} \frac{a^{-1/2} da}{\sqrt{1-\frac{a}{C}}}
    [/tex]

    Again k=1.
     
  10. Dec 17, 2009 #9
    so yeah that would actually give

    [itex]d \tau = 2 \sin^2{\phi} d \phi[/itex]

    but yeas i do want both [itex]a[/itex] and [itex]\tau[/itex] in terms of [itex]\eta[/itex]. you talk about [itex]\eta[/itex] as if it's some well known quantity, [itex]\eta=\frac{t}{a}[/itex], what does this represent and where does it come from? i haven't seen it before.

    anyway [itex]\dot{a}=\frac{da}{d \eta} \frac{d \eta}{d \tau}=\frac{da}{d \eta} \frac{1}{a}[/itex]

    so [itex]\frac{1}{a} \frac{da}{d \eta} -\frac{C}{a}+1=0[/itex]
    which rearranges to the integral you had above
    then i let [itex]\frac{a}{C}=\sin^2{\phi},da=2 \sin{\phi} \cos{\phi} d \phi, a^{\frac{1}{2}}=C^{\frac{1}{2}} \sin{\phi}[/itex]
    and get
    [itex]d \eta=\frac{C^{-1/2}C^{1/2}\sin{\phi}2\sin{\phi}\cos{\phi} d \phi}{\cos{\phi}}=2 \sin{\phi} \cos{\phi} d \phi[/itex]
    then let [itex]u=\sin{\phi},du=\cos{\phi} d \phi[/itex]
    and get
    [itex]d \eta=2 u du \Rightarrow \eta=u^2=\sin^2{\phi}=\frac{a}{C}[/itex]
    which isn't what i'm after....
     
  11. Dec 17, 2009 #10
    [itex]
    (\frac{1}{a} \frac{da}{d \eta})^2 -\frac{C}{a}+1=0
    [/itex]

    gives

    [tex]d\eta = C^{-1/2} \frac{a^{-1/2} da}{\sqrt{1-\frac{a}{C}}}[/tex]

    Which gives the solution you are afrer.

    Note that [tex]a^{-1/2} = \frac{C^{-1/2}}{sin\phi}[/tex] so everything cancels out.



    Also [itex]\eta \neq \frac{t}{a}[/itex]. The definition of eta is [itex]d\eta = \frac{a_0}{a}dt[/itex] and a is a function of t (a0 can be normalized to 1).

    Eta is the conformal time which is just a different time coordinate chosen such that it expands with the universe just like the space coodinate in Robertseon-Walker metric. It's used in cosmology because it simplifies some of the equations (as it does in this case). It's weird that you don't have it in your book, though maybe that's because it's a book on general relativity and not cosmology. In any case you can think of it purely as a mathematical parameter to arrive at the solutions you posted.
     
    Last edited: Dec 17, 2009
  12. Dec 19, 2009 #11
    hi again.

    i integrated that and got
    [itex]\eta=\frac{2 \phi}{C}[/itex] which i presumably now try and substitue back into my equations.

    so i had [itex]\phi=\sin^{-1}(\sqrt{a/C})[/itex]
    which i can put back in but then i get

    [itex]a=C \sin^2(c \eta /2)[/itex] which is close to what i want but not quite so im obviouslly screwing up somewhere but i cant see it. and how would i get the [itex]\tau[/itex] expression from that?
     
  13. Dec 19, 2009 #12
    Actually [itex]\eta=2 \phi[/itex]

    Remember that one C also come from da so all the C's cancel also.

    So [itex]a=C \sin^2(\eta /2)[/itex]

    Now recall the trigonometric identity:

    [itex]\cos(2x) = 1 - 2\sin^2(x)[/itex]

    Also since eta is defined as,

    [itex]d\eta = \frac{d\tau}{a} \Rightarrow \tau = \int_0^{\eta} a(\eta) d\eta[/itex]
     
    Last edited: Dec 19, 2009
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