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Calculate the age of the Universe at a temperature

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    I've been told to calculate the age of the Universe at [itex]T = 1 \, \text{MeV}[/itex], assuming that [itex]a(t=0)=0[/itex].

    2. Relevant equations
    Now, I've already calculated the value of [itex]H[/itex] at that temperature, which was around [itex]H(1\,\text{MeV}) \approx 0.6 \,\text{s}^{-1}[/itex]. I've also shown, that in a radiation dominated Universe, which I assume much be the case at [itex]T = 1 \, \text{MeV}[/itex], that:
    [tex]H = \frac{1}{2}t^{-1}[/tex]

    3. The attempt at a solution
    So basically, my idea was just to solve for [itex]t[/itex] in that equation, and use the value for [itex]H[/itex] I calculated, and then I end up with [itex]t = 0.85 \, \text{s}[/itex], which seems okay reasonable to me, but, my main question is the info: "assuming that [itex]a(t=0)=0[/itex]". I haven't really used that information here, so either it's just not important, or I have missed something. But what ?
     
  2. jcsd
  3. Nov 29, 2014 #2

    mfb

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    Staff: Mentor

    You cannot derive ##H = \frac{1}{2}t^{-1}## without that convention. It is needed to define where t=0 is.
     
  4. Nov 29, 2014 #3
    Well, I have shown that:

    [tex]\rho _R =\rho_{R0}\left(\frac{a_0}{a}\right)^{4}[/tex]

    So from the Friedmann equation I get:

    \begin{align}
    \frac{\dot a^{2}}{a^{2}} &= \frac{8 \pi G}{3} \frac{\rho_{R0}}{a^{4}} \nonumber \\
    &\Updownarrow \nonumber \\
    \left(a\frac{da}{dt}\right)^{2} &= \frac{8 \pi G}{3} \rho_{R0} \nonumber \\
    &\Updownarrow \nonumber \\
    a\frac{da}{dt} &= \sqrt{\frac{8 \pi G}{3}\rho_{R0}} \nonumber \\
    &\Updownarrow \nonumber \\
    a\,da &= \sqrt{\frac{8 \pi G}{3}\rho_{R0}} \,\, dt \nonumber \\
    &\Updownarrow \nonumber \\
    \int a\, da &= \sqrt{\frac{8 \pi G}{3}\rho_{R0}} \int dt \nonumber \\
    &\Updownarrow \nonumber \\
    \frac{1}{2} a^{2} &= \sqrt{\frac{8 \pi G}{3}\rho_{R0}} \, t \nonumber \\
    &\Updownarrow \nonumber \\
    a &= \sqrt[4]{\frac{32 \pi G}{3}\rho_{R0}} \, t^{1/2} \nonumber \\
    &\Updownarrow \nonumber \\
    a &\propto t^{1/2}
    \end{align}

    And from that I, again, can use the Friedmann equation, giving:

    \begin{align}
    H &= \frac{\dot a}{a} \nonumber \\
    &\Updownarrow \nonumber \\
    H &= \frac{\frac{da}{dt}}{a} \nonumber \\
    &\Updownarrow \nonumber \\
    H &= \frac{1}{t^{1/2}} \frac{d}{dt}t^{1/2} \nonumber \\
    &= \frac{1}{t^{1/2}} \left(\frac{1}{2}\frac{1}{t^{1/2}}\right) \nonumber \\
    &= \frac{1}{2} t^{-1}
    \end{align}

    That's how I got the other equation.
     
  5. Nov 29, 2014 #4

    mfb

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    The step where you perform the integration over dt and over a da should have a free integration constant. Fixing a(t=0)=0 sets this constant to zero.
     
  6. Nov 29, 2014 #5
    Hmmm, I'm not sure I follow. How does that help me ?
     
  7. Nov 30, 2014 #6

    mfb

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    Staff: Mentor

    What do you mean with "help"? You asked where you need the additional information you had, and I answered that question. The result is the same as before, but now with a correct derivation.
     
  8. Nov 30, 2014 #7
    Ah, I see what you mean now. Sorry for the confusing on my part.
    Thank you very much :)
     
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