Calculate the age of the Universe at a temperature

[Denver Dang]

Homework Statement

I've been told to calculate the age of the Universe at T = 1 \, \text{MeV}, assuming that a(t=0)=0.

Homework Equations

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

The Attempt at a Solution

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

 

[mfb]

You cannot derive $H = \frac{1}{2}t^{-1}$ without that convention. It is needed to define where t=0 is.

 

[Denver Dang]

Well, I have shown that:

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

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.

 

[mfb]

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.

 

[Denver Dang]

Hmmm, I'm not sure I follow. How does that help me ?

 

[mfb]

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.

 

[Denver Dang]

Ah, I see what you mean now. Sorry for the confusing on my part.
Thank you very much :)