Scale factor from Friedmann's equations

[TheMercury79]

If we take a flat universe dominated by radiation, the scale factor is $a(t)=t^{1/2}$
which can be derived from the first Friedmann Equation:$$(\dot a/a)^2 = \frac{8\pi G}{3c^2}\varepsilon(t)-\frac{kc^2}{R_0^2 a(t)^2}$$

But suppose I want to show this using the second Friedmann Equation
(Also known as the acceleration eqauation)$$\ddot a / a = -\frac{4\pi G}{3c^2}\varepsilon(1+ 3w)$$

Solving this second order differential equation can be tedious, but I've come up with a way for which I am unsure whether it is allowed, though it does lead to the same result.

Starting with $w=\frac{1}{3}$ for a radiation dominant universe, and the fact that the energy density in such a universe is related to scale factor by $\varepsilon_r(t) = \varepsilon_0 / a^4$, the acceleration equation can be rewritten as follows:$$\ddot a a^3 = -\frac{8\pi G}{3c^2}\varepsilon_0 = const.~~~~~~~~~~~~~~(1)$$

Like I mentioned this can be hard to solve if we consider $a(t)^3$, the scale factor as a function of time. However, since this is evalutated at a specific time, is it possible to set $a^3$ as constant because it makes evertything a whole lot easier? So that:$$\ddot a = \frac{C}{a^3}$$

And integrating both sides twice with respect to $t$ then gives $a = \frac{Ct^2}{2a^3} ==> a^4 = \frac{C}{2}t^2$
Thus $a\propto t^{1/2}$

Also we can drop the constant $\frac{C}{2}$as it will be absorbed by other constants and now we have $a(t) = t^{1/2}$

But as my question was from the beginning; Is this legitimate?
Another thing I don't like is the constant $C$ which is the constant $-\frac{8\pi G}{3c^2}\varepsilon_0$ from eq. $(1)$
The minus sign in the constant is bothering to me. It is one thing to have it in the acceleration equation, but this minus sign evidently ends up in the expression for scale factor. Because we have $a^4\propto t^2$, so the proportionality constant can't be negative

 

[George Jones]

$$\ddot a a^3 = -\frac{8\pi G}{3c^2}\varepsilon_0 = const.~~~~~~~~~~~~~~(1)$$

Like I mentioned this can be hard to solve if we consider $a(t)^3$, the scale factor as a function of time. However, since this is evalutated at a specific time, is it possible to set $a^3$ as constant because it makes evertything a whole lot easier?

No. If you evaluat $a$ at a specific time, then $\ddot a$ should be evaluated at the same time, and (1) no longer is a differential equation.

I haven't checked to see if (1) is correct. If it is, since $t$ does not appear explicitly in (1), a standard solution technique is to set $p = da/dt$. Then
$$ \frac{d^2 a}{dt^2} = \frac{dp}{dt} = \frac{dp}{da} \frac{da}{dt} = \frac{dp}{da} p .$$

 

[Orodruin]

Is this legitimate?

No. You integrated 1/a^2 as if it was constant and then concluded that a depends on t.

What you need to do is to multiply with $\dot a$ on both sides and integrate. This should give you back the firat order differential equation you have already solved.

 

[Orodruin]

I mean, it should not come as a surprise that the second equation gives you the first upon integration if they are to describe the same universe ...

 

[TheMercury79]

I mean, it should not come as a surprise that the second equation gives you the first upon integration if they are to describe the same universe ...

No it didn't. The text we use showed how if you differentiate the first you get the second. I guess the intstructions threw me off a bit. "Use the second equation directly" it said, it sounded so definite, like that's what you have to work with, nothing else. Anyway thanks. Will try and see what happens with George's method as well.

(I might also add that I knew I was off by treating a as constant, but they did it in the textbook in a previous chapter that dealt with redshift and distance, they integrated time and treated a as constant, so I thought maybe it would work here as well)

 

[Orodruin]

but they did it in the textbook in a previous chapter that dealt with redshift and distance, they integrated time and treated a as constant, so I thought maybe it would work here as well

I somehow doubt that. Can you give a more specific reference than ”the textbook”?

 

[Arman777]

The OP is talking about the derivatiion of the $1 + z = 1/a(t_e)$. In that integral we are taking $a(t_e)$ or $a(t_0)$ as constants because the time between two wave crests is too small, hence we can take then out.

In this problem we are solving for a general time.

 

[Arman777]

¨$$\ddot{a}=Ca^{-3}$$

At this point you can assume that $a(t) ∝ t^q$ where $q$ is just a number.

So we have

¨$$\ddot{a} = q(q-1) t^{q-2}$$
$$a^{-3} = t^{-3q}$$

hence
$$q (q-1)t^{q-2} = Ct^{-3q}$$

$q - 2 = -3q$
$q = 1/2$
so
$a(t) ∝t^{1/2}$