# Einstein-de Sitter - confusing constant of integration

I'm struggling to understand this derivation of the Einstein-de Sitter universe model

It starts with

$$\int R^{1/2}dR=\int H_{0}R_{0}^{3/2}dt.$$ Evaluating this integral gives$$\frac{2}{3}R^{3/2}=H_{0}R_{0}^{3/2}t+K,$$ where $K$ is a constant of integration.

$$R\left(t\right)=\frac{3}{2}R_{0}\left(H_{0}tK\right)^{2/3}.$$ Let $t=t_{0}$ when $R=R_{0}$, this becomes $$R_{0}=\frac{3}{2}R_{0}\left(H_{0}t_{0}K\right)^{2/3}.$$ Dividing the third equation by the fourth equation gives$$R\left(t\right)=R_{0}\left(\frac{t}{t_{0}}\right)^{2/3}$$

What I don't understand is how the constant of integration disappears. In the second equation it's added to the right-hand side, but in the third equation its multiplied by the right-hand side, and then cancels to give the final (correct) equation. How does that work? Apologies if I've missed the obvious.

## 1 Answer

$R=0$ at $t=0$. This is a boundary condition. Then, $K=0$ must be true. Ignore lines 3 and 4, they are wrong. Just take $K=0$ and do what you did with setting $R=R_0$ and $t=t_0$ and you've got the answer.

Edit - Response to comment:

Short answer

I guessed that you were representing the scale-factor by $R(t)$. (I normally use $a(t)$ by the way). We know that as $t$ goes to 0 (going back in time to the big bang singularity), $R$ goes to 0.

Long answer

I went back to the definition of length scales: $r=Rx$, where $r$ is the physical length and $x$ is a coordinate length [this is how $R$ is introduced]. Then I thought of the following:

• Take some coordinate system $x$, doesn't matter what it is
• Choose a value of $R$ for today (which you can always do)
• Imagine the Universe going backwards in time towards $t=0$
• As the Universe goes to $t=0$ (the big bang), the scale factor gets smaller
• At $t=0$, $r=0$ and thus we must have $R=0$.