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.


$R=0$ 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 $R=0$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.