# How do you calculate conformal time?

I've got several references that indicate that conformal time is the integral of the scale factor:

$$\eta=\int_0^t \frac{1}{a(t)}\mathrm dt.$$ So I tried to calculating this over a range of {0, present time} and immediately ran into a problem. $$a(0)$$ is zero, which means that any definite integral of the scale factor starting from the beginning of time is going to be infinite. This is probably a basic calculus question, but I'm stumped at the moment. What am I missing? How can I practically calculate the conformal time of the present time given this definition?

Integrals of the form $$\int_0^T \frac{dt}{a(t)}$$ can be convergent or divergent depending on how $$a(t)\rightarrow 0$$ as $$t\rightarrow 0$$. So one just have to calculate $$a(t)$$ and see what happens. For $$\Lambda$$CDM it converges as far as I can see when I run the equations. But for a different set of physical assumptions it may not.
Conformal time may actually lack a beginning or end depending on $$a(t)$$. In $$\Lambda$$CDM it has an endpoint, since eventually the expansion approaches exponential and the integral converges to a finite number. If $$\Lambda=0$$ it would be divergent due to slowing expansion. I'm less certain what conditions would be required to remove the beginning.
• I'm afraid I don't follow. If I had $\eta=\int_0^t \frac{f(t)}{a(t)}\mathrm dt.$, then there might be some question about whether $f(t)$ approached infinity faster than $a(t)$ approached zero, but I don't see how there's any ambiguity with just $\eta=\int_0^t \frac{1}{a(t)}\mathrm dt$. It seems like no matter what function you have for $a(t)$, if your definite integral range is {0, t}, then the result is undefined. Could you explain? – Quarkly Apr 15 at 0:22
• Suppose $a(t)=\sqrt{t}$. Then $\int 1/a(t) dt=\sqrt{t}/2 + C$ and the integral $\int_0^x 1/a(t) dt = \sqrt{x}/2$. The ting that saves us here is that $a(t)$ grows so fast near $t=0$ that the area under $1/a(t)$ is finite. – Anders Sandberg Apr 15 at 5:35