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?


1 Answer 1


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.

In any case, if the integral diverges when going to 0, then we can just define time zero to be now, and integrate closer and closer to big bang to get more and more negative conformal time.

  • $\begingroup$ 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? $\endgroup$
    – user32023
    Apr 15, 2019 at 0:22
  • $\begingroup$ 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. $\endgroup$ Apr 15, 2019 at 5:35

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