$$\left(\frac{\dot{a}}{a}\right)^2 + \frac{k}{a^2} = \frac{8 \pi G}{3} \rho$$ The Friedmann equation contains a square of the first order derivative of the scale factor $a$ with respect to time. Correspondingly, there exist two solutions to the equation: one with a $+$ and one with a $-$ sign. What is the significance of the sign?

I don't think it means time reversal symmetry, because the time $t$ is usually raised to a fractional power in the solution for constant $w$-parameter. $$a(t) = \pm\left(t/t_0\right)^m.$$

  • 1
    $\begingroup$ It has no significance. Only $a^2$ appears in the FRLW metric and therefore also in all other equations derived from it. We can just assume $a>0$. $\endgroup$ – Prahar Dec 29 '17 at 3:47
  • $\begingroup$ @Prahar It has great significance! It is the difference between an expanding and a contracting solution. $\endgroup$ – Dr. Ikjyot Singh Kohli Dec 29 '17 at 4:04
  • $\begingroup$ @Dr.IkjyotSinghKohli - I thought that has to do with whether $|a|$ is increasing or decreasing not whether its positive or negative. How can the overall sign have significance when only $a^2$ appears in the metric? $\endgroup$ – Prahar Dec 29 '17 at 4:08
  • $\begingroup$ So, as an example, let's look at expanding and contracting de Sitter universes. Very simply, an expanding de Sitter universe, we have that $\frac{3 \dot{a}}{a} = \sqrt{3 \Lambda}$, while for a contracting de Sitter universe, we have that $\frac{3 \dot{a}}{a} = -\sqrt{3 \Lambda}$. Solving the expanding case, we get that $a(t) = e^{\frac{\sqrt{\lambda } t}{\sqrt{3}}}$, solving the contracting case, we get that $a(t) = e^{-\frac{\sqrt{\lambda } t}{\sqrt{3}}}$. So, $a^2$ for the expanding is different than $a^2$ for the contracting case. @Prahar $\endgroup$ – Dr. Ikjyot Singh Kohli Dec 29 '17 at 4:37
  • $\begingroup$ @Dr.IkjyotSinghKohli - I completely agree with your example. But nothing you have said is about the OVERALL sign of $a$ which is what the original poster is asking about. You are talking about sign $a=e^{\pm \# t}$ whereas OP is asking about the sign $a = \pm e^{\# t}$. $\endgroup$ – Prahar Dec 29 '17 at 5:13

Positive gives the initial condition for an expanding universe, while negative gives the initial condition for a contracting universe.

For a barotropic fluid, we have $\rho \propto a^{-m}$, and so (assuming $k=0$ for simplicity) $$a^{m/2-1}\,da \propto -dt.$$ Doing the integral gives $$a_f^{m/2} - a_i^{m/2} \propto -\Delta t < 0,$$ i.e. the change in scale factor with time is negative.


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.