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$$\left(\frac{\dot{a}}{a}\right)^2 + \frac{k}{a^2} = \frac{8 \pi G}{3} \rho$$ 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. $$a(t) = \pm\left(t/t_0\right)^m.$$

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    $\begingroup$ 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
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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.

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