3
$\begingroup$

Let's consider a flat universe with an FRW metric with scale factor $a(t)$, with some matter content. The continuity equation $\nabla_\mu T^{\mu\nu}=0$ combined with assumptions of isotropy and homogeneity implies the equation $\frac{d\rho}{dt}=-3\frac{\dot a}{a}(\rho+p)$. Now, if we consider equations of state of the form $p=w\rho$, we can solve the above equation for $\rho$ to obtain $\rho\propto a^{-3(1+w)}$. One usually considers the cases $w=-1,0,\frac{1}{3}$, and there aren't many problems there (except $\rho=const.$ for $w=-1$). But if one puts $w<-1$, it appears that the energy density increases with the scale factor! What's worse, a somewhat straightforward calculation would imply that the scale factor decreases with time. This kind of substance violates all the energy conditions, so I'm not sure how "physical" it is. I've tried integrating the Friedmann equations for $K\neq 0$, but they seem a bit complex for an analytical solution. Does this imply that a substance with $w<-1$ can't exist in a FRW universe, or is one of my assumptions wrong ($K=0$ comes to mind)?

$\endgroup$
1
$\begingroup$

Under the assumptions that $a > 0$ and that the universe is expanding, we can derive some interesting results about the fate of such a universe.

From the Friedmann equations alone, we may derive

$$ \frac{d}{d \tau} (\rho a^3) = - P \frac{d}{d \tau} (a^3). $$

For $P = w \rho$, as long as $w \neq -1$, this yields

$$ \rho \propto \frac{1}{a^{3(1 + w)}}, $$

exactly as you stated in your question. So, yes, if the universe is expanding and $w < -1$, then the energy density does increase with time!

In the flat case with no cosmological constant, $\Lambda = K = 0$, we may integrate the first Friedmann equation $3 H^2 = 8 \pi \rho$, with this expression for $\rho$, to yield

$$a(\tau) \propto \tau^{\frac{2}{3(1+w)}},$$

exactly as in the case $w > -1.$ The Hubble parameter is thus given by

$$ H = \frac{\dot{a}}{a} = \frac{3}{3 (1 + w) \tau}. $$

In an expanding universe, we have $H > 0.$ Since $1 + w < 0$, we must have $\tau < 0$ in order for the universe to be expanding. But since $a(\tau) \propto \tau^{\frac{2}{3(1+w)}}$ and $w < -1,$ the scale factor diverges at $\tau = 0.$ So the universe will suffer a "big rip" singularity at some finite time in the future.

In a contracting universe, we have $H < 0,$ and so $\tau > 0.$ Since this is past the $\tau=0$ singularity, such a universe must have originated at a "big rip" at some finite time in the past.

So if we assume that the universe has existed for finite time, then it must be contracting (as you stated in your question), and it must have originated with a divergent scale factor. On the other hand, if we assume that the universe is expanding, then it will meet a singularity in finite time as the scale factor diverges.

$\endgroup$
  • $\begingroup$ You mentioned "But since $a(\tau) \propto \tau^{\frac{2}{3(1+w)}}$ and $w<−1$, the scale factor diverges at $\tau=0$. So the universe will suffer a "big rip" singularity at some finite time in the future." If $a(\tau)$ is divergent at $\tau=0$, why would the big rip occur at a point in the future ($\tau>0$) $\endgroup$ – IanDsouza Dec 29 '17 at 23:33
  • $\begingroup$ @IanDsouza In the first sentence of the same paragraph, I specified that (under our conventions) an expanding universe exists only in the regime $\tau < 0$. So the point $\tau = 0$ is in the future of an expanding universe. $\endgroup$ – user35736 Dec 30 '17 at 20:44
  • $\begingroup$ I see. But what is the physical significance of imposing the time to be negative? Is this just an imposition from the mathematics? $\endgroup$ – IanDsouza Dec 30 '17 at 21:02
0
$\begingroup$

What you are commenting on is something called the big rip, or phantom energy. If you have $$ \frac{d\rho}{dt} = -3\frac{\dot a}{a}(\rho + p) = 0 $$ then you have $p = -\rho$, which is the sometimes commented on as negative pressure of dark energy. However for $p = w\rho$ then $$ \frac{d\rho}{dt} = -3\frac{\dot a}{a}(1 + w)\rho $$ and we have for $\dot a/a = \sqrt{H}$ the Hubble parameter $$ \rho = \rho_0exp\left(-3\sqrt{H}(1+w)t\right) $$ for $w < -1$ something pathological happens in that the vacuum energy density grows exponentially with time.

If I put that in the FLRW equation result $$ \left(\frac{\dot a}{a}\right)^2 = H = \frac{8\pi G\rho}{3c^2} $$ and consider vacuum energy only then $a(t) = a_0exp(t\sqrt{8\pi G\rho/3c^2})$. This is the exponential acceleration of the universe. However. if I put the $w < -1$ exponent for the energy density in this equation I get an exponential growth generated by an exponential growth! The universe tears itself to pieces.

$\endgroup$
  • $\begingroup$ I don't really see where your solution for $\rho$ comes from, when I integrate the equation for $w\neq 1$, I get $\rho=\rho_0 a^{-3(1+w)}$, and the Friedmann equation $H^2=\left(\frac{\dot a}{a}\right)^2=\frac{8\pi G}{3}\rho$ integrates to a solution of the form $a(t)\propto t^{\frac{2}{3(1+w)}}$ (only the negative root though, the positive one turns out complex), so for $w<-1$ the universe shrinks instead of grows, at an accelerated rate even. Your solution would work for $H=const.$, but as the continuity and Friedmann eq. are coupled, I don't see how it can work. $\endgroup$ – blueshift Jun 6 '16 at 13:37

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.