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I have the following question for homework:

Show that the Hubble constant $H$ is time-independent in a universe in which the only contribution to energy density comes from vacuum energy.

So in this case we have $\rho = \rho_{\Lambda} = $ constant and $\rho_{\Lambda} = -p_{\Lambda}$. Then using the (simplified) Friedmann equation: $$\dot{a}^2 - \frac{8 \pi G \rho_{\Lambda}}{3}a^2 = -k$$ Dividing by $a^2$ gives $$H^2 - \frac{8 \pi G \rho_{\Lambda}}{3} = -\frac{k}{a^2}$$

But this says that $H$ depends on $a$ which isn't constant. Am I supposed to assume $k=0$? (In that case $H$ is clearly constant ...)

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3 Answers

Yes, you should probably assume $k = 0$, since otherwise the statement is not true, as you have shown.

The nature of the Friedmann equations allows us to rewrite the $k/a^2$ term as though it were itself a source of energy density $\rho_k$ varying as $a^{-2}$, complete with its own fraction $\Omega_k$ of the critical density. This is done by writing $$ \frac{8\pi G\rho_k}{3} = -\frac{kc^2}{a^2} $$ (restoring the factors of $c$ for clarity) and solving for $\rho_k$. Then $\Omega_k = \rho_k/\rho_\mathrm{crit}$. Pretty much by definition then we have $\Omega_m + \Omega_r + \Omega_\Lambda + \Omega_k \equiv 1$.

This procedure puts the equations into a "nicer" form, and so it is pretty common. Thus one often counts curvature as "energy," so saying all other energy sources vanish implies curvature vanishes as well. Ontologically one may object that curvature is intrinsic to the manifold rather than being true stress-energy, but in the Friedmann model it acts identically from a dynamical point of view.

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If you regard curvature as a form of energy density (see Chris White's post), then yes, you should set $k=0$. If not, then $H$ will only approach a constant value as $a\rightarrow \infty$: $$ H^2 \rightarrow \frac{8\pi G}{3}\!\rho_\Lambda \qquad \text{for $a\rightarrow \infty$.} $$ You can check that the general solutions of $$ \dot{a}^2 - \frac{8\pi G}{3}\!\rho_\Lambda a^2 = -kc^2, $$ are given by $$ \begin{align} a(t) &= c\sqrt{\frac{3}{8\pi G\rho_\Lambda}}\exp\left(\sqrt{\frac{8\pi G\rho_\Lambda}{3}}(t-t_\text{b})\right)\qquad &&\text{for $k=0$},\\ a(t) &= c\sqrt{\frac{3}{8\pi G\rho_\Lambda}}\cosh\left(\sqrt{\frac{8\pi G\rho_\Lambda}{3}}(t-t_\text{b})\right)\qquad &&\text{for $k=1$},\\ a(t) &= c\sqrt{\frac{3}{8\pi G\rho_\Lambda}}\sinh\left(\sqrt{\frac{8\pi G\rho_\Lambda}{3}}(t-t_\text{b})\right)\qquad &&\text{for $k=-1$}, \end{align} $$ where $t_\text{b}$ is determined by a boundary condition. Note that only $k=-1$ allows a Big Bang, since $a(t_\text{b})=0$. For $k=1$, it follows that $a(t)$ has a non-zero minimum at $t=t_\text{b}$. And for $k=0$ (a De Sitter model) we get that $a(t)\rightarrow 0$ if $(t-t_\text{b})\rightarrow -\infty$, i.e. in the infinite past.

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The formula for the Hubble Parameter is

$$ H_{a} = \text{H0}\sqrt{\text{$\Omega $R} \cdot a^{-4}+\text{$\Omega $M} \cdot a^{-3}+\text{$\Omega $K} \cdot a^{-2}+\Omega \Lambda } $$

where only

$$ \Omega \Lambda $$

is for dark Energy. The other Omegas are for radiation, matter and curvature. Reducing to dark energy reduces the formula to

$$ H_{vac} = \text{H0}\sqrt{0 \cdot a^{-4}+0 \cdot a^{-3}+0 \cdot a^{-2}+\Omega \Lambda } $$

in other words

$$ H_{vac} = \text{H0}\sqrt{\Omega \Lambda } $$

which is no longer a function of time or the scale factor, but a constant.


PS: in other modells than basic ΛCDM/FLRW, for instance quintessence or dynamic equation of state, the vacuum energy is not a constant, see http://forum.alltopic.de/viewtopic.php?p=20846#p20846

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