2
$\begingroup$

The particle horizon is given by \begin{equation} \chi_p(\eta) = \int_{t_i}^{t}\frac{dt}{a} = \int_{a_i}^{a}\frac{da}{a\dot{a}} = \int_{\ln a_i}^{\ln a} (aH)^{-1}\ d\ln a . \end{equation}

For a universe dominated by a fluid with constant equation of state $w = P/\rho$ we have, \begin{equation} (aH)^{-1} = H_0^{-1}a^{\frac{1}{2}(1 + 3w)} . \end{equation}

We can calculate the the particle horizon explicitly for this case, \begin{equation} \chi_p(a) = \frac{2H_0^{-1}}{(1 + 3w)} \left[ a^{\frac{1}{2}(1 + 3w)} - a_i^{\frac{1}{2}(1 + 3w)} \right] \equiv \eta - \eta_i \end{equation}

For familiar matter sources ($1 + 3w > 0$) the largest contribution comes from later times, i.e. \begin{equation} \eta_i = \frac{2H_0^{-1}}{(1 + 3w)}a_i^{\frac{1}{2}(1 + 3w)} \longrightarrow 0 \quad \text{for}\quad a_i\to 0,\ w > -\frac{1}{3} . \end{equation}

Therefore, the comoving horizon is finite \begin{equation} \chi_p(t) = \frac{2H_0^{-1}}{(1 + 3w)}a(t)^{\frac{1}{2}(1 + 3w)} = \frac{2}{(1 + 3w)}(aH)^{-1} . \end{equation}

Using this as motivation, Baumann and Dodelson suggest that the natural solution to the horizon problem is to conjecture a phase of decreasing Hubble radius in the early universe,

$$ \frac{d}{dt}(aH)^{-1} < 0. $$

This is cited as the solution because the Hubble sphere is growing in standard Big Bang cosmology.

Now, I don't understand why it is a natural or easy solution to consider a period of time where the Hubble radius is decreasing. Is there some key insight that I am missing? Any help would be appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

This is the motivation for inflation, which is defined as expansion where the scale factor satisfies $\ddot{a} > 0$. During inflation, the Hubble scale as measured with respect to the expansion (so-called comoving coordinates) is shrinking because the expansion proceeds at a greater rate than the proper Hubble distance.

A cosmological fluid with $p < -\rho/3$, like vacuum energy or scalar quantum fields with most of their energy stored as potential energy, are suitable sources of inflationary expansion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.