### Old and New

My previous post on Dvali and Kachru’s recent paper stimulated a lot of interest. I got into a really wonderful physics conversation with Konstantin Savvidis of the Perimeter Institute and with Sonia Paban down the hall.

In my previous post, I basically asserted that Dvali and Kachru’s proposal would not work with the sorts of parameters they envisage. But we thought it would be more fruitful to ask what range of parameters *would* be required to make it work.

Let’s expand the potential

near the saddle point, $\phi=\psi=0$,

where $\omega^2=\lambda_1(m_1^2+m_2^2)$. If the amplitude of the $\phi$ oscillations is large enough, the anharmonicities of the potential in $\phi$ might be important, but they would not significantly change the conclusion.

Write

The Hubble expansion damps these oscillations. In an inflating universe, we just get

But — and this is a crucial point — in order for the universe to be inflating, we need the $\frac{\lambda_3}{4}m_3^4$ term in the potential to dominate the energy density. If the energy stored in the $\phi$ oscillations dominates, then — instead of $p=-\rho$ and inflation— we get $|p|\ll\rho$ and ordinary matter-dominated expansion. So, regardless of how you start, the universe starts inflating only when $A_0$ satisfies

In that case, we have

As I said before, we need the oscillations to be underdamped, which means we should also require

Putting these together, we have $A_0^2 H^2 \ll \lambda_3 m_3^4$, or

Following Dvali and Kachru, let’s plug this slowly-decaying simple harmonic motion for $\phi$, $\phi=A\cos \omega t$, into the potential and treat it as a time-dependent effective potential for $\psi$ (if the amplitude of the $\psi$ motion is small enough, the back-reaction is negligible).

This is the vertically-forced harmonic oscillator. At first, I was convinced that parametric resonance would destabilize $\psi$ much earlier. But *forget* about parametric resonance. Let’s just ask: when does $\lambda_2 A(t)^2=\lambda_3 m_3^2$? Dvali and Kachru want this to happen after $N=50$ e-foldings. That is, $t=N/H$. This comes down to

Using $A_0^2\omega^2 \ll \lambda_3 m_3^4$, this implies

Let’s be as optimistic as possible, setting $m_3=M_P$ and $\omega= 1\text{TeV}$. To get $N=50$, we would still need

That’s wildly implausible.

And parametric resonance would only make things worse…

Unfortunately, the result of this little collaboration was frightfully dull (anyone still reading this post?). But it was a lot of fun to see the possibilities inherent in blogging about physics. Konstantin and I never would have hooked up, but for my post. This is why I took up blogging in the first place.

Besides, pre-blog, this depressing little calculation would just have ended up buried in my notebooks, unpublished. Now it’s on the web for all to read.

*Some* would call that progress.

**Update (9/25/2003):** Typo in penultimate formula corrected (*Thanks, Sonia!*).