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September 24, 2003

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

(1)V(ϕ,ψ)=λ 12ϕ 2((ϕm 1) 2+m 2 2)+λ 2ϕ 2ψ 2+λ 34(ψ 2m 3 2) 2 V(\phi,\psi)= \frac{\lambda_1}{2}\phi^2 \left((\phi-m_1)^2 +m_2^2 \right) +\lambda_2 \phi^2\psi^2 +\frac{\lambda_3}{4}(\psi^2-m_3^2)^2

near the saddle point, ϕ=ψ=0\phi=\psi=0,

(2)V(ϕ,ψ)λ 34m 3 4+12ω 2ϕ 2λ 32ψ 2+λ 2ϕ 2ψ 2 V(\phi,\psi)\sim \frac{\lambda_3}{4}m_3^4 +\frac{1}{2}\omega^2 \phi^2-\frac{\lambda_3}{2} \psi^2 +\lambda_2 \phi^2\psi^2

where ω 2=λ 1(m 1 2+m 2 2)\omega^2=\lambda_1(m_1^2+m_2^2). If the amplitude of the ϕ\phi oscillations is large enough, the anharmonicities of the potential in ϕ\phi might be important, but they would not significantly change the conclusion.


(3)ϕ(t)=A(t)cosωt \phi(t)=A(t)\cos \omega t

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

(4)A(t)=A 0e 32Ht A(t)= A_0 e^{-\frac{3}{2}H t}

But — and this is a crucial point — in order for the universe to be inflating, we need the λ 34m 3 4\frac{\lambda_3}{4}m_3^4 term in the potential to dominate the energy density. If the energy stored in the ϕ\phi oscillations dominates, then — instead of p=ρp=-\rho and inflation— we get |p|ρ|p|\ll\rho and ordinary matter-dominated expansion. So, regardless of how you start, the universe starts inflating only when A 0A_0 satisfies

(5)A 0ω 2λ 3m 3 4 A_0 \omega^2 \ll \lambda_3 m_3^4

In that case, we have

(6)H 2=8π3M pl 2λ 34m 3 4=2π3λ 3m 3 4M pl 2 H^2 = \frac{8\pi}{3 M_{pl}^2} \frac{\lambda_3}{4} m_3^4 = \frac{2\pi}{3} \frac{\lambda_3 m_3^4}{M_{pl}^2}

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

(7)H 2ω 2 H^2\ll \omega^2

Putting these together, we have A 0 2H 2λ 3m 3 4A_0^2 H^2 \ll \lambda_3 m_3^4, or

(8)A 0 2M pl 21 \frac{A_0^2}{M_{pl}^2}\ll 1

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

(9)V eff(ψ)=14λ 3m 3 4+12((λ 2A 2λ 3m 3 2)+λ 2A 2cos(2ωt))ψ 2 V_{\text{eff}}(\psi)=\frac{1}{4} \lambda_3 m_3^4 +\frac{1}{2} \left((\lambda_2 A^2-\lambda_3 m_3^2) +\lambda_2 A^2 \cos(2\omega t)\right)\psi^2

This is the vertically-forced harmonic oscillator. At first, I was convinced that parametric resonance would destabilize ψ\psi much earlier. But forget about parametric resonance. Let’s just ask: when does λ 2A(t) 2=λ 3m 3 2\lambda_2 A(t)^2=\lambda_3 m_3^2? Dvali and Kachru want this to happen after N=50N=50 e-foldings. That is, t=N/Ht=N/H. This comes down to

(10)λ 2=e 3Nλ 3m 3 2A 0 2 \lambda_2 = e^{3N} \textstyle{\frac{\lambda_3 m_3^2}{A_0^2}}

Using A 0 2ω 2λ 3m 3 4A_0^2\omega^2 \ll \lambda_3 m_3^4, this implies

(11)λ 2e 3Nω 2m 3 2 \lambda_2 \gg e^{3N} \textstyle{\frac{\omega^2}{m_3^2}}

Let’s be as optimistic as possible, setting m 3=M Pm_3=M_P and ω=1TeV\omega= 1\text{TeV}. To get N=50N=50, we would still need

(12)λ 210 33 \lambda_2 \gg 10^{33}

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!).

Posted by distler at September 24, 2003 5:21 PM

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