Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

[This is a version of the question that I've revised based on helpful comments from Dan.]

I haven't studied inflation at a technical level. My picture of the process is that we have an inflaton field $\phi$ which is a scalar and has a potential $V(\phi)$. Before inflation, the field starts at some $\phi_o$ that doesn't equal the value $\phi_m$ that minimizes $V$. It then rolls downhill to the minimum.

Assuming I have this right, what I don't understand is (1) why $\phi$ initially has a single value $\phi_o$ everywhere, and (2) what sets $\phi_o$. Before the onset of inflation, the universe's temperature was much higher than any scale set by $V$. When the age of the universe was on the order of the Planck time, presumably its temperature was on the Planck scale. If the temperature was that high, wouldn't thermal fluctuations cause the field to sample a wide range of values of $\phi$, contradicting #1? And if it could sample all those values, I would expect that thermodynamically, it would consist overwhelmingly of values of $\phi$ close to $\phi_m$, not some other value $\phi_o$.

share|improve this question
1  
The theory of inflation I'm familiar with doesn't have a mexican hat potential, as such a potential doesn't have the proper criterion for slow roll and thus doesn't yield the proper amount of e-folds necessary. But I do know the mexican hat potential represents the higgs potential. –  Dan Jun 7 '13 at 23:19
1  
@Dan: Thanks for your comment. Looking at this review paper, arxiv.org/abs/0705.0164 by Linde, he says on p. 9, "The first models were based on the theories with polynomial potentials, such as $V(\phi)=\pm \frac{m^2}{2} \phi^2+\frac{\lambda}{4}\phi^4$." I guess the minus sign on the first term would be a Mexican hat, but clearly things are more complicated than I thought. He starts the paper with a plain old simple harmonic oscillator potential, with the field rolling downhill toward the center. –  Ben Crowell Jun 8 '13 at 0:19
1  
True I do recall that potential, I just know that any potential needs a gently sloping initial segment (for slow roll $ V (\phi) > \dot \phi $) then a rapid decline into a harmonic oscillator state –  Dan Jun 8 '13 at 0:39
1  
Ginsburg slide 19 has a useful picture here –  twistor59 Jun 12 '13 at 7:24
1  
Still thinking... this ~ eq 56, has pretty much convinced me that the gradients fade into nothingness. Even if initial inhomogeneities mean different inflation rates in different directions, the gradients still fade and approach zero, which is what we want. –  twistor59 Jun 17 '13 at 15:56
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.