In this 2007 paper by Alan Guth discussing eternal cosmic inflation, he start's using a value $\phi$ on page 8. My understanding is that $\phi$ is the scalar field representing the dark energy of a "false vacuum" - but then why does $\phi$ appear as the dependent variable of the energy density function $V$?

  • $\begingroup$ Why is $V$ being a function of $\phi$ troubling to you? $\endgroup$
    – JamalS
    Commented Mar 26, 2014 at 11:14
  • $\begingroup$ Please don't direct-link to files... Link to the page if it's possible. $\endgroup$ Commented Mar 26, 2014 at 20:55
  • $\begingroup$ @JamalS - now that you mention it, I'm not sure. I think I was just confused about what the difference between $V$ and $\phi$ was. $\endgroup$ Commented Mar 27, 2014 at 3:56

1 Answer 1


$\phi$ in this context is typically known as the "inflaton" (a somewhat silly name, I know, but we already have quarks), it is the scalar field that drives inflation.

Any field may have a potential component, typically written as $V$. Then the Lagrangian can be written as the sum of kinetic term(s) with (the relevant covariant) derivatives and the potential term that describes how the field behaves (plus interaction terms).

For the case of the inflaton there are many models for what $V(\phi)$ should be, such as $V(\phi)=\frac14\lambda\phi^4$ as mentioned in the paper (a rather popular model right now too). That means that there is such a term in the Lagrangian.

Of course, $V(\phi)$ could be any arbitrary function of $\phi$ (of course most such functions wouldn't be applicable to inflation).

  • $\begingroup$ Ok, I think this clears it up. If I understand correctly, a newtonian analog would be $\phi$ is elevation, and $V$ is gravitational potential energy? $\endgroup$ Commented Mar 27, 2014 at 3:54
  • $\begingroup$ Yep! That's why they call one class of inflation as "slow-roll inflation" because the field needs to roll down a hill described by $V$ slowly. In fact, more generally, a potential $V$ is related to a force by $F=-\frac{dV}{dx}$. $\endgroup$
    – jazzwhiz
    Commented Mar 27, 2014 at 4:26

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