1
$\begingroup$

If we have an inflation model with potential $V(\phi) = V_0 e^{-\sqrt{\frac{2}{\lambda}} \frac{\phi}{M_p}}$, where $V_0$ and $\lambda$ are free parameters, does this lead to eternal inflation for $\lambda > 1$?

The slow roll parameter $\epsilon_V(\phi) = \frac{M_p^2}{2} (\frac{V_{'\phi}}{V})^2 = \frac{1}{\lambda}$ appears to be a constant and so for all $\lambda > 1$, $\epsilon_V(\phi) < 1$. This seems to imply that inflation never breaks down.

$\endgroup$
1
$\begingroup$

Yes, inflation does not end for single field inflation driven by a purely exponential potential. Either one interprets this form of the potential as approximating a different potential when observational scales exit the horizon, or, if taken to be exact, one must introduce some mechanism to end inflation. Non-canonical kinetic terms have been investigated for this purpose. Another thought would be to introduce a second, auxiliary field similar to hybrid inflation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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