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I am studying the Dirac-Born-Infeld (DBI) inflation model and came across this question in a past exam paper from Cambridge that considered the following Lagrangian:

$$\mathcal{L}=\sqrt{-g}A(X,\phi)$$ where $$X=\frac{1}{2}\partial^\mu\phi\partial_\mu\phi$$ and $A$ is given by:

$$A=\frac{1}{f(\phi)}\Big(1-\sqrt{1-2Xf(\phi)}\Big)-V(\phi)$$

where $f(\phi)$ is the warp factor of the throat.

I calculated the slow roll parameter which is given by:

$$\epsilon=3\frac{\frac{\gamma}{2}\dot{\phi}^2}{\frac{\gamma^2}{\gamma+1}\dot{\phi}^2+V}$$ where $$\gamma=\frac{1}{\sqrt{1-2Xf(\phi)}}$$

The problem then asks to provide a physical explanation on why inflation can be achieved even if $\epsilon_V>1$ which is different from the more conventional models seen. I can see from the $\epsilon$ that the kinetic energy times $\gamma$ must be smaller than $V$ but can't see any physical explanation for why inflation can work for $\epsilon_V>1$.

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    $\begingroup$ Please cite the paper's link. $\endgroup$
    – Avantgarde
    Jan 13 '19 at 18:50
  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Jan 13 '19 at 20:07
  • $\begingroup$ Sorry, by paper I meant a past exam paper from Cambridge. I have attached a screenshot of the question above. $\endgroup$
    – James J
    Jan 14 '19 at 10:26
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In DBI inflation, the inflaton is the scalar field parameterizing the position of a brane evolving in a warped throat geometry. Being a physical object, the brane is constrained to move at subluminal velocities, regardless of the steepness of the potential -- hence the Lorentz contraction in the kinetic term. This factor has the effect of a speed limit, allowing the field to evolve relatively slowly even in steep potentials.

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