I've found standard derivations showing how one can integrate the Friedmann equation to obtain $a(t)$ given $P_{tot}$ and $\rho_{tot}$. I'd like to know how $a(t)$ depends on the inflation potential.
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$\begingroup$ It will depend on the model you have in mind, the one I am somewhat familiar with is slow roll inflation, but I don't know enough to answer the question. I just know that the equations can be written in terms of the dynamics of the field and it's potential. $\endgroup$– TriatticusCommented Aug 14, 2018 at 17:39
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The Friedmann equation, $$H^2 = \frac{8\pi G}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$ relates the Hubble parameter to the inflaton energy density. During slow roll, $\dot{\phi}^2 \ll V$, with $H = \dot{a}/{a}$, $$\frac{{\rm d}a}{a} \propto \sqrt{V(\phi)}{\rm d}t.$$ Now, $V$ in the above is a function of $\phi$, and the integral is over time, $t$. We therefore need $\phi(t)$, which we can get via the Klein-Gordon equation, $$3H\dot{\phi} = -V'(\phi).$$
Note that it's in general not possible to get an analytic expression for $a(t)$ for an arbitrary inflaton potential, and instead you'll need to solve the coupled Friedmann and Klein-Gordon equations to obtain $a(t)$ parametrically.
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$\begingroup$ Thanks, this makes sense. Do I need to do it parametrically even for a simple potential for a massive inflaton field, say, V(\phi) = 1/2 m^2 \phi^2, if I don't want to assume slow roll so I can't neglect the derivatives of the potential? $\endgroup$– SgrACommented Aug 14, 2018 at 22:01
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$\begingroup$ If you're not assuming slow roll, you've got to solve the coupled differential equations: 1) $\ddot{\phi} + 3H\dot{\phi} + m^2\phi$ and 2) $H^2 = (8\pi G/3)(\dot{\phi}^2/2 + m^2\phi^2/2)$, with $H = \dot{a}/a$. $\endgroup$– bapowellCommented Aug 15, 2018 at 14:06
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$\begingroup$ Two questions: 1) Wouldn't there be a $-k/a^2$ term in the second equation? 2) Does this have an analytic solution? $\endgroup$– SgrACommented Aug 15, 2018 at 14:14
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$\begingroup$ 1) Why would there be? Presumably that would come from a gradient term, but the field is spatially homogeneous: perhaps you're thinking of the perturbation equation? 2) I have not idea ;)...haven't tried to solve it! $\endgroup$– bapowellCommented Aug 15, 2018 at 14:21
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$\begingroup$ 1) Isn't there a curvature term in the Friedmann equation (see for example here in eq. 4.21)? $\endgroup$– SgrACommented Aug 16, 2018 at 12:26