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I've been reading up on inflation, and have arrived at the so-called

I have to admit, I'm having trouble understanding a couple of points. First, how does ##\epsilon\ll 1## guarantee that ##H## is approximately constant (evolving slowly) during inflation? Doesn't this condition simply enforce ##\dot{H}\ll H^{2}## and not necessarily that ##\dot{H}\ll 1##? I can see in the limit that ##\epsilon\to 0## that this condition implies that ##\dot{H}=0##. Is the point that the relevant time-scale of the problem is the Hubble time ##t=H^{-1}##, and thus we want ##H## to not change appreciably over one Hubble time, such that the solution for ##H## will be well approximated by ##H\approx\text{constant}##?

As for the second condition, is this just to ensure that the equation of motion for the scalar field is dominated by the friction term ##3H\dot{\phi}## and the derivative of the potential ##V'(\phi)##? Again, how is it clear from this that ##\ddot{\phi}## is (in some sense) small during inflation.

Moving on to the "potential" slow-roll conditions, these follow from the above conditions: $$\epsilon_{V}=\frac{M_{\text{Pl}}^{2}}{2}\left(\frac{V'(\phi)}{V(\phi)}\right)^{2}\ll 1\;,\qquad\eta_{V}=M_{\text{Pl}}^{2}\frac{V''(\phi)}{V(\phi)}\ll 1$$ Analogously to the questions above, how do these condition guarantee that the potential is approximately constant during inflation? Don't they simply imply that the first and second derivatives of ##V## must be much less than ##V/M_{\text{Pl}}##? Is the point that one has ##\frac{1}{2}\dot{\phi}^{2}\ll V(\phi)##, which implies that ##p_{\phi}\approx -

\rho_{\phi}## and ##\rho_{\phi}\approx V(\phi)##, and so the continuity equation becomes ##\dot{\rho}_{\phi}\approx 0\Rightarrow V(\phi)\approx\text{constant}##?

I feel a bit dense asking this question, but I seem to be having a mental block on this.

*slow roll conditions*$$\epsilon =-\frac{\dot{H}}{H^{2}}\ll 1\; ,\qquad\eta =-\frac{\ddot{\phi}}{H\dot{\phi}}\ll 1$$I have to admit, I'm having trouble understanding a couple of points. First, how does ##\epsilon\ll 1## guarantee that ##H## is approximately constant (evolving slowly) during inflation? Doesn't this condition simply enforce ##\dot{H}\ll H^{2}## and not necessarily that ##\dot{H}\ll 1##? I can see in the limit that ##\epsilon\to 0## that this condition implies that ##\dot{H}=0##. Is the point that the relevant time-scale of the problem is the Hubble time ##t=H^{-1}##, and thus we want ##H## to not change appreciably over one Hubble time, such that the solution for ##H## will be well approximated by ##H\approx\text{constant}##?

As for the second condition, is this just to ensure that the equation of motion for the scalar field is dominated by the friction term ##3H\dot{\phi}## and the derivative of the potential ##V'(\phi)##? Again, how is it clear from this that ##\ddot{\phi}## is (in some sense) small during inflation.

Moving on to the "potential" slow-roll conditions, these follow from the above conditions: $$\epsilon_{V}=\frac{M_{\text{Pl}}^{2}}{2}\left(\frac{V'(\phi)}{V(\phi)}\right)^{2}\ll 1\;,\qquad\eta_{V}=M_{\text{Pl}}^{2}\frac{V''(\phi)}{V(\phi)}\ll 1$$ Analogously to the questions above, how do these condition guarantee that the potential is approximately constant during inflation? Don't they simply imply that the first and second derivatives of ##V## must be much less than ##V/M_{\text{Pl}}##? Is the point that one has ##\frac{1}{2}\dot{\phi}^{2}\ll V(\phi)##, which implies that ##p_{\phi}\approx -

\rho_{\phi}## and ##\rho_{\phi}\approx V(\phi)##, and so the continuity equation becomes ##\dot{\rho}_{\phi}\approx 0\Rightarrow V(\phi)\approx\text{constant}##?

I feel a bit dense asking this question, but I seem to be having a mental block on this.

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