# Troubles understanding slow-roll conditions in Inflation

• I
• "Don't panic!"

#### "Don't panic!"

I've been reading up on inflation, and have arrived at the so-called 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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Neither ##\dot H## nor ##\ddot\phi## are dimensionless. Hence, an expression such as ##\dot H \ll 1## does not make sense. Instead, ##\dot H## needs to be compared to something that carries information about a typical scale for that quantity. Since H has dimension 1/T, ##\dot H## has dimension 1/T^2 and the typical time scale is set by 1/H so it needs to be compared to ##H^2##, not to 1. The key is in your parenthesis: ”in some sense”. You can only define the ”in some sense” relaive to other quantities with the same dimensions.

Neither ##\dot H## nor ##\ddot\phi## are dimensionless. Hence, an expression such as ##\dot H \ll 1## does not make sense. Instead, ##\dot H## needs to be compared to something that carries information about a typical scale for that quantity. Since H has dimension 1/T, ##\dot H## has dimension 1/T^2 and the typical time scale is set by 1/H so it needs to be compared to ##H^2##, not to 1. The key is in your parenthesis: ”in some sense”. You can only define the ”in some sense” relaive to other quantities with the same dimensions.

Thanks for your response. I was editing my original post when your answer was posted. I'd appreciate it if you could have a look at the updated version and see if you agree with what I've written.

Why is the typical time scale set by ##1/H##? I've read that it's the time scale over which the scale factor changes appreciably. Is this why it is the relevant time scale, as it's the time-scale over which the universe changes noticeably? How can one see that this is the case? Is it simply that one can Taylor expand ##a(t)## about its value ##a(t_{1})## at some time ##t_{1}##, i.e., ##a(t_{1}+\Delta t)\approx a(t_{1})\left(1+\frac{\Delta t}{t_{H}} +\mathcal{O}\Big(\big(\frac{\Delta t}{t_{H}}\big
)^{2}\Big
)\right)##, where ##t_{H}## is the Hubble time at ##t=t_{1}##. Therefore ##a(t)## will only change appreciably from its value at ##t=t_{1}## over the time scale ##\Delta t\sim t_{H}##?

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Why is the typical time scale set by ##1/H##? I've read that it's the time scale over which the scale factor changes appreciably. Is this why it is the relevant time scale, as it's the time-scale over which the universe changes noticeably?
Basically. It's a statement that the universe will have to expand quite a lot before there is a significant change in either ##H## or ##\dot{\phi}##. And if the universe has to expand a lot before there is a significant change in ##H##, then that is an indication that the rate of expansion is nearly exponential.

Basically. It's a statement that the universe will have to expand quite a lot before there is a significant change in either ##H## or ##\dot{\phi}##. And if the universe has to expand a lot before there is a significant change in ##H##, then that is an indication that the rate of expansion is nearly exponential.

So the slow roll condition is that the change in ##H## is small per Hubble time then. Ss what I wrote in the couple of sentences at the end of my previous post correct at all (as to why the Hubble time is the appropriate time scale)?

So the slow roll condition is that the change in ##H## is small per Hubble time then. Ss what I wrote in the couple of sentences at the end of my previous post correct at all (as to why the Hubble time is the appropriate time scale)?
Not quite. ##a(t)## might change quite dramatically over that period. But it will behave close to ##a(t) = a_0 e^{H t}##. The expansion would be exactly exponential if ##H(t)## was exactly constant. The slow-roll conditions ensure that the behavior is close enough to this that it's useful to think of the expansion during inflation in terms of exponential expansion.

Mathematically the slow-roll conditions are chosen to make the equations of motion easier to solve. That requirement is ultimately why they take the precise form they do.

Not quite.

But I thought the Hubble time set the time-scale over which the scale factor changes appreciably (at least that's what I've read in a set of notes)?!

But I thought the Hubble time set the time-scale over which the scale factor changes appreciably (at least that's what I've read in a set of notes)?!
After one Hubble time with exponential expansion, ##a## increases by one factor of ##e## (2.718), i.e. nearly triples. So yes, a Hubble time is related to the amount of time for there to be a significant amount of expansion. But that amount of expansion can be pretty large depending upon the contents of the universe.

After one Hubble time with exponential expansion, ##a## increases by one factor of ##e## (2.718), i.e. nearly triples. So yes, a Hubble time is related to the amount of time for there to be a significant amount of expansion. But that amount of expansion can be pretty large depending upon the contents of the universe.

Okay, so can one say that, given the scale factor ##a(t)## at some time ##t##, then over a time interval ##\Delta t## of order the Hubble time (i.e. ##\Delta t\sim H^{-1}##) the universe will change significantly, however, for intervals much less than this (i.e. ##\Delta t\ll H^{-1}##), the scale factor at ##t+\Delta t##, ##a(t+\Delta t)## will not have changed significantly from its value at time ##t## (and thus the universe will not "look" much different than it did at time ##t##).

Okay, so can one say that, given the scale factor ##a(t)## at some time ##t##, then over a time interval ##\Delta t## of order the Hubble time (i.e. ##\Delta t\sim H^{-1}##) the universe will change significantly, however, for intervals much less than this (i.e. ##\Delta t\ll H^{-1}##), the scale factor at ##t+\Delta t##, ##a(t+\Delta t)## will not have changed significantly from its value at time ##t## (and thus the universe will not "look" much different than it did at time ##t##).
Correct, provided the ways in which the universe changes are dominated by expansion. One might imagine scenarios where other effects are dominant (e.g. gravitational collapse in a slowly-expanding universe). As inflation is inherently tied to the expansion, the time scale of expansion is the right time scale.