Undergrad Troubles understanding slow-roll conditions in Inflation

["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.

 

[Orodruin]

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.

 

["Don't panic!"]

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 $\Delta t\sim t_{H}$?

 

[kimbyd]

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.

 

["Don't panic!"]

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)?

 

[kimbyd]

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.

 

["Don't panic!"]

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)?!

 

[kimbyd]

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.

 

["Don't panic!"]

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 $t$).

 

[kimbyd]

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.