# Why does large-field inflation cause bigger gravitational waves

I'm learning about inflation and tensor perturbations during it. I've read a few times that large-field inflation is "more important" with respect to inflation as it produced bigger tensor perturbations so would be more likely to be experimentally confirmed by gravitational wave (GW) observatories. However, I haven't really been able to find any reasoning for why large field produces bigger GW. For context, I've looked at Maggiore's second textbook on GW, Mukhanov's "theory of cosmological perturbations", "inflation and the theory of cosmological perturbations" by Riotto, and "introduction to the theory of the early universe" by Gorbunov and Rubakov. Perhaps it is in one of these textbooks and I didn't see it, or perhaps I read it but didn't understand it is what implies that large field has bigger GW.

If someone knows some textbook/paper that it is explained well and could point me to it I'd appreciate it, or if someone knows the answer that'd be even better.

ETA: I tried to come up with some of my own reasoning. The only thing I got was this: I was able to show that the tensor-to-scalar ratio for both large and small field to be r=$$\frac{1}{2}\epsilon$$, where $$\epsilon$$ is the slow-roll parameter. I'm quite confident with this expression (except for maybe the exact proportionality constant, but that shouldn't matter too much) as I saw a similar one elsewhere. Then, we can express $$\epsilon$$ using its expression in terms of the potential. I'm using the reduced Planck mass as $$M^2_{PL}=\frac{1}{8\pi G}$$.

$$\epsilon = \frac{M^2_{PL}}{2}\left(\frac{V'(\phi)}{V(\phi)}\right)^2$$

For large field (taking some general $$V(\phi)=\lambda \phi^n$$)

$$\epsilon_1 = \frac{1}{2}\left(\frac{n^2M^2_{PL}}{\phi^2}\right)$$

For small field (taking some general $$V(\phi)=V_0 - g\phi^n$$

$$\epsilon_2 \approx \frac{M^2_{PL}}{2}\left(\frac{g^2n^2\phi^{2n-2}}{V_0^2}\right)$$

where I took $$V_0>>g\phi^n$$.

If we then look at the other slow-roll parameter $$\eta=M^2_{PL}\frac{V''}{V}$$. So for large field, $$epsilon \approx \eta$$ and so inflation ends when $$\epsilon$$ and $$\eta$$ are close to 1. But for small field $$\eta>\epsilon$$ and so inflation ends before $$\epsilon$$ has a chance to get close to 1 meaning the tensor-to-scalar ratio will not reach the same the magnitude and so the gravitational waves produced will be smaller??????

$$\begin{equation} N=\int\frac{d\varphi}{\sqrt{2\epsilon}}~\Longrightarrow~~~\frac{dN}{d\varphi}=\frac{1}{\sqrt{2\epsilon}}\simeq\sqrt{\frac{8}{r}}~, \end{equation}$$ then, assuming $$\epsilon$$ is nearly constant during the last $$\Delta N=50\sim 60$$ e-folds, we can write $$\frac{\Delta N}{\Delta\varphi}\simeq\sqrt{\frac{8}{r}}~\Longrightarrow~~~r\simeq 8\left(\frac{\Delta\varphi}{\Delta N}\right)^2~.$$