# Critical point of Mabuchi energy functional has zero scalar curvature

I am currently reading the proof of convergence of Kähler-Ricci flow in the case $c_1(M)=0$ from Song, Weinkove. On page 45 he defines the term Mabuchi's $K$-energy functional:

$$\frac{d}{dt}\mathrm{Mab}_{\omega_0}(\phi_t)=-\int_M\dot{\phi}_tR_{\phi_t}\omega_{\phi_t}^n$$ on the space: $$\mathrm{PSH}(M, \omega_0)=\left\{\phi\in C^\infty(M)\mid \omega_0+\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\phi>0\right\}$$ and where $\omega_{\phi_t}=\omega_0+\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\phi_t$ and $R_{\phi_t}$ is the scalar curvature of $\omega_{\phi_t}$.

Then he directly claims that if $\phi_\infty$ is a critical point of $\mathrm{Mab}_{\omega_0}$ then $\omega_\infty$ has zero scalar curvature. I do not see why is that so. I am not an expert in this realm, so I might miss some well-known facts or obvious results.

Any comment is welcome!

$\newcommand{\dd}{\partial}$It's a fundamental idiom of the calculus of variations that if $\int fg = 0$ for all $f$, then $g \equiv 0$.
The first displayed equation in your question isn't the Mabuchi energy itself, but the variation of the Mabuchi energy along an arbitrary path of smooth Kähler metrics. At a critical point, the variation vanishes (by definition) for all $\dot{\phi}_{t}$. Consequently, $$-\int_{M} \dot{\phi}_{\infty} R_{\infty}\, \omega_{\infty}^{n} = 0 \quad\text{for all \dot{\phi}_{\infty},}$$ namely, for every path through $\omega_{\infty}$ with $\dot{\omega}_{\infty} = \frac{i}{2\pi}\dd\bar{\dd} \dot{\phi}_{\infty}$. By the fundamental idiom, $R_{\infty} \equiv 0$, i.e., $\omega_{\infty}$ has vanishing scalar curvature.