这是一份warwick华威大学PX396的成功案例

$$
s_{1}=\sqrt{1-s_{3}^{2}} \sin \theta, \quad s_{2}=\sqrt{1-s_{3}^{2}} \cos \theta
$$
Then the Lagrangian becomes
$$
L=\frac{1}{2 g}\left(\partial_{\mu} \vec{s}\right)^{2}=\frac{1}{2 g}\left(\left(1-s_{3}^{2}\right)\left(\partial_{v} \theta\right)^{2}+\frac{\left(\partial_{\mu} s_{3}\right)^{2}}{1-s_{3}^{2}}\right)
$$
If we anticipate that, for small $g, s_{3} \sim \mathrm{O}(\sqrt{g})$, we can approximate $L$ by
$$
L=\frac{1}{2 g}\left[\left(\partial_{v} s_{3}\right)^{2}+\left(1-s_{3}^{2}\right)\left(\partial_{v} \theta\right)^{2}+s_{3}^{2}\left(\partial_{\mu} s_{3}\right)^{2}+\cdots\right]
$$
Finally, it is convenient to rescale $s_{3}$ to eliminate the overall factor of $g, h(r) \equiv$ $s_{3}(r) / \sqrt{g}$, so
$$
L=\frac{1}{2}\left[\left(\partial_{v} h\right)^{2}+\left(\frac{1}{g}-h^{2}\right)\left(\partial_{v} \theta\right)^{2}+g h^{2}\left(\partial_{\mu} h\right)^{2}+\cdots\right]
$$

PX396 COURSE NOTES ：

$$
m=\frac{1}{a} F(g)
$$
However, $m$ is a physical quantity that does not depend on the cutoff scheme, so
$$
\frac{\mathrm{d}}{\mathrm{d} a} m=0
$$
which implies that $a$ and $g$ must be related. We have already seen that the relation is given by asymptotic freedom in the case of the $\mathrm{O}(3)$ sigma model, and now let’s see that this determines the functional dependence of $F(g)$. Substituting the dimensional-analysis statement for $m$ into the last equation implies that
$$
-\frac{1}{a^{2}} F(g)+F^{\prime}(g) \frac{\mathrm{d} g}{\mathrm{~d} a}=0
$$
but we have determined how $g$ must depend on the ultraviolet cutoff $a$. Let’s use standard and general notation for that result, the Callan-Symanzik relation,
$$
a \frac{\mathrm{d} g}{\mathrm{~d} a}=\beta(g)
$$