Why is $B(T)\approx b(T-T_C)$ near critical point $T_C$ in Landau theory?

In Peskin&Schroeder page $270$ equation $(8.4)$ you see that they approximate the function $B(T)$ near the Curie temperature as $$B(T)\approx b(T-T_C)$$ i.e. they omit $B(T_C)$ in the Taylor expansion of $B(T)$. Similarly for $C(T)$ they just keep $C(T) \approx c.$

How come this is OK?

For those without the book in hand:

$\partial G/\partial M=H$ where $G=$ Gibbs free energy, $M=$ magnetization, $H=$ external magnetic field. For $T$ near $T_C$, $M\approx 0$ so we expand gibbs: $$G(M) = A(T)+B(T)M^2+C(T)M^4+\cdots$$ with only even powers of $M$ due to symmetry of the system under $M\rightarrow -M$.

• In Landau's theory, $B(T)$ must have different sign for $T$ between the critical point $T_c$, and $C(T)$ has the same sign. Hence, as Taylor's expansion, $B(T)$ has to consider about the first order expansion, but zero order expansion is enough for $C(T)$. Commented Feb 27, 2014 at 13:45

In Landau's theory, the order parameter $M$ should make $G(M)$ minimal. $$\frac{\partial G}{\partial M} = 2 B(T) M + 4 C(T) M^3=0$$ $$\frac{\partial^2 G}{\partial M^2} = 2 B(T) + 12 C(T) M^2 > 0$$ Hence, $M=0$ or $M = \pm M_0 = \pm \sqrt{ - \frac{B(T)}{2 C(T)}}$
When $T$ is higher than the critical point $T_c$, the groud state of system satisfies $M=0$. Hence, $M=0$ is the minimal point, and $\frac{\partial^2 G}{\partial M^2} |_{M=0} = 2 B(T) > 0$, or $B(T)|_{T>T_c}>0$.
When $T$ is lower than $T_C$, the groud state of system satisfies $M=M_0$. $$\frac{\partial^2 G}{\partial M^2} |_{M=\pm M_0} = 2 B(T) + 12 C(T) (-\frac{B(T)}{2 C(T)}) = - 4 B(T) > 0$$ Or $B(T)|_{T<T_c}<0$. And $C(T)|_{T<T_c}>0$ due to $M_0 = \sqrt{ - \frac{B(T)}{2 C(T)}}$.
Now, here is a short summary, $T>T_c$, $B(T)>0$, and $T<T_c$, $B(T)<0$, $C(T)>0$.
The simplest assumptions are, by Taylor's expansion, $B(T)=b(T-T_c)$ and $C(T)=c$.