 Solution

Let $t=(T-T_c)/T_c$, and suppose $r=r_{t}t$ with $r_t>0$ while $u>0$ and $\kappa >0$ remain finite at the transition. Critical exponents describe the leading singular powers: $M\sim (-t{)}^{\beta }$ below the transition, $\chi \sim |t{|}^{-\gamma }$, $\xi \sim |t{|}^{-\nu }$, $C_{V,s}\sim |t{|}^{-\alpha }$, $M(t=0)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$, and $G(R,t=0)\sim R^{-(D-2+\eta )}$.

The quartic equation of state $h=rM+u{M}^3$ immediately gives $M=\sqrt{-r/u}$ at zero field below the transition, so $\beta =1/2$. Differentiating with respect to the conjugate field gives ${\chi }^{-1}=r+3u{M}^2$, which is $r$ above and $2|r|$ below; therefore $\gamma =1$. At $r=0$, $M=(h/u{)}^{1/3}$ with its real sign, giving $\delta =3$. The quadratic correlation function has $\xi \propto |r{|}^{-1/2}$ and critical Ornstein--Zernike correlation function proportional to $q^{-2}$, giving $\nu =1/2$ and $\eta =0$.

Finally the minimized potential below the transition is $V_s=-r^2/(4u)$, while it is zero above. Its second temperature derivative is a finite step, so the mean-field heat-capacity critical exponent is $\alpha =0$. In particular,These are derived mean-field critical exponents. In the fluctuation-dominated critical region of a phase transition, one instead derives exponents from the scaling eigenvalues of the renormalization-group fixed point, as in the next question. A zero heat-capacity power exponent can also accompany logarithms in other theories, so specifying the power alone does not completely describe that singularity.