= Solution
At the <tricritical point>, tune $u=0$, hold $v>0$, and let $r=a t$ cross zero. The equation of state becomes $h=rM+vM^5$. At $h=0,r<0$, $M^4=-r/v$, while the inverse susceptibility on the ordered branch is $r+5vM^4=4|r|$. On the disordered side it is $r$. At $r=0$, $M=\operatorname{sgn}(h)(|h|/v)^{1/5}$. Therefore $\beta=1/4$, $\gamma=1$ and $\delta=5$.
The minimized singular <free-energy density> is
$$
f_s(r,0)=-\frac{|r|^{3/2}}{3\sqrt v}\quad(r<0),\qquad f_s(r,0)=0\quad(r>0).
$$
Two temperature derivatives give a singular <heat capacity> proportional to $|t|^{-1/2}$ on the ordered side. Consequently the <mean-field tricritical exponent calculation> gives
$$
\boxed{(\alpha,\beta,\gamma,\delta)=(\tfrac12,\tfrac14,1,5).}
$$
The displayed <scaling relations for critical exponents> hold numerically:
$$
\frac12+2\left(\frac14\right)+1=2,\qquad\frac14\cdot5=\frac14+1.
$$
These are <Landau theory> values with both relevant coefficients tuned: along a generic path with fixed $u>0$, the asymptotic transition is ordinary rather than tricritical. The sextic <upper critical dimension> is three, so fluctuations can change the powers below three dimensions, and logarithmic corrections are possible at three. The vanishing disordered mean-field singular amplitude does not invalidate the ordered-side exponent.
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