= Solution
A <phase diagram> partitions a space of thermodynamic controls into regions with distinct equilibrium phases. Its boundaries indicate <phase coexistence> or <continuous phase transitions>; a <critical point> is a termination of coexistence where the distinction between phases disappears. A concrete three-dimensional control space is $(T,\Delta,h)$ for the <Blume–Capel model>: a spin can be $0$ or $\pm1$, the exchange favors aligned nonzero spins, and $\Delta$ controls the energetic cost of nonzero spins. Varying $\Delta$ changes the balance between a continuous ordering transition and a <first-order phase transition>. Their meeting is a <tricritical point>.
Locally, the same structure is displayed by the <Landau free energy>
$$
f(M)=\frac r2M^2+\frac u4M^4+\frac v6M^6-hM,\qquad v>0,
$$
with three controls $(r,u,h)$. For $u>0$, the line $r=h=0$ consists of ordinary <critical points>. For $u<0$, equal free energies of $M=0$ and a nonzero stationary point require
$$
r+uM^2+vM^4=0,\qquad \frac r2M^2+\frac u4M^4+\frac v6M^6=0.
$$
Eliminating $r$ gives the <tricritical three-phase line>
$$
\boxed{M^2=-\frac{3u}{4v},\qquad r=\frac{3u^2}{16v},\qquad h=0.}
$$
Here $M=0$ and both signs of $M$ coexist, with a discontinuous <order parameter>. Below this line, or below $r=0$ when $u>0$, the plane $h=0$ is an ordered <phase coexistence> sheet: crossing it reverses the sign of $M$. Two <tricritical wings> also extend to nonzero $h$, separating weakly and strongly magnetized states of the same sign. Their ordinary <critical points> satisfy $f'=f''=f'''=0$. Since $f'''=6uM+20vM^3$, their <tricritical wing critical edges> are
$$
M_c^2=-\frac{3u}{10v},\qquad r_c=\frac{9u^2}{20v},\qquad h_c=\frac{6u^2}{25v}M_c.
$$
At these edges $f''''=-12u>0$, giving an ordinary quartic critical expansion around $M_c$. All these structures end at \b[the tricritical point $r=u=h=0$]. The diagrams show both the zero-field section and the genuinely three-dimensional coexistence structure.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-51-landau-phase-diagram.png]
{title=Scalar Landau phase diagram: zero-field transitions, tricritical wings and ordinary critical edges}
For the final <critical exponent> calculation, take $r=at$ with $a>0$, $t=(T-T_c)/T_c$, and hold the other controls fixed. At an ordinary <critical point>, the quartic term dominates the sextic term. For $h=0$ and $r<0$,
$$
M^2=-r/u,\qquad f_{\rm eq,s}=-r^2/(4u),
$$
whereas on the disordered side the singular contribution is zero. Two temperature derivatives give a finite jump in <heat capacity>, so $\alpha=0$. At $r=0$, the <equation of state> is $h=uM^3+O(M^5)$, so $\delta=3$.
Along the tricritical trajectory $u=0$, the nonzero minimum obeys $M^4=-r/v$ and
$$
f_{\rm eq,s}=\frac r2\sqrt{-r/v}+\frac v6(-r/v)^{3/2}
=-\frac{(-r)^{3/2}}{3\sqrt v}.
$$
Thus two temperature derivatives produce $|t|^{-1/2}$. At $r=u=0$, $h=vM^5$. Therefore the <mean-field critical exponents> are
$$
\boxed{(\alpha,\delta)_{\rm ordinary}=(0,3),\qquad(\alpha,\delta)_{\rm tricritical}=(1/2,5).}
$$
With $A$ the thermodynamic <free energy>, the physical sign is $C=-T\partial_T^2 A$, rather than the positive sign printed in the paper. This correction affects the sign of the <heat capacity>, not its <critical exponent>. The tricritical result assumes that the independent quartic control is tuned to zero; a generic trajectory through nearby ordinary critical points need not have these exponents.
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