Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 34D / iv / Solution

The expansion about the liquid-gas critical point isDifferentiating at fixed $t$ gives the requested $-9{\varphi }^2/2+O({\varphi }^3)+t[-6+O(\varphi )]$. At coexistence $\varphi =O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar{p}=1+4t-6t\varphi -3{\varphi }^3/2$.

Let $a={\varphi }_l$, $b={\varphi }_g$ for this leading cubic model. Equality of endpoint pressures gives $a^2+ab+b^2=-4t$. The Maxwell condition in pressure coordinates is ${\int }_a^b(1+\varphi ){\bar{p}}^{\prime }(\varphi )d\varphi =0$. Since the endpoint pressures agree, its first term vanishes, leavingIf $a+b\ne 0$, these two equations would imply $(a-b{)}^2=0$, contrary to distinct phases. Therefore $a=-b$, and the pressure equality gives $a^2=b^2=-4t$. HenceThus the coexistence gap has exponent one half. The printed equality ${\varphi }_l(t)=-{\varphi }_g(t)$ is a leading-order statement, not an exact identity for the full Van der Waals equation. Discarded asymmetric terms shift both phase volumes at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.