# What is the topology of a phase diagram?

Looking at various two-variable phase diagrams I was struck by that on every one I have seen so far all the phases formed simple connected regions; see, for example the phase diagrams of $$H_2O$$ or of $$Fe-Fe_3C$$ in 1. Every phase is a connected region, no two disjoint pieces correspond to the same phase. Is this really true in general and if so why?

Going further, let us consider a three-variable phase diagram, say $$p, T, y$$, where $$y$$ maybe electric or magnetic field intensity, or some other controllable extensive/intensive variable. Is it possible that now a 3d region occupied by some phase is a kind of tube that makes a half turn so that if we fix, say, $$y$$ then the resulting 2d section would have two disjoint $$p,T$$ regions of the same phase? Can such phase "tube" exist or by some thermodynamic-geometric rule is it excluded?

A possibly more difficult question if a phase can form an annular region, ie., homeomorphic to an annulus while surrounding another phase homeomorphic to a disk?

In short, what is the topology of the phase diagram and its sections?

• Could it boil down to a smart choice of variables? Think of an ising-like phase diagram (for example here). If you use the variable $x=1/M$ instead of $M$, then the ferromagnetic phase becomes disconnected. – Steven Mathey Apr 11 at 13:25

I don't have a complete answer, but I do have some thoughts. I'm going to focus on the $$T$$-$$P$$ phase diagram for a pure substance because that is what I know best.

• $$T$$ and $$P$$ are special coordinates because, at phase equilibrium they are uniform between all phases. A diagram based on $$T$$ and $$v$$ or $$P$$ and $$T$$ will not have the same topology. I mention this because it suggests (to me) that any proof of the properties you observe will in some way make use of the fact the coordinates are $$T$$ and $$P$$

• One (kind of disappointing) answer would be that this is just the way that we've defined phases: chop up the phase diagrams into regions with distinct boundaries and label each region as its own phase.

A more mathematical answer would be that, at any given $$T$$ and $$P$$, the equilibrium phase(s) are those with the lowest $$g$$ (specific Gibbs energy, also called chemical potential $$\mu$$). Imagine a plot of $$g(T,P)$$ with one series (surface) for each phase that could ever exist. For some phase X, the partial derivatives of $$g_X(T,P)$$ are

\begin{align} \left(\frac{\partial g_X}{\partial T}\right)_P = - s_X \hspace{5em} \left(\frac{\partial g_X}{\partial P}\right)_T = v_X \end{align}

• Phase boundaries occur when two of these surfaces intersect. At each such intersection, the phases have the same $$T$$, $$P$$ and $$g$$ but different $$s$$ and $$v$$, and thus the rates of change of the intersecting surfaces along the $$T$$ and $$P$$ directions are different. The surfaces therefore cross.

• Let's imagine that we are at some point $$(T_0, P_0)$$ at which increasing $$T$$ causes phase A to transform into phase B (the phase boundary between A and B at some given pressure $$P$$). At lower $$T$$, phase A has lower $$g$$, but at higher $$T$$, phase B has lower G, so it follows that

\begin{align} \left(\frac{\partial g_A}{\partial T}\right)_P &> \left(\frac{\partial g_B}{\partial T}\right)_P \\ -s_A &> -s_B \\ s_B &> s_A \end{align}

• If phases A and B can coexist again at the a higher $$T$$ labelled $$T_1$$ and the same , then they must somehow reach the same $g$ again, i.e. it must be true that

\begin{align} \int_{T_0}^{T_1} \left(\frac{\partial g_A}{\partial T}\right)_P\ \text{d}T &= \int_{T_0}^{T_1} \left(\frac{\partial g_B}{\partial T}\right)_P\ \text{d}T \\ \int_{T_0}^{T_1} -s_A(T, P_0)\ \text{d}T &= \int_{T_0}^{T_1} -s_B(T, P_0)\ \text{d}T \end{align}

• This implies that, for a second crossing, the phase which initially has higher $$s$$ must switch over to having lower $$s$$ for some period, such that the integrals cancel out and the two states again reach the same $$g$$. This strikes me as unusual, potentially impossible behaviour. A similar argument could be made by swapping $$T$$ for $$P$$ and $$-s$$ for $$v$$.

The big flaw with this mathematical approach is that it assumes that $$g_X(T,P)$$ is well-defined for every phase at every $$T$$ and $$P$$. In classical thermodynamics, this function is only well-defined in regions where X is an equilibrium phase.

• You wrote "T and P are special coordinates because, at phase equilibrium they are uniform between all phases". I think all intensive parameters would be uniform in phase equilibrium across the several phases. I do not understand the rest yet but I disagree with you that it is a matter of "chopping". Anyhow, take T and p as the independent variables, I have not seen a phase diagram yet where one phase occupies an annular region, is that possible? – hyportnex Jan 16 at 0:09
• Think of the specific volume (the inverse of the density) as an example. Liquid water and steam can coexist at the right $(T,P)$ combinations, but the only time at which they have the same density is at the critical point (when liquid and vapour become indistinguishable). The same is true for $u$, $h$, and $s$. Regarding "chopping," the suggestion isn't that it introduces new, arbitrary boundaries to the diagram; the suggestion is that the phase space is "chopped" into regions at pre-existing physical boundaries and then each of the resulting contiguous regions is labelled as a distinct phase. – user1476176 Jan 16 at 4:27
• My answer doesn't address the idea of an annular region because I honestly do not know whether it is possible. I suspect that it is not possible, but I can't make a convincing argument as to why - the argument in my answer is the best I've come up with so far. – user1476176 Jan 16 at 4:30