First-order phase transitions in Landau theory

As we have seen, Landau theory is based on the assumption that the order parameter is small near the critical point, and we have seen in the example of the Ising model how it can describe a continuous phase transition (in fact, for $t\to 0$$t\to 0$ we have $\eta \to 0$$\eta \to 0$). However, because of the symmetry properties of the Ising model we have excluded any possible cubic term; what we now want to do is to consider a more general form of ${\mathcal {L}}$${\mathcal {L}}$ which includes also a cubic term in $\eta$$\eta$, and see that this leads to the occurrence of a first-order phase transition. We have seen that since the order parameter is null for $T>{\overline {T}}$$T>{\overline {T}}$ the Landau free energy can't contain any linear term in $\eta$$\eta$. Let us therefore consider:

$${\mathcal {L}}={\frac {a}{2}}t\eta ^{2}-w\eta ^{3}+{\frac {b}{4}}\eta ^{4}-H\eta \quad \quad t={\frac {T-{\overline {T}}}{2}}$$

$${\mathcal {L}}={\frac {a}{2}}t\eta ^{2}-w\eta ^{3}+{\frac {b}{4}}\eta ^{4}-H\eta \quad \quad t={\frac {T-{\overline {T}}}{2}}$$

where we choose $w>0$$w>0$ and we have redefined $t$$t$ for convenience; as in the previous case, we must have $b>0$$b>0$ so that $\eta$$\eta$ has finite values in the equilibrium configurations. The temperature ${\overline {T}}$${\overline {T}}$ is the one at which we have the continuous transition if $w=0$$w=0$, but as we will see it doesn't have great significance now. The equilibrium configurations of the system, in absence of external fields, will be given by:

$${\frac {\partial {\mathcal {L}}}{\partial \eta }}=at\eta -3w\eta ^{2}+b\eta ^{3}=0\quad \Rightarrow \quad \eta (at-3w\eta +b\eta ^{2})=0$$

$${\frac {\partial {\mathcal {L}}}{\partial \eta }}=at\eta -3w\eta ^{2}+b\eta ^{3}=0\quad \Rightarrow \quad \eta (at-3w\eta +b\eta ^{2})=0$$

The solutions of this equation are $\eta ={\overline {\eta }}=0$$\eta ={\overline {\eta }}=0$ (disordered phase) and:

$${\overline {\eta }}_{\pm }={\frac {3w\pm {\sqrt {9w^{2}-4atb}}}{2b}}=c\pm {\sqrt {c^{2}-{\frac {at}{b}}}}$$

$${\overline {\eta }}_{\pm }={\frac {3w\pm {\sqrt {9w^{2}-4atb}}}{2b}}=c\pm {\sqrt {c^{2}-{\frac {at}{b}}}}$$

which correspond to the ordered phase, and for the sake of simplicity in the last step we have defined:

$$c:={\frac {3w}{2b}}$$

$$c:={\frac {3w}{2b}}$$

However, these two last solutions are possible only if:

$$c^{2}-{\frac {at}{b}}>0\quad \Rightarrow \quad t<t^{*}:={\frac {bc^{2}}{a}}$$

$$c^{2}-{\frac {at}{b}}>0\quad \Rightarrow \quad t<t^{*}:={\frac {bc^{2}}{a}}$$

Let us note that since $t^{*}$$t^{*}$ is positive, this will occur at temperatures higher than ${\overline {T}}$${\overline {T}}$ (because for $T={\overline {T}}$$T={\overline {T}}$ we have $t=0$$t=0$). Intuitively, since ${\mathcal {L}}\to \infty$${\mathcal {L}}\to \infty$ for $\eta \to +\infty$$\eta \to +\infty$ and ${\mathcal {L}}\to 0$${\mathcal {L}}\to 0$ for $\eta \to 0$$\eta \to 0$, we understand that ${\overline {\eta }}_{-}$${\overline {\eta }}_{-}$ will be a maximum while ${\overline {\eta }}_{+}$${\overline {\eta }}_{+}$ a minimum:

First-order transition

Same transition for lower values of the temperature

We therefore have that the introduction of the cubic term brings to an asymmetry in ${\mathcal {L}}$${\mathcal {L}}$ which leads to the formation of another minimum at $\eta ={\overline {\eta }}_{+}$$\eta ={\overline {\eta }}_{+}$ for $T<T^{*}$$T<T^{*}$.

Let us try to understand how ${\mathcal {L}}$${\mathcal {L}}$ behaves as a function of $T$$T$, by also explicitly plotting ${\mathcal {L}}$${\mathcal {L}}$ as shown in the figure above. If we start from $T>T^{*}$$T>T^{*}$, then the system will be in the disordered phase and ${\mathcal {L}}$${\mathcal {L}}$ will have only one minimum at $\eta ={\overline {\eta }}=0$$\eta ={\overline {\eta }}=0$. When $T$$T$ becomes smaller than $T^{*}$$T^{*}$ a new minimum appears at $\eta ={\overline {\eta }}_{+}$$\eta ={\overline {\eta }}_{+}$, but at the beginning we will have ${\mathcal {L}}({\overline {\eta }}_{+})>0$${\mathcal {L}}({\overline {\eta }}_{+})>0$ so this is only a local minimum (since ${\mathcal {L}}(0)=0$${\mathcal {L}}(0)=0$): in this range of temperatures the ordered phase is metastable. If we further decrease the temperature, we will reach a temperature $T^{**}$$T^{**}$ for which ${\mathcal {L}}({\overline {\eta }}_{+})=0={\mathcal {L}}(0)$${\mathcal {L}}({\overline {\eta }}_{+})=0={\mathcal {L}}(0)$: at this point the ordered and disordered phase coexist, so this is the temperature of a new transition! If we now further decrease the temperature to values lower than $T^{**}$$T^{**}$, ${\mathcal {L}}({\overline {\eta }}_{+})$${\mathcal {L}}({\overline {\eta }}_{+})$ becomes negative and so now ${\overline {\eta }}_{+}$${\overline {\eta }}_{+}$ is the global minimum of ${\mathcal {L}}$${\mathcal {L}}$: the ordered phase becomes stable and the disordered phase metastable. If now $T$$T$ becomes smaller than ${\overline {T}}$${\overline {T}}$, ${\mathcal {L}}$${\mathcal {L}}$ develops a new minimum for $\eta <0$$\eta <0$, but it is only a local minimum (the asymmetry introduced by $-w\eta ^{3}$$-w\eta ^{3}$ ensures that ${\overline {\eta }}_{+}$${\overline {\eta }}_{+}$ is always the global minimum). This means that also for ${\textstyle T<{\overline {T}}}$${\textstyle T<{\overline {T}}}$ the disordered phase with $\eta ={\overline {\eta }}_{+}$$\eta ={\overline {\eta }}_{+}$ continues to be the stable one, and so no phase transition occurs at ${\textstyle {\overline {T}}}$${\textstyle {\overline {T}}}$ any more; this is what we meant when we said that ${\overline {T}}$${\overline {T}}$ is not a relevant temperature any more. Therefore, we see that lowering the temperature of the system the value of $\eta$$\eta$ for which ${\mathcal {L}}$${\mathcal {L}}$ has a global minimum changes discontinuously from $\eta =0$$\eta =0$ to $\eta ={\overline {\eta }}_{+}$$\eta ={\overline {\eta }}_{+}$: this is a first-order transition.

As we have seen, the temperature $T^{**}$$T^{**}$ at which this first-order transition occurs is defined by two conditions: it must be a minimum of ${\mathcal {L}}$${\mathcal {L}}$ and such that the value of ${\mathcal {L}}$${\mathcal {L}}$ in that minimum is zero. Thus we can determine $T^{**}$$T^{**}$ as follows:

$${\begin{cases}{\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (at-3w\eta +b\eta ^{2})=0\\{\mathcal {L}}(\eta )=\eta ^{2}\left({\frac {a}{2}}t-w\eta +{\frac {b}{4}}\eta ^{2}\right)=0\end{cases}}\quad {\stackrel {\eta \neq 0}{\Longrightarrow }}\quad {\begin{cases}at-3w\eta +b\eta ^{2}=0\\{\frac {a}{2}}t-w\eta +{\frac {b}{4}}\eta ^{2}=0\end{cases}}$$

$${\begin{cases}{\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (at-3w\eta +b\eta ^{2})=0\\{\mathcal {L}}(\eta )=\eta ^{2}\left({\frac {a}{2}}t-w\eta +{\frac {b}{4}}\eta ^{2}\right)=0\end{cases}}\quad {\stackrel {\eta \neq 0}{\Longrightarrow }}\quad {\begin{cases}at-3w\eta +b\eta ^{2}=0\\{\frac {a}{2}}t-w\eta +{\frac {b}{4}}\eta ^{2}=0\end{cases}}$$

Solving this system for $t$$t$ and $\eta$$\eta$, we get:

$$\eta ={\overline {\eta }}^{**}=2{\frac {w}{b}}\quad \qquad t=t^{**}=2{\frac {w^{2}}{ab}}$$

$$\eta ={\overline {\eta }}^{**}=2{\frac {w}{b}}\quad \qquad t=t^{**}=2{\frac {w^{2}}{ab}}$$

Since by definition $t=(T-{\overline {T}})/2$$t=(T-{\overline {T}})/2$, we have:

$$T^{**}={\overline {T}}+4{\frac {w^{2}}{ab}}$$

$$T^{**}={\overline {T}}+4{\frac {w^{2}}{ab}}$$

Finally, we can also determine the susceptibility of the system. In the presence of an external field, the equation of state of the system is:

$${\frac {\partial {\mathcal {L}}}{\partial \eta }}=0\quad \Rightarrow \quad at\eta -3w\eta ^{2}+b\eta ^{3}=H$$

$${\frac {\partial {\mathcal {L}}}{\partial \eta }}=0\quad \Rightarrow \quad at\eta -3w\eta ^{2}+b\eta ^{3}=H$$

If we now derive both sides with respect to $H$$H$, since $\chi _{T}=\partial \eta /\partial H$$\chi _{T}=\partial \eta /\partial H$ we have:

$$at\chi _{T}-6w\eta \chi _{T}-3b\eta ^{2}\chi _{T}=1\quad \Rightarrow \quad \chi _{T}={\frac {1}{at-6w\eta +3b\eta ^{2}}}$$

$$at\chi _{T}-6w\eta \chi _{T}-3b\eta ^{2}\chi _{T}=1\quad \Rightarrow \quad \chi _{T}={\frac {1}{at-6w\eta +3b\eta ^{2}}}$$

Multicritical points in Landau theory

It is possible for a system to have more "disarranging parameters" than the sole temperature $T$$T$; let us call one such field $\Delta$$\Delta$. In this case the phase diagram of the system becomes richer, with coexistence and critical lines that intersect in points called multicritical points; one of the most common examples of a multicritical point is the tricritical point, which divides a first-order transition line from a second-order one. An example of a system of the type we are considering is the Blume-Emery-Griffiths model, which we have studied in Mean field theory for the Blume-Emery-Griffiths model. In that case the additional "disarranging field" was the concentration $x$$x$ of ${}^{3}$${}^{3}$He, and the tricritical point is the one we called $(x_{t},T_{t})$$(x_{t},T_{t})$.

Such a phenomenology can be obtained within Landau theory also with terms different from a simple cubic one; in particular, we can have first order phase transitions even when the system is invariant under parity, like in the case of the Ising model. In fact in that situation we required the coefficient of $\eta ^{4}$$\eta ^{4}$ to be always positive, but if this is not true then ${\mathcal {L}}$${\mathcal {L}}$ will be:

$${\mathcal {L}}={\frac {a}{2}}\eta ^{2}+{\frac {b}{4}}\eta ^{4}+{\frac {c}{6}}\eta ^{6}$$

$${\mathcal {L}}={\frac {a}{2}}\eta ^{2}+{\frac {b}{4}}\eta ^{4}+{\frac {c}{6}}\eta ^{6}$$

where $a$$a$, $b$$b$ and $c$$c$ are functions of $T$$T$ and $\Delta$$\Delta$, and $c$$c$ must always be positive for the stability of the system (otherwise, like in the case previously considered, the minimization of ${\mathcal {L}}$${\mathcal {L}}$ leads $\eta$$\eta$ to infinity). Now, we know that if $a$$a$ changes sign and $b$$b$ is kept positive (which can be done varying the values of $T$$T$ and $\Delta$$\Delta$ in a way such that $a$$a$ goes to zero faster than $b$$b$, depending of course on their explicit expressions) then a critical transition occurs since in this case $\eta =0$$\eta =0$ becomes a local maximum for ${\mathcal {L}}$${\mathcal {L}}$, and it develops two new global minima. Therefore, the solution of the equation $a(T,\Delta )=0$$a(T,\Delta )=0$ will give a line of critical points in $(T,\Delta )$$(T,\Delta )$ plane. However, if $b$$b$ becomes negative while $a$$a$ is still positive (which again can be done varying $T$$T$ and $\Delta$$\Delta$ so that $b$$b$ vanishes faster than $a$$a$) then something rather different happens: in this case as $b$$b$ approaches zero ${\mathcal {L}}$${\mathcal {L}}$ develops two new symmetric local minima at ${\overline {\eta }}_{\pm }$${\overline {\eta }}_{\pm }$ (similarly to the case analysed before, with the difference that now the situation is perfectly symmetric since ${\mathcal {L}}$${\mathcal {L}}$ is even) and they will become the new global minima as ${\mathcal {L}}({\overline {\eta }}_{\pm })=0$${\mathcal {L}}({\overline {\eta }}_{\pm })=0$, which happens when $b$$b$ changes sign: this way the equilibrium value of the order parameter change discontinuously from zero to a non-zero quantity so a first-order transition has indeed happened.

First-order transition with an even ${\mathcal {L}}$${\mathcal {L}}$

This means that when both $a$$a$ and $b$$b$ are null the system goes from exhibiting a continuous critical transition to a discontinuous first-order one; in other words, the tricritical point$(T_{c},\Delta _{c})$$(T_{c},\Delta _{c})$ can be determined from the solution of the equations $a(T,\Delta )=0$$a(T,\Delta )=0$ and $b(T,\Delta )=0$$b(T,\Delta )=0$.

To conclude let us consider again a system with an Ising-like Landau free energy, where $c>0$$c>0$ and $a$$a$, $b$$b$ are in general functions of the reduced temperature $t$$t$ (and also of the other "disarranging" parameter, which we now neglect). We now want to show that we can understand how the phase diagram of the system is in $(a,b)$$(a,b)$ space, i.e. that we can draw where the phase transition lines are and so we are able to visually represents where the various phases of the system are in $(a,b)$$(a,b)$ plane. First of all, we can note that when $a,b>0$$a,b>0$ the only minimum of ${\mathcal {L}}$${\mathcal {L}}$ is ${\overline {\eta }}=0$${\overline {\eta }}=0$, so the system is in the paramagnetic phase. Furthermore if $a<0$$a<0$ and $b>0$$b>0$ the system is in the magnetic phase, and a second order transition has occurred; therefore we can surely say that the half-line $(a=0,b>0)$$(a=0,b>0)$ is a second order transition line. We must thus determine where the first order transition line lies in $(a,b)$$(a,b)$ space. In order to do so, we first note that the extrema of ${\mathcal {L}}$${\mathcal {L}}$ are given by:

$$0={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+b\eta ^{2}+c\eta ^{4})\quad \Rightarrow \quad {\overline {\eta }}_{\pm }^{2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2c}}$$

$$0={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+b\eta ^{2}+c\eta ^{4})\quad \Rightarrow \quad {\overline {\eta }}_{\pm }^{2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2c}}$$

(and of course they exist only when the temperature is such that $b^{2}-4ac>0$$b^{2}-4ac>0$) and since:

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial \eta ^{2}}}_{|{\overline {\eta }}_{\pm }}=\pm {\overline {\eta }}_{\pm }^{2}\cdot 2{\sqrt {b^{2}-4ac}}$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial \eta ^{2}}}_{|{\overline {\eta }}_{\pm }}=\pm {\overline {\eta }}_{\pm }^{2}\cdot 2{\sqrt {b^{2}-4ac}}$$

we have that $\pm {\overline {\eta }}_{+}$$\pm {\overline {\eta }}_{+}$ are maxima while $\pm {\overline {\eta }}_{-}$$\pm {\overline {\eta }}_{-}$ are minima. The first order transition happens when ${\mathcal {L}}(\pm {\overline {\eta }}_{+})={\mathcal {L}}(0)=0$${\mathcal {L}}(\pm {\overline {\eta }}_{+})={\mathcal {L}}(0)=0$, so:

$${\frac {a}{2}}{\overline {\eta }}_{+}^{2}+{\frac {b}{4}}{\overline {\eta }}_{+}^{4}+{\frac {c}{6}}{\overline {\eta }}_{+}^{6}=0\quad \Rightarrow \quad {\frac {a}{2}}+{\frac {b}{4}}{\overline {\eta }}_{+}^{2}+{\frac {c}{6}}{\overline {\eta }}_{+}^{4}=0$$

$${\frac {a}{2}}{\overline {\eta }}_{+}^{2}+{\frac {b}{4}}{\overline {\eta }}_{+}^{4}+{\frac {c}{6}}{\overline {\eta }}_{+}^{6}=0\quad \Rightarrow \quad {\frac {a}{2}}+{\frac {b}{4}}{\overline {\eta }}_{+}^{2}+{\frac {c}{6}}{\overline {\eta }}_{+}^{4}=0$$

Now, from the condition $\partial {\mathcal {L}}/\partial \eta =0$$\partial {\mathcal {L}}/\partial \eta =0$ we can express ${\overline {\eta }}_{+}^{4}$${\overline {\eta }}_{+}^{4}$ as a function of ${\overline {\eta }}_{+}^{2}$${\overline {\eta }}_{+}^{2}$, and we get ${\overline {\eta }}_{+}^{4}=-(a+b{\overline {\eta }}_{+}^{2})/c$${\overline {\eta }}_{+}^{4}=-(a+b{\overline {\eta }}_{+}^{2})/c$. Substituting we get:

$${\overline {\eta }}_{+}^{2}=-4{\frac {a}{b}}$$

$${\overline {\eta }}_{+}^{2}=-4{\frac {a}{b}}$$

and substituting again in ${\overline {\eta }}_{+}^{2}=(-b+{\sqrt {b^{2}-4ac}})/(2c)$${\overline {\eta }}_{+}^{2}=(-b+{\sqrt {b^{2}-4ac}})/(2c)$ in the end we get:

$$b=-4{\sqrt {\frac {ac}{3}}}$$

$$b=-4{\sqrt {\frac {ac}{3}}}$$

so the first order transition line is a parabola in $(a,b)$$(a,b)$ plane (in particular it will lie in the fourth quadrant). In the end the situation is as follows:

Phase diagram of the system in $(a,b)$$(a,b)$space

As we can see the tricritical point of the system, being the point that divides the first-order from the second-order transition line, is the origin $(0,0)$$(0,0)$ of the parameter space.

We conclude by noting that in such situations strange things can happen; in particular we can show that if we move through the tricritical point along the $a$$a$ axis (thus keeping $b=0$$b=0$) the critical exponents of the system change from the "trivial" ones predicted by mean field theories. In fact if $b=0$$b=0$ then the Landau free energy is ${\mathcal {L}}=a\eta ^{2}/2+c\eta ^{6}/6$${\mathcal {L}}=a\eta ^{2}/2+c\eta ^{6}/6$ and its minima are given by:

$$0={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+c\eta ^{4})$$

$$0={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+c\eta ^{4})$$

Excluding the case $\eta =0$$\eta =0$ and supposing that $a\sim t$$a\sim t$ we get:

$$\eta \sim \left(-{\frac {a}{c}}\right)^{1/4}\sim (-t)^{1/4}\quad \Rightarrow \quad \beta ={\frac {1}{4}}$$

$$\eta \sim \left(-{\frac {a}{c}}\right)^{1/4}\sim (-t)^{1/4}\quad \Rightarrow \quad \beta ={\frac {1}{4}}$$

Furthermore, the state equation of the system in the presence of an external field $h$$h$ will be:

$$h={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+c\eta ^{4})$$

$$h={\frac {\partial {\mathcal {L}}}{\partial \eta }}=\eta (a+c\eta ^{4})$$

and at the critical temperature $a=0$$a=0$, so that in the end:

$$h=cm^{5}\quad \Rightarrow \quad \delta =5$$

$$h=cm^{5}\quad \Rightarrow \quad \delta =5$$

Analogous computations for the other critical exponents give $\gamma =1$$\gamma =1$ and $\alpha =1/2$$\alpha =1/2$.

We therefore see that the critical exponents do indeed change if the system passes through its tricritical point in the phase diagram.