# Is this even a phase transition?

I'm trying to understand how individual contributions of single particles can add up to something one would call a phase transition. For this I consider a model of a system of $N$ one-dimensional particles in a potential well like this:

$$U(x_i)=\begin{cases} 0&\text{if }|x|<d\\ U_{II}&\text{if }d<|x|<L\\ \infty&\text{if }|x|>L \end{cases}.$$

Here $x_i$ is position of $i$th particle, $U_{II}>0$ is height of potential barrier, $d$ is width of potential well and $L>d$ is the width of the total motion area. There's no interaction between the particles, only with the external potential $U$.

Now I calculate how average kinetic energy $\left\langle T\right\rangle$ depends on average total energy per particle. This should be proportional to temperature, since both inside the well and outside of it I have an ideal or almost ideal gas. Here's what I get for $\langle T\rangle$:

$$\langle T\rangle=E-U_{II}\frac{\Delta t_{II}}{\Delta t_I+\Delta t_{II}}=\begin{cases}\frac{(L-d)U_{II}}{L+d\left(1-\sqrt{1-\frac{U_{II}}{E}}\right)}+E&\text{if }E>U_{II}\\ E&\text{if }E<U_{II}\end{cases},$$

where $\Delta t_{I,II}$ are times spent in the areas of $U=0$ and $U=U_{II}$ respectively per period of motion. Here's what it looks like for $N=1,2,3,4,5,10$ (yes, the last one is $10$ not $6$), $L=30$, $d=0.1$ and $U_{II}=7$: I calculate it as

$$\langle T\rangle=\int_0^\infty dE_1\int_0^\infty dE_2\cdots\int_0^\infty dE_N\left(\frac1N\sum_{i=1}^N\langle T_i\rangle\delta(NE-E_1-E_2-\cdots-E_N)\right)$$

using Monte-Carlo integration. Here $\langle T_i\rangle$ is kinetic energy of $i$th particle averaged in time, and $\langle T\rangle$ is $\langle T_i\rangle$ averaged over all particles and possible combinations of energies of each particle $E_i$ such that total energy of the system is $EN$. The discontinuity of the curve for one particle is due to the particle going out of the potential well ($E\gtrsim U_{II}$): near this energy the particle most of the time moves very slowly above the barrier of $U=U_{II}$, thus its average kinetic energy is small.

And here's how what I suppose to be heat capacity, namely $\frac{dE}{d\langle T\rangle}$, looks (the noise is due to derivatives of Monte-Carlo-integrated function) This seems to resemble the peak of heat capacity of a second order transition, c.f. these plots for the Ising model simulations.

But my doubt is that as I increase number of particles, the peak appears to decrease its amplitude and widen. This shouldn't happen for a phase transition, which is normally a many-particle effect. So my question is: is what I described above even a phase transition at all? Or will the peak of heat capacity actually drop to zero once I take the limit of $N\to\infty$? Or am I completely wrong about what I think is heat capacity or temperature for this system?

• I don't understand your system. Are the particles interacting, and if so in what way? What is causing the huge discontinuity in the one-particle E vs T relation? But all this aside, just from the shown graphs I would agree with you that the change in heat capacity looks more like a one-particle effect that might vanish in the thermodynamic limit, rather than a phase transition. – Rococo Dec 26 '16 at 21:52
• Is there a missing equality in your equation for <T>. It seems to include both continuous integrals and discrete summations? – JMLCarter Dec 27 '16 at 2:48
• Can you identify params in your first expression too? – JMLCarter Dec 27 '16 at 3:01
• @Rococo, JMLCarter: I've edited the question to address the comments. – Ruslan Dec 27 '16 at 6:07

Let's calculate the partition function for $N$ particles:

\begin{align} Z(\beta)&=\left(\int_{-\infty}^{\infty}\exp\left(-\beta\frac{p^2}{2m}\right)\mathrm dp\int_{-L/2}^{L/2}\exp\left(-\beta \begin{cases} 0&\text{if }|x|<d/2\\ U_{II}&\text{if }d<|x|<L/2 \end{cases} \right)\mathrm dx\right)^N=\\ &=\left(\frac{2\pi m}\beta\right)^{N/2}\left(d+e^{-\beta U_{II}}(L-d)\right)^N. \end{align}

Calculating average energy per particle, we have (setting $k_B=1$):

\begin{align}\langle E\rangle&=-\frac1N\frac {\mathrm d\ln Z}{\mathrm d\beta}=\\ &=\frac1{2\beta}+\frac{(L-d)U_{II}}{de^{\beta U_{II}}+L-d}=\\ &=\frac T2+\frac{(L-d)U_{II}}{de^{U_{II}/T}+L-d}. \end{align}

Now if we plot $T(\langle E\rangle)$ dependence, superimposing it on the picture in the OP, we'll get (dashed line here): We can see that

1. The curves from the OP don't appear to converge to it.
2. The dependence is the same regardless of $N$

This gives two conclusions:

1. The calculations in the OP don't represent equilibrium state.
2. There's no phase transition anyway, since in the thermodynamic limit the dependence of $\langle E\rangle$ on $T$ doesn't have singularities (it remains the same regardless of number of particles). The nonlinearity is just an increase in heat capacity for some tempearture, but the heat capacity per particle always remains finite.

I believe that by restricting yourself to an ensemble of non-interacting particles, you have a priori forbidden a phase change from occurring, in that you have essentially required the particles to remain an ideal gas at all temperatures. After all, any other phase requires some interaction to constrain the particles' motion and energy. The discontinuity you're getting for low $N$ is a numerical artifact that should indeed be smoothed out at higher particle multiplicity.