The entropic pressure of a lattice polygon

Let ${c}_{n}(\vec{r})$ be the number of self-avoiding walks in Z² of length n steps from the origin to the lattice point $\vec{r}$.

Similarly, let ${c}_{n}(r)={\sum }_{r=\vert \vec{r}\vert }{c}_{n}(\vec{r})$ be the number of self-avoiding walks of length n from the origin to lattice points a distance $\vert \vec{r}\vert =r$ on a spherical shell from the origin. Since the endpoints of walks counted by c_n(r) ends on a spherical shell a distance r from the origin, one should have

$$\begin{eqnarray} &&{c}_{n}(r)\simeq {A}_{0}\hspace{0.167em} {r}^{d-1}\hspace{0.167em} {c}_{n}(\vec{r}) \end{eqnarray} \tag{ 8 }$$

in d dimensions.

One may similarly consider walks in a half-space instead. Let ${c}_{n}^{+}(\vec{r};L)$ be the number of walks from the origin in a half-space defined by the (d − 1)-dimensional hyperplane L, of length n and ending in the vertex $\vec{r}$. Put ${c}_{n}^{+}(r;L)={\sum }_{r=\vert \vec{r}\vert }{c}_{n}^{+}(\vec{r};L)$, the number of walks from the origin in a half-space defined by L, of length n and with endpoint a distance r from the origin. Since these walks end on a hemispherical shell of radius r from the origin, one should have

$$\begin{eqnarray} &&{c}_{n}^{+}(r;L)\simeq {A}_{1}\hspace{0.167em} {r}^{d-1}\hspace{0.167em} {c}_{n}^{+}(\vec{r};L) \end{eqnarray} \tag{ 9 }$$

in d dimensions.

The average distance of the endpoint of a self-avoiding walk of length n from the origin is denoted by 〈r〉_n, and this introduces a metric in the model of self-avoiding walks. It is expected that 〈r〉_n ≃ C₀ n^ν, where ν is the metric exponent. In two dimensions the exact value is $\nu =\frac{3}{4}$ [25, 26].

In terms of c_n(r) the average distance of the endpoint of the walk from the origin may be calculated from

$$\begin{eqnarray} &&\left \langle r\right \rangle _{n}=\frac{\mathop{\sum }_{r\geq 0}r \hspace{0.167em} {c}_{n}(r)}{{c}_{n}}\simeq {C}_{0}\hspace{0.167em} {n}^{\nu }. \end{eqnarray} \tag{ 10 }$$

This gives a way for estimating the scaling of c_n(r).

For fixed n, c_n(r) should decay as r increases with length scale n^ν. That is, one may guess that

$$\begin{eqnarray} &&{c}_{n}(r)={B}_{0}\hspace{0.167em} {r}^{x}\hspace{0.167em} {\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}{c}_{n} \end{eqnarray} \tag{ 11 }$$

where x is an exponent that must be determined. This assumption may be used to determine the scaling of 〈r〉_n. Substituting this assumption in equation (10) and approximating the summation by an integral and using equation (11) gives

$$\begin{eqnarray*} &&\fl \left \langle r\right \rangle _{n}\simeq \int \nolimits _{0}^{n}{B}_{0}\hspace{0.167em} {r}^{1+x}{\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}\hspace{0.167em} \mathrm{d}r\simeq {B}_{0}\left ({C}_{0}{n}^{\nu }\right )^{x+2}\int \nolimits _{0}^{\infty }\left [\frac{r}{{C}_{0}{n}^{\nu }}\right ]^{x+1}{\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}\frac{\mathrm{d}r}{{C}_{0}{n}^{\nu }} \end{eqnarray*}$$

assuming that n ≫ n^ν (since $\nu =\frac{3}{4}$ in two dimensions, this is true for large n). The last integral is a constant, so one concludes that

$$\begin{eqnarray} &&\left \langle r\right \rangle _{n}\sim {n}^{\nu (x+2)}. \end{eqnarray} \tag{ 12 }$$

Since 〈r〉_n ≃ C₀ n^ν, it follows that x =− 1 in equation (11).

Taking the above together shows that c_n(r) scales as

$$\begin{eqnarray} &&{c}_{n}(r)\simeq {B}_{0}\hspace{0.167em} {r}^{-1}\hspace{0.167em} {\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}\hspace{0.167em} {c}_{n}. \end{eqnarray} \tag{ 13 }$$

This result gives the scaling of ${c}_{n}(\vec{r})$ as well by equation (8):

$$\begin{eqnarray} &&{c}_{n}(\vec{r})\simeq \frac{{r}^{1-d}}{{A}_{0}}\hspace{0.167em} {c}_{n}(r)\simeq \frac{{B}_{0}}{{A}_{0}{r}^{d}}\hspace{0.167em} {\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}{c}_{n}=A \hspace{0.167em} {r}^{-d}{\mathrm{e}}^{-r/{C}_{0}{n}^{\nu }}{c}_{n} \end{eqnarray} \tag{ 14 }$$

where $r=\vert \vec{r}\vert$ and A is a constant.

The above arguments apply mutatis mutandis to ${c}_{n}^{+}(\vec{r})$, with c_n in the above replaced by ${c}_{n}^{+}$. This follows in particular because the scaling of the average distance of the endpoint of half-space walks is given by

$$\begin{eqnarray} &&\left \langle r\right \rangle _{n}^{+}=\frac{\mathop{\sum }_{r\geq 0}r \hspace{0.167em} c_{n}^{+}(r)}{{c}_{n}^{+}}\simeq {C}_{1}\hspace{0.167em} {n}^{\nu }. \end{eqnarray} \tag{ 15 }$$

The rest of the argument is the same. That is, one finally obtains

$$\begin{eqnarray} &&{c}_{n}^{+}(\vec{r},L)\simeq \frac{{r}^{1-d}}{{A}_{1}}\hspace{0.167em} {c}_{n}^{+}(r)\simeq \frac{{B}_{1}}{{A}_{1}{r}^{d}}\hspace{0.167em} {\mathrm{e}}^{-r/{C}_{1}{n}^{\nu }}{c}_{n}^{+}=B \hspace{0.167em} {r}^{-d}{\mathrm{e}}^{-r/{C}_{1}{n}^{\nu }}{c}_{n}^{+}(L) \end{eqnarray} \tag{ 16 }$$

where $r=\vert \vec{r}\vert$ and B is a constant.

To determine a scaling form for the pressure due to a rooted lattice polygon, it will be necessary to determine a scaling relation for lattice polygons passing through a lattice point $\vec{r}$.

Denote the number of rooted lattice polygons of length n passing through the point $\vec{r}$ by ${p}_{n}(\vec{r})$. Approximate this by considering a pair of walks which step from the origin to $\vec{r}$. If the walks avoid one another, then their union is a rooted lattice polygon passing through $\vec{r}$. If they do not avoid one another, then an upper bound on ${p}_{n}(\vec{r})$ is obtained.

However, if the two walks are confined to two different half-spaces defined by a (d − 1)-dimensional hyperplane L passing through the origin and $\vec{r}$, then they avoid one another, and one may estimate

$$\begin{eqnarray} &&{p}_{n}(\vec{r})\gtrsim \sum _{k=0}^{n}{c}_{k}^{+}(\vec{r},L)\hspace{0.167em} {c}_{n-k}^{+}(\vec{r},L). \end{eqnarray} \tag{ 17 }$$

This may be approximated by using equation (16) as estimates for ${c}_{n}^{+}(\vec{r},L)$. This gives

$$\begin{eqnarray} &&{p}_{n}(\vec{r})\gtrsim {B}^{2}\sum _{k=0}^{n}{r}^{-2 d}{\mathrm{e}}^{-r/{C}_{1}{k}^{\nu }}{\mathrm{e}}^{-r/{C}_{1}(n-k)^{\nu }}{c}_{k}^{+}(L)\hspace{0.167em} {c}_{n-k}^{+}(L). \end{eqnarray} \tag{ 18 }$$

Next, rescale $r=\vert \vec{r}\vert$ by its mean $\left \langle r\right \rangle _{n}^{+}\simeq {C}_{1}\hspace{0.167em} {n}^{\nu }$: that is, choose $r=a \hspace{0.167em} \left \langle r\right \rangle _{n}^{+}$ in the above, for a > 0 in equation (18). This estimates the number of polygons through the point $\vec{r}=\left (a \hspace{0.167em} \left \langle r\right \rangle _{n}^{+}\right )\hspace{0.167em} (\vec{r}/\vert \vec{r}\vert )$. Denote this by ${\widehat{p}}_{n}(a,L)$ (that is, ${\widehat{p}}_{n}(a,L)$ is the number of polygons of length n passing through a point fixed on L a distance $\rho =a\left \langle r\right \rangle _{n}^{+}$ from the origin). This gives

$$\begin{eqnarray} &&{\widehat{p}}_{n}(a,L)\gtrsim {B}^{2}\sum _{k=0}^{n}\left (a \hspace{0.167em} {C}_{1}\hspace{0.167em} {n}^{\nu }\right )^{-2 d}{\mathrm{e}}^{-a(n/k)^{\nu }}{\mathrm{e}}^{-a(n/(n-k))^{\nu }}{c}_{k}^{+}(L)\hspace{0.167em} {c}_{n-k}^{+}(L). \end{eqnarray} \tag{ 19 }$$

The scaling of ${c}_{n}^{+}(L)$ is given in equation (6), and is asymptotically independent of L. Thus, the above may be averaged over L (that is, over all directions through the origin) so that the number of polygons passing through a point a distance $\rho =a\left \langle r\right \rangle _{n}^{+}$ from the origin is approximated by

$$\begin{eqnarray} &&\fl {\widehat{p}}_{n}(a)\gtrsim \frac{{B}^{2}({A}^{\prime})^{2}{\mu }^{n}}{\left (a{C}_{1}{n}^{\nu }\right )^{2 d}}\int \nolimits _{0}^{n}{k}^{{\gamma }_{1}-1}\hspace{0.167em} (n-k)^{{\gamma }_{1}-1}\hspace{0.167em} {\mathrm{e}}^{-a(n/k)^{\nu }}{\mathrm{e}}^{-a(n/(n-k))^{\nu }}\hspace{0.167em} \mathrm{d}k\nonumber\\ &&=~\frac{{B}^{2}({A}^{\prime})^{2}{\mu }^{n}}{\left (a{C}_{1}{n}^{\nu }\right )^{2 d}}\hspace{0.167em} {n}^{2{\gamma }_{1}-1}\hspace{0.167em} \int \nolimits _{0}^{1}{x}^{{\gamma }_{1}-1}(1-x)^{{\gamma }_{1}-1}{\mathrm{e}}^{-a/{x}^{\nu }}\hspace{0.167em} {\mathrm{e}}^{-a/(1-x)^{\nu }}\hspace{0.167em} \mathrm{d}x \end{eqnarray} \tag{ 20 }$$

where the summation is now approximated by an integral.

Denote the integral above by g(a):

$$\begin{eqnarray} &&g(a)=\int \nolimits _{0}^{1}{x}^{{\gamma }_{1}-1}(1-x)^{{\gamma }_{1}-1}{\mathrm{e}}^{-a/{x}^{\nu }}\hspace{0.167em} {\mathrm{e}}^{-a/(1-x)^{\nu }}\hspace{0.167em} \mathrm{d}x. \end{eqnarray} \tag{ 21 }$$

Put ${C}_{2}={B}^{2}({A}^{\prime})^{2}/{C}_{1}^{2 d}$. Then the above reduces to

$$\begin{eqnarray} &&{\widehat{p}}_{n}(a)\gtrsim {C}_{2}\hspace{0.167em} {\mu }^{n}\hspace{0.167em} {a}^{-2 d}g(a)\hspace{0.167em} {n}^{2{\gamma }_{1}-1-2 d \nu }. \end{eqnarray} \tag{ 22 }$$

Notice that the dependence on direction has disappeared, so that ${\widehat{p}}_{n}(a)$ is the approximate number of rooted polygons of length n, passing through a particular point a distance equal to $\rho =a\left \langle r\right \rangle _{n}^{+}\simeq a \hspace{0.167em} {C}_{1}\hspace{0.167em} {n}^{\nu }$ from the origin.

If d = 2, then ${\gamma }_{1}=\frac{6 1}{6 4}$ and $\nu =\frac{3}{4}$, and the a- and n-dependence of ${\widehat{p}}_{n}(a)$ becomes

$$\begin{eqnarray} &&{\widehat{p}}_{n}(a)\gtrsim \frac{{C}_{2}\hspace{0.167em} {\mu }^{n}g(a)}{{a}^{4}}\hspace{0.167em} {n}^{-6 7/3 2}. \end{eqnarray} \tag{ 23 }$$

It remains to determine the pressure. A plot of g(a)/a⁴ is given in figure 4 on a log–log scale.

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By equation (3) the pressure due to rooted polygons at a point a distance $\rho =a \hspace{0.167em} \left \langle r\right \rangle _{n}^{+}\simeq a \hspace{0.167em} {C}_{1}\hspace{0.167em} {n}^{\nu }$ from the origin is approximately given by

$$\begin{eqnarray} &&{\mathbf{P}}_{n}(a)=-\log \left (1-\frac{{\widehat{p}}_{n}(a)}{{p}_{n}}\right ), \end{eqnarray} \tag{ 24 }$$

where ${\widehat{p}}_{n}(a)$ is the number of polygons passing through a point a distance $\rho =a\left \langle r\right \rangle _{n}^{+}$ from the origin.

The scaling of this is obtained by using equations (7) and (22). In particular, for a and n large this becomes

$$\begin{eqnarray} &&{\mathbf{P}}_{n}(a)\simeq -\log \left (1-\frac{{C}_{2}g(a)}{B{a}^{2 d}}{n}^{2{\gamma }_{1}-1-2 d \nu -{\alpha }_{s}+2}\right ). \end{eqnarray} \tag{ 25 }$$

It is readily checked that the power of n in the above is negative and that g(a) quickly approaches zero with increasing a. Thus, for large n the logarithm can be expanded. Keeping only the leading term shows that

$$\begin{eqnarray} &&{\mathbf{P}}_{n}(a)\simeq \frac{{C}_{2}\hspace{0.167em} g(a)}{B{a}^{2 d}}{n}^{2{\gamma }_{1}-1-2 d \nu -{\alpha }_{s}+2}. \end{eqnarray} \tag{ 26 }$$

If d = 2, then substitution of the exact values of the exponents and combining the constants above give

$$\begin{eqnarray} &&{\mathbf{P}}_{n}(a)\simeq \frac{C g(a)}{{a}^{4}}{n}^{-1 9/3 2}, \end{eqnarray} \tag{ 27 }$$

for some constant C.

This prediction may be tested numerically by plotting [n^19/32 P_n(a)] against $[\vert \vec{r}\vert /{n}^{3/4}]$. That is, rescaling length in the lattice by a factor n^ν and the pressure by n^19/32 should collapse data for a range of choices of $\vec{r}$ and n to a single universal curve which is only a function of the parameter a. The shape of this universal curve is given by a rescaling of g(a)/a⁴.

A unit mass test particle placed near the origin in the lattice will experience a average pressure gradient in the vicinity of the polygon. If the test particle can move freely without friction, then it will accelerate from high to low pressure.

Thus, assume that a particle is placed near the polygon, and that it will be accelerating in a frictionless environment down the pressure gradient. Assume in addition that temperature is constant (that is, the polygon is in contact with a large heat bath keeping its temperature fixed). This last assumption is important since the particle will drain energy from the polygon, cooling it down, in the absence of a heat bath.

Suppose that the particle will move along the X-axis. The pressure gradient between lattice sites (x,0) and (x + 1,0) is ΔP_n(x,0) = P_n(x + 1,0) − P_n(x,0). Assuming that the acceleration of the particle from (x,0) to (x + 1,0) is constant, it will be given by

$$\begin{eqnarray} &&\frac{\Delta {v}_{n}(x,0)}{\Delta t}={P}_{n}(x,0)-{P}_{n}(x+1,0), \end{eqnarray} \tag{ 28 }$$

where Δv_n(x,0) = v_n(x + 1,0) − v_n(x,0) and v_n(x,0) is the speed of the particle at the point (x,0), and Δt is the time interval it takes for the particle to move from (x,0) to (x + 1,0).

Notice that Δv_n(x,0)/Δt = (Δv_n(x,0)/Δx)(Δx/Δt) ≈ v_n(x,0) Δv_n(x,0)/Δx since Δx/Δt ≈ v_n(x,0). That is, if ΔP_n(x,0) = P_n(x + 1,0) − P_n(x,0), then equation (28) becomes

$$\begin{eqnarray} &&{v}_{n}(x,0)\hspace{0.167em} \Delta {v}_{n}(x,0)\approx -\Delta {P}_{n}(x,0) \end{eqnarray} \tag{ 29 }$$

where Δx = 1 was used and where ΔP_n(x,0) = P_n(x + 1,0) − P_n(x,0) is the pressure drop from (x,0) to (x + 1,0).

Equation (29) may be approximated by a differential equation in x. Integrating this from y to x gives

$$\begin{eqnarray} &&\frac{1}{2}{v}_{n}^{2}(z,0)\vert _{y}^{x}=-{P}_{n}(z,0)\vert _{y}^{x}. \end{eqnarray} \tag{ 30 }$$

This becomes

$$\begin{eqnarray} &&\frac{1}{2}\left ({v}_{n}^{2}(x,0)-{v}_{n}^{2}(y,0)\right )={P}_{n}(y,0)-{P}_{n}(x,0). \end{eqnarray} \tag{ 31 }$$

Observe that the left-hand side is the difference in kinetic energy of a unit mass particle, and the right-hand side is the drop in pressure. For example, putting y = 1 gives

$$\begin{eqnarray} &&{v}_{n}(x,0)=\sqrt{{v}_{n}^{2}(1,0)+2 \hspace{0.167em} \left ({P}_{n}(1,0)-{P}_{n}(x,0)\right )}. \end{eqnarray} \tag{ 32 }$$

That is, if the particle is put with zero velocity at (1,0), then its velocity at (x,0) is

$$\begin{eqnarray} &&{v}_{n}(x,0)=\sqrt{2 \hspace{0.167em} \left ({P}_{n}(1,0)-{P}_{n}(x,0)\right )}. \end{eqnarray} \tag{ 33 }$$

If x is large (say x > n) then P_n(x,0) = 0. This gives the terminal velocity of the particle once it has moved far from the polygon:

$$\begin{eqnarray} &&{v}_{n}^{\mathrm{ter}}=\sqrt{2 \hspace{0.167em} {P}_{n}(1,0)} \end{eqnarray} \tag{ 34 }$$

for a particle released at (1,0) in rest.

Notice that the rescaling of P_n(a) in equation (26) is rotationally symmetric about the origin in the square lattice (independent of direction). Hence, the arguments above generalize to particles accelerating in other directions from the origin—it is therefore appropriate to consider a particle moving freely along the X-axis, without loss of generality. For example, the terminal velocity will only be dependent on the distance the particle is released from the origin, and not on the direction of its trajectory.

The pressure as a function of (x,0) for 1 ≤ x ≤ 30 is plotted for values of n ranging from n = 10 to 200 in figure 5. As expected, the pressure approaches zero with increasing distance from the origin—for polygons of length n the pressure at lattice points (x,y) with |x| > n/2 or |y| > n/2 is necessarily zero, since a rooted polygon cannot pass through such points.

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The scaling of the pressure can be uncovered by considering equation (27). That is, if distance is rescaled by n^−3/4, then pressure should be rescaled by n^19/32. This implies that plotting n^19/32P_n(x,0) against x/n^3/4 should collapse the data in figure 5 onto a single curve. This is illustrated in figure 6.

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The data in figure 6 includes points values of n∈{10,20,50,100,150,200,250} for 1 ≤ x ≤ 29, and show that the scaling of P_n(x,0) is well described by equation (27). In fact, the approximation made in equation (17), and carried through equation (23)–(27), apparently captures the dominant contributions to the pressure.

Plotting the data in figure 6 on a log–log scale gives figure 7. This plot shows that there are two scaling regimes. The first is the data for large n (and x small compared to n). The rescaled pressure is large and decays slowly with increasing n. The other regime is for large x (and n small compared to x). In this regime the pressure decays quickly with distance. Observe that the shape of this data is a scaled representation of the plot of g(a)/a⁴ on a log–log scale in figure 4. This is not accidental, as seen in equation (27).

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The scaling of P_n(a) in equation (27) predicts that the scaling of the pressure is isotropic—that is, the same in all direction. Above, the scaling along a lattice axis was examined, in this section other directions will be considered.

Consider first the pressure P_n(x,x) at lattice points {(x,x)|1 ≤ x ≤ 29} along the diagonal direction in the lattice. The pressure is a function of the geometric distance $\Vert (x,x)\Vert _{2}=\sqrt{2}\hspace{0.167em} \vert x\vert$ from the origin (that is, distance is measured using the L₂-norm). Thus, rescaling distance in this case by $\sqrt{2}\hspace{0.167em} {n}^{-3/4}$ should again collapse the rescaled pressure n^19/32P_n(x,x) to a single curve. This is seen in figure 8 for values of n ranging from 10 to 250.

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Comparison to figure 6 shows that the data scales similarly, and may in fact be plotted on the same set of axes. This is done in figure 9—that is, the data of figures 6 and 8 are plotted on the same axes and scale. This shows that the scaling along the diagonal is the same as along the X-axis.

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The data in figure 9 is numerical confirmation that one may expect the same scaling of the pressure in any direction from the origin. This may be made even more clear by plotting all the data for points (x,y) with 1 ≤ x ≤ 29 and 1 ≤ y ≤ 29. That is, plot the rescaled pressures n^19/32P_n(x,y) against n^−3/4||(x,y)||₂ for 4 ≤ n ≤ 250, 1 ≤ x ≤ 29 and 1 ≤ y ≤ 29 on the same graph. The result is shown in figure 10. The data collapse to the same curve.

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The terminal velocity of a unit mass test particle accelerating without dissipation of energy at constant temperature when released at (1,0) is given by equation (34). Intermediate velocities as a function of position are given by equation (33).

Velocity along the X-axis as a function of position is plotted in figure 11. Increasing the size of the polygon tends to decrease the rate of acceleration and lower intermediate velocities. However, the terminal velocity seems to be quite insensitive to the length of the polygon, and in each case levels off close to 1.33.

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The velocity can be rescaled by rescaling length by n^ν. In figure 12 this is done for all the data in figure 11 and also for data from n = 250. With increasing n the data start to accumulate on a rescaled velocity curve. This rescaling is compounded by the release of the particle at (1,0), since the starting point is also rescaled. By instead choosing the origin at x = 2, and then rescaling, the curve in figure 12 becomes better defined in figure 13, with data points clustering more tightly over the wide range of n-values used.

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