Correlations and temperature

It is obvious that for cold temperatures that the angles are to be small and therefore that the energetics is given by a quadratic in the shape of $-\theta^2$ multiplied by some constant. For these reasons, and the symmetry of the problem, we can now see that the distribution for the angles is given by the Boltzmann formula making for $P(\theta) = e^{-\theta^2/T}$ . Of course there are multiplying factors but of no importance to the fact we are trying to solve.

Taking the average the full range from $-\infty$ to $+\infty$ and knowing that the errors due to this range extension are very small, the average angle can be solved for

$\left<\theta\right> = \frac{\int\limits_{-\infty}^{+\infty}\theta^2 e^{-\theta^2/T}d\theta}{\int\limits_{-\infty}^{+\infty} e^{-\theta^2/T}d\theta} = \frac{T}{2}$ .

Then so, the angle of each move is proportional to $\sqrt{T}$ , the total links to move an angle so that we are in anti-alignment with the starting rod must be proportional to the inverse $1/\sqrt{T}$ . The answer is now easy to see.

A given state of the infinite stack of rotating charges is just one configuration out of the infinite set of possible configurations of the angles $\left\{\ldots,\theta_{-1},\mbox{ }\theta_0,\mbox{ }\theta_1,\ldots\right\}$ .

Of the many possibilities, the stack will tend to occupy the configurations that have the most favorable free energies, i.e. $\displaystyle P(\vec{\theta}) \sim e^{-F(\vec{\theta})/k_{B}T}$ . As $T \rightarrow 0$ , lower and lower energy states are occupied until only the lowest possible energy state can be occupied.

The free energy $F$ of two interacting charges $i$ and $i+1$ is given by $\displaystyle \frac{kq_{i}q_{i+1}}{r_{i,i+1}}$ .

For the stack, every charge contributes $\displaystyle -2k\frac{q^2}{r_{i,i+1}}$ (one interaction for each of its two neighbors).

The distance between two adjacent charges is approximately $\displaystyle\sqrt{1+R^2\theta^2}$ and so the free energy per charge site is given by

$\begin{aligned} \displaystyle F_{site} &= -2k\frac{q^2}{\sqrt{1+(R\theta)^2}}\\ &\approx -2kq^2\left(1-R^2\theta^2\right)\end{aligned}$

The probability of a particular configuration is then given by

$\begin{aligned} P(\vec{\theta}) &\sim \exp -\frac{F_{site}}{k_B T} \\ &= \displaystyle \exp 2kq^2\frac{1-R^2\theta^2}{k_B T}\mbox{ (for small }\theta)\\ &\sim \exp -2kq^2 \frac{R^2\theta^2}{k_B T} \end{aligned}$

Therefore, the probability of the angle between adjacent charges has the general form $P(\theta) \sim e^{-A\theta^2/T}$ .

At $T=0$ the probability becomes zero for all angles except for $\theta = 0$ , as stated in the problem description. At temperatures slightly above zero, other configurations open up to the system but the system is still cut off from $\theta^2$ much greater than $T$ because the probability falls off dramatically for $\theta$ much beyond $k_BT$ . In other words, the partition function is dominated by the region where $\theta \sim \sqrt{T}$ . Therefore, the size of $\theta_{site}$ scales as $\sqrt{T}$ .

The correlation length $L$ is defined as the number of rods in a half turn of the helical stack (a shift of $\pi$ ), i.e.

$\begin{aligned} L &\approx \pi/\theta_{site} \\ &\sim 1/\sqrt{T} \\ &=\boxed{ \displaystyle\frac{1}{\sqrt{T-T_c}}} \end{aligned}$ .

Therefore, the exponent $\nu$ is $\displaystyle\boxed{\frac12}$ .

At T the avarage angle between the two adjacent rods is θ, then the potential energy E=-kq^2/sqrt(1+(θR)^2). At T=0, E(0)=-kq^2.

For θ <<1, E= -kq^2 + [k(qR)^2/2] θ^2, Compared with the mininum energy at absolute 0, T= E-E(0) = C θ^2, where C is a constant = [k(qR)^2/2].

Hence θ ∝ T^(1/2). L ∝ 1/θ ∝ T^(-1/2). Namely, ν = 1/2

Suppose the diffence in angle \theta between two consecutive rods is \alpha (alpha is small because T is small), the speed of each charge is v. According to kinetic theory of gases: \frac{m * v^2}{2} = \frac{3 * k_B * T}{2}

Hence m * v^2 = 3 * k_B * T

Because \alpha is small, we suppose the distance between two consecutive charges is d, which is the distance between the planes containing the orbit of each charge. Therefore, the attractive force between two consecutive charges is F_0 = \frac{k * q^2}{d^2}

The projection of this force into the plane containing the orbit of the charge: F p = F 0 * \sin \gamma where \gamma is the angle between the vertical axis and the straight line connecting two consecutive charges. Since \alpha is small, \sin \gamma = \frac{r * \alpha}{d}.
Therefore, F_p = \frac{k * q^2 * r * \alpha}{d^3}

The total force on each charge is: F = 2 * F_p * \sin \frac{\alpha}{2}

Since \alpha is small, F \approx F_p * \alpha = \frac{k * q^2 * r * \alpha^2}{d^3}

According to Newton Law II, F = \frac{m * v^2}{r} = \frac{3 * k_B * T}{r}

Therefore \alpha^2 = \frac{3 * k_B * T * d^3}{k * q^2 * r^2}

Therefore L = \frac{2 * \pi}{\alpha} = 2 * \pi * q * r * sprt{\frac{k}{3 * k_B * T * d^3}}

Therefore L \sim T^(-0.5) or v = 0.5

The equipartition theorem states that the average kinetic energy is $E \propto kT$ , where $k$ is Boltzmann's constant. We can therefore say that the total energy per charge ABOVE the ground state energy is also $\propto kT$ . Since $T$ is very small the charges will in general be almost aligned. The potential energy between a pair of charges that is exactly aligned is given by

$E=-\frac{Kq^2}{1}$

while the potential energy for a pair that is almost aligned is

$E=-\frac{Kq^2}{(1+(R\Delta \theta)^2)^{1/2}}$

where $\Delta \theta$ is the misalignment angle between the charges. $\Delta \theta$ is small, and so for a misaligned pair the potential energy above the ground state is

$\Delta E=\frac{Kq^2 R^2}{2} (\Delta \theta)^2\propto kT.$

If the Nth particle is oppositely aligned from the 0th particle, then $\Delta \theta=\pi/N$ . Substituting this into the expression above yields $N\propto T^{-1/2}$ which gives $\nu=1/2$ .