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The Critical Scaling of the
Helicity Modulus of the O(3)
classical Heisenberg ferromagnet

Robert G. Brown


Acknowledgements: This work was done with my good friend and colleague, Dr. Mikael Ciftan. We gratefully acknowledge the support of the Army Research Office.

Classical Heisenberg ferromagnet (CHF) (the ${\cal O}(3)$ model on a 3$d$ simple cubic lattice with periodic boundary conditions) in zero external field:

{\cal H} = - \sum_{a < b}^{nn} J_{ab} {\bf S}_a \cdot {\bf S}_b +
{\cal H}_{\rm bath}(\{{\bf S}_i\},T)

To compute and ``measure'' (with Monte Carlo) the critical exponents of the model, in particular the critical exponent $\nu_E$ of the Helicity Modulus $\Upsilon(t)$.


Review of Theory

In the last expression, $\Upsilon(L) \approx L^{-2\beta/\nu}$. $2\beta/\nu$ term from $m^2$ clearly dominant ($\eta$ is very small, $\sim 0.01$, for this model, while $\beta \approx 1/3$ and $\nu
\approx 2/3$). The helicity modulus should vanish sharply near $T_c$ according to Landau theory.


We cannot directly measure the free energy density $dF/dV$. We can directly measure the enthalpy density $E$. Following an identical argument:

\Delta E(\Theta) \approx \frac{1}{2} \Upsilon_E(T)(\nabla \phi)^2
\end{displaymath} (9)

where $\Delta E(\Theta)$ is the change in internal energy caused by twisting the boundary conditions through the angle $\Theta \le \pi/2$ with either helicity. From this obvious substitutions yield:
\Upsilon_E(T) = \frac{2L^2}{\Theta^2} \Delta E(\Theta)
\end{displaymath} (10)

With a page or two of algebra we can show that:

\Upsilon_E(t) \sim t^{v_E} \sim t^{-\phi}
\end{displaymath} (11)

with the critical exponent
v_E = -\phi = 1 - 2\nu - \alpha
\end{displaymath} (12)

This is what we wish to measure, in part to invert this equation and deduce the values of $\nu$ and $\alpha$.

Note that as before, if we make the finite size scaling hypothesis we will actually measure:

\Upsilon(L) \sim (L)^{-v_E/\nu} \sim (L)^{2 - \frac{1 - \alpha}{\nu}}
\end{displaymath} (13)

-v_E/\nu = 2 - \frac{1-\alpha}{\nu}
\end{displaymath} (14)

The enthalpy helicity should thus diverge at $T_c$.

It is easy to show that:

$\displaystyle d-2 -v_E/\nu$ $\textstyle =$ $\displaystyle d - \frac{1-\alpha}{\nu}$ (15)
$\displaystyle 1 - v_E/\nu$ $\textstyle =$ $\displaystyle \frac{1}{\nu}$ (16)
$\displaystyle \nu$ $\textstyle =$ $\displaystyle \frac{1}{1 - \frac{v_E}{\nu}}$ (17)

where the second step uses ``hyperscaling'' (widely believed but by no means proven for this model) to eliminate $\alpha$ for $d = 3$. With this we can compute $\alpha$ and $\nu$ given $-v_E/\nu$ and possibly check hyperscaling.

Measuring $\Upsilon(T,L)$ with Monte Carlo


Figure 1: E per spin as function of interlayer twist angle $\Delta \phi $ for $L = 64$ (in progress). This is fit to obtain the helicity.

Figure 2: $\Upsilon (T)$ for $L = 8$ and $L = 16$, evaluated at low precision to get trend only.

Figure 3: The helicities for various $L$ at $T_c$. The nonlinear least squares fit of this yields $x/\nu = -v_e/\nu $.

Best result to date: $x/\nu = 0.353 \pm 0.02$


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Robert G. Brown 2003-06-02