Dhrubaditya Mitra

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Can classical non-equilibrium systems be diamagnetic ?

A well-known result in classical physics, the Bohr-van Leeuwen theorem, states that many body systems obeying classical laws of physics cannot be either diamagnetic or paramagentic. There are at least two ways to understand this theorem.

Consider charged particles moving in a domain with random velocities and at equilibrium with a thermal bath. Ignore the particle-particle interaction, hence the partiition function of the system becomes a product of single particle partition functions. The equation of motion of a single charged particle can be written as \begin{eqnarray} \dot{{\bf x}} &=& {\bf v} \\ \dot{{\bf v}} &=& -\gamma_1 ({\bf u} - {\bf v}) + \gamma_2 ({\bf v} \times {\bf B}) \end{eqnarray} where $\gamma_1 = 6\pi\mu a/m$ and $\gamma_2 = (q/m)$ with $m$ the mass of the particle, which is assumed to be a sphere, $a$ its radius, and $q$ its charge. The velocity ${\bf u}$ is the velocity of a the background fluid which constitutes the thermal bath, hence it obeys the fluctuation-dissiaption relation. In other words, ${\bf u}$ is a random, Gaussian, delta-correlated in time process with zero mean and its variance given by $k_{\rm B} T$ where $T$ is the temperature and $k_{\rm B}$ is the Boltzmann constant. Under the influence of an external magnetic field, the charged particles will execute circular motions as shown in the picture below. If we consider one such orbit as sketch inside the domain below we can calculate the magnetic field due to the circular motion of the charged particles by the Biot-Savart formula \begin{equation} {\bf B}_{\rm in} = \frac{\mu_0}{4\pi} \sum_{{\rm j}=1,N}\frac{q{\bf v}_{\rm j}\times {\hat r}_{\rm j}}{r^3_{\rm j}} \end{equation} Here we have ignored relativistic effects (otherwise we would have to use the much more cumbersome Lennert-Wichert potential). As all the particles, irrespective of their initial velocities, move in the same direction (clockwise in the figure below) we get an net contribution to the induced magnetic field ${\bf B}_{\rm in}$ in a direction which is into the plane of the figure. This is opposite to the direction of ${\bf B}$ which is the external imposed magnetic field. Hence we obtain diamagnetism. The fallacy of this argument is realised by looking at the boundary of the domain, which is assumed to be perfectly reflecting. Here, as shown in the figure we get a net currect which is opposite in sign the current induced inside the domain. It can be shown that the magentic field due to this two currents exactly cancel. Hence the induced magnetic field should be zero. Thus classical physics gives neither diamagnetism nor paramagentism.

A beautiful pedagogic discussion from this aspect of the problem can be found in the book Surprises in theoretical physics by Rudolf Peirls.

The second argument is even simpler. It is to notice that magnetic forces do no work. Hence they do not contribute to the Hamiltonian of the system of classical charged particles. At equilibrium, all the statistical properties of the system is determined by its partition function which is given by the Gibbs rule, \begin{equation} {\mathcal Z} = \sum_{{\rm all}\;{\rm states}}\exp[-\beta {\mathcal H}] \end{equation} As the external magnetic field does not contribute to the hamiltonian ${\mathcal H}$ the induced magnetic field must be zero. This argument is even more powerful because it works even when there is particle-particle charged interaction. But this works only in equilbrium systems. A paedagogic discussion from this aspect can be found in Chapter 34 of the second volume of Feynmann Lectures on Physics.

In recent years an interesting attempt has been made to circumvent the first argument by putting the particles in the surface of a sphere which has no boundaries. Numerical solutions of the system of many particles showed that diamagnetism is possible. But recasting the problem in a Fokker-Planck picture and then solving the Fokker-Planck equation (under assumption of stationarity and some other simplifying ones) shows that no diamagnetism is observed. A more recent work on this problem by N. Kumar has demonstrated an interesting asepct. In this work, Kumar has assumed that the charged particles violate fluctuation-dissipation relation. Insted of assuming a delta-correlated velocity, he added to it a correlted part (in time) and showed that diamagnetism and paramagnetism are both possible if fluctuation-dissipation realtion is not true. This begs the question, what happens if ${\bf u}$ is a turbulent solution of the Navier-Stokes equation ?

In that case we solve the following set of equations: \begin{eqnarray} \partial_t {\bf u} + {\bf u}\cdot\nabla{\bf u} &=& \nu\nabla^2{\bf u} -\nabla p + {\bf J}\times{\bf B} + {\bf f}\\ \nabla\cdot{\bf u} &=& 0 \\ \partial_t {\bf B} &=& \nabla\times\left({\bf u}\times{\bf B} - \eta{\bf J} \right) \end{eqnarray} in a periodic box. Where $\rho = 1$ and $\mu_0 = 1$ has been assumed, and ${\bf J} = \nabla\times{\bf B}$. The external force ${\bf f}$ is assumed to be non-helical limited to small wave-numbers. In addition to the equations of the charged particles written down before. The numerical problem then is to calculate the induced magnetic field, as given by the Biot-Savart formula, once the system of particles have reached statistically stationary state. Whether we obtain diamagnetism or paramagnetism or neither is the main question.

Note here, that the velocity field ${\bf u}$ no longer satisfies the fluctuation-dissipation relation, so if Kumar's latest results apply here we would obtain diamagnetism (or paramagnetism). If we assume (as is reasonable) that the system of charged particles reach a statistically stationary state then the question of dia (or para) magnetism is that the following quantity, \begin{equation} {\bf M}_{\rm in} \equiv \int \frac{{\bf v}\times\hat{r}}{r^3} \mathcal{P}({\bf r},{\bf v}) d^drd^dv \end{equation} is zero, positive or negative. The probability distribution function in phase space, $\mathcal{P}({\bf r},{\bf v})$, is the quantity of interest.

A different quantity is the PDF of two particle distribution in phase space, $\mathcal{P}_2({\bf r}_1,{\bf r}_2,{\bf v}_1,{\bf v}_2)$ which is the joint probability to find a two charged particles $1$ and $2$ at specific points in phase space. The particle-particle correlation function in real space, \begin{equation} \Phi_2({\bf r}_1,{\bf r}_2) \equiv \int \mathcal{P}_2d^dv_1d^dv_2 \end{equation} should be a function of $\ell \equiv \mid{\bf r}_1 - {\bf r}_2 \mid$ alone due to homogeneity and isotropy in the absence of the external magnetic field. This is the quantity that has been calculated by H. Homann, J. Bec, H. Fichtner, and R. Grauer to study clustering and shown to behave like, \begin{equation} \Phi_2(\ell) \sim \ell^{D_2} \end{equation} with $D_2$ as a function of the Stokes number (${\rm St} \equiv \gamma_1 \tau_{\rm K}$) is shown in Figure 5 of their paper.

To begin with we should calculate the following quantities:

  1. Add external magnetic field to the spectral code. In our spectral code, this may be calculated in the following manner. In the equation for the magnetic field the cross-product ${\bf u} \times {\bf B}$ is calculated in real space. It is best to add a constant magnetic field the transformed-to-real-space ${\bf B}$ at this point. Similarly, one should add the same constant magnetic field to the magnetic field in the term ${\bf J}\times{\bf B}$ at the same time. While calculating the trajectories of the charged particles one must add the same magnetic field. In other words, as soon as the magnetic field is transformed to real space add a constant part to it. Before fourier transforming to Fourier space, this constant component may even be subtracted out.
  2. Let them reach a non-equilibrium statistically stationary state (NESS). It is possibly best to use exactly the same dimensionless parameters as used by Homann et al.
  3. The PDF of speeds. This would presumably be Maxwellian. In the presence of of the external magnetic field, the three compoments of velocities may have different distribution. In that case we should calculate the in-plane and parallel to the magnetic field PDFs.
  4. The auto-correlation function of ${\bf u}$ as sampled by the charged particles. Kumar's work assumes that it would have exponential in time behavior.
  5. The induced magnetic moment. Once we get a value, we should reverse the magnetic field and see if we see an exactly opposite value. Can we understand the diamagnetism in terms of Kumar's work.
  6. How does this induced magnetic field (if it is non-zero) depend on the clustering ?

Dimensionless parameters
Let us start by non-dimensionalizing the equations of evolution. To solve for the flow, we use equations of isothermal magnetohydrodynamics (IMHD). The non-dimensional form of IMHD are: \begin{eqnarray} \partial_t {\bf u} + {\bf u}\cdot\nabla{\bf u} &=& \frac{1}{{\rm Re}}\nabla^2{\bf u} - \frac{1}{{\rm Ma}}\nabla p + \frac{1}{\rm M_{\rm A}}{\bf J}\times{\bf B} + {\bf f}\\ \partial_t \rho + \nabla\cdot(\rho{\bf u}) &=& 0 \\ \partial_t {\bf B} &=& \nabla\times\left({\bf u}\times{\bf B} - \frac{1}{{\rm Re}_{\rm M}}{\bf J} \right) \end{eqnarray} where ${\rm Re} = \frac{u_{\rm rms}}{k_{\rm f}\nu}$ is the Reynolds number, ${\rm Re}_{\rm M} = \frac{u_{\rm rms}}{k_{\rm f}\eta}$ is the magnetic Reynolds number, ${\rm Ma} = \frac{u_{\rm rms}}{c_{\rm s}}$ is the Mach number, ${\rm M}_{\rm A} = \frac{u_{\rm rms}}{c_{\rm A}}$ is the Alfvenic Mach number, $c_{\rm s}$ is the speed of sound, $v_{\rm A}$ is the characteristic Alfvenic speed, $u_{\rm rms}$ is the root-mean-square velocity of turbulence. The characteristic length scale is set by $\ell = 1/k_{\rm f}$ the wavenumber corresponding to the forcing and the characteristic time scale is set by $\tau_{\rm L} = 1/u_{\rm rms}k_{\rm f}$. Now, let us use the same length and time scales to non-dimensionalize the equations of motion of the charged particles. As the charged particles feel a drag forces their typical speeds would be comparable to the flow speeds. Hence we use $u_{\rm rms}$ to set the velocity scale. The non-dimensionalized equations are: \begin{eqnarray} \dot{{\bf x}} &=& {\bf v} \\ \dot{{\bf v}} &=& \frac{1}{{\rm St}}({\bf u} - {\bf v}) + \frac{1}{{\rm Lo}}( -{\bf u}\times{\bf B} + {\bf v} \times {\bf B}) \end{eqnarray} where ${\rm St} \equiv \frac{\tau_{\rm p}}{\tau_{\rm L}}$, $ {\rm Lo} \equiv \frac{1}{\omega_{\rm c}\tau_{\rm L}}$, $ \tau_{\rm p} \equiv \frac{m_{\rm p}}{6\pi\mu a_{\rm p}}$ and $ \omega_{\rm c} \equiv \frac{q v_{\rm A}}{m_{\rm p}}$. For computational convenience we divide the particles by their radii $a_{\rm p}$. They are assumed spherical, hence their mass $m_{\rm p} = \rho_{\ast}\frac{4}{3}\pi a_{\rm p}^3$ with $\rho_{\ast}$ is the material density of the charged particles. To begin with it may be best to start with particles whose ${\rm St}$ and ${\rm Lo}$ are both order unity. This implies that we have to choose \begin{equation} \tau_{\rm p}\omega_{\rm c} \approx 1 \end{equation} In other words, \begin{equation} \frac{q v_{\rm A}}{6\pi\mu a_{\rm p}} \approx 1 \end{equation} We can do this by choosing the charge of the dust particle accordingly. We know from experience that the magnetic energy that develops in the types of simulations we are running is roughly the same order as the kinetic energy, hence $v_{\rm A} \approx u_{\rm rms}$.
Preliminary Results
We have started running simulations in resolution of $256^3$, with the following parameters: \begin{eqnarray} {\rm Re}={\rm Rm}= 77.5, \frac{v_{\rm A}}{B_{\rm eq}} = 0.50 \\ {\rm St}= 0.45, 1.8, 4.0, 7.1, 28.6 \\ {\rm Lo}= 2.4, 0.3, 0.08, 0.04, 0.005 \end{eqnarray} The Alfven velocity ($v_{\rm A}$) is the given by the magnitude of the external field imposed on the system. The equipartition magnetic field $B_{\rm eq} \equiv u_{\rm rms}$ which is the root-mean-square value of the turbulent velociy. We have a milliion particles in the system with five different values of the Stokes and Lorentz number. The induced magnetic moment is given by \begin{equation} M_i \equiv \epsilon_{ijk}\langle x_j v_k \rangle \end{equation} where the symbol $\langle \cdot \rangle$ denotes averaging over all the particles (irrespective of their Stokes and Lorenz number) in the box. The result is shown below :
The magnetic moment is normalized by $M_0 \equiv \frac{u^2_{\rm rms}}{\omega_{\rm c}}$. The black line is for an imposed magnetic field along the positive $z$ direction and the red line is for the negative $z$ direction. This gives rise to several intriguing questions:
Computational Resources

Last modified: Thu Nov 5 18:21:54 CET 2015