Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 69 (Magnetohydrodynamics and Turbulence)

Statistical homogeneity makes the two-point correlation depend on the displacement $\mathbf{u}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Substituting the Fourier transforms and changing variables from $(\mathbf{r},{\mathbf{r}}^{\prime })$ to $(\mathbf{u},{\mathbf{r}}^{\prime })$ givesThe last integral is a Dirac delta function, soThe plus sign occurs because neither Fourier transform has undergone complex conjugation. For a real magnetic field, the version with ${\hat{B}}_j({\mathbf{k}}^{\prime }{)}^{\ast }$ instead has $\delta (\mathbf{k}-{\mathbf{k}}^{\prime })$. Homogeneity determines dependence on the displacement vector; dependence only on its norm needs statistical isotropy as well.

Statistical isotropy and statistical reflection symmetry allow the spectral tensor to contain only ${\delta }_{ij}$ and $k_i{k}_j$:The solenoidal magnetic-field constraint is essential here. Taking its Fourier transform gives $k_i{\hat{B}}_i=0$, hence $k_i{\hat{C}}_{ij}=0$ and $A+D{k}^2=0$. Thus the spectral tensor is a scalar multiple of the transverse projector of a vector field. Its trace is $2A$, so the prescribed normalization givesThis is the solenoidal isotropic spectral tensor. Without the solenoidal condition, isotropy and reflection symmetry would leave two scalar functions. Without reflection symmetry, an antisymmetric term proportional to $i{ϵ}_{ijm}{k}_m$ could additionally encode magnetic helicity.

Write the transverse displacement as $\mathbf{s}$ and take the common line-of-sight interval to be $[0,L]$. The rotation measure is a line integral of the magnetic field, so its correlation isSubstitute the Fourier inversion of the solenoidal isotropic spectral tensor. Keeping the finite observation length gives the exact windowed expressionThe allowed extension of the longitudinal displacement integral to the whole real line is the long-path projection of a magnetic correlation. It applies when $L$ is much larger than the correlation length: equivalently, the squared window becomes $2\pi L\delta (k_z)$ inside the integral. The Dirac delta function sets $k_z=0$, leavingStatistical isotropy makes this a function of $s=|\mathbf{s}|$. The displayed two-dimensional relation is the intended long-path approximation; the preceding windowed formula accounts for finite-length edge effects.

Invert the two-dimensional Fourier transform from part (c):In polar coordinates, the angular integral is $2\pi J_0(ks)$, where $J_0$ is a Bessel function of the first kind. The Fourier-Hankel normalization for an isotropic spectrum therefore yieldsThis Hankel inversion of a rotation-measure correlation recovers the scalar spectral tensor coefficient. To obtain the shell-integrated magnetic energy spectrum, take the trace of the three-dimensional tensor:With magnetic energy density $B^2/(8\pi )$ in Gaussian units, the one-dimensional convention ${\int }_0^{\infty }{E}_B(k)dk=\mathbb{E}|\mathbf{B}{|}^2/(8\pi )$ gives $\boxed{E_B(k)=k^{2}H(k)/(8{\pi }^3)}$. A spectrum normalized instead to $\mathbb{E}|\mathbf{B}{|}^2/2$ has coefficient $k^{2}H(k)/(2{\pi }^2)$; the physical normalization must be stated.

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.

Let $b_l=\delta B_l/B_0$. The nonlinear term has two perpendicular spatial derivatives, so its rate relative to the fluctuation amplitude isThe wave period from part (a), using $k\sim k_{\perp }$, is ${\tau }_w\sim l_{\parallel }{l}_{\perp }/(v_A{d}_i)$. A weak wave cascade requires that one interaction change a packet only slightly:For interactions with decorrelated phases, successive fractional changes accumulate as a random walk. An order-one change needs $N{\chi }^2\sim 1$, givingConstant energy flux per unit mass then gives $ϵ\sim v_A^2{b}_l^2/{\tau }_l\sim v_A^3{d}_i{b}_l^4{l}_{\parallel }/l_{\perp }^3$, henceThe weak electron-magnetohydrodynamic cascade scaling uses both locality in scale and the decorrelation assumption; a small amplitude by itself does not specify the cascade time.

For an isotropic weak electron-magnetohydrodynamic cascade, set $l_{\parallel }\sim l_{\perp }\sim l$. Part (b) gives $\delta B_l\propto l^{1/2}$. With the one-dimensional shell convention for the magnetic energy spectrum, fluctuations on scale $l\sim k^{-1}$ satisfy $\delta B_l^2\sim k{E}_B(k)$ up to normalization constants. ThereforeThis is the shell-integrated spectrum. The three-dimensional spectral tensor coefficient has two additional powers of $k^{-1}$ and scales as $H(k)\propto k^{-4}$. These exponents apply within the assumed local, weak cascade range.

Critical balance makes the nonlinear interaction time comparable to the wave period:An order-one interaction transfers energy in ${\tau }_l\sim {\tau }_{\mathrm{nl}}$, so constant energy flux gives $ϵ\sim v_A^3{d}_i{b}_l^3/l_{\perp }^2$. Thus the critically balanced electron-magnetohydrodynamic cascade hasFor the corresponding perpendicular magnetic energy spectrum, $\delta B_l^2\sim k_{\perp }{E}_B(k_{\perp })$ gives $E_B(k_{\perp })\propto k_{\perp }^{-7/3}$. The increasing ratio $l_{\parallel }/l_{\perp }\propto l_{\perp }^{-2/3}$ describes progressively more anisotropic fluctuations at smaller perpendicular scales, within the range in which the model and critical balance assumptions hold.

The strain is an Ornstein-Uhlenbeck process. Multiplying its equation by the integrating factor $e^{t/\tau }$ and using the zero initial condition givesThe second expression is a stochastic integral against Brownian motion. A deterministic linear functional of Gaussian white noise is Gaussian, and its expected value is zero. For $t\ge t^{\prime }$, the noise covariance givesFor $t^{\prime }\gg \tau$, the initial-condition term is negligible. The stationary autocorrelation is thereforeIts normalized exponential decay has correlation time $\tau$. The zero-start Ornstein-Uhlenbeck covariance also shows how stationarity is approached.

Differentiate the product of Dirac delta functions defining $\tilde{P}$, regarding $B$ and $\sigma$ as independent density arguments. The random continuity equation isFor $0<s<t$, the causal functional derivatives follow from the explicit strain solution and $\tilde{B}(t)=B_0\exp [{\int }_0^t\tilde{\sigma }(u)du]$:Consequently,In the Furutsu–Novikov formula, the covariance $\kappa \delta (t-s)$ samples the upper endpoint of the causal time integral. Symmetric regularization of the white noise gives half the mass there. The $B$ response vanishes as $s\uparrow t$, while the strain response tends to one, so $\mathbb{E}[f(t)\tilde{P}(t)]=-(\kappa /2){\partial }_{\sigma }P$. Averaging the continuity equation yields the Fokker-Planck equationThere is no direct diffusion term in $B$: its equation contains the time integral of colored noise, whereas the strain equation receives Gaussian white noise. The same result follows from the Itô diffusion with drift $(\sigma B,-\sigma /\tau )$ and diffusion vector $(0,\sqrt{\kappa })$.

Multiply the Fokker-Planck equation by $B^n$ and integrate over $B>0$ and all $\sigma$, assuming the required boundary terms vanish. Integration by parts givesThe mixed moment is not determined by $\mathbb{E}[B^n]$: the current strain and the accumulated magnetic field are correlated. Zero mean strain does not imply $\mathbb{E}[\sigma B^n]=0$, so this is an unclosed moment equation.

Keep the strain dependence by defining $P_n(\sigma ,t)={\int }_0^{\infty }{B}^{n}P(B,\sigma ,t)dB$. The magnetic drift term satisfiesThe strain derivatives commute with the $B$ integral. HenceThis weighted density obeys a tilted Ornstein-Uhlenbeck Fokker-Planck equation. Its integral over $\sigma$ is the desired moment, while retaining $\sigma$ makes the evolution closed.

For a separated weighted density, write $P_n=e^{\gamma t}p(\sigma )$. Its eigenvalue equation isSet $p=\psi \exp [-{\sigma }^2/(2\kappa \tau )]$. Differentiation eliminates the first derivative of $\psi$, giving the tilted Ornstein-Uhlenbeck oscillator transformationComplete the square and introduce the dimensionless variable $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$. The equation becomesComparing with the quantum harmonic oscillator gives $2E=1+\kappa n^2{\tau }^3-2\gamma \tau$. Its energies $E_m=m+1/2$ therefore correspond to ${\gamma }_m=\kappa n^2{\tau }^2/2-m/\tau$. The dominant nonnegative separated mode is the ground state, which is nodeless. ConsequentlyFor an independent check, $\log [\tilde{B}(t)/B_0]={\int }_0^t\tilde{\sigma }(s)ds$ is a centered Gaussian random variable with varianceThe exponential moment of a Gaussian linear functional gives $\mathbb{E}[B^n]=B_0^n\exp [n^{2}V(t)/2]$, whose long-time growth rate is the same ${\gamma }_n$. This exact Ornstein-Uhlenbeck multiplicative amplification formula also distinguishes the finite-time transient from the asymptotic exponential law. A general transient can contain sign-changing excited eigenfunctions in its expansion while the total weighted density remains nonnegative.

Assume statistically steady, three-dimensional turbulence at large Reynolds number, with local statistical homogeneity and statistical isotropy at scales small compared with the forcing scale $L$. The Kolmogorov 1941 theory additionally assumes a local energy cascade, constant mean energy flux per unit mass $ϵ$, and negligible viscous dissipation within the inertial range. Its dimensional form neglects corrections from internal intermittency.

At separation $l$, let $\delta u_l$ be a typical velocity increment. The eddy turnover time is ${\tau }_l\sim l/\delta u_l$. Constant transfer rate then gives $ϵ\sim \delta u_l^2/{\tau }_l\sim \delta u_l^3/l$, henceHere $E(k)$ is the one-dimensional turbulent energy spectrum, obtained from $\delta u_l^2\sim kE(k)$ with $k\sim l^{-1}$. The scaling holds only forThe Kolmogorov length scale follows by setting the scale-dependent Reynolds number $\delta u_{l}l/\nu$ to one. The associated Kolmogorov microscales give ${\tau }_{\eta }=(\nu /ϵ{)}^{1/2}$. At larger scales the forcing matters; at smaller scales kinematic viscosity matters.

Take a positive initial separation $0<l_0\ll \eta$. In the dissipation range, the velocity field is smooth, so the difference of the two Lagrangian trajectories obeys the linearized separation equation $\dot{\mathbf{l}}\simeq (\nabla \mathbf{u})\mathbf{l}$. Chaotic stretching gives a positive Lyapunov exponent of order ${\tau }_{\eta }^{-1}$ and a typical separationThis needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.

Once $l$ lies in the inertial range, the Kolmogorov 1941 theory gives a relative speed of order $(ϵl{)}^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim (ϵl{)}^{1/3}$ integrates toAfter loss of the entry-scale memory, the statistical Richardson pair dispersion law is $\mathbb{E}[l^2]\sim g_Rϵ(t-t_1{)}^3$. Thus the characteristic separation grows as $(t-t_1{)}^{3/2}$ while $\eta \ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.

For $l\gg L$, the particle velocities become approximately independent. If their Lagrangian velocity autocorrelations have a finite Lagrangian integral time $T_L\sim L/U$, the diffusive large-scale pair dispersion has effective relative diffusivity of order $UL$. After a further velocity-decorrelation transient,This is the long-time mechanism of Taylor turbulent dispersion applied to the difference of two decorrelated trajectories. The root mean square separation eventually loses memory of the $L^2$ term.

Reaching $l=L$ at the end of the inertial range takesOrder-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer eddy turnover time.

The three-regime turbulent pair-separation model in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.