Part 3, Chapter 2

MHD Equations

Ideal and resistive magnetohydrodynamics derived from two-fluid theory

2.1 The MHD Approximation

MHD treats plasma as a single conducting fluid. This is obtained from the two-fluid equations by summing over species and making several key approximations:

  • Quasi-neutrality: ne = Zni = n (charge separation negligible on scales L much greater than the Debye length)
  • Low frequency: omega much less than Omegaci (slower than ion cyclotron frequency)
  • Large scales: L much greater than rhoi, lambdaD, c/omegapi
  • Non-relativistic: u much less than c (displacement current negligible)
  • Small electron mass: me/mi much less than 1 (electron inertia negligible)

2.2 Mass Density and Center-of-Mass Velocity

We define the single-fluid (MHD) variables by summing over the two species:

Step 1: Mass density

$$\rho = m_i n_i + m_e n_e = m_i n + m_e n \approx m_i n$$

Since me/mi is approximately 1/1836, the electron mass contribution is negligible.

Step 2: Center-of-mass velocity

$$\rho\mathbf{v} = m_i n_i \mathbf{u}_i + m_e n_e \mathbf{u}_e \approx m_i n \mathbf{u}_i$$

So the MHD velocity is essentially the ion velocity: v approximately equals ui.

Step 3: Current density

$$\mathbf{J} = en(\mathbf{u}_i - \mathbf{u}_e)$$

This allows us to express the electron velocity as ue = ui - J/(en) = v - J/(en).

Step 4: Total pressure

$$p = p_i + p_e = n_i k_B T_i + n_e k_B T_e = n k_B(T_i + T_e)$$

In MHD we work with a single scalar pressure that is the sum of electron and ion pressures.

2.3 MHD Continuity Equation

Add the two species continuity equations, weighted by their masses:

Step 1: Ion continuity

$$\frac{\partial(m_i n_i)}{\partial t} + \nabla\cdot(m_i n_i \mathbf{u}_i) = 0$$

Step 2: Electron continuity

$$\frac{\partial(m_e n_e)}{\partial t} + \nabla\cdot(m_e n_e \mathbf{u}_e) = 0$$

Step 3: Sum and use definitions

Adding these and using rho = mini + mene and rho v = miniui + meneue:

Result: MHD Mass Continuity

$$\boxed{\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0}$$

2.4 MHD Momentum Equation

Add the ion and electron momentum equations. The collision terms cancel (Rei = -Rie by Newton's third law).

Step 1: Write out both momentum equations

$$m_i n\left(\frac{\partial \mathbf{u}_i}{\partial t} + \mathbf{u}_i\cdot\nabla\mathbf{u}_i\right) = en(\mathbf{E}+\mathbf{u}_i\times\mathbf{B}) - \nabla p_i + \mathbf{R}_{ie}$$
$$m_e n\left(\frac{\partial \mathbf{u}_e}{\partial t} + \mathbf{u}_e\cdot\nabla\mathbf{u}_e\right) = -en(\mathbf{E}+\mathbf{u}_e\times\mathbf{B}) - \nabla p_e + \mathbf{R}_{ei}$$

Step 2: Add the equations

On the left side, we neglect electron inertia (me much less than mi), giving rho dv/dt. On the right side:

$$en(\mathbf{u}_i - \mathbf{u}_e)\times\mathbf{B} = \mathbf{J}\times\mathbf{B}$$

The electric field terms en(E) - en(E) cancel due to quasi-neutrality. The collision terms cancel: Rie + Rei = 0.

Step 3: Combine pressures

$$-\nabla p_i - \nabla p_e = -\nabla(p_i + p_e) = -\nabla p$$

Result: MHD Momentum Equation

$$\boxed{\rho\frac{d\mathbf{v}}{dt} = \rho\left(\frac{\partial\mathbf{v}}{\partial t} + \mathbf{v}\cdot\nabla\mathbf{v}\right) = \mathbf{J}\times\mathbf{B} - \nabla p}$$

The MHD momentum equation has only the J x B force and pressure gradient -- the electric field has been eliminated! This is a profound simplification.

2.5 Ohm's Law from Electron Momentum

The MHD Ohm's law is derived from the electron momentum equation alone (not the sum). Since me is small, the electron inertia is negligible.

Step 1: Electron momentum with negligible inertia

$$0 \approx -en(\mathbf{E}+\mathbf{u}_e\times\mathbf{B}) - \nabla p_e + \mathbf{R}_{ei}$$

We set the left side to zero because men due/dt is negligible compared to the electromagnetic forces.

Step 2: Express ue in terms of MHD variables

Using ue = v - J/(en):

$$\mathbf{u}_e\times\mathbf{B} = \mathbf{v}\times\mathbf{B} - \frac{\mathbf{J}\times\mathbf{B}}{en}$$

Step 3: Model the friction force

The electron-ion friction is proportional to the relative drift:

$$\mathbf{R}_{ei} = -m_e n \nu_{ei}(\mathbf{u}_e - \mathbf{u}_i) = m_e n\nu_{ei}\frac{\mathbf{J}}{en} = \frac{m_e\nu_{ei}}{ne^2}\mathbf{J} \equiv \eta_{res}\mathbf{J}$$

Here eta is the Spitzer resistivity, arising from electron-ion collisions.

Step 4: Solve for E

Rearranging the electron momentum equation:

$$\mathbf{E} + \mathbf{v}\times\mathbf{B} = \eta\mathbf{J} + \frac{1}{en}\left(\mathbf{J}\times\mathbf{B} - \nabla p_e\right)$$

This is the generalized Ohm's law. The J x B/(en) term is the Hall term, and the pressure gradient term drives diamagnetic effects.

Ideal MHD Ohm's Law (drop Hall, pressure, resistive terms)

$$\boxed{\mathbf{E} + \mathbf{v}\times\mathbf{B} = 0}$$

Resistive MHD Ohm's Law (keep only resistivity)

$$\boxed{\mathbf{E} + \mathbf{v}\times\mathbf{B} = \eta\mathbf{J}}$$

2.6 The Induction Equation

We derive the evolution equation for B by combining Faraday's law with Ohm's law.

Step 1: Start from Faraday's law

$$\frac{\partial \mathbf{B}}{\partial t} = -\nabla\times\mathbf{E}$$

Step 2: Substitute resistive Ohm's law

From E = -v x B + eta J:

$$\frac{\partial \mathbf{B}}{\partial t} = -\nabla\times(-\mathbf{v}\times\mathbf{B} + \eta\mathbf{J})$$
$$= \nabla\times(\mathbf{v}\times\mathbf{B}) - \nabla\times(\eta\mathbf{J})$$

Step 3: Use Ampere's law to eliminate J

In the MHD limit (no displacement current): J = (nabla x B)/mu0. Assuming uniform resistivity:

$$\nabla\times(\eta\mathbf{J}) = \frac{\eta}{\mu_0}\nabla\times(\nabla\times\mathbf{B}) = \frac{\eta}{\mu_0}\left[\nabla(\nabla\cdot\mathbf{B}) - \nabla^2\mathbf{B}\right]$$

Since div B = 0 (always), the first term vanishes:

$$= -\frac{\eta}{\mu_0}\nabla^2\mathbf{B}$$

Result: Resistive Induction Equation

$$\boxed{\frac{\partial\mathbf{B}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{B}) + \frac{\eta}{\mu_0}\nabla^2\mathbf{B}}$$

The first term is advection (field frozen to fluid) and the second is diffusion (field slips through fluid).

Ideal limit (eta = 0):

$$\frac{\partial\mathbf{B}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{B})$$

This implies the frozen-in flux theorem: magnetic field lines are "frozen" into the fluid.

2.7 Magnetic Reynolds Number and Flux Freezing

The ratio of the advection term to the diffusion term defines the magnetic Reynolds number:

Dimensional analysis

$$|\nabla\times(\mathbf{v}\times\mathbf{B})| \sim \frac{vB}{L}, \qquad \left|\frac{\eta}{\mu_0}\nabla^2\mathbf{B}\right| \sim \frac{\eta B}{\mu_0 L^2}$$
$$\boxed{R_m = \frac{\mu_0 v L}{\eta} = \frac{\text{advection rate}}{\text{diffusion rate}}}$$

Rm much greater than 1

Ideal MHD valid. Field frozen to plasma. Typical of astrophysical plasmas (Rm ~ 1010 in the solar corona).

Rm ~ 1 or less

Resistive effects important. Field diffuses through plasma. Laboratory plasmas, reconnection layers.

Flux Freezing Theorem (Alfven's theorem)

When Rm is much greater than 1, the magnetic flux through any closed loop moving with the fluid is conserved:

$$\Phi = \int_S \mathbf{B}\cdot d\mathbf{A} = \text{const for ideal MHD}$$

Proof sketch: The rate of change of flux through a surface S(t) moving with velocity v is:

$$\frac{d\Phi}{dt} = \int_S \frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{A} + \oint_C (\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$$

Using Stokes' theorem on the line integral and substituting the ideal induction equation:

$$= \int_S \left[\frac{\partial\mathbf{B}}{\partial t} - \nabla\times(\mathbf{v}\times\mathbf{B})\right]\cdot d\mathbf{A} = 0$$

Magnetic diffusion time

When v = 0 (static plasma), the induction equation becomes a diffusion equation:

$$\frac{\partial\mathbf{B}}{\partial t} = \frac{\eta}{\mu_0}\nabla^2\mathbf{B} \qquad\Rightarrow\qquad \tau_d = \frac{\mu_0 L^2}{\eta}$$

For a tokamak with L ~ 1 m and eta ~ 10-7 Omega m: taud ~ 10 seconds. For the Sun with L ~ 109 m: taud ~ 1010 years.

2.8 MHD Force Balance: Magnetic Pressure and Tension

The Lorentz force J x B can be decomposed into magnetic pressure and tension:

Step 1: Use Ampere's law

$$\mathbf{J}\times\mathbf{B} = \frac{1}{\mu_0}(\nabla\times\mathbf{B})\times\mathbf{B}$$

Step 2: Apply the vector identity

$$(\nabla\times\mathbf{B})\times\mathbf{B} = (\mathbf{B}\cdot\nabla)\mathbf{B} - \nabla\left(\frac{B^2}{2}\right)$$

Result

$$\boxed{\mathbf{J}\times\mathbf{B} = -\nabla\left(\frac{B^2}{2\mu_0}\right) + \frac{(\mathbf{B}\cdot\nabla)\mathbf{B}}{\mu_0}}$$

Magnetic Pressure

$$p_B = \frac{B^2}{2\mu_0}$$

Acts isotropically, pushes perpendicular to field lines. Analogous to gas pressure.

Magnetic Tension

$$\frac{(\mathbf{B}\cdot\nabla)\mathbf{B}}{\mu_0} = \frac{B^2}{\mu_0 R_c}\hat{n}$$

Acts along curved field lines toward the center of curvature Rc. Like tension in a rubber band.

2.9 The Complete Ideal MHD System

Collecting all the equations together (with adiabatic closure), the full ideal MHD system is:

$$\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) = 0 \qquad\text{(mass)}$$
$$\rho\frac{d\mathbf{v}}{dt} = \mathbf{J}\times\mathbf{B} - \nabla p = \frac{1}{\mu_0}(\nabla\times\mathbf{B})\times\mathbf{B} - \nabla p \qquad\text{(momentum)}$$
$$\frac{d}{dt}\left(\frac{p}{\rho^\gamma}\right) = 0 \qquad\text{(energy/adiabatic)}$$
$$\frac{\partial\mathbf{B}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{B}) \qquad\text{(induction)}$$
$$\nabla\cdot\mathbf{B} = 0 \qquad\text{(solenoidal constraint)}$$

Counting: 8 equations (1 scalar + 3 vector + 1 scalar + 3 vector) for 8 unknowns (rho, vx, vy, vz, p, Bx, By, Bz). The system is closed. The div B = 0 constraint is maintained automatically if satisfied initially.

Interactive Simulation

Flux Freezing: 1D Magnetic Field Advection

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Key Takeaways

  • -- MHD is derived from two-fluid equations by summing over species and exploiting me much less than mi
  • -- The MHD momentum equation contains only J x B and pressure gradient (no explicit E field)
  • -- Ohm's law comes from the electron momentum equation with negligible electron inertia
  • -- The induction equation governs B evolution: advection (frozen flux) vs diffusion
  • -- Magnetic Reynolds number Rm = mu0vL/eta determines the regime
  • -- J x B decomposes into magnetic pressure B2/(2mu0) and magnetic tension (B dot nabla)B/mu0
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