Reconnection rate and Ohm's law¶

Advanced level

This page builds on The flux function ψ and X-points, O-points and islands.

The flux function page showed that \(E_M = \partial\psi/\partial t\). This page asks two questions: does that hold everywhere, and if the in-plane field vanishes at the X-line, how can \(E_M\) be nonzero there?

Two different statements¶

1. Faraday's law (always true in 2D). From \(\mathbf{E} = -\partial_t\mathbf{A} - \nabla\phi\), the \(M\) component is

\[ E_M = -\frac{\partial A_M}{\partial t} - \underbrace{\frac{\partial\phi}{\partial M}}_{=\,0} = \frac{\partial\psi}{\partial t}, \]

using \(\partial_M = 0\) and \(\psi = -A_M\). No assumption about the plasma enters. It holds at every point of the L–N plane, X-line or not.

2. Ohm's law (a plasma closure). This determines what \(E_M\) actually is. With ideal Ohm's law, \(\mathbf{E} + \mathbf{v}\times\mathbf{B} = 0\):

\[ E_M = -(\mathbf{v}\times\mathbf{B})_M = -(v_N B_L - v_L B_N) = -\mathbf{v}\cdot\nabla\psi , \]

so combining with Faraday's law,

\[ \frac{\partial\psi}{\partial t} + \mathbf{v}\cdot\nabla\psi = 0 . \]

\(\psi\) is carried with the flow. Each field line keeps its \(\psi\) label as it moves, so the flux between two moving field lines is conserved. This is the 2D version of \(BA = \text{const}\). Away from the X-line, a nonzero \(E_M\) just means field lines are drifting past a fixed point; no flux is reconnected.

The puzzle at the X-line¶

From the definition of \(\psi\),

\[ \nabla\psi = \left(\frac{\partial\psi}{\partial L}, \frac{\partial\psi}{\partial N}\right) = (-B_N,\; B_L). \]

This is the in-plane field rotated by 90°, so \(|\nabla\psi| = |\mathbf{B}_{\text{in-plane}}|\), and \(\nabla\psi\perp\mathbf{B}\) everywhere (which is why B runs along contours of \(\psi\)). At the X-line, \(B_L = B_N = 0\), so \(\nabla\psi = 0\). Ideal Ohm's law would then force

\[ E_M = -\mathbf{v}\cdot\nabla\psi = 0 \quad\text{at the X-line}. \]

But observed and simulated \(E_M\) at the X-line is not zero. Is the framework inconsistent?

Resolution

No. This is a proof by contradiction that ideal Ohm's law fails at the X-line. If it held, \(\psi_X\) would never change and no flux could ever be reconnected. A nonzero \(E_M\) at the X-line means some non-ideal term must be nonzero there.

The generalized Ohm's law¶

For the electron fluid,

\[ \mathbf{E} + \mathbf{v}_e\times\mathbf{B} = \mathbf{R}, \qquad \mathbf{R} = -\frac{\nabla\cdot\mathsf{P}_e}{ne} - \frac{m_e}{e}\frac{d\mathbf{v}_e}{dt} + \eta\mathbf{J}. \]

Taking the \(M\) component and combining with Faraday's law:

\[ \frac{\partial\psi}{\partial t} + \mathbf{v}_e\cdot\nabla\psi = R_M . \]

Away from the diffusion region, \(R_M\approx 0\): \(\psi\) is carried with the electrons, and field lines keep their labels.
At the X-line, \(\nabla\psi = 0\), so the advection term drops out, leaving

\[ \boxed{\;\frac{\partial\psi_X}{\partial t} = E_M\big|_X = R_M\big|_X\;} \]

The reconnection rate equals the non-ideal term at the X-line. In collisionless plasmas \(\eta\) is negligible, so electron inertia and the off-diagonal (nongyrotropic) parts of the electron pressure tensor supply \(R_M\).

Why the electron velocity, not the bulk velocity?

Ions decouple from the field at the ion inertial length \(d_i\). Between \(d_i\) and the electron inertial length \(d_e\), the field is frozen to the electrons, and the Hall term \(\mathbf{J}\times\mathbf{B}/(ne)\) accounts for the difference between \(\mathbf{v}\) and \(\mathbf{v}_e\). The Hall term is proportional to \(\mathbf{B}\), so it also vanishes at the X-line and cannot support \(E_M\) there. Only the electron pressure and inertia terms, acting on the \(d_e\) scale, can.

Why this is the reconnection rate¶

It is frame-independent. If the X-line moves with velocity \(\mathbf{u}\), the value of \(\psi\) following it changes as \(d\psi_X/dt = \partial_t\psi + \mathbf{u}\cdot\nabla\psi = \partial_t\psi\), because \(\nabla\psi = 0\) there.
It measures flux transfer. \(\psi_X\) labels the separatrix. The flux between the separatrix and a reference line far upstream is \(\psi_{\text{up}} - \psi_X\). When \(\psi_X\) changes, flux moves from the inflow (unreconnected) region to the outflow (reconnected) region:

\[ E_M\big|_X = \text{flux reconnected per unit time per unit length in } M . \]

The steady-state special case¶

In steady state, \(\partial_t\mathbf{B} = 0\), so \(\nabla\times\mathbf{E} = 0\). In 2D that forces \(E_M\) to be uniform, the same everywhere, including the inflow region where ideal Ohm's law holds:

\[ E_M = -v_N B_L . \]

That is why the reconnection rate can be estimated from the inflow speed and upstream field. When reconnection is time-dependent, the value at the X-line is the one that defines the rate.

What MMS sees

In an electron diffusion region (EDR) encounter, MMS measures \(\mathbf{E} + \mathbf{v}_e\times\mathbf{B} \ne 0\) at the X-line, together with a positive \(\mathbf{J}\cdot(\mathbf{E} + \mathbf{v}_e\times\mathbf{B})\): energy being dissipated in the electron frame.

Check yourself

In the inflow region, \(B_L = 20\) nT and the inflow speed is \(v_N = 40\) km/s towards the sheet. The upstream Alfvén speed is 400 km/s. Estimate \(E_M\) and the normalised reconnection rate.

Answer

\(|E_M| = |v_N| B_L = (4\times10^4)(2\times10^{-8}) = 0.8\) mV/m. Normalised rate \(= v_N/v_A = 0.1\), the canonical fast-reconnection value.

Practice problems¶

Try each problem before opening its solution. Click an option to check a concept question; type a number and press Check for a numerical one (forms like 6.2e7 or 6.2x10^7 both work).

Concept questions¶

Problem 1

In collisionless reconnection, which term of the generalised Ohm's law supports \(E_M\) exactly at the X-line?

The electron pressure tensor and electron inertia
The Hall term \(\mathbf{J}\times\mathbf{B}/ne\)
\(\mathbf{v}\times\mathbf{B}\)
Resistivity \(\eta\mathbf{J}\)

The in-plane field vanishes at the X-line, so \(\mathbf{v}\times\mathbf{B}\) and the Hall term vanish there. Collisions are negligible, which leaves the electron pressure tensor and inertia.

Problem 2

In steady 2D reconnection, how does \(E_M\) vary across the L–N plane?

It is uniform
It peaks at the X-line and vanishes elsewhere
It vanishes at the X-line
It grows with distance from the X-line

Steady state means \(\partial_t\mathbf{B} = 0\), so \(\nabla\times\mathbf{E} = 0\); in 2D that forces \(E_M\) to be constant.

Show that…¶

Problem 3: ψ is carried with the flow

Combine \(E_M = \partial_t\psi\) with ideal Ohm's law to show that \(\partial_t\psi + \mathbf{v}\cdot\nabla\psi = 0\).

Hint

Compute \(E_M = -(\mathbf{v}\times\mathbf{B})_M\) and use \(B_L = \partial_N\psi\), \(B_N = -\partial_L\psi\).

Proof

\(-(\mathbf{v}\times\mathbf{B})_M = -(v_NB_L - v_LB_N) = -(v_N\partial_N\psi + v_L\partial_L\psi) = -\mathbf{v}\cdot\nabla\psi\). Setting this equal to \(\partial_t\psi\) gives the result: \(\psi\) is constant following the flow.

Problem 4: the Hall term vanishes at the X-line

Without a guide field, show that the \(M\) component of the Hall term \(\mathbf{J}\times\mathbf{B}/(ne)\) vanishes at the X-line.

Hint

At the X-line \(B_L = B_N = 0\), and with no guide field \(B_M = 0\) too.

Proof

\(\mathbf{B} = 0\) at the X-line, so \(\mathbf{J}\times\mathbf{B} = 0\) there, whatever \(\mathbf{J}\) is. The Hall term cannot support \(E_M\).

Numbers¶

Problem 5

Reconnection proceeds at a normalised rate of 0.1 with \(B = 10\) nT and \(v_A = 500\) km/s upstream. What is \(E_M\), in mV/m?

Solution

\(E_M = 0.1\,v_AB = 0.1\times5\times10^5\times10^{-8} = 5\times10^{-4}\) V/m \(= 0.5\) mV/m.

Problem 6

How much flux per unit length in \(M\) does that rate reconnect in one minute, in Wb/m?

Solution

\(E_M\,\Delta t = 5\times10^{-4}\times60 = 0.03\) Wb/m.

Problem 7

In the inflow, \(B_L = 20\) nT and \(v_N = 40\) km/s. What is \(|E_M|\), in mV/m?

Solution

\(|E_M| = v_NB_L = 4\times10^4\times2\times10^{-8} = 8\times10^{-4}\) V/m \(= 0.8\) mV/m.