The flux function ψ¶
Advanced level
This page assumes vector calculus (Stokes' theorem, line integrals) and is meant for students who have finished the Plasma Basics and Magnetic Reconnection sections.
In two dimensions, the whole magnetic field can be described by a single scalar, the flux function \(\psi\). It is the most useful tool for picturing reconnection.
Coordinates¶
Reconnection studies use LMN coordinates, a right-handed system aligned with the current sheet:
- \(L\): along the reconnecting field (the outflow direction),
- \(M\): along the current (out of the reconnection plane),
- \(N\): normal to the current sheet (the inflow direction).
They match the \((x, y, z)\) of the Harris sheet page with \(L = x\), \(M = y\), \(N = z\). A 2D model assumes nothing varies along \(M\): \(\partial/\partial M = 0\).
Defining ψ¶
With \(\partial_M = 0\), the divergence-free condition becomes
That is precisely the condition for a scalar \(\psi\) to exist with
In terms of the vector potential of the previous section, \(\psi = -A_M\).
The same idea as a stream function
For 2D incompressible flow, \(\nabla\cdot\mathbf{v} = 0\) gives \(v_x = \partial_y\psi\), \(v_y = -\partial_x\psi\). Streamlines are contours of \(\psi\), and the volume flux between two streamlines is \(\Delta\psi\). The magnetic version is a direct copy.
Field lines are contours of ψ¶
Along a field line, \(d\mathbf{l}\parallel\mathbf{B}\). Then
because \(dL/dN = B_L/B_N\). So ψ is constant along every field line. To draw field lines, you just draw contours of \(\psi\).
Why Δψ is the magnetic flux¶
Magnetic flux is \(\int\mathbf{B}\cdot d\mathbf{a}\). The surface to picture is a ribbon: take any curve \(C\) in the L–N plane from \(P_0\) to \(P\) and sweep it a length \(\Delta M\) along \(M\).
For a small step \(d\mathbf{l} = (dL, dN)\) along \(C\), the ribbon's area element is
Then
Step by step: the cross product and the dot product
Along \(C\), a small step is \(d\mathbf{l} = dL\,\hat{\mathbf{L}} + dN\,\hat{\mathbf{N}}\). For a right-handed \((L, M, N)\) system the cyclic order is \(L\to M\to N\to L\), so \(\hat{\mathbf{M}}\times\hat{\mathbf{N}} = \hat{\mathbf{L}}\) and \(\hat{\mathbf{M}}\times\hat{\mathbf{L}} = -\hat{\mathbf{N}}\). Then \(\hat{\mathbf{M}}\times d\mathbf{l} = dN\,\hat{\mathbf{L}} - dL\,\hat{\mathbf{N}}\), and dotting with \(\mathbf{B} = B_L\hat{\mathbf{L}} + B_M\hat{\mathbf{M}} + B_N\hat{\mathbf{N}}\) gives \(B_L\,dN - B_N\,dL\).
Geometrically, \((dN, -dL)\) is the step \((dL, dN)\) rotated by 90°: the in-plane normal to \(C\). The dot product keeps only the part of \(\mathbf{B}\) crossing the curve. A guide field \(B_M\) runs parallel to the ribbon and contributes nothing.
The integrand is a total differential, so only the endpoints matter:
\(\psi(P) - \psi(P_0)\) is the magnetic flux per unit length in \(M\) between the field lines through \(P_0\) and \(P\).
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Why only the endpoints matter¶
Parametrize \(C\) as \(\mathbf{r}(t) = (L(t), N(t))\), \(t\) from 0 to 1. Along \(C\), \(\psi\) is an ordinary function of one variable, and by the chain rule \(d\psi/dt = \partial_L\psi\,\dot L + \partial_N\psi\,\dot N\), exactly the integrand. So
This is the gradient theorem, \(\int_C\nabla\psi\cdot d\mathbf{r} = \psi(\text{end}) - \psi(\text{start})\): the ordinary fundamental theorem of calculus, applied along the curve.
Not every integrand works this way. \(\int L\,dN\) depends on the path; around a closed loop it gives the enclosed area. A form \(a\,dL + b\,dN\) is a total differential exactly when \(\partial a/\partial N = \partial b/\partial L\). Here \(a = -B_N\), \(b = B_L\), so the condition is
The flux integral is path-independent because \(\nabla\cdot\mathbf{B} = 0\), the same reason \(\psi\) exists at all.
Three stages, each dropping a dimension
- Surface integral (2D). The ribbon needs two coordinates: \(s\) along \(C\) and \(m\) along \(M\). \(\Phi = \int_0^{\Delta M}\!\int_C \mathbf{B}\cdot(\hat{\mathbf{M}}\times\hat{\mathbf{t}})\,ds\,dm\).
- Line integral (1D). Nothing depends on \(m\), so the \(m\)-integral is just a factor \(\Delta M\): \(\Phi = \Delta M\int_C (B_L\,dN - B_N\,dL)\).
- Endpoint values (0D). The integrand is \(d\psi\), so \(\Phi = \Delta M\,[\psi(P) - \psi(P_0)]\), just as \(\int_1^3 2x\,dx = 9 - 1\).
The first drop comes from symmetry in \(M\); the second from the fundamental theorem of calculus. In the figure, the shaded ribbon is stage 1, the curve \(C\) is stage 2, and the two dots are stage 3.
The same result using Stokes' theorem¶
Since \(\mathbf{B} = \nabla\times\mathbf{A}\), \(\int_{\text{ribbon}}\mathbf{B}\cdot d\mathbf{a} = \oint_{\partial(\text{ribbon})}\mathbf{A}\cdot d\mathbf{l}\), where the loop is the ribbon's edge, traversed by the right-hand rule relative to \(d\mathbf{a}\propto\hat{\mathbf{M}}\times\hat{\mathbf{t}}\). On each edge, \(d\mathbf{l}\) is just the small step along that edge:
| Edge | Path | \(d\mathbf{l}\) | Contribution |
|---|---|---|---|
| 1 | At \(P_0\), from \(m=0\) to \(\Delta M\) | \(+\hat{\mathbf{M}}\,dm\) | \(+A_M(P_0)\,\Delta M\) |
| 2 | Back copy of \(C\), \(P_0\to P\) | step along \(C\) | \(+\int_C\mathbf{A}_{\text{in-plane}}\cdot d\mathbf{l}\) |
| 3 | At \(P\), from \(m=\Delta M\) back to 0 | \(-\hat{\mathbf{M}}\,dm\) | \(-A_M(P)\,\Delta M\) |
| 4 | Front copy of \(C\), \(P\to P_0\) | reversed step | \(-\int_C\mathbf{A}_{\text{in-plane}}\cdot d\mathbf{l}\) |
Edges 2 and 4 cover the same curve in opposite directions with identical \(\mathbf{A}\), so they cancel. Only the edges along \(M\) survive:
using \(\psi = -A_M\). This matches the direct surface integral, sign included. \(A_M\) carries the flux because it is the only component of \(\mathbf{A}\) that produces \((B_L, B_N)\).
It doesn't depend on the curve¶
Two curves \(C\) and \(C'\) from \(P_0\) to \(P\), together with their ribbons, enclose a closed surface. \(\nabla\cdot\mathbf{B} = 0\) means zero net flux out of it, so both ribbons carry the same flux. Every field line that crosses one curve crosses the other.
Field strength from contour spacing¶
\(\nabla\psi = (\partial_L\psi, \partial_N\psi) = (-B_N, B_L)\) is the in-plane field rotated by 90°, so
Draw contours at equal steps \(\Delta\psi\). Neighbouring contours a perpendicular distance \(\Delta s_\perp\) apart satisfy \(|\mathbf{B}_{\text{in-plane}}|\,\Delta s_\perp\approx\Delta\psi = \text{const}\). Close contours mean strong field. Where the field is nearly along \(L\) this reads \(B_L\,\Delta n\approx\Delta\psi\). It is the 2D version of \(BA = \text{const}\), with \(\Delta n\) playing the role of the tube's cross-section per unit length in \(M\).
Current and electric field¶
The out-of-plane current is the Laplacian of ψ:
In 2D, a gauge change \(\nabla\chi(L,N,t)\) has no \(M\) component, so \(\nabla\phi\) has no \(M\) component either and
holds exactly. The only gauge freedom left in \(\psi\) is an additive constant, which affects neither \(\mathbf{B}\) nor \(E_M\).
The reconnection rate
The flux reconnected per unit time (per unit length in \(M\)) is the rate at which \(\psi\) changes at the X-point:
The out-of-plane electric field at the X-point is the reconnection rate. This is how simulations and spacecraft measure it. But ideal MHD says \(E_M = 0\) wherever \(\nabla\psi = 0\), including the X-point. Why that isn't a contradiction is the subject of Reconnection rate and Ohm's law.
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 a field-line plot with contours of \(\psi\) drawn at equal steps, where is the in-plane field strongest?
- Where the contours are closest together
- Where the contours are furthest apart
- Where the contours are most curved
- It cannot be told from the plot
\(|\mathbf{B}_{\text{in-plane}}| = |\nabla\psi|\approx\Delta\psi/\Delta s_\perp\): close contours mean a steep \(\psi\) and a strong field.
Problem 2
What does the difference \(\psi(P)-\psi(P_0)\) measure?
- The magnetic flux per unit length in \(M\) between the field lines through \(P_0\) and \(P\)
- The current between \(P_0\) and \(P\)
- The length of the field line
- The energy stored between the points
Integrating \(\mathbf{B}\cdot d\mathbf{a}\) over a ribbon from \(P_0\) to \(P\) gives \(\Delta M[\psi(P)-\psi(P_0)]\).
Show that…¶
Problem 3: ψ is constant along field lines
Using \(B_L = \partial_N\psi\) and \(B_N = -\partial_L\psi\), show that \(\mathbf{B}\cdot\nabla\psi = 0\).
Hint
Write out \(\mathbf{B}\cdot\nabla\psi = B_L\partial_L\psi + B_N\partial_N\psi\).
Proof
\(B_L\partial_L\psi + B_N\partial_N\psi = \partial_N\psi\,\partial_L\psi - \partial_L\psi\,\partial_N\psi = 0\). So \(\psi\) does not change along \(\mathbf{B}\): field lines are contours of \(\psi\).
Problem 4: the current is the Laplacian
Show that \(\mu_0J_M = \nabla^2\psi\).
Hint
\(\mu_0J_M = \partial_NB_L - \partial_LB_N\); substitute the definitions of \(B_L\) and \(B_N\).
Proof
\(\partial_N(\partial_N\psi) - \partial_L(-\partial_L\psi) = \partial_N^2\psi + \partial_L^2\psi = \nabla^2\psi\).
Numbers¶
Problem 5
Field-line contours are drawn every \(\Delta\psi = 4\times10^{-5}\) Wb/m. In the inflow region they are 2 km apart. What is \(B_L\), in nT?
Solution
\(B_L\approx\Delta\psi/\Delta n = 4\times10^{-5}/2000 = 2\times10^{-8}\) T \(= 20\) nT.
Problem 6
For a Harris sheet, \(\psi = B_0\lambda\ln\cosh(z/\lambda)\). With \(B_0 = 20\) nT and \(\lambda = 1000\) km, how much flux per unit length lies between \(z = 0\) and \(z = \lambda\), in Wb/m?
Solution
\(\Delta\psi = B_0\lambda\ln\cosh1 = 2\times10^{-8}\times10^6\times0.434 = 8.7\times10^{-3}\) Wb/m.