Vector and scalar potentials¶
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.
Reconnection theory is written almost entirely in terms of potentials. This page shows where they come from and why they are not extra assumptions, only another way of writing two of Maxwell's equations.
Why B = ∇ × A¶
It comes from \(\nabla\cdot\mathbf{B} = 0\), the fact that there are no magnetic monopoles.
For any smooth vector field \(\mathbf{A}\), the divergence of a curl vanishes:
So writing \(\mathbf{B} = \nabla\times\mathbf{A}\) guarantees \(\nabla\cdot\mathbf{B} = 0\). The useful direction is the converse: any divergence-free field can be written as a curl. This is the Poincaré lemma, valid in a simply connected region (one with no holes running through it). You can even build \(\mathbf{A}\) explicitly:
and checking that \(\nabla\times\mathbf{A} = \mathbf{B}\) uses only \(\nabla\cdot\mathbf{B} = 0\).
Why E = −∂A/∂t − ∇φ¶
Substitute \(\mathbf{B} = \nabla\times\mathbf{A}\) into Faraday's law:
A field with zero curl is the gradient of a scalar. Writing that scalar as \(-\phi\):
Where does ∇φ come from?¶
Let \(\mathbf{F} \equiv \mathbf{E} + \partial_t\mathbf{A}\), so \(\nabla\times\mathbf{F} = 0\). Define a scalar by a line integral from a fixed reference point \(\mathbf{r}_0\):
The path doesn't matter. Two different paths from \(\mathbf{r}_0\) to \(\mathbf{r}\) form a closed loop. By Stokes' theorem,
so both paths give the same value, and \(\phi\) is a function of position alone.
Then \(\mathbf{F} = -\nabla\phi\). Extend the path by a small step \(d\mathbf{r}\). Over a short enough segment \(\mathbf{F}\) is essentially constant, so
But for any smooth scalar the chain rule gives the total differential
Comparing, \(\nabla\phi\cdot d\mathbf{r} = -\mathbf{F}\cdot d\mathbf{r}\) for every \(d\mathbf{r}\). Choosing \(d\mathbf{r}\) along \(x\), then \(y\), then \(z\) matches the components one at a time, so \(\nabla\phi = -\mathbf{F}\).
The logical order
\(d\phi = -\mathbf{F}\cdot d\mathbf{r}\) comes first, from the definition of \(\phi\), and \(\mathbf{F} = -\nabla\phi\) follows. The reverse also works: starting from \(\mathbf{F} = -\nabla\phi\), the chain rule gives \(d\phi = -\mathbf{F}\cdot d\mathbf{r}\). The two statements are equivalent.
This is the familiar electrostatics argument. When \(\partial_t\mathbf{B} = 0\), \(\nabla\times\mathbf{E} = 0\) and \(\mathbf{E} = -\nabla\phi\). With time-varying fields, \(\mathbf{E}\) itself is no longer curl-free, but \(\mathbf{E} + \partial_t\mathbf{A}\) is.
Two versions of the same lemma¶
| Condition | Consequence |
|---|---|
| \(\nabla\times\mathbf{F} = 0\) | \(\mathbf{F} = -\nabla\phi\) (scalar potential) |
| \(\nabla\cdot\mathbf{B} = 0\) | \(\mathbf{B} = \nabla\times\mathbf{A}\) (vector potential) |
In each case, the vanishing curl (or divergence) makes an integral independent of the path (or surface).
The region must be simply connected
If the region has a hole, for example the space around a wire with the wire excluded, a loop around the hole doesn't bound a surface inside the region. Stokes' theorem can't be applied, \(\oint\mathbf{F}\cdot d\mathbf{l}\) can be nonzero even though \(\nabla\times\mathbf{F} = 0\), and \(\phi\) becomes multivalued. The field of a straight wire is the standard example: it is curl-free outside the wire, yet \(\oint\mathbf{B}\cdot d\mathbf{l} = \mu_0 I \neq 0\).
Gauge freedom¶
\(\mathbf{A}\) is not unique. Adding a gradient, \(\mathbf{A}\to\mathbf{A}' = \mathbf{A} + \nabla\chi\), leaves \(\mathbf{B}\) unchanged because \(\nabla\times\nabla\chi = 0\).
\(\mathbf{E}\) is physical too, so it must not change either. Expanding \(\mathbf{E} = -\partial_t\mathbf{A}' - \nabla\phi'\):
which equals the original \(\mathbf{E} = -\partial_t\mathbf{A} - \nabla\phi\) only if \(\phi' = \phi - \partial_t\chi\). A gauge transformation acts on both potentials together:
and leaves both \(\mathbf{B}\) and \(\mathbf{E}\) unchanged.
A common mistake
Starting from \(\mathbf{B} = \nabla\times(\mathbf{A}+\nabla\chi)\), it is tempting to write \(\nabla\times(\mathbf{E} + \partial_t\mathbf{A} + \nabla\chi) = 0\) and then \(\mathbf{E} + \partial_t\mathbf{A} + \nabla\chi = 0\). Two steps are wrong:
- The time derivative also acts on \(\nabla\chi\), so the term is \(\nabla(\partial_t\chi)\), not \(\nabla\chi\). And since the curl of any gradient is zero, it drops out of the curl anyway.
- A zero curl means the field is a gradient, not that it is zero. Setting the bracket to zero is the special choice \(\phi = 0\), not a consequence.
Check yourself
Show that \(\nabla\cdot(\nabla\times\mathbf{A}) = 0\) by writing out the components.
Answer
\(\partial_x(\partial_y A_z - \partial_z A_y) + \partial_y(\partial_z A_x - \partial_x A_z) + \partial_z(\partial_x A_y - \partial_y A_x)\). Every mixed second derivative appears twice with opposite signs, and they cancel because partial derivatives commute.
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
Under the gauge transformation \(\mathbf{A}\to\mathbf{A}+\nabla\chi\), how must \(\phi\) change so that \(\mathbf{E}\) stays the same?
- \(\phi\to\phi-\partial\chi/\partial t\)
- \(\phi\to\phi+\partial\chi/\partial t\)
- \(\phi\to\phi+\chi\)
- \(\phi\) does not change
\(\mathbf{E} = -\partial_t\mathbf{A}-\nabla\phi\) picks up \(-\nabla(\partial_t\chi)\) from \(\mathbf{A}\), which \(\phi\to\phi-\partial_t\chi\) cancels.
Problem 2
A vector field has zero curl everywhere in a simply connected region. What can you conclude?
- It is the gradient of a scalar
- It is zero
- It is the curl of another field
- Nothing
Zero curl makes line integrals path-independent (Stokes), so the field is the gradient of a potential. It need not be zero.
Problem 3
Outside a straight wire, \(\nabla\times\mathbf{B} = 0\), yet \(\oint\mathbf{B}\cdot d\mathbf{l} = \mu_0I\neq0\) around the wire. Why?
- The region outside a wire is not simply connected
- Ampère's law fails there
- The field is not static
- The loop is too large
A loop around the wire cannot be shrunk to a point without crossing the wire, so Stokes' theorem cannot be used on a surface inside the region.
Show that…¶
Problem 4: curl of a gradient
Show that \(\nabla\times\nabla\chi = 0\) by writing out one component.
Hint
Look at the \(x\) component: \(\partial_y(\partial_z\chi) - \partial_z(\partial_y\chi)\).
Proof
The \(x\) component is \(\partial_y\partial_z\chi - \partial_z\partial_y\chi = 0\) because partial derivatives commute; the same holds for \(y\) and \(z\).
Problem 5: a uniform field
Show that \(\mathbf{A} = \tfrac12\mathbf{B}_0\times\mathbf{r}\) gives a uniform field \(\mathbf{B}_0\).
Hint
Use \(\nabla\times(\mathbf{a}\times\mathbf{b}) = \mathbf{a}(\nabla\cdot\mathbf{b}) - (\mathbf{a}\cdot\nabla)\mathbf{b}\) for constant \(\mathbf{a}\), with \(\nabla\cdot\mathbf{r} = 3\).
Proof
\(\nabla\times(\mathbf{B}_0\times\mathbf{r}) = \mathbf{B}_0(\nabla\cdot\mathbf{r}) - (\mathbf{B}_0\cdot\nabla)\mathbf{r} = 3\mathbf{B}_0 - \mathbf{B}_0 = 2\mathbf{B}_0\). Halving gives \(\mathbf{B}_0\).
Numbers¶
Problem 6
A uniform field \(B = 2\) T points along \(z\). Using \(\mathbf{A} = \tfrac12\mathbf{B}\times\mathbf{r}\), compute \(\oint\mathbf{A}\cdot d\mathbf{l}\) around a circle of radius 0.5 m in the \(xy\) plane (in Wb).
Solution
On the circle \(|\mathbf{A}| = \tfrac12Br\), tangent to it, so \(\oint\mathbf{A}\cdot d\mathbf{l} = \tfrac12Br\cdot2\pi r = \pi r^2B = 1.57\) Wb: the flux, as Stokes requires.
Problem 7
For the same \(\mathbf{A}\), what is \(|\mathbf{A}|\) at a distance 1 m from the \(z\) axis, in T·m?
Solution
\(|\mathbf{A}| = \tfrac12Br = \tfrac12\times2\times1 = 1\) T·m.