Most discussions of the magnetic vector potential defined through $\mathbf{B}=\nabla\times\mathbf{A}$ are only for working with static electric fields (for example, Griffiths:

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If we instead require that

$$\nabla\times\mathbf{B}=\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})\ -\nabla^2\mathbf{A} = \mu_0\mathbf{J} +\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

using the full form of Ampere's law, then we can still have the freedom to choose an $\mathbf{A}$ with zero divergence, giving us a form of Poisson's equation:

$$\nabla^2\mathbf{A} = -\mu_0\mathbf{J} -\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

Which can be solved in principle using analogies to the electric scalar potential $V$.

Why is this form of the magnetic vector potential in its full generality seldom seen? Is there a contradiction which prevents its use?

  • $\begingroup$ Turn ahead to Section 10.1 in your copy of Griffiths, and look for the phrase "Coulomb gauge". $\endgroup$ Commented Jan 10, 2022 at 14:17
  • $\begingroup$ BTW your curl of the curl you wrote down is in correct $\endgroup$ Commented Jan 10, 2022 at 14:28
  • 1
    $\begingroup$ "Why is this form of the magnetic vector potential in its full generality seldom seen?" Why should choosing a specific gauge ($\nabla\cdot A=0$) be considered "full generality"? $\endgroup$
    – oliver
    Commented Jan 10, 2022 at 14:32
  • $\begingroup$ to add on to Oliver. All gauges are all equally correct and all predict the same E and B fields $\endgroup$ Commented Jan 10, 2022 at 14:53

1 Answer 1


Yes it can be used,

Given,$ \nabla × A = B$


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

$\nabla × E = -\frac{\partial (\nabla × A) }{\partial t}$

$\nabla × E = -\nabla × \frac{\partial A }{\partial t}$

$\nabla × E + \nabla × \frac{\partial A }{\partial t} = 0$

$\nabla × (E + \frac{\partial A }{\partial t})= 0$

Because the curl of this quantity is zero, it can be written as the gradient of a scalar function ( or the negative of a gradient of a scalar function, which is used to match the definition of electrostatic potential)

$E + \frac{\partial A }{\partial t} = -\nabla V$

$E = -\nabla V -\frac{\partial A }{\partial t} $

From here we can substitute the definitions of E into the other maxwell equations

This will obtain 2 equations that interlink A and V in terms of the source terms $\rho$ and J.

using the "lorenz" gauge choice ( not lorentz) we can decouple these and solve easier( or using the coulomb gauge)

This is called the potential formulation of maxwells equations

  • $\begingroup$ Also, when learning the potential formulation it is important you actually understand WHY we can prescribe the divergence of A to follow specific gauges. Look at my answer here ; physics.stackexchange.com/questions/682626/… $\endgroup$ Commented Jan 10, 2022 at 14:50

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