How can I derive that $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}$$ where $\phi$ is the scalar potential and $\vec{A}$ the vector potential?


3 Answers 3


$\def\vA{{\vec{A}}}$ $\def\vB{{\vec{B}}}$ $\def\vD{{\vec{D}}}$ $\def\vE{{\vec{E}}}$ $\def\vH{{\vec{H}}}$ $\def\vS{{\vec{S}}}$ $\def\eps{\varepsilon}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$

Faraday's law: $$\rot(\vE)+\dot\vB=0$$ Source-less B-field: $$\div(\vB)=0$$ From this in a simply connected domain there follows the existence of a vector potential $\vA$ with $$\vB = \rot\vA$$ Therewith, Faraday's law reads $$\rot(\vE+\dot{\vA})=0$$ The curl-freeness of the vector field $\vE+\dot{\vA}$ (in a simply connected domain) implies the existence of a scalar potential with $$\vE+\dot{\vA} = -\grad\varphi$$ and that is your formula. You see $\varphi$ is just defined in the way you wrote it down.

The purpose of introducing $\vA$ is to solve $\div\vB=0$ and the purpose of introducing $\varphi$ is to solve Faraday's law. In most cases the only equation which remains to be solved is Ampre's law $$ \rot\vH = \vS + \dot\vD $$ which reads with our new independent variables $\varphi$ and $\vA$ as $$ \rot(\mu\rot \vA) = (\kappa + \eps\partial_t)(-\grad\varphi - \dot\vA). $$ Maybe, one has also a pre-defined space-charge density $\rho$. That would imply the equation $$ \rho = \div \vD = \div (\eps(-\grad\varphi - \dot\vA)) $$ For constant $\eps$ you have $$ -\frac{\rho}{\eps} = \Delta\varphi + \div \dot\vA $$ Now, you have some degrees of freedom in the choice of $\vA$. If $\vA$ is a vector potential for $\vB$ then for any smooth scalar function $\varphi'$ also $\vA':=\vA+\grad\varphi'$ is a vector potential for $\vB$ since $\rot\grad=0$. One possible choice is $\div\vA = 0$. Therewith, the equation for the space charge reads just $$ -\frac{\rho}{\eps} = \Delta\varphi $$ which we know from electro-statics. The nice thing of $\div\vA=0$ is that the above equation decouples from the magnetics. (But only if $\rho$ assumed to be predefined.) So one can solve the problem staggered.

The condition $\div\vA=0$ can always be satisfied. If we initially have a vector potential $\vA'$ with $\div\vA'\neq 0$ then we define $\vA := \vA'+\grad\varphi'$ such that $$ 0 = \div \vA = \div(\vA' + \grad\varphi')= \div\vA' + \Delta \varphi' $$ To find the function $\varphi'$ for the modification of $\vA'$ we just have to solve the poisson equation $$ \Delta\varphi' = -\div\vA' $$ for $\varphi'$. The coice $\div A=0$ is the so called Coulomb gauging.

A field is determined by its curl and sources. Under the assumption that we have already the required boundary conditions for $\varphi$ and $\vA$ then the fixation of the divergence for the vector potential (i.e., the gauging) determines the potentials uniquely. Beside the Coulomb gauging there are other forms of useful gauging (e.g., Lorentz-gauging).


The equation $$\vec{E}=-\vec{\nabla}\phi$$ only holds when there is no changing magnetic field. As you might know, a changing magnetic field gives rise to an electric field. The equation governing this is $$\vec{\nabla}\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$. This is known as Faraday's law of induction. Remembering that $\vec{B}=\vec{\nabla}\times\vec{A}$, we see that $$\vec{\nabla}\times \vec{E}=-\vec{\nabla}\times\frac{\partial \vec{A}}{\partial t}$$ This tells you that $$\vec{E}=-\frac{\partial\vec{A}}{\partial t} + \text{an irrotational term (curl=0)}$$ From vector calculus we know that $\vec{\nabla}\phi$ for a scalar $\phi$ has zero curl, so this fits nicely with the first equation, which holds when there is no changing magnetic field. All in all, we find $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial\vec{A}}{\partial t}$$ This equation holds in general.


In that form, there isn't really much to prove. The full statement, is of course, one of existence:

For each electric field configuration $\mathbf{E}(\mathbf{r},t)$, there exist a scalar function $\phi$ and a vector field $\mathbf{A}$ such that $$\mathbf{E}(\mathbf{r},t)=-\nabla \phi -\frac\partial{\partial t}\mathbf{A}.\tag1$$

In this form, which ignores the accompanying magnetic field, you can simply set $$ \begin{align} \phi&=0\\\mathbf{A}&=-\int_{t_0}^t\mathbf{E}(\mathbf{r},\tau)\text d\tau +\mathbf{A}_0(\mathbf{r}) \end{align} $$ and these functions will satisfy that equation, with the additional freedom of choice of the constant vector field $\mathbf{A}_0$

Of course, this is essentially the easy way out: it is valid, but we have ignored the primary role that we want $\mathbf{A}$ to play in the magnetic side of the story, and if charges are present it will leave us in a very uncomfortable gauge.

The problem is, then, to also address the second half of the statement, which goes

and also $$\mathbf{B}=\nabla\times \mathbf{A},\tag2$$ where $\mathbf{B}$ is the corresponding magnetic field of the configuration.

Now, of course, you have more work to do, because $\mathbf{B}$ is fixed but the choice above already gives an expression for $$ \nabla \times \mathbf{A} =-\int_{t_0}^t\nabla \times\mathbf{E}(\mathbf{r},\tau)\text d\tau +\nabla \times\mathbf{A}_0(\mathbf{r}). $$ Fortunately, by Faraday's law this simplifies somewhat to $$ \nabla \times \mathbf{A} =+\int_{t_0}^t \frac{\partial \mathbf{B}}{\partial t}(\mathbf{r},t) \text d\tau +\nabla \times\mathbf{A}_0(\mathbf{r}) =\mathbf{B}(\mathbf{r},t)-\mathbf{B}(\mathbf{r},t_0)+\nabla \times\mathbf{A}_0(\mathbf{r}). $$

Thus, for the full result to hold, we now only need to find an integration constant $\mathbf{A}_0(\mathbf{r})$ which will satisfy the statement (2) at a particular time $t_0$, $$\mathbf{B}(\mathbf{r},t_0)=\nabla\times \mathbf{A}(\mathbf{r},t_0),$$ and it will hold for all other times.

You might have noticed that I have been happily evading any vector calculus, but unfortunately my luck runs out at this point. I have a known vector field $\mathbf{B}(\mathbf{r},t_0)$, and I need to prove the existence of a second vector field $\mathbf{A}(\mathbf{r},t_0)$ which will curl to $\mathbf{B}$.

This is not automatic, of course, and as usual it depends on the magnetic Gauss law, $\nabla\cdot\mathbf{B}=0$. There are a number of ways to prove this theorem, and I won't go in depth into them; I'll just mention that every solution can be expressed in the form $$ \mathbf{A}(\mathbf{r},t_0)=\int\frac{\nabla\times\mathbf{B}(\mathbf{r}',t_0)}{|\mathbf{r}-\mathbf{r}'|}\text d\mathbf{r}' +\nabla\psi(\mathbf{r}), $$ as long as the magnetic field is reasonably bounded, and where $\psi(\mathbf{r})$ is an arbitrary function. For a proof, one needs to take the curl of the above expression with due attention to the singularities, which will return the magnetic field.

This gives, then the full vector potential that will give the correct electric and magnetic fields: $$ \mathbf{A}=-\int_{t_0}^t\mathbf{E}(\mathbf{r},\tau)\text d\tau +\int\frac{\nabla\times\mathbf{B}(\mathbf{r}',t_0)}{|\mathbf{r}-\mathbf{r}'|}\text d\mathbf{r}'. $$ I will agree that this is a slightly unsatisfactory answer, in that the immensely-useful scalar potential is now gone. The reason for this is that you still have the full freedom of gauge choice to change both potentials in tandem whilst still retaining the same force fields. To prove existence, one simply needs to exhibit a solution, and that's it; after that, it's simply a matter of transforming to a more convenient gauge.


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