# Deriving electric field

Can anyone explain where the following expression for the electric field vector comes from? $$\mathbf E(\mathbf r,t) = -\nabla \phi(\mathbf r,t) - \frac{\partial}{\partial t}\mathbf A(\mathbf r,t)$$

where $\phi$ and $\mathbf A$ are the scalar and vector potentials, respectively.

Presumably it can it be derived from Maxwell's Equations?

It requires 2 of Maxwell's laws:

• the Maxwell-Faraday equation, $$\frac{\partial\mathbf B}{\partial t}=-\nabla\times\mathbf E$$
• the divergence condition $$\nabla\cdot\mathbf B=0$$

Then it requires 2 vector calculus identities:

• divergence of the curl is identically zero $$\nabla\cdot\nabla\times\mathbf A=0$$
• curl of the gradient is identically zero $$\nabla\times\nabla\phi=0$$ (if you haven't proved these yourself already, go ahead and do so).

By applying these two identities in the appropriate places and working through some simple calculus, you can work out your relation.

I will expand on Kyle Kanos's answer by deriving the actual relation and commenting on it slightly. Two of Maxwell's equations, namely the homogeneous ones, $\nabla \cdot \mathbf{B} = 0$ and $\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0$, can be automatically solved if one always writes the electric and magnetic fields, $\mathbf{E}$ and $\mathbf{B}$ as derived objects that come from other fields, $\mathbf{A}$ and $\phi$, called "potentials". If one picks an arbitrary vector field $\mathbf{A}$ and writes the $\mathbf{B}$ field as $\mathbf{B} := \nabla \times \mathbf{A}$ then this automatically solves Gauss's Law, since the divergence of a curl is always zero. On the other hand, if one also writes the electric field as $\mathbf{E} := -(\nabla \phi + \frac{\partial \mathbf{A}}{\partial t})$ then Faraday's Law is automatically also satisfied since:

$$\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = -\nabla \times (\nabla \phi + \frac{\partial \mathbf{A}}{\partial t}) + \nabla \times \frac{\partial \mathbf{A}}{\partial t} = 0$$

where the last equatlity holds since the curl of a gradient is always 0. Note that you can not choose to write the electric field in a way totally independent from the way you write the magnetic field since Faraday's Law relates the two of them even in the absence of sources, that is why you get the $\frac{\partial \mathbf{A}}{\partial t}$ term in your equation. I would not say that the expression is "derived" from Maxwell's equations, it is more a convenient definition of the electric field that, coupled with a related convenient definition of the magnetic field automatically satisfy two of Maxwell's equations.

The fields, $\phi$ and $\mathbf{A}$, are originated because of the sources in the electromagnetic theory. The other two answers by Ignicio and Kyle Kanos do not provide the uniqueness of those definitions of the electric and magnetic fields.

Without the sources, both $\mathbf{E}$ and $\mathbf{B}$ could be derived from scalar fields, e.g. $\mathbf{B}=-\nabla \psi$, because Maxwell's equations become symmetric for electric and magnetic fields in the sourceless case.

If you only consider the Gauss Law of electrostatic, $$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$ one can derive the electric field from a scalar potential since the force, $\mathbf{F}=q\mathbf{E}$, is conservative, i.e., $\nabla \times \mathbf{F}=0$ which means the work is independent from the path. This will give us $$\mathbf{E}= - \nabla \phi$$ uniquely for this electrostatic case.

On the other hand, we know that, at least if fields are not changing with time, $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$$ This equation guarantees the Magnetic field can not be derived from a scalar potential since the curl of such a field would vanish. So, the magnetic field can be derived uinquely from $$\tag{1} \mathbf{B}=\nabla \times \mathbf{A} - \nabla \psi$$ As you may notice in electrodynamics, the second term of Eq.(1) is not present because of the absence of the magnetic monopoles. Therefore, one can safely write $\psi=0$. Nevertheless, for permanent magnets, one can still use $\psi \neq 0$, and if there are no currents, $\mathbf{A}=0$. See, solutions for Biot-Savart law.

If we consider the time dependence of the sources, then the electric field becomes $$\mathbf{E} = - \nabla \phi - \frac{\partial \mathbf{A}}{\partial t}$$ because Faraday law, $\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0$ as given in the answers of Kyle Kanos and Ignacio in detail.

On the other hand, Ampere's law do not provide a term like $\frac{\partial \mathbf{C}}{\partial t}$ for the magnetic field, as if $\mathbf{E} = \nabla \times \mathbf{C} + \cdots$, because there does not exist any magnetic currents as we know so far.

In conclusion, the definitions of the electric and magnetic fields in terms of scalar and vector potential are completely derived from the empirical knowledge of the sources. Without sources, the definitions would not be unique at all.