# 7.6: Appendix - Review of Free Electromagnetic Field


Here we review the derivation of the vector potential for the plane wave in free space. We begin with Maxwell’s equations (SI):

\begin{align} \overline {\nabla} \cdot \overline {B} &= 0 \label{6.78} \\[4pt] \overline {\nabla} \cdot \overline {E} &= \rho / \varepsilon _ {0} \label{6.79} \\[4pt] \overline {\nabla} \times \overline {E} &= - \dfrac {\partial \overline {B}} {\partial t} \label{6.80} \\[4pt] \overline {\nabla} \times \overline {B} &= \mu _ {0} \overline {J} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial \overline {E}} {\partial t} \label{6.81} \end{align}

Here the variables are: $$\overline {E}$$, electric field; $$\overline {B}$$, magnetic field; $$\overline {J}$$, current density; $$\rho$$, charge density; $$\mathcal {E} _ {0}$$, electrical permittivity; $$\mu _ {0}$$, magnetic permittivity. We are interested in describing $$\overline {E}$$ and $$\overline {B}$$ in terms of a vector and scalar potential, $$\overline {A}$$ and $$\varphi$$.

Next, let’s review some basic properties of vectors and scalars. Generally, vector field $$\overline {F}$$ assigns a vector to each point in space. The divergence of the field

$\overline {\nabla} \cdot \overline {F} = \dfrac {\partial F _ {x}} {\partial x} + \dfrac {\partial F _ {y}} {\partial y} + \dfrac {\partial F _ {z}} {\partial z} \label{6.82}$

is a scalar. For a scalar field $$\phi$$, the gradient

$\nabla \phi = \dfrac {\partial \phi} {\partial x} \hat {x} + \dfrac {\partial \phi} {\partial y} \hat {y} + \dfrac {\partial \phi} {\partial z} \hat {z} \label{6.83}$

is a vector for the rate of change at one point in space. Here

$\hat {x}^{2} + \hat {y}^{2} + \hat {z}^{2} = \hat {r}^{2}$

are unit vectors. Also, the curl

$\overline {\nabla} \times \overline {F} = \left| \begin{array} {l l l} {\hat {x}} & {\hat {y}} & {\hat {z}} \\ {\dfrac {\partial} {\partial x}} & {\dfrac {\partial} {\partial y}} & {\dfrac {\partial} {\partial z}} \\ {F _ {x}} & {F _ {y}} & {F _ {z}} \end{array} \right|$

is a vector whose $$x$$, $$y$$, and $$z$$ components are the circulation of the field about that component. Some useful identities from vector calculus that we will use are

\begin{align} \overline {\nabla} \cdot ( \overline {\nabla} \times \overline {F} ) &= 0 \label{6.85} \\[4pt] \nabla \times ( \nabla \phi ) &= 0 \label{6.86} \\[4pt] \nabla \times ( \overline {\nabla} \times \overline {F} ) &= \overline {\nabla} ( \overline {\nabla} \cdot \overline {F} ) - \overline {\nabla}^{2} \overline {F} \label{6.87} \end{align}

## Gauge Transforms

We now introduce a vector potential $$\overline {A} ( \overline {r} , t )$$ and a scalar potential $$\varphi ( \overline {r} , t )$$, which we will relate to $$\overline {E}$$ and $$\overline {B}$$. Since

$\overline {\nabla} \cdot \overline {B} = 0$

and

$\overline {\nabla} ( \overline {\nabla} \times \overline {A} ) = 0,$

we can immediately relate the vector potential and magnetic field

$\overline {B} = \overline {\nabla} \times \overline {A} \label{6.88}$

Inserting this into Equation \ref{6.80} and rewriting, we can relate the electric field and vector potential:

$\overline {\nabla} \times \left[ \overline {E} + \dfrac {\partial \overline {A}} {\partial t} \right] = 0 \label{6.89}$

Comparing Equations \ref{6.89} and \ref{6.86} allows us to state that a scalar product exists with

$\overline {E} = \dfrac {\partial \overline {A}} {\partial t} - \nabla \varphi \label{6.90}$

So summarizing our results, we see that the potentials $$\overline {A}$$ and $$\varphi$$ determine the fields $$\overline {B}$$ and $$\overline {E}$$:

\begin{align} \overline {B} ( \overline {r} , t ) &= \overline {\nabla} \times \overline {A} ( \overline {r} , t ) \label{6.91} \\[4pt] \overline {E} ( \overline {r} , t ) &= - \overline {\nabla} \varphi ( \overline {r} , t ) - \dfrac {\partial} {\partial t} \overline {A} ( \overline {r} , t ) \label{6.92} \end{align}

We are interested in determining the classical wave equation for $$\overline {A}$$ and $$\varphi$$. Using Equation \ref{6.91}, differentiating Equation \ref{6.92}, and substituting into Equation \ref{6.81}, we obtain

$\overline {\nabla} \times ( \overline {\nabla} \times \overline {A} ) + \varepsilon _ {0} \mu _ {0} \left( \dfrac {\partial^{2} \overline {A}} {\partial t^{2}} + \overline {\nabla} \dfrac {\partial \varphi} {\partial t} \right) = \mu _ {0} \overline {J} \label{6.93}$

Using Equation \ref{6.87},

$\left[ - \overline {\nabla}^{2} \overline {A} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial^{2} \overline {A}} {\partial t^{2}} \right] + \overline {\nabla} \left( \overline {\nabla} \cdot \overline {A} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial \varphi} {\partial t} \right) = \overline {\mu} _ {0} \overline {J} \label{6.94}$

From Equation \ref{6.90}, we have

$\overline {\nabla} \cdot \overline {E} = - \dfrac {\partial \overline {\nabla} \cdot \overline {A}} {\partial t} - \overline {\nabla}^{2} \varphi \label{6.95}$

and using Equation \ref{6.79},

$\dfrac {- \partial \overline {V} \cdot \overline {A}} {\partial t} - \overline {\nabla}^{2} \varphi = \rho / \varepsilon _ {0} \label{6.96}$

Notice from Equations \ref{6.91} and \ref{6.92} that we only need to specify four field components ($$A_{x}, A_{y}, A_{z}, \varphi$$ to determine all six $$\bar{E}$$ and $$\bar{B}$$ components. But $$\bar{E}$$ and $$\bar{B}$$ do not uniquely determine $$\bar{A}$$ and $$\varphi$$. So we can construct $$\bar{A}$$ and $$\varphi$$ in any number of ways without changing $$\bar{E}$$ and $$\bar{B}$$. Notice that if we change $$\bar{A}$$ by adding $$\bar{\nabla} \chi$$ where $$\chi$$ is any function of $$\bar{r}$$ and $$t$$ this will not change $$\bar{B} \quad(\nabla \times(\nabla \cdot B)=0)$$. It will change $$E$$ by $$\left(-\frac{\partial}{\partial t} \bar{\nabla} \chi\right)$$, but we can change $$\varphi$$ to $$\varphi^{\prime}=\varphi-(\partial \chi / \partial t)$$. Then $$\bar{E}$$ and $$\bar{B}$$ will both be unchanged. This property of changing representation (gauge) without changing $$\bar{E}$$ and $$\bar{B}$$ is gauge invariance. We can define a gauge transformation with

$\bar{A}^{\prime}(\bar{r}, t)=\bar{A}(\bar{r}, t)+\bar{\nabla} \cdot \chi(\bar{r}, t) \label{6.97}$

$\varphi^{\prime}(\bar{r}, t)=\varphi(\bar{r}, t)-\frac{\partial}{\partial t} \chi(\bar{r}, t) \label{6.98}$

Up to this point, $$A^{\prime} \text {and} \varphi^{\prime}$$ are undetermined. Let’s choose a $$\chi$$ such that:

$\overline {\nabla} \cdot \overline {A} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial \varphi} {\partial t} = 0 \label{6.99}$

which is known as the Lorentz condition. Then from Equation \ref{6.93}:

$- \nabla^{2} \overline {A} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial^{2} \overline {A}} {\partial t^{2}} = \mu _ {0} \overline {J} \label{6.100}$

The right hand side of this equation can be set to zero when no currents are present. From Equation \ref{6.96}, we have:

$\varepsilon _ {0} \mu _ {0} \dfrac {\partial^{2} \varphi} {\partial t^{2}} - \nabla^{2} \varphi = \dfrac {\rho} {\varepsilon _ {0}} \label{6.101}$

Equations \ref{6.100} and \ref{6.101} are wave equations for $$\overline {A}$$ and $$\varphi$$. Within the Lorentz gauge, we can still arbitrarily add another $$\chi$$; it must only satisfy Equation \ref{6.99}. If we substitute Equations \ref{6.97} and \ref{6.98} into Equation \ref{6.101}, we see

$\nabla^{2} \chi - \varepsilon _ {0} \mu _ {0} \dfrac {\partial^{2} \chi} {\partial t^{2}} = 0 \label{6.102}$

So we can make further choices/constraints on $$\bar{A} \text {and} \varphi$$ as long as it obeys Equation \ref{6.102}. We now choose $$\varphi=0$$, the Coulomb gauge, and from Equation \ref{6.99} we see

$\overline {\nabla} \cdot \overline {A} = 0 \label{6.103}$

So the wave equation for our vector potential when the field is far currents ($$J= 0$$) is

$- \overline {\nabla}^{2} \overline {A} + \varepsilon _ {0} \mu _ {0} \dfrac {\partial^{2} \overline {A}} {\partial t^{2}} = 0 \label{6.104}$

The solutions to this equation are plane waves:

$\overline {A} = \overline {A} _ {0} \sin ( \omega t - \overline {k} \cdot \overline {r} + \alpha ) \label{6.105}$

where $$\alpha$$ is a phase. $$k$$ is the wave vector which points along the direction of propagation and has a magnitude

$k^{2} = \omega^{2} \mu _ {0} \varepsilon _ {0} = \omega^{2} / c^{2} \label{6.106}$

Since $$\overline {\nabla} \cdot \overline {A} = 0$$ (Equation \ref{6.103}), then

$- \overline {k} \cdot \overline {A} _ {0} \cos ( \omega t - \overline {k} \cdot \overline {r} + \alpha ) = 0$

and

$\overline {k} \cdot \overline {A} _ {0} = 0 \label{6.107}$

So the direction of the vector potential is perpendicular to the direction of wave propagation ($$\overline {k} \perp \overline {A _ {0}}$$). From Equations \ref{6.91} and \ref{6.92}, we see that for $$\varphi = 0$$:

\begin{align} \overline {E} &= - \dfrac {\partial \overline {A}} {\partial t} \\[4pt] &= - \omega \overline {A} _ {0} \cos ( \omega t - \overline {k} \cdot \overline {r} + \alpha ) \label{6.108} \\[4pt] \overline {B} &= \overline {\nabla} \times \overline {A} \\[4pt] &= - \left( \overline {k} \times \overline {A} _ {0} \right) \cos ( \omega t - \overline {k} \cdot \overline {r} + \alpha ) \label{6.109} \end{align}

Here the electric field is parallel with the vector potential, and the magnetic field is perpendicular to the electric field and the direction of propagation ($$\overline {k} \perp \overline {E} \perp \overline {B}$$). The Poynting vector describing the direction of energy propagation is

$\overline {S} = \varepsilon _ {0} c^{2} ( \overline {E} \times \overline {B} )$

and its average value, the intensity, is

$I = \langle S \rangle = \dfrac {1} {2} \varepsilon _ {0} c E _ {0}^{2}.$

This page titled 7.6: Appendix - Review of Free Electromagnetic Field is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Andrei Tokmakoff via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.