7.8: Appendix- Electromagnetic Fields and Potentials

Wave Equations and Electromagnetic Fields 

\( \let\bibref\eqref %make-command_bibref-equivalent-to-eqref\)Here we review the derivation of the vector potential, electric field, and magnetic field for electromagnetic plane waves propagating in free space. The starting point for all discussions of electromagnetic waves is Maxwell’s equations in vacuum, also known as the microscopic form, which are in SI units:

\[\begin{gather}
\nabla \cdot \boldsymbol{\mathcal{B}} = 0 \label{EMField01}\\
\nabla \cdot \boldsymbol{\mathcal{E}} = \frac{\rho}{\varepsilon_0} \label{EMField02} \\ 
\nabla \times \boldsymbol{\mathcal{E}} = - \frac{\partial \boldsymbol{\mathcal{B}}}{\partial t} \label{EMField03} \\
\nabla \times \boldsymbol{\mathcal{B}} = \mu_{0} \mathbf{J} + \varepsilon_0 \mu_0 \frac{\partial \boldsymbol{\mathcal{E}}}{\partial t} \label{EMField04}\end{gather}\]

The four vector variables here are \(\boldsymbol{\mathcal{E}}\), the electric field (\(\text{V} \, \text{m}^{- 1})\); \(\boldsymbol{\mathcal{B}}\), the magnetic flux density (\(\text{C} \, \text{m}^{- 2})\); \(\mathbf{J}\), the current density (\(\text{A} \, \text{m}^{- 2})\); \(\boldsymbol{\rho}\), the charge density (\(\text{C} \, \text{m}^{- 3}\)). The fundamental constants are \(\varepsilon_0\), the electrical vacuum permittivity (\(8.85419 \times 10^{- 12} \, \text{F} \, \text{m}^{- 1}\)), and \(\mu_0\), the magnetic vacuum permittivity (\(1.256637 \times 10^{- 6} \, \text{N} \, \text{A}^{- 2}\)), which are related to the speed of light in a vacuum by

\[c = \frac{1}{\sqrt {\varepsilon_0 \mu_0}} \label{EMField05} \]

In vacuum, the magnetic field \(\mathbf{H}\) (\(\text{A} \, \text{m}^{- 2}\)) is directly proportional to \(\boldsymbol{\mathcal{B}}\): \(\boldsymbol{\mathcal{B}} = \mu_0 \mathbf{H}\). We are interested in describing \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) in terms of \(\mathbf{A}\), the magnetic vector potential (\(\text{V} \, \text{s} \, \text{m}^{- 1}\)), and \(\varphi\), the scalar electrostatic potential (\(\text{J} \, \text{c}^{- 1}\)).

To start, let’s review some basic properties of vector and scalar fields. A field \(\mathbf{F} (\mathbf{r})\) associates a physical property to different points in space, and may be time dependent. A scalar field assigns a number to each point in space, whereas a vector field assigns a vector. The divergence of a vector field

\[\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \label{EMField06} \]

is a scalar. For a scalar field \(\phi\), the gradient

\[\nabla \phi = \frac{\partial \phi}{\partial x}\hat x + \frac{\partial \phi}{\partial y}\hat y + \frac{\partial \phi}{\partial z}\hat z \label{EMField07} \]

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

\[\nabla \times \mathbf{F} = \left| {\begin{array}{*{20}{c}}
 \hat x & \hat y & \hat z \\
 \frac{\partial}{\partial x} & \frac{\partial}{\partial y}& \frac{\partial}{\partial z} \\
 F_x & F_y & F_z
\end{array}} \right| \label{EMField08} \]

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

\[\nabla \cdot \left(\nabla \times \mathbf{F} \right) = 0 \label{EMField09} \]\[\nabla \times \left(\nabla \phi \right) = 0 \label{EMField10} \]\[\nabla \times \left(\nabla \times \mathbf{F} \right) = \nabla \left(\nabla \cdot \mathbf{F} \right) - \nabla^2 \mathbf{F} \label{EMField11} \]

To obtain the wave equation for the electric field, we can substitute eq. (\ref{EMField02}) and (\ref{EMField03}) in eq. (\ref{EMField11}) with \(\mathbf{F} = \boldsymbol{\mathcal{E}}\) to obtain:

\[ - \frac{\partial (\nabla \times \boldsymbol{\mathcal{B}})}{\partial t} = \frac{\nabla \rho}{\varepsilon_0} - \nabla^2 \mathcal{E} \label{EMField12} \]

Substituting eq. (\ref{EMField04}) into this expression gives

\[ \nabla^2 \boldsymbol{\mathcal{E}} - \varepsilon_0 \mu_0 \frac{\partial^2 \boldsymbol{\mathcal{E}}}{\partial t^2} = \frac{\nabla \rho}{\varepsilon_0} + \mu _{0} \frac{\partial \mathbf{J}}{\partial t} \label{EMField13} \]

This is the inhomogeneous wave equation for the electric field. In the absence of charges and currents, the terms on the right-hand side are zero, and we obtain the wave equation for the free electric field:

\[ \nabla^2 \boldsymbol{\mathcal{E}} - \varepsilon_0 \mu_0 \frac{\partial^2 \boldsymbol{\mathcal{E}}}{\partial t^2} = 0 \label{EMField14} \]

This form of a wave equation has as solutions functions of the form \(\mathcal{E}(x - vt)\) that propagate in the \(x\) direction with a velocity of magnitude \(v = (\varepsilon_0 \mu_0)^{- 1 /2}\). In vacuum that velocity is the speed of light, \(c\).

\[c = \frac{1}{\sqrt {\varepsilon_0 \mu_0}} \label{EMField15} \]

In spherical coordinates, solutions to eq. (\ref{EMField14}) take the form \( \frac{1}{r} \mathcal{E} \left(r - vt \right) \).

To obtain the plane wave solutions to eq. (\ref{EMField14}), we posit solutions with a form that separates the spatial and temporal variables:

\[ \boldsymbol{\mathcal{E}} (\mathbf{r},t) = \boldsymbol{\mathcal{E}} (\mathbf{r}) e^{\mathrm{i} \omega t} \label{EMField16} \]

The real electric field is understood to be obtained from \(\operatorname{Re} \left[ \boldsymbol{\mathcal{E}}(\mathbf{r}, t) \right]\). Substituting (\ref{EMField16}) into (\ref{EMField14}) we obtain an equation for the spatial part of the field:

\[ \nabla^2 \boldsymbol{\mathcal{E}} (\mathbf{r}) - k^2 \boldsymbol{\mathcal{E}} (\mathbf{r}) = 0 \label{EMField17} \]

where \(k^2 = {\omega^2 c^2}\). This equation has solutions of the form \(\boldsymbol{\mathcal{E}} (\mathbf{r}) = \boldsymbol{\mathcal{E}}_0 e^{- \mathrm{i}\mathbf{k} \cdot \mathbf{r}}\), which have a periodicity in space governed by the wavevector \(\mathbf{k}\). The solution allows positive and negative values of the wavevector but requires that \(\mathbf{k}\) and frequency are related by the dispersion relationship

\[| \mathbf{k} | = \frac{\omega}{c} = \frac{2\pi}{\lambda} \label{EMField18} \]

where \(\lambda\) is the wavelength of light.

With these results for \(\boldsymbol{\mathcal{E}} (\mathbf{r})\) one finds that the resulting solutions to the wave-equation (\ref{EMField14}) have the sinusoidal form

\[\boldsymbol{\mathcal{E}} (\mathbf{r}, t) = \tilde{\boldsymbol{\mathcal{E}}} _0 e^{\mathrm{i} \left(\mathbf{k} \cdot \mathbf{r} - \omega t \right) } \label{EMField18.5}\]

with the complex amplitude \( \tilde{\boldsymbol{\mathcal{E}}}_0\). The real part is calculated as:

\[\text{Re}[\boldsymbol{\mathcal{E}} (\mathbf{r}, t)]= \frac{1}{2}\tilde{\boldsymbol{\mathcal{E}}} _0 e^{\mathrm{i} \left(\mathbf{k} \cdot \mathbf{r} - \omega t \right) } + \frac{1}{2}\tilde{\boldsymbol{\mathcal{E}}} _0^* e^{- \mathrm{i}\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right)} \label{EMField19} \]

Alternatively, we can express this as:

\[ \boldsymbol{\mathcal{E}} (\mathbf{r}, t) = {\boldsymbol{\mathcal{E}}_0}\cos \left(\mathbf{k} \cdot \mathbf{r} - \omega t - \alpha \right) \label{EMField20} \]

Here the phase factor \(\alpha\) is added to illustrate that sine function solutions (\(\alpha = \pi /2\)) are equally valid. This is possible because one can add an arbitrary phase factor \(e^{- \mathrm{i}\alpha}\) to eq. (\ref{EMField16}), which simply carries through.

Equations (\ref{EMField19}) and (\ref{EMField20}) describe a monochromatic plane wave of frequency \(\omega\) traveling in the direction of the wavevector \(\mathbf{k}\) at the speed of light. For any value of \(k\), there are positive and negative solutions, for which the sinusoidal argument can take the form (\(\mathrm{i} \mathbf{k} \cdot \mathbf{r} - \mathrm{i} \omega t\)) to describe propagation in the \(\mathbf{k}\) direction and (\(- \mathrm{i} \mathbf{k} \cdot \mathbf{r} - \mathrm{i} \omega t\)) for backward propagating in the \(- \mathbf{k}\) direction. Additionally, Maxwell's equation with zero charge density, \(\nabla \cdot \boldsymbol{\mathcal{E}} = 0\), dictates that these waves satisfy \(\mathbf{k} \cdot \boldsymbol{\mathcal{E}}_0 = 0\), so the electric fields are oriented perpendicular to the direction of travel, i.e. transverse waves.

A similar approach using the other two Maxwell equations can be used to obtain the wave equation for the magnetic field. For a free electromagnetic wave, we find

\[ \nabla^2 \boldsymbol{\mathcal{B}} - \varepsilon_0 \mu_0 \frac{\partial^2 \boldsymbol{\mathcal{B}}} {\partial t^2} = 0 \label{EMField21} \]

This is the same wave equation as that for the electric field and therefore must have the same solutions:

\[\boldsymbol{\mathcal{B}} (\mathbf{r}, t) = {\boldsymbol{\mathcal{B}}_0} e^{\mathrm{i}\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right)} + \boldsymbol{\mathcal{B}}_0^* e^{- \mathrm{i}\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right)} \label{EMField22} \]

or

\[ \boldsymbol{\mathcal{B}} (\mathbf{r}, t) = \boldsymbol{\mathcal{B}}_0 \cos \left(\mathbf{k} \cdot \mathbf{r} - \omega t - \alpha \right) \nonumber \]

There are similar constraints on the constants \(\mathcal{B}_0\), \(\mathbf{k}\), and \(\omega\).

Maxwell’s equations (\ref{EMField03}) and (\ref{EMField04}) dictate that \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) are not independent of each other. If we substitute the solutions eq. (\ref{EMField19}) and (\ref{EMField22}) into these equations, we find that the frequency wavevector and phase of the waves must be the same for \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) and that

\[\mathbf{k} \times \boldsymbol{\mathcal{E}}_0 = \omega \boldsymbol{\mathcal{B}}_0 \label{EMField24} \]\[\mathbf{k} \times \boldsymbol{\mathcal{B}}_0 = - \mu _0 \varepsilon _0 \omega \boldsymbol{\mathcal{E}}_0 \label{EMField23} \]

Thus, the orientation of the magnetic field is perpendicular to the electric field and the direction of propagation \(\left(\mathbf{k} \, \bot \, \boldsymbol{\mathcal{E}} \, \bot \, \boldsymbol{\mathcal{B}} \right)\). These expressions hold for all plane waves. Taking the absolute value of eq. (\ref{EMField24}) and using eq. (\ref{EMField15}) and (\ref{EMField18}), we find that the magnitude of the magnetic and electric fields are related by the speed of light:

\[\frac{\mathcal{E}_0}{\mathcal{B}_0} = c \label{EMField25} \]

Separately, one can show that the Poynting vector describing the wave’s energy flux is \(\mathbf{S} = \varepsilon_0 c^2 (\boldsymbol{\mathcal{E}} \times \boldsymbol{\mathcal{B}})\). \(\mathbf{S}\) is parallel to \(\mathbf{k}\) and gives the rate of energy passage through a given area per unit time (\(\text{W} \, \text{cm}^{-2}\)). One calculates the intensity or irradiance as the value obtained by averaging \(\mathbf{S}\) over one cycle of the field: \(I = \left\langle S \right\rangle = \frac{1}{2} \varepsilon_0 c\mathcal{E} _0^2\).

Vector Potential

We now introduce the magnetic vector potential \( \mathbf{A} (\mathbf{r}, t ) \) and a electrostatic scalar potential \(\varphi (\mathbf{r}, t )\), which we will relate to \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\). We start by recognizing that any vector field \(F\) can be written in terms of a vector potential and scalar potential according to

\[ \mathbf{F} = \nabla \Phi + \nabla \times \mathbf{A} \label{EMField26} \]

Since the magnetic field has no divergence, \(\nabla \cdot \boldsymbol{\mathcal{B}} = 0 \), and eq. (\ref{EMField09}) indicates that \(\nabla \cdot \left(\nabla \times \mathbf{A} \right) = 0 \), we can relate the vector potential and magnetic field through

\[ \boldsymbol{\mathcal{B}} = \nabla \times \mathbf{A} \label{EMField27} \]

Inserting this into eq. (\ref{EMField03}) and rewriting, we can relate the electric field and vector potential:

\[\nabla \times \left[ \boldsymbol{\mathcal{E}} + \frac{\partial \mathbf{A}}{\partial t} \right] = 0 \label{EMField28} \]

Comparing eqs. (\ref{EMField28}) and (\ref{EMField10}) allows us to state that a scalar product exists such that

\[ \boldsymbol{\mathcal{E}} = - \frac{ \partial \mathbf{A}}{\partial t} - \nabla \varphi \label{EMField29} \]

This equation shows that, when the vector potential is independent of time, we recover the expected relationship between the electric field and the gradient of the electrostatic potential: \(\boldsymbol{\mathcal{E}} = - \nabla \varphi\). Summarizing our results, we see that the potentials \(\mathbf{A}\) and \(\varphi\) determine the fields \(\boldsymbol{\mathcal{B}}\) and \(\boldsymbol{\mathcal{E}}\):

\[ \boldsymbol{\mathcal{B}} (\mathbf{r}, t) = \nabla \times \mathbf{A} (\mathbf{r}, t) \label{EMField30} \]\[ \boldsymbol{\mathcal{E}} (\mathbf{r}, t) = - \nabla \varphi (\mathbf{r}, t) - \frac{\partial}{\partial t} \mathbf{A} (\mathbf{r}, t) \label{EMField31} \]

We are interested in determining the classical wave equationS for \(\mathbf{A}\) and \(\varphi\). Using eq. (\ref{EMField30}), differentiating eq. (\ref{EMField31}), and substituting into eq. (\ref{EMField04}), we obtain

\[\nabla \times \left(\nabla \times \mathbf{A} \right) + \varepsilon_0 \mu_0 \left(\frac{\partial^2 \mathbf{A}}{\partial t^2} + \nabla \frac{\partial \varphi}{\partial t} \right) = \mu_0 \mathbf{J} \label{EMField32} \]

Using eq. (\ref{EMField11}),

\[\left[ - \nabla^2 \mathbf{A} + \varepsilon_0 \mu_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} \right] + \nabla \left(\nabla \cdot \mathbf{A} + \varepsilon_0 \mu_0 \frac{\partial \varphi}{\partial t} \right) = \mu_0 \mathbf{J} \label{EMField33} \]

From eq. (\ref{EMField29}), we have

\[\nabla \cdot \boldsymbol{\mathcal{E}} = - \frac{\partial \nabla \cdot \mathbf{A}}{\partial t} - \nabla^2 \varphi \label{EMField34} \]

and using eq. (\ref{EMField02}),

\[\frac{- \partial \nabla \cdot \mathbf{A}}{\partial t} - \nabla^2 \varphi = \frac{\rho}{\varepsilon_0} \label{EMField35} \]

To continue, we must specify the divergence of \(A\) in order to uniquely specify the vector potential. Notice from eqs. (\ref{EMField30}) and (\ref{EMField31}) that we only need to specify four field components (\(A_x, \, A_y, \, A_z, \, \varphi\)) to determine all six \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) components. But \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) do not uniquely determine \(\mathbf{A}\) and \(\varphi\). So we can construct \(\mathbf{A}\) and \(\varphi\) in any number of ways without changing \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\). Notice that if we change \(\mathbf{A}\) by adding \(\nabla \chi\), where \(\chi\) is any scalar function of \(\mathbf{r}\) and \(t\), \(\boldsymbol{\mathcal{B}}\) will not change (\(\nabla \times \left(\nabla \cdot \boldsymbol{\mathcal{B}} \right) = 0 \)). Adding \(\nabla \chi\) will change \(\boldsymbol{\mathcal{E}}\) by \( ( - \tfrac{\partial}{\partial t} \nabla \chi)\), but we can change \(\varphi\) to \(\varphi ' = \varphi - \left(\partial \chi/ \partial t \right)\). Then \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) will both be unchanged. This property of changing representation (changing the gauge) without changing \(\boldsymbol{\mathcal{E}}\) and \(\boldsymbol{\mathcal{B}}\) is known as gauge invariance. We can define a gauge transformation with

\[\mathbf{A} \! ' (\mathbf{r}, t) = \mathbf{A} (\mathbf{r}, t) + \nabla \chi (\mathbf{r}, t) \label{EMField36} \]

\[\varphi '(\mathbf{r}, t) = \varphi (\mathbf{r}, t) - \frac{\partial}{\partial t}\chi (\mathbf{r}, t) \label{EMField37} \]

Up to this point, \(\mathbf{A} \! '\) and \(\varphi '\) are undetermined. Let’s choose a \( \chi \) such that:

\[\nabla \cdot \mathbf{A} + \varepsilon_0 \mu_0 \frac{\partial \varphi}{\partial t} = 0 \label{EMField38} \]

This is known as the Lorentz condition (gauge). It is selected so that when we insert it into eq. (\ref{EMField32}), the second term of the left hand side vanishes, and we are left with

\[- \nabla^2 \mathbf{A} + \varepsilon_0 \mu_0 \frac{\partial^2 \mathbf{A}} {\partial t^2} = \mu_0 \mathbf{J} \label{EMField39} \]

Similarly, inserting eq. (\ref{EMField38}) into eq. (\ref{EMField34}) gives

\[\nabla^2 \varphi - \varepsilon_0 \mu_0 \frac{\partial^2 \varphi}{\partial t^2} = -\frac{\rho}{\varepsilon_0} \label{EMField40} \]Equations (\ref{EMField39}) and (\ref{EMField40}) are inhomogeneous wave equations for \(\mathbf{A}\) and \(\varphi\), with the current density and charge density as source terms, so this gauge is useful when describing time-varying charges. Within the Lorentz gauge, we can still arbitrarily add another \(\chi\); it must only satisfy eq. (\ref{EMField38}). If we substitute eqs. (\ref{EMField36}) and (\ref{EMField37}) into eq. (\ref{EMField40}), we see that \(\chi\) follows the wave equation

\[\nabla ^{2}\chi - \varepsilon_0 \mu_0 \frac{\partial^2 \chi}{\partial t^2} = 0 \label{EMField41} \]

So we can make further choices or add constraints to \(\mathbf{A}\) and \(\varphi\), as long as they obey eq. (\ref{EMField41}).

In the Coulomb gauge, we set

\[\nabla \cdot \mathbf{A} = 0 \label{EMField42} \]

which is selected so that insertion into eq. (\ref{EMField34}) gives the Poisson equation for the scalar potential: \(\nabla^2 \varphi =- \rho /\varepsilon_0\). This means treating static charge distributions is straightforward. A special case of the Coulomb gauge is the Gibbs gauge, which sets \(\varphi=0\) and can be used for free electromagnetic waves in the absence of charges and currents. The challenge for general solutions in the Coulomb gauge is that the vector potential wave equation has a more complicated form. From eq. (\ref{EMField32}):
\[\nabla ^2\mathbf{A} -\varepsilon_0 \mu _0 \frac{\partial ^2\mathbf{A}}{\partial t^2} = -\mu_0\,\mathbf{J}+\varepsilon _0 \mu _0 \nabla \kern-.3em \left( \frac{\partial \varphi }{\partial t} \right)\label{EMField43} \]

One can simplify this equation further by recognizing that the currents can be separated into a sum of longitudinal and transverse contributions, \( \mathbf{J}=\mathbf{J}_{\parallel}+\mathbf{J}_{\bot } \), which correspond to the components that are curl free \(\left(\nabla \times \mathbf{J}_{\parallel}=0 \right)\) and divergence free \(\left(\nabla \cdot \mathbf{J}_{\bot }\right)\). Jackson shows how the second term on the right hand side of eq. (\ref{EMField43}) is equal to \(\mu_0 \mathbf{J}_{\parallel}\),\(\bibref{Cite1} %in-text-citation\) meaning that the source term for the wave equation of \(\mathbf{A}\) purely arises from the transverse current:

\[ \nabla^2 \mathbf{A}-\varepsilon_0 \mu_0 \frac{ \partial^2 \mathbf{A} }{ \partial t^2 }=-\mu_0 \mathbf{J}_\bot \]

Indeed, we will see that using the Coulomb gauge results in electric fields that are purely transverse.

In isolation, the vector potential is far from any currents, and the wave equation for the free vector potential becomes:

\[ \nabla^2 \mathbf{A} - \varepsilon_0 \mu_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = 0 \label{EMField43.5} \]

Noting \(\varepsilon_0 \mu_0 = c^2\), we see this is an equation for a wave propagating at the speed of light. As described in the previous section, we look for harmonic plane wave solutions to this equation and find

\[\mathbf{A} (r,t) = \tilde{\mathbf{A}}_0 e^{\mathrm{i} \left(\mathbf{k} \cdot \mathbf{r} - \omega t \right)} \label{EMField44} \]

where \(\tilde{\mathbf{A}}_0\) is a complex vector amplitude. Alternatively, we can write a solution in real form:

\[ \mathbf{A} = \mathbf{A}_0 \sin \left(\omega t - \mathbf{k} \cdot \mathbf{r} + \alpha \right) \label{EMField45} \]

Here \(\alpha\) is an arbitrary constant phase, so a cosine solution is obtained by setting \(\alpha = \pi /2\) or, equivalently, setting \( \tilde{\mathbf{A}}_0 = \mathbf{A}_0 e^{\mathrm{i} \alpha} \). As before, \(\mathbf{k}\) is the wave vector which points along the direction of propagation and has a magnitude

\[ k^2 = \omega^2 \mu_0 \varepsilon_0 = \frac{\omega^2}{c^2} \label{EMField46} \]

which allows for positive and negative wavenumbers. Since \(\nabla \cdot \mathbf{A} = 0\) (see eq. (\ref{EMField42})), \(- \mathbf{k} \cdot {\mathbf{A}_0}\cos \left(\omega t - \mathbf{k} \cdot \mathbf{r} + \alpha \right) = 0\). Therefore,

\[ \mathbf{k} \cdot \mathbf{A}_0 = 0 \label{EMField47} \]

This indicates that the vector potential is oriented perpendicular to the direction of wave propagation (\(\mathbf{k} \, \bot \, \mathbf{A}_0\)).

Finally, we can obtain the electric and magnetic field from eqs. (\ref{EMField30}) and (\ref{EMField31}) with \(\nabla \varphi = 0\)

\[ \boldsymbol{\mathcal{E}} = - \frac{\partial \mathbf{A}}{\partial t} \label{EMField48} \]\[ \boldsymbol{\mathcal{B}} = \nabla \times \mathbf{A} \label{EMField49} \]

The expressions corresponding to the real form of \(\mathbf{A}\) in eq. (\ref{EMField45}) are

\[ \boldsymbol{\mathcal{E}} = - \omega \mathbf{A}_0 \cos \left(\omega t - \mathbf{k} \cdot \mathbf{r} + \alpha \right) \label{EMField50} \]\[ \boldsymbol{\mathcal{B}} = - \left(\mathbf{k} \times \mathbf{A}_0 \right)\cos \left(\omega t - \mathbf{k} \cdot \mathbf{r} + \alpha \right) \label{EMField51} \]

and the expressions corresponding to the complex form in eq. (\ref{EMField44}) are

\[ \tilde{\mathcal{E}} = -\mathrm{i}\omega {\mathbf{\tilde{A}}_0}\, e^{\mathrm{i}(k\cdot r-\omega t)}=-\mathrm{i}\omega \mathbf{\tilde{A}} \label{EMField52} \]

\[ \tilde{\mathcal{B}} =-\mathrm{i} \left( \mathbf{k}\times\mathbf{A}_{0} \right) e^{\mathrm{i}(k\cdot r-\omega t)}=-\mathrm{i} \left( \mathbf{k}\times \mathbf{\tilde{A}} \right) \label{EMField53} \]

From eq. (\ref{EMField50}), we observe that the electric field is parallel with the vector potential (\(\boldsymbol{\mathcal{E}} \parallel \! \mathbf{A}_0 \)), and the magnetic field is perpendicular to the electric field and the direction of propagation \(\left(\mathbf{k} \, \bot \, \boldsymbol{\mathcal{E}} \, \bot \, \boldsymbol{\mathcal{B}} \right)\), consistent to what we observed in the previous section.

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References

\[\tag {1} \label {Cite1}\]

Jackson, J. D. Classical Electrodynamics, 2nd ed.; Wiley, 1975.