For the force $\mathbf{F}$ on a particle with charge $q$ that moves with velocity $\mathbf{v}$ through the electromagnetic field $\left(\mathbf{E}, \mathbf{B}\right)$, the following holds
This force is called the Lorentz force.
The Maxwell equations for the electromagnetic field (EMF) $\left(\mathbf{E}, \mathbf{B}\right)$ ($\rho$ is the charge density and $\mathbf{j}$ is the current density) are
\[ \begin{align} \nabla\cdot\mathbf{E} &= 4\pi\rho\tag{3.2}\label{eq:mwg_1},\\ \nabla\cdot\mathbf{B} &= 0\tag{3.3}\label{eq:mwg_2},\\ \nabla\times\mathbf{E} + \frac{1}{c}\frac{\partial\mathbf{B}}{\partial t} &= \mathbf{0}\tag{3.4}\label{eq:mwg_3},\\ \nabla\times\mathbf{B} - \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t} &= \frac{4\pi}{c}\mathbf{j}\tag{3.5}\label{eq:mwg_4}. \end{align} \]
Charges and currents are therefore sources of the electromagnetic field. Equations (3.1) - (3.5) are the basis of electrodynamics, from which the theory follows. The Maxwell equations have a memorable structure: they each consist of a differential equation for the divergent and rotational parts of the fields $\mathbf{E}, \mathbf{B}$. For each of the fields $\mathbf{E}$ and $\mathbf{B}$ there is a homogeneous and an inhomogeneous equation. With Gauss's and Stokes's theorems, the integral forms of the Maxwell equations result:
\[ \begin{align} \int_{\partial V}^{}\mathbf{E}\cdot d\mathbf{n} &= 4\pi Q\\ \int_{\partial V}\mathbf{B}\cdot d\mathbf{n} &= 0\\ \int_{\partial A}^{}\mathbf{E}\cdot d\mathbf{s} + \frac{1}{c}\frac{\partial }{\partial t}\int_{A}\mathbf{B}\cdot d\mathbf{n} &= 0\\ \int_{\partial A}^{}\mathbf{B}\cdot d\mathbf{s} - \frac{1}{c}\frac{\partial}{\partial t}\int_{A}^{}\mathbf{E}\cdot d\mathbf{n} &= \frac{4\pi}{c}I \end{align} \]
Here $Q$ is the charge in $V$ and $I$ is the current passing through $A$. The corresponding experimental findings that led to the formulation of the equations are the following:
Coulomb's Law
There are no magnetic monopoles.
Faraday's law of induction
Ampère's law and Maxwell's displacement current
Writing the Maxwell equations component by component, one obtains eight equations for the six components of the electromagnetic field. The question therefore arises whether the Maxwell equations are overdetermined and have no non-trivial solutions. This section shows that this is not the case.
Because of Eq. (3.3), there exists a vector field $\mathbf{A}$ with
\[ \begin{align} \mathbf{B} = \nabla\times\mathbf{A}, \tag{3.10}\label{eq:b_vom_pot} \end{align} \]
the field $\mathbf{A}$ is called the vector potential. Thereby Eq. (3.3) is satisfied by virtue of Eq. (B.48). Because of the third Maxwell equation, Eq. (3.4), one has
\[ \begin{align} \nabla\times\left(\mathbf{E} + \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}\right) = \mathbf{0}. \end{align} \]
So there exists a potential $\phi$ with
\[ \begin{align} \mathbf{E} = -\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t},\tag{3.12}\label{eq:e_vom_pot} \end{align} \]
$\phi$ is called the scalar potential. Thereby Eq. (3.4) is satisfied. The definitions of the potentials are therefore obtained from the homogeneous Maxwell equations. Substituting the previous definitions into Eq. (3.5), one obtains with Eq. (B.54)
\[ \begin{align} -\Delta\mathbf{A} + \nabla\left(\nabla\cdot\mathbf{A}\right) + \frac{1}{c}\nabla\frac{\partial\phi}{\partial t} + \frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2} = \frac{4\pi}{c}\mathbf{j}. \end{align} \]
If one adds a conservative field $\nabla\phi'$ to $\mathbf{A}$, then $\mathbf{B}$ does not change. With the replacement
\[ \begin{align} \phi\to\phi - \frac{1}{c}\frac{\partial\phi'}{\partial t} \end{align} \] $\mathbf{E}$ does not change either.
This is called the gauge invariance of the Maxwell equations. The gauge can be fixed by any linear relationship between the potentials; with the Lorenz gauge
\[ \begin{align} \nabla\cdot\mathbf{A} + \frac{1}{c}\frac{\partial\phi}{\partial t} = 0\tag{3.15}\label{eq:lorenz-eichung} \end{align} \]
becomes the fourth Maxwell equation
\[ \begin{align} \Delta \mathbf{A} - \frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2} = -\frac{4\pi}{c}\mathbf{j}.\tag{3.16}\label{eq:ed_pot_1} \end{align} \]
The first Maxwell equation becomes
\[ \begin{align} -\Delta\phi - \frac{1}{c}\frac{\partial}{\partial t}\nabla\cdot\mathbf{A} = 4\pi\rho\Leftrightarrow\Delta\phi - \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = -4\pi\rho.\tag{3.17}\label{eq:ed_pot_2} \end{align} \]
This gives four linear, independent differential equations for $\phi$, $\mathbf{A}$, which replace the Maxwell equations. The EMF $\left(\mathbf{E}, \mathbf{B}\right)$ can be determined from this using the equations (3.12) and (3.10).
If one forms the partial time derivative of Eq. (3.2), multiplies the divergence of Eq. (3.5) by $c$, and adds the two resulting equations, one obtains
\[ \begin{align} \nabla\cdot\frac{\partial\mathbf{E}}{\partial t} - \nabla\cdot\frac{\partial\mathbf E}{\partial t} &= 4\pi\frac{\partial\rho}{\partial t} + 4\pi\nabla\cdot\mathbf{j}\nonumber \end{align} \]
\[ \begin{align} \Leftrightarrow \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{j} &= 0. \end{align} \]
This is the charge continuity equation.
Now the Lagrange and Hamilton functions of a particle with mass $m$ and charge $q$ in the electromagnetic field $\left(\mathbf{E}, \mathbf{B}\right)$ are to be derived. With the ansatz for the potential $U = U\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right)$
\[ \begin{align} U\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right) = q\phi\left(\mathbf{r}, t\right) - \frac{q}{c}\mathbf{A}\left(\mathbf{r}, t\right)\cdot\newdot{\mathbf{r}}, \tag{3.19}\label{eq:ed_ansatz_kraft} \end{align} \]
it follows from Eq. (2.61)
\[ \begin{align} \mathbf{F}\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right)&\stackrel{\href{ch-40-vector-analysis.html#eq:diff_op_rule_6}{\text{Eq. (B.52)}}}{=} -q\nabla\phi + \frac{q}{c}\left(\newdot{\mathbf{r}}\cdot\nabla\right)\mathbf{A} + \frac{q}{c}\newdot{\mathbf{r}}\times\mathbf{B} - \frac{q}{c}\md{}\mathbf{A} = q\left(-\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t} + \frac{1}{c}\newdot{\mathbf{r}}\times\mathbf{B}\right)\nonumber\\ &= q\left(\mathbf{E} + \frac{\newdot{\mathbf{r}}}{c}\times\mathbf{B}\right). \end{align} \]
This yields the Lorentz force according to Eq. (3.1); this shows the correctness of the ansatz Eq. (3.19). Thus the Lagrange function of a particle in the EMF is
\[ \begin{align} L\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right) &= \frac{m}{2}\newdot{\mathbf{r}}^2 - q\phi\left(\mathbf{r}, t\right) + \frac{q}{c}\newdot{\mathbf{r}}\cdot\mathbf{A}\left(\mathbf{r}, t\right) \end{align} \]
From this it follows for the canonical momenta
\[ \begin{align} p_i = \frac{\partial L}{\partial\newdot{x}_i} = m\newdot{x}_i + \frac{q}{c}A_i\Rightarrow \newdot{x}_i = \frac{1}{m}\left(p_i - \frac{q}{c}A_i\right). \end{align} \]
For the Hamilton function $H = H\left(\mathbf{r}, \mathbf{p}, t\right)$, one therefore has
\[ \begin{align} H\left(\mathbf{r}, \mathbf{p}, t\right) &= \frac{1}{m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\cdot\mathbf{p} - \frac{m}{2}\frac{1}{m^2}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi\left(\mathbf{r}, t\right) - \frac{q}{ cm}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\cdot\mathbf{A}\left(\mathbf{r}, t\right)\nonumber\\ &= \frac{1}{m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\left(\mathbf{p} - \frac{1}{2}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right) - \frac{q}{c}\mathbf{A}\right) + q\phi\left(\mathbf{r}, t\right) = \frac{1}{2m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi\left(\mathbf{r}, t\right).\tag{3.23}\label{eq:hamilton_funktion_teilchen_emf} \end{align} \]
In a vacuum, the Maxwell equations read
\[ \begin{align} \nabla\cdot\mathbf{E} &= 0, \tag{3.24}\label{eq:mwg_1_vak}\\ \nabla\cdot\mathbf{B} &= 0, \tag{3.25}\label{eq:mwg_2_vak}\\ \nabla\times\mathbf{E} + \frac{1}{c}\frac{\partial\mathbf{B}}{\partial t} &= \mathbf{0}, \tag{3.26}\label{eq:mwg_3_vak}\\ \nabla\times\mathbf{B} - \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t} &= \mathbf{0}.\tag{3.27}\label{eq:mwg_4_vak} \end{align} \]
If one applies $\nabla\times $ to Eqs. (3.26) and (3.27), one obtains
\[ \begin{align} - \Delta\mathbf{E} + \frac{1}{c}\frac{\partial}{\partial t}\nabla\times\mathbf{B} &= \mathbf{0}\Leftrightarrow\Delta\mathbf{E} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\mathbf{E} = \mathbf{0},\\ - \Delta\mathbf{B} - \frac{1}{c}\frac{\partial}{\partial t}\nabla\times\mathbf{E} &= \mathbf{0}\Leftrightarrow\Delta\mathbf{B} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\mathbf{B} = \mathbf{0}, \end{align} \]
i.e. wave equations for the EMF $\left(\mathbf{E}, \mathbf{B}\right)$. One now makes the ansätze
\[ \begin{align} \mathbf{E} = \mathbf{E}_0e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)}, & {} & \mathbf{B} = \mathbf{B}_0e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t + \varphi\right)}. \end{align} \]
More general waves can be obtained by means of the Fourier transform from the superposition of such plane waves. One immediately sees that electromagnetic waves in a vacuum are dispersion-free and have phase velocity $c$. One is interested in the ratio of the amplitudes $\mathbf{E}_0$, $\mathbf{B}_0$, as well as the phase shift $\varphi$. From Eqs. (3.24) and (3.25) it follows that $\mathbf{E}_0, \mathbf{B}_0\perp\mathbf{k}$. Electromagnetic waves are therefore transverse waves. One further finds, with Eq. (B.51),
\[ \begin{align} \nabla\times\mathbf{E} &= -i\mathbf{E}\times\mathbf{k}. \end{align} \]
Moreover, one has
\[ \begin{align} \nabla\times\mathbf{B} = -i\mathbf{B}\times\mathbf{k}, & {} & \frac{\partial\mathbf{E}}{\partial t} = -i\omega\mathbf{E}, & {} & \frac{\partial\mathbf{B}}{\partial t} = -i\omega\mathbf{B}. \end{align} \]
Substituting this into Eqs. (3.26) - (3.27), one obtains
\[ \begin{align} - i\mathbf{E}\times\mathbf{k} - \frac{1}{c}i\omega\mathbf{B} &= \mathbf{0}\tag{3.33}\label{eq:ed_deriv_1},\\ - i\mathbf{B}\times\mathbf{k} + \frac{1}{c}i\omega\mathbf{E} &= \mathbf{0}\tag{3.34}\label{eq:ed_deriv_2}. \end{align} \]
With $c = \frac{\omega}{k}$ it follows
\[ \begin{align} \mathbf{E}\times\mathbf{k} &= -k\mathbf{B}. \end{align} \]
From this it follows that
\[ \begin{align} \mathbf{E}\cdot\mathbf{B} = 0, \varphi = 0. \end{align} \]
Since $\mathbf{E}\perp\mathbf{k}$,
\[ \begin{align} E = B \end{align} \]
and thus
\[ \begin{align} E_0 &= B_0. \end{align} \]
To determine the energy of the electromagnetic field, consider a system of $N$ charges $q_i$ at locations $\mathbf{r}_i$. The work that the electromagnetic field does on this system is, according to Eq. (3.1), given by
\[ \begin{align} W &= \sum_{i = 1}^{N}\int\mathbf{F}\left(\mathbf{r}_i\right)\cdot\frac{d\mathbf{r}_i}{dt}dt = \sum_{i = 1}^{N}\int q_i\left(\mathbf{E}\left(\mathbf{r}_i\right) + \frac{1}{c}\frac{d\mathbf{r}_i}{dt}\times\mathbf{B}\right)\cdot\frac{d\mathbf{r}_i}{dt}dt\nonumber\\ &= \sum_{i = 1}^{N}\int q_i\mathbf{E}\left(\mathbf{r}_i\right)\cdot\frac{d\mathbf{r}_i}{dt}dt. \end{align} \]
It follows
\[ \begin{align} \frac{dW}{dt} = \sum_{i = 1}^{N}q_i\mathbf{E}\left(\mathbf{r}_i\right)\cdot\frac{d\mathbf{r}_i}{dt}. \end{align} \]
Here, one substitutes the current density
\[ \begin{align} \mathbf{j}\left(\mathbf{r}, t\right) = \sum_{i = 1}^{N}q_i\mathbf{v}_i\delta\left(\mathbf{r} - \mathbf{r}_i\right) \end{align} \]
to obtain
\[ \begin{align} \frac{dW}{dt} = \int_{\mathbb{R}^3}\mathbf{j}\cdot\mathbf{E}d^3r \end{align} \]
This can be transformed using the Maxwell equations,
\[ \begin{align} \mathbf{j}\cdot\mathbf{E} &= \frac{c}{4\pi}\mathbf{E}\cdot\nabla\times\mathbf{B} - \frac{1}{4\pi}\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t}. \end{align} \]
With Eq. (B.55), one obtains
\[ \begin{align} \mathbf{j}\cdot\mathbf{E} &= -\frac{c}{4\pi}\nabla\cdot\left(\mathbf{E}\times\mathbf{B}\right) + \frac{c}{4\pi}\mathbf{B}\cdot\nabla\times\mathbf{E} - \frac{1}{8\pi}\frac{\partial\mathbf{E}^2}{\partial t} = -\frac{c}{4\pi}\nabla\cdot\left(\mathbf{E}\times\mathbf{B}\right) - \frac{1}{8\pi}\frac{\partial}{\partial t}\left(\mathbf{E}^2 + \mathbf{B}^2\right). \end{align} \]
Now define the energy density $w$ of the electromagnetic field by
as well as the radiation flux density as
\[ \begin{align} \mathbf{S}\left(\mathbf{r}, t\right) \coloneqq \frac{c}{4\pi}\left(\mathbf{E}\times\mathbf{B}\right). \end{align} \]
It thus follows
Eq. (3.47) is the Poynting theorem. This can still be illustrated. One integrates Eq. (3.47) over a time-independent volume $V$:
\[ \begin{align} \frac{\partial}{\partial t}\int_{V}w\left(\mathbf{r}, t\right)d^3r + \int_{\partial V}\mathbf{S}\cdot d\mathbf{n} = -\frac{dW}{dt}. \end{align} \]
If Eq. (3.45) is the energy density of the electromagnetic field, then
\[ \begin{align} U = \int_{V}w\left(\mathbf{r}, t\right)d^3r \end{align} \]
is the energy of the same. It thus follows
\[ \begin{align} \frac{dU}{dt} + \frac{dW}{dt} = -\int_{\partial V}\mathbf{S}\cdot d\mathbf{n}. \end{align} \]
Radiation means energy transport through electromagnetic waves. The physical quantities that describe radiation are called radiation quantities.
| Radiation quantity | SI unit | Definition |
|---|---|---|
| radiant energy | J | the energy transferred to a system in a specific time interval |
| spectral radiant energy | J/m | radiant energy per wavelength |
| radiant flux | W | the temporal rate at which energy is transferred by radiation |
| spectral radiant flux | W/m | radiant flux per wavelength |
| radiant flux density | W/m$^2$ | in general a vector quantity, radiant energy that passes through a certain area per time |
| spectral radiant flux density | W/m$^3$ | radiant flux density per wavelength |
| radiance | W/m$^2$ | radiant flux density per solid angle |
| spectral radiance | W/m$^3$ | radiance per wavelength |
| energy density | J/m$^3$ | energy present per volume in the form of electromagnetic waves |
| spectral energy density | J/m$^4$ | energy density per wavelength |
The spectral quantities are to be understood as follows: one imagines a device that can measure the underlying quantity filtered by wavelength. For example, one measures a radiant flux $\phi$ acting on a system, but the measurement is cut off at a wavelength $\lambda > 0$. In this way one defines the function $\varphi\left(\lambda\right)$. The derivative of this function $\phi_\lambda \coloneqq \frac{d\varphi}{d\lambda}$ with respect to $\lambda$ is called the spectral radiant flux, since this quantity permits a statement about the spectral distribution of the energy. One has
\[ \begin{align} \phi = \int_{0}^{\infty}\phi_\lambda d\lambda. \end{align} \]
Spectral radiation quantities can also be defined by differentiation with respect to the frequency. The word spectral is sometimes omitted.
The Maxwell equations yield the phase velocity $c$ for electromagnetic waves in a vacuum in every IS. This contradicts the Galilean transformation, which is therefore incorrect and must be replaced by the Lorentz transformation; the latter must contain the Galilean transformation as a limiting case for velocities $v\ll c$. This is the statement of the special theory of relativity.
An element $\left(x^{\left(\alpha\right)}\right)$ of the spacetime is defined by a 4-vector
\[ \begin{align} \left(x^{\left(\alpha\right)}\right) \coloneqq\left(\begin{array}{c} ct\\ x\\ y\\ z \end{array}\right), \end{align} \]
spacetime is the set of all 4-vectors. $x, y, z$ are Cartesian coordinates in an IS, $t$ is the time coordinate determined by an arbitrary time zero point $t_0$. Greek indices are always supposed to run from zero to three. One further defines an event as a measurable physical process without temporal and spatial extension. Let two events be denoted by the indices 1 and 2, then they can be defined by their corresponding space-time coordinates:
\[ \begin{align} \left(x_1^{\left(\alpha\right)}\right) &= \left(\begin{array}{c} ct_1\\ x_1\\ y_1\\ z_1 \end{array}\right), \nonumber\\ \left(x_2^{\left(\alpha\right)}\right) &= \left(\begin{array}{c} ct_2\\ x_2\\ y_2\\ z_2 \end{array}\right). \end{align} \]
One defines the distance $s_{1, 2}^2$ of the two events in spacetime by
\[ \begin{align} s_{1, 2}^2 \coloneqq c^2\left(t_2 - t_1\right)^2 - \left(x_2 - x_1\right)^2 - \left(y_2 - y_1\right)^2 - \left(z_2 - z_1\right)^2. \end{align} \]
Now define an IS' whose axes are parallel to those of IS and which moves with constant velocity $\mathbf{v} = v\mathbf{e}_x$ relative to IS. At time $t = t' = 0$, the origins of the two coordinate systems are at the same place. Experimentally one finds
\[ \begin{align} s_{1, 2}^2 &= s_{1, 2}'^2. \end{align} \]
For a photon (or an electromagnetic wavefront) that moves with $c$ in every IS, one has
\[ \begin{align} s_{1, 2}^2 &= s_{1, 2}'^2 = 0, \end{align} \]
so the statement is clear. For a particle moving uniformly in a straight line,
\[ \begin{align} s_{1, 2}^2 &= \left(c^2 - v^2\right)\left(t_2 - t_1\right) = \left(c^2 - v'^2\right)\left(t_2' - t_1'\right) = s_{1, 2}'^2 \end{align} \]
is to be verified experimentally. One further defines the matrix $\eta$ by
\[ \begin{align} \eta \coloneqq\left(\begin{array}{cccc} 1&0&0&0\\ 0& -1&0&0\\ 0&0& -1&0\\ 0&0&0& -1 \end{array}\right). \end{align} \]
This allows the distance between two events $ds$ to be developed linearly:
\[ \begin{align} ds^2 &= c^2dt^2 - dx^2 - dy^2 - dz^2 = \sum_{\alpha = 0}^{3}\sum_{\beta = 0}^{3}\eta_{\alpha, \beta}dx^{\left(\alpha\right)} dx^{\left(\beta\right)} = :\eta_{\alpha, \beta}dx^{\left(\alpha\right)} dx^{\left(\beta\right)} \end{align} \]
Here, the Einstein summation convention was introduced: one sums over two identical indices, one of which is at the bottom and the other at the top. For the Lorentz transformation, one now makes the ansatz
\[ \begin{align} x'^{\left(\alpha\right)} = \Lambda_\beta^{\left(\alpha\right)} x^{\left(\beta\right)} + b^{\left(\alpha\right)}. \end{align} \]
From this it follows
\[ \begin{align} dx'^{\left(\alpha\right)} = \Lambda_\beta^{\left(\alpha\right)} dx^{\left(\beta\right)}. \end{align} \]
Now one requires $ds^2 = ds'^2$ and obtains
\[ \begin{align} ds'^2 &= \eta_{\alpha, \beta}dx'^{\left(\alpha\right)} dx'^{\left(\beta\right)} = \eta_{\alpha, \beta}dx'^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\delta\right)} = \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)} dx^{\left(\gamma\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\delta\right)}\nonumber\\ &= \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\gamma\right)} dx^{\left(\delta\right)} = \eta_{\gamma.\delta}dx^{\left(\gamma\right)} dx^{\left(\delta\right)} = ds^2. \end{align} \]
Since this is to hold for all $dx$ and $dx'$, one has
\[ \begin{align} \eta_{\gamma, \delta} &= \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)}. \end{align} \]
This means
\[ \begin{align} \eta_{\gamma, \delta} &= \sum_{\alpha, \beta = 0}^{3}\eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)} = \sum_{\alpha = 0}^{3}\Lambda_\gamma^{\left(\alpha\right)}\sum_{\beta = 0}^{3}\eta_{\alpha, \beta}\Lambda_\delta^{\left(\beta\right)} = \sum_{\alpha = 0}^{3}\left(\Lambda_\alpha^{\left(\gamma\right)}\right)^T\sum_{\beta = 0}^{3}\eta_{\alpha, \beta}\Lambda_\delta^{\left(\beta\right)} = \left(\Lambda^T\eta\Lambda\right)_{\gamma, \delta}, \end{align} \]
so
\[ \begin{align} \Lambda^T\eta\Lambda = \eta.\tag{3.65}\label{eq:bed_srt} \end{align} \]
For the arrangement of the two coordinate systems IS and IS' described above, $\mathbf{b} = \mathbf{0}$. Furthermore, the assumptions $y' = y$ and $z' = z$ are reasonable. The transformation between $\left(x, t\right)$ and $\left(x', t'\right)$ cannot depend on the $y$- or $z$-coordinates, since a rotation of both coordinate systems about the $x$-axis must not lead to any change. The transformation matrix $\Lambda$ therefore has the form
\[ \begin{align} \Lambda = \left(\begin{array}{cccc} \Lambda_0^{(0)}&\Lambda_1^{(0)}&0&0\\ \Lambda_0^{\left(1\right)}&\Lambda_1^{\left(1\right)}&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right). \end{align} \]
Furthermore, for the origin of IS' one has
\[ \begin{align} x^{\left(0\right)} = ct&\leftrightarrow x'^{\left(0\right)} = ct',\\ x^{\left(1\right)} = vt&\leftrightarrow x'^{\left(1\right)} = 0,\\ x^{\left(2\right)} = 0&\leftrightarrow x'^{\left(2\right)} = 0,\\ x^{\left(3\right)} = 0&\leftrightarrow x'^{\left(3\right)} = 0. \end{align} \]
From Eq. (3.65), it follows
\[ \begin{align} \left(\begin{array}{cc} \Lambda_0^{(0)}&\Lambda_0^{\left(1\right)}\\ \Lambda_1^{(0)}&\Lambda_1^{\left(1\right)} \end{array}\right)\left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right)\left(\begin{array}{cc} \Lambda_0^{(0)}&\Lambda_1^{(0)}\\ \Lambda_0^{\left(1\right)}&\Lambda_1^{\left(1\right)} \end{array}\right) &= \left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right)\nonumber\\ \Rightarrow\left(\begin{array}{cc} \Lambda_0^{(0)}& -\Lambda_0^{\left(1\right)}\\ \Lambda_1^{(0)}& -\Lambda_1^{\left(1\right)} \end{array}\right)\left(\begin{array}{cc} \Lambda_0^{(0)}&\Lambda_1^{(0)}\\ \Lambda_0^{\left(1\right)}&\Lambda_1^{\left(1\right)} \end{array}\right) &= \left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right)\nonumber\\ \Rightarrow\left(\begin{array}{cc} \left(\Lambda_0^{(0)}\right)^2 - \left(\Lambda_0^{\left(1\right)}\right)^2&\Lambda_0^{(0)}\Lambda_1^{(0)} - \Lambda_0^{\left(1\right)}\Lambda_1^{\left(1\right)}\\ \Lambda_0^{(0)}\Lambda_1^{(0)} - \Lambda_0^{\left(1\right)}\Lambda_1^{\left(1\right)}&\left(\Lambda_0^{\left(1\right)}\right)^2 - \left(\Lambda_1^{\left(1\right)}\right)^2 \end{array}\right) &= \left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right). \end{align} \]
From this follow the three conditions
\[ \begin{align} \left(\Lambda_0^{(0)}\right)^2 - \left(\Lambda_0^{\left(1\right)}\right)^2 &= 1,\\ \left(\Lambda_0^{\left(1\right)}\right)^2 - \left(\Lambda_1^{\left(1\right)}\right)^2 &= -1,\\ \Lambda_0^{(0)}\Lambda_1^{(0)} - \Lambda_0^{\left(1\right)}\Lambda_1^{\left(1\right)} &= 0. \end{align} \]
For two real numbers $\psi, \varphi$, one can make the ansatz
\[ \begin{align} \Lambda_0^{\left(1\right)} &= -\sinh\left(\psi\right),\\ \Lambda_1^{(0)} &= -\sinh\left(\varphi\right). \end{align} \]
Thus one has
\[ \begin{align} \Lambda_0^{(0)} &= \pm\cosh\left(\psi\right), \Lambda_1^{\left(1\right)} &= \pm\cosh\left(\varphi\right). \end{align} \]
For $\psi, \varphi\to0$, the identity transformation should result, so one excludes the minus sign at this point. The third condition is
\[ \begin{align} \cosh\left(\psi\right)\sinh\left(\varphi\right) = \cosh\left(\varphi\right)\sinh\left(\psi\right), \end{align} \]
from this it follows that $\psi = \varphi$. Thus one has
\[ \begin{align} \left(\begin{array}{cc} \Lambda_0^{(0)}&\Lambda_1^{(0)}\\ \Lambda_0^{\left(1\right)}&\Lambda_1^{\left(1\right)} \end{array}\right) &= \left(\begin{array}{cc} \cosh\left(\psi\right)& -\sinh\left(\psi\right)\\ - \sinh\left(\psi\right)&\cosh\left(\psi\right) \end{array}\right). \end{align} \]
It thus holds
\[ \begin{align} x'^{\left(1\right)} = 0 &= -\sinh\left(\psi\right)ct + \cosh\left(\psi\right)x^{\left(1\right)} = \left[-\sinh\left(\psi\right)c + \cosh\left(\psi\right)v\right]t. \end{align} \]
From this it follows
\[ \begin{align} \tanh\left(\psi\right) = \frac{v}{c}\Rightarrow\psi = \text{artanh}\left(\frac{v}{c}\right). \end{align} \]
$\psi$ is called the rapidity. One further defines
\[ \begin{align} \gamma &\coloneqq \cosh\left(\psi\right) = \frac{1}{\sqrt{1 - \tanh\left(\psi\right)^2}} = \frac{1}{\sqrt{1 - v^2/c^2}}, \end{align} \]
then
\[ \begin{align} \sinh\left(\psi\right) = \gamma\frac{v}{c}. \end{align} \]
In this case, the Lorentz transformation is thus given by
\[ \begin{align} \left(\begin{array}{c} ct'\\ x'\\ y'\\ z' \end{array}\right) &= \left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\left(\begin{array}{c} ct\\ x\\ y\\ z \end{array}\right)\tag{3.84}\label{eq:lorentztrafo_spec} \end{align} \]
The potential equations (3.16) and (3.17), together with the Lorenz gauge condition (3.15), are summarized here once more:
\[ \begin{align} \left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi &= -4\pi\rho,\\ \left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\mathbf{A} &= -4\pi\mathbf{j},\\ \frac{1}{c}\frac{\partial\phi}{\partial t} + \nabla\cdot\mathbf{A} &= 0. \end{align} \]
Now define the four-vector
\[ \begin{align} \left(A^{\left(\alpha\right)}\right) \coloneqq\left(\begin{array}{c} \phi\\ A_x\\ A_y\\ A_z \end{array}\right) \end{align} \]
One further defines
\[ \begin{align} \left(j^{\left(\alpha\right)}\right) &\coloneqq \left(\begin{array}{c} \rho\\ j_x\\ j_y\\ j_z \end{array}\right) \end{align} \]
As shorthand notation for partial derivatives, one defines
\[ \begin{align} \partial_0 &= \partial^{(0)} \coloneqq\frac{1}{c}\frac{\partial}{\partial t},\\ \partial^{(i)} &\coloneqq \frac{\partial}{\partial x_i} = -\frac{\partial}{\partial x^{(i)}} = -\partial_i \end{align} \]
for $1\leq i\leq 3$. With the d'Alembert operator
\[ \begin{align} \Box \coloneqq\partial_\beta\partial^{\left(\beta\right)} = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \Delta \end{align} \]
this can be written as
\[ \begin{align} \Box A^{\left(\alpha\right)} = \frac{4\pi}{c}j^{\left(\alpha\right)} \end{align} \]
The Lorenz gauge condition is therefore written as
\[ \begin{align} \partial_\alpha A^{\left(\alpha\right)} = 0. \end{align} \]
One defines the field strength tensor $F^{\left(\alpha, \beta\right)}$ by
\[ \begin{align} F^{\left(\alpha, \beta\right)} &= \partial^{\left(\alpha\right)} A^{\left(\beta\right)} - \partial^{\left(\beta\right)} A^{\left(\alpha\right)}. \end{align} \]
This matrix is antisymmetric:
\[ \begin{align} F^{\left(\beta, \alpha\right)} &= \partial^{\left(\beta\right)} A^{\left(\alpha\right)} - \partial^{\left(\alpha\right)} A^{\left(\beta\right)} = -\left(\partial^{\left(\alpha\right)} A^{\left(\beta\right)} - \partial^{\left(\beta\right)} A^{\left(\alpha\right)}\right) = -F^{\left(\alpha, \beta\right)} \end{align} \]
At this point, recall
\[ \begin{align} \mathbf{E} = -\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t},& {} & \mathbf{B} = \nabla\times\mathbf{A} \end{align} \]
It follows that
\[ \begin{align} F^{\left(\alpha, \beta\right)} &= \left(\begin{array}{cccc} 0&\frac{1}{c}\frac{\partial A_x}{\partial t} + \frac{\partial\phi}{\partial x}&\frac{1}{c}\frac{\partial A_y}{\partial t} + \frac{\partial\phi}{\partial y}&\frac{1}{c}\frac{\partial A_z}{\partial t} + \frac{\partial\phi}{\partial z}\\ - \frac{\partial\phi}{\partial x} - \frac{1}{c}\frac{\partial A_x}{\partial t}&0&\frac{\partial}{\partial y} - \frac{\partial}{\partial x}&\frac{\partial}{\partial z} - \frac{\partial}{\partial x}\\ - \frac{\partial\phi}{\partial y} - \frac{1}{c}\frac{\partial A_y}{\partial t}&\frac{\partial}{\partial x} - \frac{\partial}{\partial y}&0&\frac{\partial}{\partial z} - \frac{\partial}{\partial y}\\ - \frac{\partial\phi}{\partial z} - \frac{1}{c}\frac{\partial A_z}{\partial t}&\frac{\partial}{\partial x} - \frac{\partial}{\partial z}&\frac{\partial}{\partial y} - \frac{\partial}{\partial z}&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x& -E_y& -E_z\\ E_x&0& -B_z&B_y\\ E_y&B_z&0& -B_x\\ E_z& -B_y&B_x&0 \end{array}\right). \end{align} \]
Consider again the two inertial systems IS and IS', already used in Sect. 3.5. Let the electromagnetic fields in these systems be denoted by $\left(\mathbf{E}, \mathbf{B}\right)$ and $\left(\mathbf{E'}, \mathbf{B'}\right)$. The field-strength tensors transform according to Eq. (3.84) as
\[ \begin{align} F' &= \Lambda F\Lambda^T = \Lambda\left(\begin{array}{cccc} 0& -E_x& -E_y& -E_z\\ E_x&0& -B_z&B_y\\ E_y&B_z&0& -B_x\\ E_z& -B_y&B_x&0 \end{array}\right)\left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\left(\begin{array}{cccc} \gamma E_x\frac{v}{c}& -\gamma E_x& -E_y& -E_z\\ \gamma E_x& -\gamma E_x\frac{v}{c}& -B_z&B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -\gamma^2E_x + \gamma^2E_x\frac{v^2}{c^2}& -\gamma E_y + B_z\gamma\frac{v}{c}& -E_z\gamma - B_y\gamma\frac{v}{c}\\ - \gamma^2E_x\frac{v^2}{c^2} + \gamma^2E_x&0&E_y\gamma\frac{v}{c} - B_z\gamma&E_z\gamma\frac{v}{c} + \gamma B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x& -\gamma E_y + B_z\gamma\frac{v}{c}& -E_z\gamma - B_y\gamma\frac{v}{c}\\ E_x&0&E_y\gamma\frac{v}{c} - B_z\gamma&E_z\gamma\frac{v}{c} + \gamma B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x'& -E_y'& -E_z'\\ E_x'&0& -B_z'&B_y'\\ E_y'&B_z'&0& -B_x'\\ E_z'& -B_y'&B_x'&0 \end{array}\right). \end{align} \]
From this, one obtains
\[ \begin{align} E_x' = E_x, & {} & E_y' = \gamma\left(E_y - B_z\frac{v}{c}\right), & {} & E_z' = \gamma\left(E_z + B_y\frac{v}{c}\right),\\ B_x' = B_x, & {} & B_y' = \gamma\left(B_y + E_z\frac{v}{c}\right), & {} & B_z' = \gamma\left(B_z - E_y\frac{v}{c}\right). \end{align} \]
This can be generalized in vector form to
\[ \begin{align} \mathbf{E}_\parallel' &= \mathbf{E}_\parallel,\\ \mathbf{B}_\parallel' &= \mathbf{B}_\parallel,\\ \mathbf{E}_\perp' &= \gamma\left(\mathbf{E}_\perp + \frac{\mathbf{v}}{c}\times\mathbf{B}\right),\\ \mathbf{B}_\perp' &= \gamma\left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c}\times\mathbf{E}\right), \tag{3.105}\label{eq:trafo_emf_srt_4} \end{align} \]
where parallel and perpendicular components are understood with respect to $\mathbf{v}$. The inverse transformation is obtained with $\mathbf{v}\to\mathbf{v'}$:
\[ \begin{align} \mathbf{E}_\parallel = \mathbf{E'}_\parallel, & {} & \mathbf{B}_\parallel = \mathbf{B'}_\parallel,\\ \mathbf{E}_\perp = \gamma\left(\mathbf{E'}_\perp - \frac{\mathbf{v}}{c}\times\mathbf{B'}\right), & {} & \mathbf{B}_\perp = \gamma\left(\mathbf{B'}_\perp + \frac{\mathbf{v}}{c}\times\mathbf{E'}\right). \end{align} \]