3 Electrodynamics

3.1 Lorentz force

For the force $\mathbf{F}$ on a particle with charge $q$, which moves with the speed $\mathbf{v}$ through the electromagnetic field $\left(\mathbf{E}, \mathbf{B}\right)$, applies

This force is called Lorentz force.

3.2 Maxwell equations

The Maxwell equations (MWGen) for the electromagnetic field (EMF) $\left(\mathbf{E}, \mathbf{B}\right)$ ($\rho$ is the charge density and $\mathbf{j}$ is the current density) are

Charges and currents are therefore sources of the electromagnetic field. The equations (3.1) - (3.5) are the basis of the ED, from which the theory follows. The MWGen have a memorable structure: They each consist of a differential equation for the divergent or rotational portion 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 Maxwell's equations result:

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

3.2.1 Formulation about potentials

If you write down the MWGen component by component, you get eight equations for the six components of the electromagnetic field. The question therefore arises as to whether the MWGen are overdetermined and do not have non-trivial solutions. This section shows that this is not the case.

Because of Eq. (3.3) 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. This means that Eq. (3.3) according to Eq. (B.48) is met. Because of the third Maxwell equation Eq. (3.4) applies

\[ \begin{align} \nabla\times\left(\mathbf{E} + \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}\right) = \mathbf{0}. \end{align} \]

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. So Eq. (3.4) fulfilled. The definitions of the potentials are therefore obtained from the homogeneous Maxwell equations. If you put the previous specifications in Eq. (3.5), one obtains with Eq. (B.54)

\[ \begin{align} -\Delta\mathbf{A} + \nabla\left(\nabla\cdot\mathbf{A}\right) + \frac{1}{c}\nabla\frac{\partial\phi}{\partial t} + \frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2} = \frac{4\pi}{c}\mathbf{j}. \end{align} \]

This is called the gauge invariance of the MWGen. The calibration can be determined by any linear relationship between the potentials, with the Lorenz calibration

\[ \begin{align} \nabla\cdot\mathbf{A} + \frac{1}{c}\frac{\partial\phi}{\partial t} = 0\tag{3.15}\label{eq:lorenz-eichung} \end{align} \]

becomes the fourth MWG

\[ \begin{align} \Delta \mathbf{A} - \frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2} = -\frac{4\pi}{c}\mathbf{j}.\tag{3.16}\label{eq:ed_pot_1} \end{align} \]

The first MWG will be closed

\[ \begin{align} -\Delta\phi - \frac{1}{c}\frac{\partial}{\partial t}\nabla\cdot\mathbf{A} = 4\pi\rho\Leftrightarrow\Delta\phi - \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = -4\pi\rho.\tag{3.17}\label{eq:ed_pot_2} \end{align} \]

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).

3.2.2 Continuity equation of charge

If one forms the partial time derivative of Eq. (3.2), multiplies the divergence of Eq. (3.5) with $c$ and adding the two resulting equations, you get

\[ \begin{align} \nabla\cdot\frac{\partial\mathbf{E}}{\partial t} - \nabla\cdot\frac{\partial\mathbf E}{\partial t} &= 4\pi\frac{\partial\rho}{\partial t} + 4\pi\nabla\cdot\mathbf{j}\nonumber \end{align} \]

This is the charge continuity equation.

3.2.3 Lagrange and Hamilton functions of a particle in the EMF

Now the Lagrange and Hamilton functions of a particle with mass $m$ and charge $q$ in the electromagnetic field $\left(\mathbf{E}, \mathbf{B}\right)$ should be derived. If one sets 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)

\[ \begin{align} \mathbf{F}\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right)&\stackrel{\href{ch-40-vector-analysis.html#eq:diff_op_rule_6}{\text{Glg. (B.52)}}}{=} -q\nabla\phi + \frac{q}{c}\left(\newdot{\mathbf{r}}\cdot\nabla\right)\mathbf{A} + \frac{q}{c}\newdot{\mathbf{r}}\times\mathbf{B} - \frac{q}{c}\md{}\mathbf{A} = q\left(-\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t} + \frac{1}{c}\newdot{\mathbf{r}}\times\mathbf{B}\right)\nonumber\\ &= q\left(\mathbf{E} + \frac{\newdot{\mathbf{r}}}{c}\times\mathbf{B}\right). \end{align} \]

This results in the Lorentz force according to Eq. (3.1), this shows the correctness of the approach Eq. (D.173). So the Lagrange function of a particle in the EMF is

\[ \begin{align} L\left(\mathbf{r}, \newdot{\mathbf{r}}, t\right) &= \frac{m}{2}\newdot{\mathbf{r}}^2 - q\phi\left(\mathbf{r}, t\right) + \frac{q}{c}\newdot{\mathbf{r}}\cdot\mathbf{A}\left(\mathbf{r}, t\right) \end{align} \]

This follows for the canonical impulses

\[ \begin{align} p_i = \frac{\partial L}{\partial\newdot{x}_i} = m\newdot{x}_i + \frac{q}{c}A_i\Rightarrow \newdot{x}_i = \frac{1}{m}\left(p_i - \frac{q}{c}A_i\right). \end{align} \]

For the Hamilton function $H = H\left(\mathbf{r}, \mathbf{p}, t\right)$ therefore applies

\[ \begin{align} H\left(\mathbf{r}, \mathbf{p}, t\right) &= \frac{1}{m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\cdot\mathbf{p} - \frac{m}{2}\frac{1}{m^2}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi\left(\mathbf{r}, t\right) - \frac{q}{ cm}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\cdot\mathbf{A}\left(\mathbf{r}, t\right)\nonumber\\ &= \frac{1}{m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)\left(\mathbf{p} - \frac{1}{2}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right) - \frac{q}{c}\mathbf{A}\right) + q\phi\left(\mathbf{r}, t\right) = \frac{1}{2m}\left(\mathbf{p} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi\left(\mathbf{r}, t\right).\tag{3.23}\label{eq:hamilton_funktion_teilchen_emf} \end{align} \]

3.3 Electromagnetic waves

In a vacuum the MWGen is

\[ \begin{align} \nabla\cdot\mathbf{E} &= 0, \tag{3.24}\label{eq:mwg_1_vak}\\ \nabla\cdot\mathbf{B} &= 0, \tag{3.25}\label{eq:mwg_2_vak}\\ \nabla\times\mathbf{E} + \frac{1}{c}\frac{\partial\mathbf{B}}{\partial t} &= \mathbf{0}, \tag{3.26}\label{eq:mwg_3_vak}\\ \nabla\times\mathbf{B} - \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t} &= \mathbf{0}.\tag{3.27}\label{eq:mwg_4_vak} \end{align} \]

If you apply $\nabla\times $ to the equations (3.26) and (3.27), you get

\[ \begin{align} - \Delta\mathbf{E} + \frac{1}{c}\frac{\partial}{\partial t}\nabla\times\mathbf{B} &= \mathbf{0}\Leftrightarrow\Delta\mathbf{E} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\mathbf{E} = \mathbf{0},\\ - \Delta\mathbf{B} - \frac{1}{c}\frac{\partial}{\partial t}\nabla\times\mathbf{E} &= \mathbf{0}\Leftrightarrow\Delta\mathbf{B} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\mathbf{B} = \mathbf{0}, \end{align} \]

so wave equations for the EMF $\left(\mathbf{E}, \mathbf{B}\right)$. Approaches are now being made

\[ \begin{align} \mathbf{E} = \mathbf{E}_0e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)}, & {} & \mathbf{B} = \mathbf{B}_0e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t + \varphi\right)}. \end{align} \]

More general waves can be obtained with the FT 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 the equations (3.24) and (3.25) it follows $\mathbf{E}_0, \mathbf{B}_0\perp\mathbf{k}$. Electromagnetic waves are therefore transverse waves. One continues with Eq. (B.51)

\[ \begin{align} \nabla\times\mathbf{E} &= -i\mathbf{E}\times\mathbf{k}. \end{align} \]

Also apply

\[ \begin{align} \nabla\times\mathbf{B} = -i\mathbf{B}\times\mathbf{k}, & {} & \frac{\partial\mathbf{E}}{\partial t} = -i\omega\mathbf{E}, & {} & \frac{\partial\mathbf{B}}{\partial t} = -i\omega\mathbf{B}. \end{align} \]

If you insert this into the equations (3.26) - (3.27), you get

\[ \begin{align} - i\mathbf{E}\times\mathbf{k} - \frac{1}{c}i\omega\mathbf{B} &= \mathbf{0}\tag{3.33}\label{eq:ed_deriv_1},\\ - i\mathbf{B}\times\mathbf{k} + \frac{1}{c}i\omega\mathbf{E} &= \mathbf{0}\tag{3.34}\label{eq:ed_deriv_2}. \end{align} \]

With $c = \frac{\omega}{k}$ it follows

\[ \begin{align} \mathbf{E}\times\mathbf{k} &= -k\mathbf{B}. \end{align} \]

Follow from this

\[ \begin{align} \mathbf{E}\cdot\mathbf{B} = 0, \varphi = 0. \end{align} \]

Since $\mathbf{E}\perp\mathbf{k}$,

\[ \begin{align} E = B \end{align} \]

and thus

\[ \begin{align} E_0 &= B_0. \end{align} \]

3.4 Energy of the electromagnetic field

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 you can see the current density

\[ \begin{align} \mathbf{j}\left(\mathbf{r}, t\right) = \sum_{i = 1}^{N}q_i\mathbf{v}_i\delta\left(\mathbf{r} - \mathbf{r}_i\right) \end{align} \]

insert:

\[ \begin{align} \frac{dW}{dt} = \int_{\mathbb{R}^3}\mathbf{j}\cdot\mathbf{E}d^3r \end{align} \]

This can be transformed using Maxwell's 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) is obtained

\[ \begin{align} \mathbf{j}\cdot\mathbf{E} &= -\frac{c}{4\pi}\nabla\cdot\left(\mathbf{E}\times\mathbf{B}\right) + \frac{c}{4\pi}\mathbf{B}\cdot\nabla\times\mathbf{E} - \frac{1}{8\pi}\frac{\partial\mathbf{E}^2}{\partial t} = -\frac{c}{4\pi}\nabla\cdot\left(\mathbf{E}\times\mathbf{B}\right) - \frac{1}{8\pi}\frac{\partial}{\partial t}\left(\mathbf{E}^2 + \mathbf{B}^2\right). \end{align} \]

Now define the energy density $w$ of the electromagnetic field by

as well as the radiation flux density as

This follows

Eq. (3.47) is the Poynting theorem. This can still be illustrated. One integrates Eq. (3.47) over a time-independent volume $V$:

\[ \begin{align} \frac{\partial}{\partial t}\int_{V}w\left(\mathbf{r}, t\right)d^3r + \int_{\partial V}\mathbf{S}\cdot d\mathbf{n} = -\frac{dW}{dt}. \end{align} \]

Is Eq. (9.33) is the energy density of the electromagnetic field, so is

\[ \begin{align} U = \int_{V}w\left(\mathbf{r}, t\right)d^3r \end{align} \]

the energy of the same. Thus follows

\[ \begin{align} \frac{dU}{dt} + \frac{dW}{dt} = -\int_{\partial V}\mathbf{S}\cdot d\mathbf{n}. \end{align} \]

3.4.1 Radiation quantities

Radiation means energy transport through electromagnetic waves. The physical quantities that describe radiation are called radiation quantities.

Summary of definitions of radiation quantities.

Radiation size	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
Radiation 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$^2$	i. 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
Strahldichte	W/m$^2$	Strahlungsflussdichte pro Raumwinkel
spectral radiance	W/m$^3$	radiance per wavelength
Energiedichte	J/m$^3$	Energie, die pro Volumen in Form elektromagnetischer Wellen vorhanden ist
spectral energy density	J/m$^4$	energy density per wavelength

The spectral quantities are to be understood as follows: Imagine a device that can measure the underlying quantity filtered by wavelength. For example, you measure a radiation flux $\phi$ that acts on a system, but the measurement stops at a wavelength $\lambda > 0$. This is how you define the function $\varphi\left(\lambda\right)$. The derivative of this function $\phi_\lambda \coloneqq \frac{d\varphi}{d\lambda}$ after $\lambda$ is called spectral radiation flux, since this quantity allows a statement about the spectral distribution of the energy. It applies

\[ \begin{align} \phi = \int_{0}^{\infty}\phi_\lambda d\lambda. \end{align} \]

Spectral radiation quantities can also be defined by deriving them from the frequency. The word spectral is sometimes omitted.

3.5 Special theory of relativity

The MWGen gives the phase velocity $c$ for electromagnetic waves in vacuum in every IS. This is a contradiction to the Galileo transformation. This is therefore incorrect and should be replaced by the Lorentz transformation, which must contain the Galilei 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 a constant speed $\mathbf{v} = v\mathbf{e}_x$ relative to IS. At time $t = t' = 0$ the origins of the two KS 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) moving in every IS with $c$, holds

\[ \begin{align} s_{1, 2}^2 &= s_{1, 2}'^2 = 0, \end{align} \]

so the statement is clear. For a particle moving in a straight line and uniformly

\[ \begin{align} s_{1, 2}^2 &= \left(c^2 - v^2\right)\left(t_2 - t_1\right) = \left(c^2 - v'^2\right)\left(t_2' - t_1'\right) = s_{1, 2}'^2 \end{align} \]

to be verified experimentally. Now we further define 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's sum convention was introduced: Two identical indices, one of which is at the bottom and the other at the top, are summed. Now we start with the Lorentz transformation

\[ \begin{align} x'^{\left(\alpha\right)} = \Lambda_\beta^{\left(\alpha\right)} x^{\left(\beta\right)} + b^{\left(\alpha\right)}. \end{align} \]

From this it follows

\[ \begin{align} dx'^{\left(\alpha\right)} = \Lambda_\beta^{\left(\alpha\right)} dx^{\left(\beta\right)}. \end{align} \]

Now you ask $ds^2 = ds'^2$ and get

\[ \begin{align} ds'^2 &= \eta_{\alpha, \beta}dx'^{\left(\alpha\right)} dx'^{\left(\beta\right)} = \eta_{\alpha, \beta}dx'^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\delta\right)} = \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)} dx^{\left(\gamma\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\delta\right)}\nonumber\\ &= \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)} dx^{\left(\gamma\right)} dx^{\left(\delta\right)} = \eta_{\gamma.\delta}dx^{\left(\gamma\right)} dx^{\left(\delta\right)} = ds^2. \end{align} \]

Since this is supposed to apply to all $dx$ and $dx'$,

\[ \begin{align} \eta_{\gamma, \delta} &= \eta_{\alpha, \beta}\Lambda_\gamma^{\left(\alpha\right)}\Lambda_\delta^{\left(\beta\right)}. \end{align} \]

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 two KS IS and IS' described above, $\mathbf{b} = \mathbf{0}.$ Furthermore, the assumptions $y' = y$ and $z' = z$ make sense. The transformation between $\left(x, t\right)$ and $\left(x', t'\right)$ cannot depend on the y or z coordinates, since rotating both CS around the x axis must not lead to any change. So the transformation matrix $\Lambda$ 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, the origin of IS'

\[ \begin{align} x^{\left(0\right)} = ct&\leftrightarrow x'^{\left(0\right)} = ct',\\ x^{\left(1\right)} = vt&\leftrightarrow x'^{\left(1\right)} = 0,\\ x^{\left(2\right)} = 0&\leftrightarrow x'^{\left(2\right)} = 0,\\ x^{\left(3\right)} = 0&\leftrightarrow x'^{\left(3\right)} = 0. \end{align} \]

From Eq. (3.65) 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

\[ \begin{align} \left(\Lambda_0^{(0)}\right)^2 - \left(\Lambda_0^{\left(1\right)}\right)^2 &= 1,\\ \left(\Lambda_0^{\left(1\right)}\right)^2 - \left(\Lambda_1^{\left(1\right)}\right)^2 &= -1,\\ \Lambda_0^{(0)}\Lambda_1^{(0)} - \Lambda_0^{\left(1\right)}\Lambda_1^{\left(1\right)} &= 0. \end{align} \]

One can assume $\psi, \varphi$ for two real numbers

\[ \begin{align} \Lambda_0^{\left(1\right)} &= -\sinh\left(\psi\right),\\ \Lambda_1^{(0)} &= -\sinh\left(\varphi\right). \end{align} \]

So they apply

\[ \begin{align} \Lambda_0^{(0)} &= \pm\cosh\left(\psi\right), \Lambda_1^{\left(1\right)} &= \pm\cosh\left(\varphi\right). \end{align} \]

The identical transformation should result for $\psi, \varphi\to0$, so exclude 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 $\psi = \varphi$. So it applies

\[ \begin{align} \left(\begin{array}{cc} \Lambda_0^{(0)}&\Lambda_1^{(0)}\\ \Lambda_0^{\left(1\right)}&\Lambda_1^{\left(1\right)} \end{array}\right) &= \left(\begin{array}{cc} \cosh\left(\psi\right)& -\sinh\left(\psi\right)\\ - \sinh\left(\psi\right)&\cosh\left(\psi\right) \end{array}\right). \end{align} \]

Therefore applies

\[ \begin{align} x'^{\left(1\right)} = 0 &= -\sinh\left(\psi\right)ct + \cosh\left(\psi\right)x^{\left(1\right)} = \left[-\sinh\left(\psi\right)c + \cosh\left(\psi\right)v\right]t. \end{align} \]

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 Rapidity. One continues to define

\[ \begin{align} \gamma &\coloneqq \cosh\left(\psi\right) = \frac{1}{\sqrt{1 - \tanh\left(\psi\right)^2}} = \frac{1}{\sqrt{1 - v^2/c^2}}, \end{align} \]

then applies

\[ \begin{align} \sinh\left(\psi\right) = \gamma\frac{v}{c}. \end{align} \]

In this case the Lorentz transformation is complete

\[ \begin{align} \left(\begin{array}{c} ct'\\ x'\\ y'\\ z' \end{array}\right) &= \left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\left(\begin{array}{c} ct\\ x\\ y\\ z \end{array}\right)\tag{3.84}\label{eq:lorentztrafo_spec} \end{align} \]

set.

3.6 Relativistic treatment of the electromagnetic field

The equations of the potentials (3.16) and (3.17) as well as the Lorenz calibration (3.15) are summarized again here:

\[ \begin{align} \left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi &= -4\pi\rho,\\ \left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\mathbf{A} &= -4\pi\mathbf{j},\\ \frac{1}{c}\frac{\partial\phi}{\partial t} + \nabla\cdot\mathbf{A} &= 0. \end{align} \]

Now the 4-vector

\[ \begin{align} \left(A^{\left(\alpha\right)}\right) \coloneqq\left(\begin{array}{c} \phi\\ A_x\\ A_y\\ A_z \end{array}\right) \end{align} \]

defined. Define further

\[ \begin{align} \left(j^{\left(\alpha\right)}\right) &\coloneqq \left(\begin{array}{c} \rho\\ j_x\\ j_y\\ j_z \end{array}\right) \end{align} \]

It is defined as a shorthand notation for partial derivatives

\[ \begin{align} \partial_0 &= \partial^{(0)} \coloneqq\frac{1}{c}\frac{\partial}{\partial t},\\ \partial^{(i)} &\coloneqq \frac{\partial}{\partial x_i} = -\frac{\partial}{\partial x^{(i)}} = -\partial_i \end{align} \]

for $1\leq i\leq 3$. With the d'Alembert operator

\[ \begin{align} \Box \coloneqq\partial_\beta\partial^{\left(\beta\right)} = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \Delta \end{align} \]

you can do this as

\[ \begin{align} \Box A^{\left(\alpha\right)} = \frac{4\pi}{c}j^{\left(\alpha\right)} \end{align} \]

note down. The Lorenz calibration 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 it is still on

\[ \begin{align} \mathbf{E} = -\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t},& {} & \mathbf{B} = \nabla\times\mathbf{A} \end{align} \]

remembered. It applies

\[ \begin{align} F^{\left(\alpha, \beta\right)} &= \left(\begin{array}{cccc} 0&\frac{1}{c}\frac{\partial A_x}{\partial t} + \frac{\partial\phi}{\partial x}&\frac{1}{c}\frac{\partial A_y}{\partial t} + \frac{\partial\phi}{\partial y}&\frac{1}{c}\frac{\partial A_z}{\partial t} + \frac{\partial\phi}{\partial z}\\ - \frac{\partial\phi}{\partial x} - \frac{1}{c}\frac{\partial A_x}{\partial t}&0&\frac{\partial}{\partial y} - \frac{\partial}{\partial x}&\frac{\partial}{\partial z} - \frac{\partial}{\partial x}\\ - \frac{\partial\phi}{\partial y} - \frac{1}{c}\frac{\partial A_y}{\partial t}&\frac{\partial}{\partial x} - \frac{\partial}{\partial y}&0&\frac{\partial}{\partial z} - \frac{\partial}{\partial y}\\ - \frac{\partial\phi}{\partial z} - \frac{1}{c}\frac{\partial A_z}{\partial t}&\frac{\partial}{\partial x} - \frac{\partial}{\partial z}&\frac{\partial}{\partial y} - \frac{\partial}{\partial z}&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x& -E_y& -E_z\\ E_x&0& -B_z&B_y\\ E_y&B_z&0& -B_x\\ E_z& -B_y&B_x&0 \end{array}\right). \end{align} \]

Consider again the two KS IS and IS', which were already worked with in Sect. 3.5. The electromagnetic fields in the systems are denoted by $\left(\mathbf{E}, \mathbf{B}\right)$ and $\left(\mathbf{E'}, \mathbf{B'}\right)$. The field strength tensors transform with Eq. (3.84) like

\[ \begin{align} F' &= \Lambda F\Lambda^T = \Lambda\left(\begin{array}{cccc} 0& -E_x& -E_y& -E_z\\ E_x&0& -B_z&B_y\\ E_y&B_z&0& -B_x\\ E_z& -B_y&B_x&0 \end{array}\right)\left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} \gamma& -\gamma\frac{v}{c}&0&0\\ - \gamma\frac{v}{c}&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)\left(\begin{array}{cccc} \gamma E_x\frac{v}{c}& -\gamma E_x& -E_y& -E_z\\ \gamma E_x& -\gamma E_x\frac{v}{c}& -B_z&B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -\gamma^2E_x + \gamma^2E_x\frac{v^2}{c^2}& -\gamma E_y + B_z\gamma\frac{v}{c}& -E_z\gamma - B_y\gamma\frac{v}{c}\\ - \gamma^2E_x\frac{v^2}{c^2} + \gamma^2E_x&0&E_y\gamma\frac{v}{c} - B_z\gamma&E_z\gamma\frac{v}{c} + \gamma B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x& -\gamma E_y + B_z\gamma\frac{v}{c}& -E_z\gamma - B_y\gamma\frac{v}{c}\\ E_x&0&E_y\gamma\frac{v}{c} - B_z\gamma&E_z\gamma\frac{v}{c} + \gamma B_y\\ \gamma E_x - \gamma B_z\frac{v}{c}& -\gamma E_y\frac{v}{c} + \gamma B_z&0& -B_x\\ \gamma E_z + \gamma B_y\frac{v}{c}& -\gamma E_z\frac{v}{c} - \gamma B_y&B_x&0 \end{array}\right)\nonumber\\ &= \left(\begin{array}{cccc} 0& -E_x'& -E_y'& -E_z'\\ E_x'&0& -B_z'&B_y'\\ E_y'&B_z'&0& -B_x'\\ E_z'& -B_y'&B_x'&0 \end{array}\right). \end{align} \]

Follow from this

\[ \begin{align} E_x' = E_x, & {} & E_y' = \gamma\left(E_y - B_z\frac{v}{c}\right), & {} & E_z' = \gamma\left(E_z + B_y\frac{v}{c}\right),\\ B_x' = B_x, & {} & B_y' = \gamma\left(B_y + E_z\frac{v}{c}\right), & {} & B_z' = \gamma\left(B_z - E_y\frac{v}{c}\right). \end{align} \]

This can be generalized vectorially 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 components parallel and perpendicular to $\mathbf{v}$ are meant. The inverse transformation is obtained with $\mathbf{v}\to\mathbf{v'}$

\[ \begin{align} \mathbf{E}_\parallel = \mathbf{E'}_\parallel, & {} & \mathbf{B}_\parallel = \mathbf{B'}_\parallel,\\ \mathbf{E}_\perp = \gamma\left(\mathbf{E'}_\perp - \frac{\mathbf{v}}{c}\times\mathbf{B'}\right), & {} & \mathbf{B}_\perp = \gamma\left(\mathbf{B'}_\perp + \frac{\mathbf{v}}{c}\times\mathbf{E'}\right). \end{align} \]