Chapter 58Natural Units and the Four-Vector Formulation of Electromagnetism — Preparing for Particle Physics

58.1 Natural Units $c=\hbar=1$ — Removing Units to Make Equations Easier to Read

58.1.1 Starting from the “eV” of high school

In the atomic-physics unit of high school physics, we used the electronvolt $\mathrm{eV}$, rather than the joule $\mathrm{J}$, as the unit of energy. The kinetic energy that one electron (the magnitude of its electric charge is $e=1.602\,176\,634\times10^{-19}\ \mathrm{C}$) gains when it is accelerated through a potential difference of $1\ \mathrm{V}$ is $1\ \mathrm{eV}$; that is,

This follows from $W=qV$ (Section 36.1). The energy needed to pull the electron of a hydrogen atom away from the atom is $13.6\ \mathrm{eV}$, so in the world of atoms the $\mathrm{eV}$ is a unit of “just the right size.” Energies in the world of atomic nuclei are expressed in $\mathrm{MeV}$ (megaelectronvolts, $10^6\ \mathrm{eV}$), and energies in the world of elementary particles in $\mathrm{GeV}$ (gigaelectronvolts, $10^9\ \mathrm{eV}$) or $\mathrm{TeV}$ (teraelectronvolts, $10^{12}\ \mathrm{eV}$).

Next, recall $E=mc^2$ (the rest energy), which we learned in Chapter 57. A particle of mass $m$ has energy $mc^2$ even when it is at rest. Mass can therefore also be measured using $\mathrm{eV}$, in the unit “$\mathrm{eV}/c^2$.” The mass of the electron is $m_{\mathrm e}c^2=0.511\ \mathrm{MeV}$, that is, $m_{\mathrm e}=0.511\ \mathrm{MeV}/c^2$, and the mass of the proton is $m_{\mathrm p}=938.272\ \mathrm{MeV}/c^2\approx0.938\ \mathrm{GeV}/c^2$. Likewise, for momentum it is convenient to use the unit $\mathrm{eV}/c$, because of $E=pc$ (for a massless particle) and $E^2=p^2c^2+m^2c^4$ (Section 57.5). In this way, mass, momentum, and energy can all be expressed in units of $\mathrm{eV}$ multiplied by a power of $c$.

In that case, writing “$/c^2$” or “$/c$” every time is a nuisance. If we simply decide to “measure in units of the speed of light $c$” and set $c=1$, then mass, momentum, and energy can all be expressed in the same $\mathrm{eV}$, and $E^2=p^2c^2+m^2c^4$ becomes $E^2=p^2+m^2$. This is the first step in the idea of natural units.

58.1.2 Setting $c=1$ and $\hbar=1$

A unit in which the speed of light is $1$ has in fact long been used in astronomy. A “light-year” is the distance light travels in one year—a length fixed by a time. In the same way, if we decide to measure lengths by “the time light takes to travel,” length and time become the same unit. Since the speed of light in vacuum has the exact (defined) value $c=299\,792\,458\ \mathrm{m/s}$, the time light takes to travel $1\ \mathrm{m}$ is

To set $c=1$ means to “regard $1\ \mathrm{m}$ and $3.3356\ \mathrm{ns}$ as the same unit of length.” Already in Part VIII we have used $x^0=ct$, the time multiplied by $c$, as the “time coordinate” (Section 54.5). That was a partial implementation of the idea of setting $c=1$.

There is one more constant, the Planck constant. The energy of a photon (a particle of light) can be written $E=h\nu=\hbar\omega$ using the frequency $\nu$ and the angular frequency $\omega=2\pi\nu$ ($\hbar\equiv h/(2\pi)$ is read “h-bar” and is called the reduced Planck constant; $h=6.626\,070\,15\times10^{-34}\ \mathrm{J\,s}$ is an exact (defined) value). Setting $\hbar=1$ gives $E=\omega$; that is, “energy” and “angular frequency (the inverse of a time)” become the same unit.

It is easiest to think in terms of the number of units. In SI units, mechanical quantities are expressed with three independent units: length (m), mass (kg), and time (s). Setting $c=1$ makes length and time the same unit, and the number of independent units drops to two. If we set $\hbar=1$ as well, then, since the unit of $\hbar$ is $\mathrm{J\,s}=\mathrm{kg\,m^2/s}$, mass (= energy) and time are linked, and only one independent unit remains—energy (Figure 58.1).

58.1.3 Conversion using $\hbar c=197.327\ \mathrm{MeV\,fm}$

We want to convert the results calculated in natural units back to $\mathrm{m}$, $\mathrm{s}$, and $\mathrm{kg}$ in the end and compare them with experiment. The bridge for that is the product $\hbar c$ of $\hbar$ and $c$. Since $\hbar=h/(2\pi)=1.054\,571\,817\times10^{-34}\ \mathrm{J\,s}$ and $c=299\,792\,458\ \mathrm{m/s}$, we have

The unit of length in the world of atomic nuclei is $1\ \mathrm{fm}=10^{-15}\ \mathrm{m}$ (femtometer; also called the “fermi”; the radius of the proton is about $0.84\ \mathrm{fm}$), and the unit of energy is $\mathrm{MeV}$, so rewriting Eq. \eqref{eq:58-hbarc-eVm} gives

($1.973\,270\times10^{-7}\ \mathrm{eV\,m}=1.973\,270\times10^{-7}\times10^{-6}\ \mathrm{MeV}\times10^{15}\ \mathrm{fm}=197.327\ \mathrm{MeV\,fm}$.) In natural units $\hbar c=1$, so Eq. \eqref{eq:58-hbarc} says “$1=197.327\ \mathrm{MeV\,fm}$”; that is, it gives the conversion table

A length $\ell$ and an energy $E$ are related by $E=\hbar c/\ell$. The shorter the length, the larger the corresponding energy—this is one of the reasons why it is said in elementary particle physics that “we probe the small world with high energy” (Figure 58.2). The correspondence between length and energy here arises naturally from the uncertainty principle of quantum mechanics: when a particle is confined to a region of size about $\ell$, the magnitude of its momentum is about $\hbar/\ell$ (quantum mechanics is not treated in this book).

Note how the points of Figure 58.2 are to be read. (i) The visible-light point corresponds to a photon of energy $2.5\ \mathrm{eV}$ (wavelength $\lambda=2\pi\hbar c/E\approx500\ \mathrm{nm}$), for which $\ell=\hbar c/E\approx79\ \mathrm{nm}$; this is $1/(2\pi)$ of the wavelength (this corresponds to our having written $E=2\pi\hbar c/\lambda$ in Example 58.2(c)). (ii) The hydrogen-atom point lies at $\hbar c/a_0=\alpha m_{\mathrm e}c^2\approx3.7\ \mathrm{keV}$, for the Bohr radius $a_0=\hbar/(\alpha m_{\mathrm e}c)$. This is a different quantity from the ionization energy of the hydrogen atom, $13.6\ \mathrm{eV}=\alpha^2m_{\mathrm e}c^2/2$, and is $2/\alpha\approx274$ times as large. The “typical energy of a particle confined in a region of size $\ell$” and the “binding energy within that region” are not the same.

58.1.4 Restoring the units — recovery by dimensional analysis

Let us review the procedure for restoring $c$ and $\hbar$ in an equation written in natural units. The principle is simple; it is only the requirement that “every term in a physical equation has the same dimension” (the dimensional analysis of Section 1.3). In SI units, $[\hbar]=\mathrm{kg\,m^2/s}$ and $[c]=\mathrm{m/s}$, so once a mass $m$ is given, the following quantities can be formed from these (using $\mathrm{J}=\mathrm{kg\,m^2/s^2}$):

(Length: $\dfrac{\mathrm{kg\,m^2/s}}{\mathrm{kg}\cdot\mathrm{m/s}}=\mathrm{m}$; time: $\dfrac{\mathrm{kg\,m^2/s}}{\mathrm{kg\,m^2/s^2}}=\mathrm{s}$.)

58.1.5 $k_{\mathrm B}=1$ and notes on the units of electromagnetism

If we also set the Boltzmann constant $k_{\mathrm B}=1.380\,649\times10^{-23}\ \mathrm{J/K}$, which appears in thermodynamics and statistical mechanics (Part IV and Part VII), equal to 1, the temperature $T$ can be measured in units of energy. Since $k_{\mathrm B}=8.617\,333\times10^{-5}\ \mathrm{eV/K}$, $1\ \mathrm{eV}$ converted to a temperature corresponds to $1.160\,452\times10^{4}\ \mathrm{K}$.

Let us also prepare for one more topic: the treatment of the “lifetime” of short-lived particles. For a particle that exists for only a very short time, the shorter the lifetime $\tau$, the more the value of the particle's energy (mass) is blurred, by about $\Gamma\sim\hbar/\tau$ (this is the uncertainty principle of quantum mechanics; it is not derived in this book). The width $\Gamma$ of this blurring is called the decay width, and the larger $\Gamma$ is, the faster the particle decays. The mean lifetime is $\tau=\hbar/\Gamma$, which in natural units can be written $\tau=1/\Gamma$.

58.2 The Four-Gradient $\partial_\mu$ and the d'Alembertian $\Box$ — Lorentz Invariance of the Wave Equation

In the previous section we sorted out units. From here on, we rewrite the electromagnetism of Part V in a form whose shape does not change under Lorentz transformations (the covariant form). The equations of electromagnetism contain many partial derivatives $\partial/\partial t$, $\partial/\partial x$, …. So we first examine how these partial derivatives transform when the coordinates are changed by a Lorentz transformation.

58.2.1 The four-gradient $\partial_\mu$

We arrange the partial derivatives of a scalar field $\psi(t,x,y,z)$ (a field whose value at each point is determined by a single number, such as temperature or electric potential) as partial derivatives with respect to the four coordinates $x^\mu=(ct,x,y,z)$.

Since $x^0=ct$, we have $\dfrac{\partial}{\partial x^0}=\dfrac1c\dfrac{\partial}{\partial t}$. The $\nabla=(\partial_x,\partial_y,\partial_z)$ (nabla, or del) used in Part V is itself the spatial part of the four-gradient, and the four-gradient may be thought of as “the gradient including the time derivative” (University Mathematics, Section 18.4). Below we abbreviate $\partial_t=\partial/\partial t$.

Now, how does $\partial_\mu$ change under a Lorentz transformation? We find the partial derivative $\partial'_\mu=\partial/\partial x'^\mu$ written in the coordinates $x'^\mu$ of S′ by partial differentiation of a composite function (the chain rule; University Mathematics, Section 6.4). Regarding $\psi$ as a function of the coordinates $x^\nu$ of S, and noting that $x^\nu$ is a function of $x'^\mu$, we obtain

For a transformation in the $x$ direction, from the inverse transformation $x^0=\gamma(x'^0+\beta x'^1)$, $x^1=\gamma(x'^1+\beta x'^0)$, $x^2=x'^2$, $x^3=x'^3$ we get $\partial x^0/\partial x'^0=\gamma$, $\partial x^1/\partial x'^0=\gamma\beta$, $\partial x^0/\partial x'^1=\gamma\beta$, $\partial x^1/\partial x'^1=\gamma$, so

Let us compare this with $a'^0=\gamma(a^0-\beta a^1)$, $a'^1=\gamma(a^1-\beta a^0)$. Since the lowered components are $a_0=a^0,\ a_1=-a^1$, we get $a'_0=\gamma(a_0+\beta a_1)$, $a'_1=\gamma(a_1+\beta a_0)$, which has the same form as Eq. \eqref{eq:58-partial-boost}. That is, $\partial_\mu$ transforms as a four-vector with a lower index. In general we can state the following ($(\Lambda^{-1})^\nu{}_\mu$ is an entry of the inverse matrix of $\Lambda$).

58.2.2 The d'Alembertian $\Box$

Let us write the combination “second time derivative $-$ second space derivative” that appears in the equation of a wave in terms of the four-gradient. Forming the contraction (inner product) of $\partial_\mu$ and $\partial^\mu$, we obtain

(since $\partial^i=-\partial_i$, the spatial components carry a $-$ sign. $\nabla^2=\nabla\cdot\nabla$ is the Laplacian; University Mathematics, Section 19.3).

The d'Alembertian is the “spacetime version” of the Laplacian $\nabla^2$. Just as the Laplacian does not change its form under rotations of space (transformations that rotate the coordinate axes), the d'Alembertian does not change its form under Lorentz transformations.

This theorem is connected with the invariance of the speed of light. The wave equation in vacuum, with $\psi$ a scalar field (or any component of an electromagnetic wave), is

(Section 11.2, Section 42.3). When the coordinates are replaced by a Lorentz transformation, Theorem 58.2 gives $\Box'\psi=0$; that is, in frame S′ as well, $\dfrac1{c^2}\pdiff{^2\psi}{t'^2}=\nabla'^2\psi$ holds, a wave equation of the same form with the same $c$. That the speed of light (electromagnetic waves) is $c$ in any inertial frame of reference comes out automatically from the form of the equation.

58.2.3 Plane waves and the wave four-vector

The most basic solution of the wave equation is the plane wave. In frame S, we write a plane wave with angular frequency $\omega$ and wave vector $\bm k$ as $\psi=\cos(\omega t-\bm k\cdot\bm x)$ (following the convention of Part II, we take the phase in the form $\omega t-kx$; Section 11.3). The four components obtained by lining up $\omega/c$ and $\bm k$,

are called the wave four-vector. Then the phase can be written with the four-dimensional inner product $k\cdot x=k_\mu x^\mu=k^0x^0-\bm k\cdot\bm x=\omega t-\bm k\cdot\bm x$: $\psi=\cos(k\cdot x)$.

The phase $\omega t-\bm k\cdot\bm x$ is the quantity that determines at which time and place the crests and troughs of the wave are located, and since the event “there is a crest” is a fact independent of the observer, the value of the phase must be the same in every inertial frame (the phase is Lorentz invariant). Since $k_\mu x^\mu$ is invariant for an arbitrary four-position $x^\mu$, $k^\mu$ is a four-vector that transforms in the same way as $x^\mu$.

We substitute the plane wave into the wave equation, Eq. \eqref{eq:58-wave}. To simplify the calculation, we use the complex form $\psi=\exp(-ik\cdot x)=\exp[-i(\omega t-\bm k\cdot\bm x)]$ (we write the exponential function as $\exp$ to avoid confusion with the elementary charge $e$; this has the same form as $\exp[i(kx-\omega t)]$ in Part II, and the sign of its phase is the reverse of the form $\omega t-kx$, but the real part $\cos(\omega t-\bm k\cdot\bm x)=\cos(\bm k\cdot\bm x-\omega t)$ is the same; since the wave equation is linear in $\psi$, the real part and the imaginary part are each again a solution). Since $\partial_\mu\exp(-ik_\nu x^\nu)=-ik_\mu\exp(-ik\cdot x)$,

Hence the condition for $\Box\psi=0$ is $k_\mu k^\mu=0$, that is,

(this is the dispersion relation $\omega=ck$ of Section 42.4). That the “length” of the wave four-vector is zero (lightlike, or null) expresses the fact that the wave travels at the speed of light.

58.3 The Four-Current and Conservation of Charge

The sources of the electromagnetic field are charges and currents. In Part V we treated the charge density $\rho\ [\mathrm{C/m^3}]$ (the electric charge per unit volume) and the current density $\bm j\ [\mathrm{A/m^2}]$ (the electric charge passing through a unit area per unit time; its direction is the direction of the flow of charge) as separate quantities (Sections 37.1–37.2). However, to a moving observer, a charge that was at rest moves and becomes a “current,” and a wire that was carrying a current may look as if it carries “charge.” $\rho$ and $\bm j$ ought to mix with each other, just as time and space do. Let us find out how they mix.

58.3.1 Why do the charge density and the current density form a single set?

Suppose that a collection of charged particles (it may be an electron beam, or the electrons inside a metal) is moving with a uniform velocity $\bm v$. Let $\rho_0$ be the charge density seen by an observer who moves together with this collection of particles (the rest frame). In the rest frame the charges are at rest, so there is no current.

What happens in the laboratory frame (the frame in which this collection appears to move with velocity $\bm v$)? Here we use two facts. (i) Charge is Lorentz invariant; that is, the charge of a particle does not change no matter how fast the particle moves (the independence of charge from speed is consistent both with precise experiments on the electrical neutrality of atoms and molecules and with analyses of charged particles moving inside accelerators. For example, the speeds of the electrons and protons inside an atom vary widely, yet the observation that the atom as a whole is electrically neutral supports the conclusion that the magnitudes of the electron charge and the proton charge are equal irrespective of their motion). (ii) The length of a moving object along its direction of motion contracts to $1/\gamma$ (Lorentz contraction, Section 55.1), so the volume that holds the same number of particles is $1/\gamma$ of that in the rest frame. Since the charge is unchanged while the volume becomes $1/\gamma$, the charge density becomes $\gamma$ times larger:

The current density is the charge density multiplied by the velocity (Chapter 37), so $\bm j=\rho\bm v=\gamma\rho_0\bm v$. If we now line up $c\rho$ and $\bm j$ as four components, we obtain

Here $u^\mu=\gamma(c,\bm v)$ is the four-velocity (Section 57.4). Since $\rho_0$ is a Lorentz-invariant quantity (the charge density in the rest frame), $\rho_0u^\mu$ transforms as a four-vector.

Since $j^\mu$ is a four-vector, the transformation law of Chapter 57, $j'^0=\gamma(j^0-\beta j^1)$, $j'^1=\gamma(j^1-\beta j^0)$, $j'^2=j^2$, $j'^3=j^3$, gives the charge density and the current density in the frame S′, which moves at speed $v=\beta c$ in the $x$ direction:

For example, when a current $j_x\neq0$ flows in a wire that is electrically neutral ($\rho=0$) in the frame S, in the frame S′ we have $\rho'=-\gamma\beta j_x/c\neq0$; that is, the wire appears to be charged. This fact is the essence of why the magnetic field is the “relativistic face of the electric field” (Section 58.7).

58.3.2 The covariant form of conservation of charge

Conservation of charge (the continuity equation, Section 37.2) states that the increase of the electric charge in a region equals the current that flows in through its boundary:

Using the four-current and the four-gradient, the left-hand side is exactly $\partial_\mu j^\mu$. Indeed, $\partial_0j^0=\dfrac1c\pdiff{(c\rho)}{t}=\pdiff{\rho}{t}$ and $\partial_ij^i=\pdiff{j_x}{x}+\pdiff{j_y}{y}+\pdiff{j_z}{z}=\nabla\cdot\bm j$, so

holds. By Theorem 58.1, the four-divergence of a four-vector is Lorentz invariant, so this equation says that the law “charge is conserved” holds in the same form in every inertial frame. If $\partial_\mu j^\mu$ is 0 in one inertial frame, it is 0 in all inertial frames.

58.4 The Vector Potential and the Four-Potential

Next, let us move on to the electric field $\bm E$ and the magnetic flux density $\bm B$. If we tried to combine $\bm E,\bm B$ into a four-vector right away, we would find that $\bm E$ and $\bm B$ have three components each, six in all, which do not fit into the four components of a four-vector. So we take the following order: first we combine the “potentials” that lie behind them—the electric potential $\varphi$ and the vector potential $\bm A$—into a four-vector, and then we produce $\bm E,\bm B$ from it by differentiation.

58.4.1 Review of the electric potential and the vector potential

In Part V we learned two tools for expressing the electromagnetic field in terms of potentials.

• Electric potential $\varphi\ [\mathrm{V}]$: the electrostatic field is the field of a conservative force with $\rot\bm E=\bm0$, so it can be written as $\bm E=-\nabla\varphi$ (Section 36.1, Section 4.4. The fact that “a vector field whose curl is zero can be written as a gradient” is treated in University Mathematics, Section 20.3).
• Vector potential $\bm A\ [\mathrm{V\,s/m}=\mathrm{T\,m}=\mathrm{Wb/m}]$: the magnetic field satisfies $\operatorname{div}\bm B=0$ (Law 38.3). Since “a vector field whose divergence is zero can be written as the curl of some $\bm A$,” we may set $\bm B=\rot\bm A$ (Section 38.5. For a steady current, $\bm A(\bm r)=\dfrac{\mu_0}{4\pi}\iiint\dfrac{\bm j(\bm r')}{\abs{\bm r-\bm r'}}\dd V'$). Conversely, if we set $\bm B=\rot\bm A$, then $\operatorname{div}\bm B=0$ holds automatically, because “the divergence of a curl is $0$” (University Mathematics, Section 19.5).

When the electric field varies with time, the electrostatic relation $\bm E=-\nabla\varphi$ must be modified. Substituting $\bm B=\rot\bm A$ into Faraday's law of induction $\rot\bm E=-\partial_t\bm B$ (Section 40.2) gives

(The order of partial differentiation can be exchanged: $\partial_t\rot\bm A=\rot\,\partial_t\bm A$. This is Schwarz's theorem, University Mathematics, Section 6.3.) The vector field $\bm E+\partial_t\bm A$, whose curl is zero, can be written as the gradient of some function (with a $-$ sign attached), so if we call that function the electric potential $\varphi$, we obtain $\bm E+\partial_t\bm A=-\nabla\varphi$. In this way we arrive at the following.

The verification of the second half is easy: $\operatorname{div}\bm B=\operatorname{div}(\rot\bm A)=0$ and $\rot\bm E=-\rot(\nabla\varphi)-\partial_t(\rot\bm A)=-\bm 0-\partial_t\bm B$ (“the curl of a gradient is $0$”). Two of Maxwell's four equations are, so to speak, “digested” by the introduction of the potentials, and the remaining two (Gauss's law and the Ampère–Maxwell law) become the equations that the potentials satisfy (Section 58.6).

58.4.2 The four-potential

Let us check the units. From the second term of $\bm E=-\nabla\varphi-\partial_t\bm A$, we have $[\bm A]=[E]\cdot\mathrm{s}=\mathrm{V/m}\cdot\mathrm{s}=\mathrm{V\,s/m}$. On the other hand, the electric potential has $[\varphi]=\mathrm{V}$, so $\varphi/c$, obtained by dividing $\varphi$ by the speed of light $c\ [\mathrm{m/s}]$, has the unit $\mathrm{V\,s/m}$, exactly the same dimension as $\bm A$. So, just as with $x^\mu=(ct,\bm x)$ and $j^\mu=(c\rho,\bm j)$, we line up $\varphi/c$ as the time component and $\bm A$ as the spatial components.

That this $A^\mu$ indeed transforms as a four-vector (that is, that $\varphi$ and $\bm A$ mix in the same way as $\rho$ and $\bm j$) will be verified in Section 58.6.3: after deriving $\Box A^\mu=\mu_0j^\mu$, we use the fact that $j^\mu$ on the right-hand side is a four-vector. Here we simply write down in advance the result for a transformation in the $x$ direction:

58.4.3 Gauge transformations

The potentials have freedom in how they are chosen. For an arbitrary function $\chi(t,\bm x)$, if we simultaneously replace $\bm A$ and $\varphi$ by

then $\bm E,\bm B$ do not change. Indeed, $\bm B\to\rot\bm A+\rot\nabla\chi=\bm B$ (“the curl of a gradient is $0$”), and

(the last two terms cancel each other when the order of partial differentiation is exchanged). This replacement is called a gauge transformation. Using $\partial^\mu\chi=\left(\dfrac1c\partial_t\chi,\ -\nabla\chi\right)$, Eq. \eqref{eq:58-gauge} can be combined into a single equation in four-vector form:

(time component: $\varphi/c\to\varphi/c-\dfrac1c\partial_t\chi$; spatial component: $\bm A\to\bm A+\nabla\chi$). The electric and magnetic fields are the observed physical quantities, and the potentials are “tools” for calculating them, so it is quite acceptable for there to be freedom in how the tools are chosen (the Lorenz condition of Section 58.6 is a way of using this freedom skillfully). Incidentally, in quantum mechanics it is known that the potentials affect the waves of electrons even in regions where the electric and magnetic fields are zero (the Aharonov–Bohm effect), so the role of the potentials becomes more essential.

58.5 The Field-Strength Tensor $F^{\mu\nu}$ — Combining the Electric and Magnetic Fields into One

From the four-potential $A^\mu$, let us construct a single quantity that generates both the electric field and the magnetic flux density. Looking at Eq. \eqref{eq:58-EB-potential}, we see that $\bm E$ is a combination of the spatial derivative of $\varphi$ and the time derivative of $\bm A$, while $\bm B$ is a combination of spatial derivatives (the curl) of $\bm A$. Each of $\bm E,\bm B$ has the form “the difference of two terms, each obtained by differentiating a component of $A$ once.” So we differentiate the four-potential with the four-gradient and take the difference between the result and the same expression with the indices exchanged.

58.5.1 The definition of the field-strength tensor

An antisymmetric tensor has $4\times3/2=6$ independent components, which exactly matches the three components of $\bm E$ plus the three components of $\bm B$. This is why the electric and magnetic fields appear as “six components of a single quantity.” Moreover, $F^{\mu\nu}$ is unchanged by the gauge transformation, Eq. \eqref{eq:58-gauge-cov} ($A^\nu\to A^\nu-\partial^\nu\chi$):

(the result is the same if the order of the partial differentiations is exchanged). This is consistent with the fact that the observed $\bm E,\bm B$ do not change under a gauge transformation.

58.5.2 Computing the components

We compute the components of the definition, Eq. \eqref{eq:58-Fmunu}, one by one, using Eqs. \eqref{eq:58-partial-lower} and \eqref{eq:58-Amu} (note that $\partial^0=\partial_0=\dfrac1c\partial_t$, $\partial^i=-\partial_i$, $A^0=\varphi/c$, $A^i=A_i$).

First, the time–space components $F^{0i}$ ($i=1,2,3$):

(In the last step we used $E_i=-\partial_i\varphi-\partial_tA_i$.) Next, the space–space components $F^{ij}$. For example, $F^{12}$ is

Similarly, $F^{23}=-\partial_yA_z+\partial_zA_y=-B_x$ and $F^{31}=-\partial_zA_x+\partial_xA_z=-B_y$. Using the antisymmetry $F^{i0}=-F^{0i}$ and $F^{ji}=-F^{ij}$, we collect all the components into a $4\times4$ matrix, as follows (the rows are labeled by the first index $\mu$ and the columns by the second index $\nu$).

The components of the tensor with lowered indices, $F_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}F^{\alpha\beta}$, do not change sign when the index being lowered is $0$, and change sign when it is $i$. The time–space components $F_{0i}$ contain just one spatial index, so the sign changes, whereas $F_{ij}$ contains two, so it does not: $F_{0i}=-F^{0i}=E_i/c$ and $F_{ij}=F^{ij}$. In other words, the matrix becomes the one obtained by the replacement $\bm E\to-\bm E$.

Since both $\partial^\mu$ (Theorem 58.1) and $A^\nu$ (that it is a four-vector will be verified in Section 58.6.3) are four-vectors, $F^{\mu\nu}$ is a second-rank tensor for which each of the two indices transforms with $\Lambda$ under a Lorentz transformation:

($\mathsf T$ denotes the transpose. The reason we transpose the right-hand $\Lambda$ and multiply it from the right is to line up the order of the matrix product when we sum $\Lambda^\nu{}_\beta$ over the second index $\beta$.) Since $\Lambda$ is a matrix that mixes time and space, the components of $F$, that is, the components of $\bm E$ and of $\bm B$, get mixed with one another. This is the subject of Section 58.7.

58.5.3 Invariants

From $F^{\mu\nu}$ we can build quantities that do not change when the coordinate system is changed (Lorentz invariants). One of them is $F_{\mu\nu}F^{\mu\nu}$, in which all the indices are contracted.

Here $E^2=\abs{\bm E}^2=\bm E\cdot\bm E$ and $B^2=\abs{\bm B}^2=\bm B\cdot\bm B$ (we use this notation from here on in this chapter).

This invariant has a physical meaning. If $E^2-c^2B^2\gt0$ (the electric field dominates) in one inertial frame, then $E^2-c^2B^2\gt0$ in every other inertial frame as well. Therefore, a field in which the electric field dominates cannot appear as a purely magnetic field to any moving observer, and vice versa. Also, for an electromagnetic wave, $E=cB$ and $\bm E\perp\bm B$ (Section 42.4), so both invariants are $0$, and an electromagnetic wave remains one with $E=cB$ and $\bm E\perp\bm B$ as seen from any inertial frame.

58.6 The Covariant Form of Maxwell's Equations and the Lorenz Condition

The groundwork is laid. At the end of Part V (Section 42.2), electromagnetism was condensed into four equations, Maxwell's equations (Gauss's law, Gauss's law for magnetism, Faraday's law, and the Ampère–Maxwell law). When these are rewritten in terms of the four-current $j^\mu$ and the field-strength tensor $F^{\mu\nu}$, they become just two equations.

58.6.1 The equation with charges and currents as sources, $\partial_\mu F^{\mu\nu}=\mu_0j^\nu$

Let us check, by writing out the components, that this equation is nothing other than Gauss's law and the Ampère–Maxwell law.

There is one more important consequence. Applying $\partial_\nu$ to both sides of Eq. \eqref{eq:58-maxwell-inhom}, the right-hand side becomes $\mu_0\partial_\nu j^\nu$ and the left-hand side becomes $\partial_\nu\partial_\mu F^{\mu\nu}$. The order of partial differentiation can be interchanged ($\partial_\nu\partial_\mu$ is symmetric in $\mu,\nu$), but $F^{\mu\nu}$ is antisymmetric, so

Since this quantity equals its own negative, $\partial_\nu\partial_\mu F^{\mu\nu}=0$, and therefore $\partial_\nu j^\nu=0$ (Eq. \eqref{eq:58-continuity-cov}) follows. Maxwell's equations have conservation of charge “built in.” It was precisely for the sake of this consistency that Maxwell added the displacement current (Section 42.1).

58.6.2 The two equations without sources

The remaining two, $\operatorname{div}\bm B=0$ and Faraday's law, hold automatically once the potentials are introduced (Theorem 58.3). These too can be summarized in the language of tensors as follows. For the lower-index components $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, taking the sum of three terms with cyclically permuted indices gives

In fact, substituting $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and so on into the left-hand side, we get

(when the order of partial differentiation is interchanged, the six terms cancel exactly two at a time). Eq. \eqref{eq:58-maxwell-hom} is trivial ($0=0$) when two of the indices $(\lambda,\mu,\nu)$ are the same, and is meaningful only when the three are all different. There are four such triples: $(1,2,3),\ (0,2,3),\ (0,3,1),\ (0,1,2)$.

• $(\lambda,\mu,\nu)=(1,2,3)$: Using $F_{23}=F^{23}=-B_x$ and so on, $\partial_1F_{23}+\partial_2F_{31}+\partial_3F_{12}=-(\partial_xB_x+\partial_yB_y+\partial_zB_z)=0$. That is, $\operatorname{div}\bm B=0$ (Gauss's law for magnetism).
• $(0,1,2)$: Using $F_{12}=-B_z,\ F_{20}=-E_y/c,\ F_{01}=E_x/c$, we get $\partial_0F_{12}+\partial_1F_{20}+\partial_2F_{01}=-\dfrac1c\left[\pdiff{B_z}{t}+\left(\pdiff{E_y}{x}-\pdiff{E_x}{y}\right)\right]=0$. That is, $(\rot\bm E)_z=-\partial_tB_z$. $(0,2,3)$ and $(0,3,1)$ similarly give the $x$ component and the $y$ component, and together they yield $\rot\bm E=-\partial_t\bm B$ (Faraday's law).

58.6.3 The Lorenz condition and $\Box A^\mu=\mu_0j^\mu$

Substituting $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$ into Eq. \eqref{eq:58-maxwell-inhom} gives an equation for $A^\mu$ alone:

(here $\partial_\mu\partial^\nu A^\mu=\partial^\nu\partial_\mu A^\mu$, by interchanging the order of partial differentiation). The second term, $\partial_\mu A^\mu$, is a nuisance. So we eliminate it by using the freedom of the gauge transformation of Section 58.4. Under the gauge transformation $A^\mu\to A^\mu-\partial^\mu\chi$, $\partial_\mu A^\mu$ changes into $\partial_\mu A^\mu-\Box\chi$, so if we choose a $\chi$ satisfying $\Box\chi=\partial_\mu A^\mu$ (this is a kind of wave equation, so a solution always exists), the new $A^\mu$ satisfies the following condition.

Since the left-hand side of this condition is a four-divergence, its form does not change under Lorentz transformations ($\partial_\mu A^\mu=0$ holds in every inertial frame). Note also that the Lorenz condition does not fix the gauge completely. Even if we perform a gauge transformation $A^\mu\to A^\mu-\partial^\mu\chi$ with a $\chi$ satisfying $\Box\chi=0$, we get $\partial_\mu A^\mu\to\partial_\mu A^\mu-\Box\chi=\partial_\mu A^\mu$, and the condition is preserved (residual gauge freedom). Under the Lorenz condition, $\partial^\nu(\partial_\mu A^\mu)=0$, and the following equation is obtained.

(The equation for $\mu=0$ is $\Box(\varphi/c)=\mu_0c\rho$; we multiplied both sides by $c$ and used $\mu_0c^2=1/\varepsilon_0$.) The equations for the potentials are such that the four components of $\varphi$ and $\bm A$ are each, independently, a wave equation with source $j^\mu$. In a steady state, where the time variation is ignored, $\Box\to-\nabla^2$, so we return to $\nabla^2\varphi=-\rho/\varepsilon_0$ (Poisson's equation, Section 36.7) and $\nabla^2\bm A=-\mu_0\bm j$ (the equation satisfied by $\bm A=\frac{\mu_0}{4\pi}\iiint\frac{\bm j}{R}\dd V'$ of Chapter 38). In source-free vacuum ($j^\mu=0$), $\Box A^\mu=0$, which is the wave equation for electromagnetic waves (Section 42.3), and the wave travels at the speed of light $c$.

$A^\mu$ is a four-vector. Here we carry out the check that was postponed in Sections 58.4.2 and 58.5.2. In the Lorenz gauge, $\Box A^\mu=\mu_0j^\mu$ holds, $\Box$ is Lorentz invariant (Theorem 58.2), and $j^\mu$ is a four-vector (Definition 58.4). So let us construct $\tilde A'^\mu=\Lambda^\mu{}_\nu A^\nu$, the solution $A^\mu(x)$ in frame S carried over to frame S′ by the transformation law of a four-vector. Since $\Lambda$ is constant, $\Box'\tilde A'^\mu=\Lambda^\mu{}_\nu\Box A^\nu=\mu_0\Lambda^\mu{}_\nu j^\nu=\mu_0j'^\mu$, so an equation of the same form holds in S′ as well. The Lorenz condition too is preserved, $\partial'_\mu\tilde A'^\mu=\partial_\mu A^\mu=0$, because the four-divergence is invariant (Theorem 58.1). The boundary condition “no radiation coming in from the past” is also Lorentz invariant, so the retarded solution obtained by solving the equation independently in S′ coincides with $\tilde A'^\mu$. Thus we see that $A^\mu$ transforms as a four-vector, just like $x^\mu$ and $j^\mu$. For general potentials not in the Lorenz gauge, the potential in S′ may differ from the result of the transformation law by a gauge transformation $A^\mu\to A^\mu-\partial^\mu\chi$, but $F^{\mu\nu}$ (and hence $\bm E,\bm B$) does not change (Section 58.5.1).

The solution in the presence of sources is given by the retarded potential, which expresses the fact that information about the source propagates at the speed of light $c$: $\varphi(\bm r,t)=\dfrac1{4\pi\varepsilon_0}\iiint\dfrac{\rho(\bm r',\,t-\abs{\bm r-\bm r'}/c)}{\abs{\bm r-\bm r'}}\dd V'$ (not derived in this book; for details see textbooks such as Jackson's). The electric potential at the position $\bm r$ at time $t$ is determined by the charge distribution that the source had at time $t-\abs{\bm r-\bm r'}/c$ (earlier by the time it takes light to travel from the source to $\bm r$).

58.7 Transformation of Electric and Magnetic Fields — They Mix for Different Observers

The central result of this chapter is the transformation law of the electric and magnetic fields, which we derive next. Using the fact that $F^{\mu\nu}$ is a second-rank tensor (Eq. \eqref{eq:58-Ftrans}), the $\bm E',\bm B'$ seen by a moving observer can be found by a mechanical calculation.

58.7.1 Derivation of the transformation law

Consider the frame S′ that moves in the $x$ direction at speed $v=\beta c$ relative to the inertial frame S. The entries of the transformation matrix $\Lambda$ are $\Lambda^0{}_0=\Lambda^1{}_1=\gamma$, $\Lambda^0{}_1=\Lambda^1{}_0=-\beta\gamma$, $\Lambda^2{}_2=\Lambda^3{}_3=1$, and all the others are $0$. We compute the components of $F'^{\mu\nu}=\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta F^{\alpha\beta}$ in Eq. \eqref{eq:58-Ftrans} one at a time. Since the only indices that appear in rows $0,1$ of $\Lambda$ are $0,1$, the sums are simple.

That the vector form agrees with Eq. \eqref{eq:58-EBtrans} can be checked by computing $\bm v\times\bm B=(0,\ -vB_z,\ vB_y)$ and $\bm v\times\bm E=(0,\ -vE_z,\ vE_y)$. Eq. \eqref{eq:58-EBtrans-vec} is easy to remember. The dimensions also match: $vB\ [\mathrm{m/s\cdot T}=\mathrm{V/m}]$ has the same dimension as $E$, and $vE/c^2\ [\mathrm{(m/s)(V/m)/(m/s)^2}=\mathrm{V\,s/m^2}=\mathrm{T}]$ has the same dimension as $B$.

Let us use Eq. \eqref{eq:58-EBtrans} to verify the invariance of $\bm E\cdot\bm B$, which we postponed in Section 58.5.

58.7.2 A space containing only an electric field, as seen by a moving observer

Suppose that in the frame S there is only an electric field $\bm E=(0,E_y,0)$ (that is, $\bm B=\bm0$). An example is the uniform electric field between the plates of a parallel-plate capacitor. Setting $B_y=B_z=E_z=0$ in Eq. \eqref{eq:58-EBtrans}, we find in the frame S′

This is shown in Figure 58.4. Although there was no magnetic field in the frame S, the moving observer observes a magnetic field $B'_z\neq0$. It is not zero even for $v\ll c$. Moreover, the combination of the two quantities $E'_y$ and $cB'_z$, $E'^2-c^2B'^2=\gamma^2E_y^2(1-\beta^2)=E_y^2$, is constant, independent of $\beta$ (Theorem 58.4).

58.7.3 A charge near a current-carrying wire — the magnetic force as the relativistic face of the electric force

As you learned in high school physics, a charge moving parallel to the current near a current-carrying wire is attracted toward the wire (parallel currents in the same direction attract each other; Section 39.2). This force is the magnetic force $q\bm v\times\bm B$. However, an observer moving together with this charge would observe that the charge is at rest ($\bm v=\bm0$), so the magnetic force is zero, and yet the charge begins to be drawn toward the wire. What exerts the force? The answer is: “To the moving observer, the wire is electrically charged.”

Let us calculate with a simple model (Figure 58.5). We represent the wire along the $x$ axis by a row of positive charges moving in the $+x$ direction at speed $u$ (with linear density $+\lambda$ in the laboratory frame S) and a row of stationary negative charges (linear density $-\lambda$). The wire as a whole is neutral, and the current is $I=\lambda u$ (in a real wire, the carriers that move are electrons (negative charges). In that case the signs of all the charges in the wire are reversed, so the direction of the force acting on $q$ is simply reversed (pointing away from the wire), and the conclusion that the magnetic force and the electric force agree is the same. The speed of the charge $q$ may differ from the carrier speed, but then the calculation becomes a little more complicated). At the point $(0,r,0)$, a distance $r$ from the wire, a point charge $q\ (\gt0)$ is moving at the same speed $u$ in the $+x$ direction.

We carry out this calculation in the laboratory frame S and in the frame S′ that moves together with $q$ (moving in the $+x$ direction at speed $u$ relative to S, with $\beta=u/c$), and compare the results.

In the end, the force that we called the “magnetic force” in S is observed as an “electric force” in S′. The magnetic field and the electric field are “two faces of the same electromagnetic field” that can be interchanged by the motion of the observer. One may say that the magnetic force emerges when a relativistic effect (the change of charge density due to Lorentz contraction) is added to the electric force. From this viewpoint, “Coulomb's law” and the “force between currents,” which we learned in high school as separate laws, can be understood from a single law.

58.8 The Covariant Lorentz Force

Let us also rewrite the force that the electromagnetic field exerts on a charge, namely the Lorentz force $\bm F=q(\bm E+\bm v\times\bm B)$ (Section 39.4), in covariant form. In high school we learned the force $q\bm v\times\bm B$ that a magnetic field exerts on a charge and the force $q\bm E$ that an electric field exerts on a charge as separate laws. With the field-strength tensor $F^{\mu\nu}$, these combine into a single equation.

58.8.1 Equation of motion $\dd p^\mu/\dd\tau=qF^{\mu\nu}u_\nu$

We describe the motion of a charged particle (charge $q$, mass $m$) in terms of the particle's proper time $\tau$ (Section 56.5). The four-velocity is $u^\mu=\gamma(c,\bm v)$, the four-momentum is $p^\mu=mu^\mu=(E/c,\ \bm p)$ with $\bm p=\gamma m\bm v$ and $E=\gamma mc^2$ (Sections 57.4–57.5), and $\dd\tau=\dd t/\gamma$.

Let us verify in components that this equation contains the familiar Lorentz force.

Why this form? The left-hand side $\dd p^\mu/\dd\tau$ is the four-vector $p^\mu$ differentiated with respect to the invariant $\tau$, so it is a four-vector. Therefore the right-hand side must also be a four-vector. Among the four-vectors that can be built from the electromagnetic field $F^{\mu\nu}$ and the particle's four-velocity $u_\nu$ and that are linear in both $F$ and $u$, the simplest is $F^{\mu\nu}u_\nu$. Checking in components, we find that this is exactly the Lorentz force.

This equation has two meanings. First, since $F^{\mu\nu}$ is a second-rank tensor and $u_\nu$ is a (lower-index) four-vector, the right-hand side of Eq. \eqref{eq:58-lorentz-force} is a four-vector that transforms in the same way as $p^\mu$. Since both sides transform in the same way, the same form $\dd p'^\mu/\dd\tau=qF'^{\mu\nu}u'_\nu$ holds in frame S′ as well, and this equation has the same form in every inertial frame. Second, the components of $F'^{\mu\nu}$ in frame S′ give the electric field and the magnetic flux density $\bm E',\bm B'$ measured in frame S′. For the force written as $q(\bm E+\bm v\times\bm B)$ in frame S to be written in the same form $q(\bm E'+\bm v'\times\bm B')$ in frame S′ as well, $\bm E',\bm B'$ must transform as in Theorem 58.5. Conversely, if $F^{\mu\nu}$ is a tensor, this is satisfied automatically. The agreement of the magnetic force and the electric force that we saw in the example of “a charge near a wire” in Section 58.7 is precisely a manifestation of this consistency.

Let us make one more consistency check. As we saw in Chapter 57, the “length” of the four-velocity must always satisfy $u_\mu u^\mu=c^2$. From Eq. \eqref{eq:58-lorentz-force} and $p^\mu=mu^\mu$,

The last quantity, $u_\mu F^{\mu\nu}u_\nu$, is $0$ because it is the antisymmetric $F^{\mu\nu}$ contracted with the symmetric $u_\mu u_\nu$ (it changes sign under the exchange $\mu\leftrightarrow\nu$, so it equals the negative of itself; this is the same argument as for $\partial_\nu\partial_\mu F^{\mu\nu}=0$ in Section 58.6). Even when the particle is acted on by a force from the electromagnetic field, $u_\mu u^\mu=c^2$ is preserved.

58.8.2 Relativistic cyclotron motion

Let us treat a charged particle moving in a uniform magnetic flux density $\bm B$ (no electric field) relativistically. In nonrelativistic mechanics, a particle entering perpendicular to the magnetic field with speed $v$ undergoes circular motion with radius $r=mv/(qB)$ and angular frequency $\omega_{\mathrm c}=qB/m$ (Section 39.6). In relativity, $\dd E/\dd t=q\bm E\cdot\bm v=0$ implies that $E=\gamma mc^2$ is constant, that is, $\gamma$ and the speed $v$ are constant. The equation of motion is $\dd\bm p/\dd t=q\bm v\times\bm B$, and the magnitude of $\bm p=\gamma m\bm v$ is constant; only its direction rotates. The magnitude of the force is $qvB$ ($\bm v\perp\bm B$), and when a vector of magnitude $p$ is rotated with angular velocity $\omega$, its rate of change is $\omega p$, so $\omega p=qvB$, that is,

This is the nonrelativistic formula with the mass $m$ replaced by $\gamma m$. However, the relation $r=p/(qB)$, “the radius of the trajectory is the momentum $p$ divided by $qB$,” holds unchanged in relativity. For a particle whose charge is the elementary charge $e$, expressing the momentum in $\mathrm{GeV}/c$, we obtain the following practical formula from $pc=ecBr$.

This figure is the map that unites the many laws learned separately in Part V—Coulomb's law, the Biot–Savart law, Ampère's law, Faraday's law, and the Lorentz force—under a single principle, Lorentz invariance.

58.9 The Relativistic Lagrangian of a Free Particle

In Part VI (Chapters 43–46), we rewrote mechanics in the form “fix the Lagrangian $L=T-U$, and the motion is determined so that the action $S=\int L\,\dd t$ is stationary ($\delta S=0$)” (Hamilton's principle). This formalism can be used in special relativity just as it stands. All we need is to find an action that does not change under the Lorentz transformation.

58.9.1 The action and Lagrangian of a free particle

The action $S$ must be a quantity that takes the same value no matter which inertial frame it is computed in (a Lorentz invariant). Otherwise, the “principle that determines the motion” would change from one inertial frame to another. Among the Lorentz invariants that can be built from the trajectory that a free particle of mass $m$ traces out in spacetime (the world line, Section 56.3), the simplest is the proper time $\tau=\int\dd\tau$ measured along the world line (Section 56.5). We take as the action of a free particle the proper time multiplied by the constant $-mc^2$, which adjusts the dimension to that of action $[\mathrm{J\,s}]$:

(the sign of the coefficient was chosen so that the nonrelativistic limit becomes $L=T$, as shown below). Using the fact that the proper time of a moving particle is $\dd\tau=\dd t/\gamma=\sqrt{1-v^2/c^2}\,\dd t$ (Chapter 56), we obtain

which is of the form $S=\int L\,\dd t$. Let us also check the dimensions: $\mathrm{J\,s}$, obtained by multiplying $mc^2\ [\mathrm J]$ by $\dd\tau\ [\mathrm s]$, is the dimension of action. Also, the reason for using the proper time is that, among the Lorentz invariants determined by the world line alone, there is none simpler than this (the length of the world line is nothing other than the proper time multiplied by $c$).

Let us check the nonrelativistic limit. For $v\ll c$, applying $\sqrt{1-x}=1-\dfrac x2-\dfrac{x^2}8-\cdots$ (Taylor expansion, University Mathematics, Section 4.2) with $x=v^2/c^2$, we obtain

The first term $-mc^2$ is a constant, and because it enters Lagrange's equations of motion only through differentiation, it does not affect the motion (adding a constant does not change the meaning of $L$). The second term corresponds to the kinetic energy $\frac12mv^2$ we have used until now. The third and later terms are relativistic corrections (Figure 58.7).

58.9.2 Momentum and the Hamiltonian

Apply the definition of the generalized momentum $p=\partial L/\partial\dot q$ (Section 43.2) to Eq. \eqref{eq:58-L-free}. In one dimension, with $q=x$ and $\dot x=v$, differentiating $L=-mc^2(1-v^2/c^2)^{1/2}$ with respect to $v$ (differentiating a composite function) gives

which agrees with the relativistic momentum $\gamma m\bm v$ obtained from the four-momentum in Chapter 57 (in three dimensions as well, $\bm p=\gamma m\bm v$). Next, computing the Hamiltonian (Section 45.1) $H=pv-L$, we obtain

(we used $1/\gamma=\gamma(1-\beta^2)$). The Hamiltonian has turned out to be the total energy $E=\gamma mc^2$. Furthermore, to write $H$ as a function of $p$ rather than $v$ (this is the meaning of the Legendre transformation), we can use the relation $E^2=p^2c^2+m^2c^4$ from Chapter 57, which gives:

The Hamiltonian $H=\sqrt{p^2c^2+m^2c^4}$ becomes $H\approx mc^2+p^2/(2m)$ for $v\ll c$, returning to the case $U=0$ of the nonrelativistic $H=T+U$ (Section 45.3) (the first term $mc^2$ is a constant).

For a free particle, Lagrange's equation of motion $\dfrac{\dd}{\dd t}\pdiff{L}{v}=\pdiff{L}{x}$ has $\partial L/\partial x=0$, so $\dd(\gamma mv)/\dd t=0$; that is, the momentum is constant. When there is an external force ($L=-mc^2/\gamma-U(x)$), it reads $\dd(\gamma mv)/\dd t=-\partial U/\partial x$, which is the relativistic version of “$\dd\bm p/\dd t=\bm F$” ($\bm p=\gamma m\bm v$).

58.9.3 A particle in an electromagnetic field (Advanced)

The Lagrangian of a particle of charge $q$ in an electromagnetic field is the free-particle Lagrangian plus an interaction term with the potentials:

In terms of the four-potential, the action can be written as $S=-mc^2\int\dd\tau-q\int A_\mu\dd x^\mu$: since $A_\mu\dd x^\mu=(\varphi/c)(c\,\dd t)-\bm A\cdot\dd\bm x=(\varphi-\bm A\cdot\bm v)\dd t$, we have $-qA_\mu\dd x^\mu=(-q\varphi+q\bm v\cdot\bm A)\dd t$, which agrees with Eq. \eqref{eq:58-L-em}. Since $A_\mu\dd x^\mu$ is the inner product of four-vectors and hence Lorentz invariant, this action is also invariant.

Let us check, for the $x$ component, that the Lorentz force emerges from this Lagrangian (here $\varphi$ and $\bm A$ are the values evaluated at the particle's position $\bm x(t)$).

For Eq. \eqref{eq:58-L-em}, the canonical momentum is $\bm P=\partial L/\partial\bm v=\gamma m\bm v+q\bm A$ (different from the mechanical momentum $\gamma m\bm v$), and the Hamiltonian is $H=\bm P\cdot\bm v-L=\gamma mc^2+q\varphi=\sqrt{c^2(\bm P-q\bm A)^2+m^2c^4}+q\varphi$. That is, the interaction with the electromagnetic field is introduced simply by replacing, in the free-particle $H=\sqrt{p^2c^2+m^2c^4}$, $\bm p\to\bm P-q\bm A$ and $E\to H-q\varphi$. This replacement (minimal coupling) is used just as it is when treating an electron in an electromagnetic field in relativistic quantum mechanics, which we touch on in the next section.

58.10 From $E^2=p^2c^2+m^2c^4$ to Relativistic Quantum Mechanics — A Look Back at the Whole Book

58.10.1 From $E^2=p^2c^2+m^2c^4$ to relativistic quantum mechanics

Several of the equations in this chapter lead directly into quantum mechanics. Let us close by previewing that direction in just a few words (this book does not treat quantum mechanics itself).

The relation $E^2=p^2c^2+m^2c^4$ of Chapter 57 becomes $E^2=p^2+m^2$ in natural units, and $p_\mu p^\mu=m^2c^2$ when written with the four-momentum. In quantum mechanics, energy and momentum are replaced by operators (rules that act on a function to produce another function; for example, $\partial_t$ itself) ($E\to i\hbar\,\partial_t$, $\bm p\to-i\hbar\nabla$; written in four-vector form, $p^\mu\to i\hbar\,\partial^\mu$). Substituting these into $E^2=p^2c^2+m^2c^4$, letting the result act on a wave function $\psi$, and dividing by $\hbar^2c^2$, we obtain

This equation is the Klein–Gordon equation. Here $\Box$ is exactly the d'Alembertian defined in this chapter (Definition 58.3), and when the mass is zero the equation returns to the wave equation $\Box\psi=0$. In natural units it can be written neatly as $(\Box+m^2)\psi=0$. From the same starting point one constructs the Dirac equation for particles that, like the electron, have spin (the intrinsic angular momentum that a particle has, as if it were rotating about its own axis; $\hbar/2$ for the electron), and it predicts the existence of antiparticles (particles with the same mass and opposite charge; the positron for the electron). Furthermore, an electron in an electromagnetic field is treated by the replacement corresponding to the minimal coupling of Section 58.9 ($\bm p\to\bm P-q\bm A$, $E\to H-q\varphi$). In this way, the four-potential of this chapter and the way of writing equations in natural units become, just as they are, the common language of particle physics. We leave the detailed meanings of operators, spin, antiparticles, and so on to the quantum mechanics materials; here it is enough to know their names.

You can experience what lies beyond with simulators. The Relativistic Quantum Mechanics Simulator (Relativistic-QM-simulator) places side by side, and compares, the dispersion relations of the Schrödinger equation, which starts from $E=p^2/2m$, and of the Klein–Gordon and Dirac equations, which start from $E^2=p^2c^2+m^2c^4$, and shows how a particle in a box and a free electron gas (from metals to white dwarfs) become “relativistic.” The Elementary Particle Simulator (elementary-particles-simulator) takes the “correspondence between length and energy” seen in Figure 58.2 of this chapter as a yardstick, and lays out side by side the hierarchy that continues through atoms, nuclei, and quarks, together with gravity, the electromagnetic force, the weak force, and the strong force. The full-fledged study that follows—the kinematics of scattering and decay with four-momentum, field theory written in natural units, and the Standard Model of particle physics—is covered in the companion volume “Particle Physics” (AD 09).

58.10.2 Looking back at the whole book

Now that we have finished Part VIII, let us look back over this book (Parts I–VIII, 58 chapters in all). The following table is an overview of the contents of the eight parts and of how they connect.

Following the rightmost column of the table, we see that this book is not “eight separate subjects” but is tied together by one thick thread.

• From the equation of motion to the principle of action. In Part I we solved Newton's equation of motion as a differential equation; in Part VI we rewrote it as the stationarity condition $\delta S=0$ on the action; and in Part VIII we rewrote it once more as the relativistic action $S=-mc^2\int\dd\tau$. The Hamiltonian $H$ is the energy itself, and in quantum mechanics it becomes an operator.
• From the wave equation to the speed of light, and then to spacetime. The wave equation of Part II became the equation of electromagnetic waves in Part V, where we derived that the speed of the wave is $c=1/\sqrt{\varepsilon_0\mu_0}$. From the postulate that $c$ is the same in every inertial frame arose the Lorentz transformation and the idea of spacetime in Part VIII, and in this chapter we arrived at the four-dimensional wave equation $\Box A^\mu=\mu_0j^\mu$.
• The micro and the macro. The quantities treated by the thermodynamics of Part IV (temperature, entropy, free energy) were derived from the motion of molecules in the statistical mechanics of Part VII. Black-body radiation (Chapter 53) showed that light behaves as quanta (photons), and this is the entrance to quantum mechanics.
• The strength of materials. The stress, strain, and fracture mechanics of Part III extend the balance of forces of Part I and form a foundation of materials science. The energy release rate and Griffith theory use the idea of free energy from thermodynamics.

The physical constants used in this book are collected in Appendix B (appB_constants.html), and the solutions to the exercises in Appendix A (appA_solutions.html). If, after reading the whole book, you want to review according to your goals, we recommend the following orders.

• To consolidate mechanics and analytical mechanics: Part I (Mechanics) → Part VI (Analytical Mechanics) → Part VIII (Special Relativity, especially Section 58.9 of Chapter 58).
• To connect electromagnetism and relativity: Part II (Waves) → Part V (Electromagnetism) → Part VIII (Chapters 54–58).
• With a focus on heat and statistics: Part IV (Thermodynamics) → Part VII (Statistical Mechanics) → Chapter 53 (Photons and Phonons).
• With a focus on the strength of materials: Part III (Mechanics of Materials and Fracture Mechanics) → Part IV (Thermodynamics, surface energy and free energy) → Part VII (Statistical Mechanics; Weibull statistics is in Chapter 22).

58.11 Summary and Exercises

58.11.1 Summary

• Natural units: If we set $c=\hbar=1$ (and $k_{\mathrm B}=1$), mass, momentum, energy, and temperature are expressed in $\mathrm{eV}$, and length and time in $\mathrm{eV}^{-1}$ (Definition 58.1, Table 58.1). The relation $\hbar c=197.327\ \mathrm{MeV\,fm}$ converts between $\mathrm{fm}$ and $\mathrm{MeV}$ (Eq. \eqref{eq:58-hbarc}), and units are restored by dimensional analysis (Formula 58.1). $E^2=p^2c^2+m^2c^4$ becomes $E^2=p^2+m^2$.
• Four-gradient and d'Alembertian: $\partial_\mu=(\frac1c\partial_t,\nabla)$ transforms as a four-vector with a lower index (Theorem 58.1), and $\Box=\partial_\mu\partial^\mu=\frac1{c^2}\partial_t^2-\nabla^2$ is Lorentz invariant (Theorem 58.2). The form of the wave equation $\Box\psi=0$ is unchanged by the Lorentz transformation, which resolves the problem of the speed of light that arose when this form broke down under the Galilean transformation. The wave four-vector of a plane wave is $k^\mu=(\omega/c,\bm k)$, with $k_\mu k^\mu=0$.
• Four-current $j^\mu=(c\rho,\bm j)=\rho_0u^\mu$. Conservation of charge is $\partial_\mu j^\mu=0$ (Eq. \eqref{eq:58-continuity-cov}).
• Four-potential $A^\mu=(\varphi/c,\bm A)$ (Definition 58.5). $\bm E=-\nabla\varphi-\partial_t\bm A$ and $\bm B=\rot\bm A$ (Theorem 58.3). Under the gauge transformation $A^\mu\to A^\mu-\partial^\mu\chi$, $\bm E,\bm B$ are invariant.
• Field-strength tensor $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$: an antisymmetric $4\times4$ matrix, with $F^{0i}=-E_i/c$, $F^{12}=-B_z$, and so on (Eq. \eqref{eq:58-Fmatrix}). The invariants are $E^2-c^2B^2$ and $\bm E\cdot\bm B$ (Theorem 58.4).
• Maxwell's equations reduce to two: $\partial_\mu F^{\mu\nu}=\mu_0j^\nu$ (Gauss's law and the Ampère–Maxwell law, Law 58.1) and $\partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0$ ($\operatorname{div}\bm B=0$ and Faraday's law). Under the Lorenz condition $\partial_\mu A^\mu=0$, we have $\Box A^\mu=\mu_0j^\mu$.
• Transformation of the electric and magnetic fields (Theorem 58.5): Under a Lorentz transformation in the $x$ direction, $E_x,B_x$ are invariant, while the perpendicular components mix, as in $E'_y=\gamma(E_y-vB_z)$. For an observer moving along with the charge, the magnetic force on a charge near a current-carrying wire is an electric force due to the electric charging of the wire.
• Covariant Lorentz force $\dd p^\mu/\dd\tau=qF^{\mu\nu}u_\nu$ (Law 58.2). The radius of the trajectory of a charged particle of momentum $p$ is $r=p/(qB)$, and $p\ [\mathrm{GeV}/c]=0.2998\,B\ [\mathrm{T}]\,r\ [\mathrm{m}]$ (Formula 58.2).
• Lagrangian of a free particle $L=-mc^2/\gamma$ (Formula 58.3) gives $\bm p=\gamma m\bm v$ and $H=\gamma mc^2=\sqrt{p^2c^2+m^2c^4}$. For $v\ll c$, $L\approx-mc^2+\frac12mv^2$. In an electromagnetic field, $L=-mc^2/\gamma-q\varphi+q\bm v\cdot\bm A$.
• Applying $p^\mu\to i\hbar\partial^\mu$ to $E^2=p^2c^2+m^2c^4$ gives the Klein–Gordon equation $(\Box+m^2c^2/\hbar^2)\psi=0$.

Related simulator: Relativistic Quantum Mechanics Simulator (dispersion relations of the equations that start from $E^2=p^2c^2+m^2c^4$, a particle in a box, and the relativistic treatment of the free electron gas), Elementary Particle Simulator (the hierarchy from atoms to elementary particles seen on scales of length and energy, and the four forces).

58.11.2 Exercises

58.11.3 References