(a) Show $\vec E\!\cdot\!\vec B$ is Lorentz-invariant. (b) Frame-independence of $\vec E\perp\vec B$. (c) For free plane waves, $\vec E\perp\vec B$. (d) $\vec E\times\vec B = A\vec k$ with $A\neq 0$.

(a) $\vec E\!\cdot\!\vec B$ is Lorentz-invariant

The dual field strength tensor is $\tilde F^{\mu\nu} = \tfrac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$. Form the scalar contraction

$$F_{\mu\nu}\tilde F^{\mu\nu} = \tfrac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} = -(4/c)\vec E\!\cdot\!\vec B.$$

(The factor follows from expanding $F$ in terms of $\vec E,\vec B$ components.) Since $\epsilon^{\mu\nu\rho\sigma}$ is a Lorentz pseudotensor and $F$ is a tensor, the contraction is a Lorentz pseudoscalar — identical numerical value in every (proper, orthochronous) Lorentz frame. Hence $\vec E\!\cdot\!\vec B$ is invariant.

(b) Orthogonality is frame-independent

$\vec E\perp\vec B \iff \vec E\!\cdot\!\vec B = 0$. By (a), if this vanishes in one frame, it vanishes in all. Orthogonality of $\vec E$ and $\vec B$ is a frame-independent statement.

(c) Apply to free plane waves

By Problem 1.121, $\vec E\!\cdot\!\vec B = 0$ for any free EM plane wave (in any frame). Alternative direct derivation: $\vec E\!\cdot\!\vec B = (\vec k\varepsilon^0 - \omega\vec\varepsilon)\!\cdot\!(-\vec k\times\vec\varepsilon) = -\varepsilon^0\vec k\!\cdot\!(\vec k\times\vec\varepsilon) + \omega\vec\varepsilon\!\cdot\!(\vec k\times\vec\varepsilon) = 0 + 0 = 0$ by triple-product identities.

(d) $\vec E\times\vec B \propto \vec k$

Substitute the plane-wave forms:

$$\vec E\times\vec B = (\vec k\varepsilon^0 - \omega\vec\varepsilon)\times(-\vec k\times\vec\varepsilon)\cdot(-i)^2 e^{2ik\cdot x}.$$

Apply BAC-CAB: $\vec A\times(\vec B\times\vec C) = \vec B(\vec A\!\cdot\!\vec C) - \vec C(\vec A\!\cdot\!\vec B)$.

Term 1: $\vec k\varepsilon^0\times(-\vec k\times\vec\varepsilon) = -\varepsilon^0[\vec k(\vec k\!\cdot\!\vec\varepsilon) - \vec\varepsilon(\vec k\!\cdot\!\vec k)] = -\varepsilon^0[(\omega\varepsilon^0/c)\vec k - |\vec k|^2\vec\varepsilon]$.

Term 2: $-\omega\vec\varepsilon\times(-\vec k\times\vec\varepsilon) = \omega[\vec k|\vec\varepsilon|^2 - \vec\varepsilon(\vec k\!\cdot\!\vec\varepsilon)] = \omega[\vec k|\vec\varepsilon|^2 - \vec\varepsilon(\omega\varepsilon^0/c)]$.

Sum: the $\vec\varepsilon$-coefficient is $\varepsilon^0|\vec k|^2 - \omega^2\varepsilon^0/c = \varepsilon^0(|\vec k|^2 - (\omega/c)^2) = 0$ by null dispersion. The $\vec k$-coefficient survives: $-(\omega/c)(\varepsilon^0)^2 + \omega|\vec\varepsilon|^2 = \omega[|\vec\varepsilon|^2 - (\varepsilon^0)^2/c]$. With careful unit choice ($c=1$): $\omega(\vec\varepsilon\!\cdot\!\vec\varepsilon - (\varepsilon^0)^2) = -\omega\varepsilon_\mu\varepsilon^\mu$. For physical (spacelike) polarisations $\varepsilon^2 < 0$, so $A = -\omega\varepsilon^2 > 0$.

$$\boxed{\;\vec E\times\vec B = A\vec k,\quad A \neq 0.\;}$$

Poynting vector $\vec S = \vec E\times\vec B/\mu_0\propto\vec k$ — energy streams in the propagation direction, as physical intuition demands.

$\vec B = B\hat z$, no electric field. Electron has $\vec u = u\hat x$ at $t=0$. Find the trajectory.

Constants of motion

The Lorentz force $\vec F = -e\vec v\times\vec B$ is perpendicular to $\vec v$, so it does no work: $\vec F\!\cdot\!\vec v = 0$. Therefore the particle's energy — and hence $\gamma$ and $|\vec v|$ — are conserved.

Equations of motion

$$\frac{d\vec p}{dt} = -e\vec v\times\vec B,\qquad \vec p = \gamma m_0\vec v.$$

Since $\gamma$ is constant, this is equivalent to $\gamma m_0\,d\vec v/dt = -e\vec v\times\vec B$. With $\vec B = B\hat z$:

$$\gamma m_0\dot v_x = -eB\,v_y,\qquad \gamma m_0\dot v_y = +eB\,v_x.$$

(Negative charge $-e$ flipped the standard sign.) Define

$$\omega_c \equiv \frac{eB}{\gamma m_0}\quad\text{(relativistic cyclotron frequency)}.$$

Solution

Coupled linear system. Decouple by introducing $w = v_x + i v_y$: $\dot w = i\omega_c w$, so $w(t) = u e^{i\omega_c t}$ (with $w(0) = u$ from initial $\vec v(0) = u\hat x$).

$v_x = u\cos\omega_c t$, $v_y = u\sin\omega_c t$. Integrating:

$$\boxed{\;\vec r(t) = R(\sin\omega_c t,\;1 - \cos\omega_c t,\;0),\quad R = u/\omega_c = \gamma m_0 u/(eB) = p/(eB).\;}$$

Circle of radius $R$, centred at $(0, R, 0)$, counterclockwise viewed from $+\hat z$.

The universal radius formula

$$\boxed{\;R = p/(eB).\;}$$

Holds at all energies, the foundational formula for magnetic-spectrometer design and synchrotrons. Numerical mnemonic: $R[\text{m}] = p[\text{GeV}/c]/(0.3\,B[\text{T}])$ — e.g., a 1 GeV proton in $B = 1$ T orbits at $R \approx 3.3$ m. The LHC dipoles ($B = 8.3$ T) bend 7 TeV protons at $R \approx 2.8$ km — matching the actual machine radius.

Newtonian limit

$\gamma\to 1$: $\omega_c\to eB/m_0$, the classical cyclotron frequency. Independent of $u$ (Newtonian!) but acquires energy-dependence relativistically through $\gamma$ — the reason classical cyclotrons stop working above ~10 MeV and synchrotrons (variable $\omega$) take over.

In $K$: $\vec E = (cB,0,0)$, $\vec B = (B,0,0)$. In $K'$: $\vec E' = (0,2cB,cB)$ and $B'_x = 0$. Find $B'_y$, $B'_z$.

The two field invariants

Recall

$$I_1 = |\vec E|^2 - c^2|\vec B|^2 = -\tfrac{c^2}{2}F_{\mu\nu}F^{\mu\nu},\qquad I_2 = \vec E\!\cdot\!\vec B = -\tfrac{c}{4}F_{\mu\nu}\tilde F^{\mu\nu}.$$

Both are Lorentz scalars (resp. pseudoscalars) — same value in $K$ and $K'$.

Invariants in $K$

$\vec E = (cB, 0, 0)$, $\vec B = (B, 0, 0)$ (parallel, $|\vec E| = c|\vec B|$):

$$I_1 = c^2 B^2 - c^2 B^2 = 0,\qquad I_2 = (cB)(B) = c B^2.$$

Set up equations in $K'$

$\vec E' = (0, 2cB, cB)$, $B'_x = 0$, so $\vec B' = (0, B'_y, B'_z)$. Match invariants:

$$I_1: \quad |\vec E'|^2 - c^2|\vec B'|^2 = 5c^2 B^2 - c^2(B'_y{}^2 + B'_z{}^2) = 0 \Rightarrow B'_y{}^2 + B'_z{}^2 = 5 B^2.$$

$$I_2: \quad \vec E'\!\cdot\!\vec B' = 2cB\cdot B'_y + cB\cdot B'_z = cB^2 \Rightarrow 2B'_y + B'_z = B.$$

Solve the system

From $I_2$: $B'_z = B - 2B'_y$. Substitute into $I_1$: $B'_y{}^2 + (B - 2B'_y)^2 = 5B^2$, expand: $5B'_y{}^2 - 4B\,B'_y + B^2 = 5B^2$, so $5B'_y{}^2 - 4B\,B'_y - 4B^2 = 0$.

Quadratic formula: $B'_y = (4B\pm\sqrt{16B^2 + 80B^2})/10 = (4B\pm\sqrt{96}\,B)/10 = (4\pm 4\sqrt 6)\,B/10 = 2(1\pm\sqrt 6)\,B/5$.

$$\boxed{\;B'_y = \frac{2(1\pm\sqrt 6)}{5}B,\quad B'_z = \frac{1\mp 4\sqrt 6}{5}B.\;}$$

The $\pm$ pair

The two solutions correspond to two distinct points on the same Lorentz orbit — reached by boosts that differ in the relative orientation of $\vec v$ and the $(yz)$-plane decomposition. Invariants alone do not uniquely determine $\vec B'$; the boost direction provides the missing information.

$\vec E = (\alpha,-\alpha,0)$, $\vec B = (0,0,2\alpha/c)$. In another frame: $\vec E' = (0,0,2\alpha)$, $\vec B' = (B'_x,\alpha/c,B'_z)$. Find $B'_x,B'_z$.

Invariants in $S$ (natural units $c=1$)

$\vec E = (\alpha, -\alpha, 0)$: $|\vec E|^2 = 2\alpha^2$. $\vec B = (0, 0, 2\alpha)$: $|\vec B|^2 = 4\alpha^2$ (in these natural units, treat $c$ as 1 then restore).

$$I_1 = |\vec E|^2 - c^2|\vec B|^2 = 2\alpha^2 - 4\alpha^2 = -2\alpha^2\quad\text{(magnetic-like)}.$$

$$I_2 = \vec E\!\cdot\!\vec B = 0\cdot\alpha + 0\cdot(-\alpha) + 0\cdot 0 = 0.$$

Apply to $S'$

$\vec E' = (0, 0, 2\alpha)$, $\vec B' = (B'_x, \alpha/c, B'_z)$.

$I_2 = \vec E'\!\cdot\!\vec B' = 0\cdot B'_x + 0\cdot(\alpha/c) + 2\alpha\cdot B'_z = 2\alpha B'_z$. Set equal to 0: $B'_z = 0$.

$I_1 = |\vec E'|^2 - c^2|\vec B'|^2 = 4\alpha^2 - c^2(B'_x{}^2 + \alpha^2/c^2 + 0) = 4\alpha^2 - c^2 B'_x{}^2 - \alpha^2 = 3\alpha^2 - c^2 B'_x{}^2$. Set equal to $-2\alpha^2$: $c^2 B'_x{}^2 = 5\alpha^2$.

$$\boxed{\;B'_x = \pm\sqrt 5\,\alpha/c,\quad B'_z = 0.\;}$$

Magnetic-like Lorentz class

$I_1 < 0$ means the field is in the magnetic-like Lorentz class — there exists a frame where $\vec E = 0$ but $\vec B \neq 0$. (Magnetic-like fields cannot be Lorentz-transformed to pure electric.) The same machinery extends to monopole + dipole configurations (Problem 1.141).

$\vec E = (0,\beta,-\beta)$, $\vec B = (2\beta,0,0)$. In $K'$: $\vec E' = (2\beta,0,0)$, $\vec B' = (B'_x,B'_y,\beta/c)$. Find $B'_x,B'_y$.

Invariants in $K$

$\vec E = (0, \beta, -\beta)$, $\vec B = (2\beta, 0, 0)$. In natural units ($c=1$):

$$|\vec E|^2 = 2\beta^2,\quad |\vec B|^2 = 4\beta^2,\quad I_1 = 2\beta^2 - 4\beta^2 = -2\beta^2.$$

$$\vec E\!\cdot\!\vec B = 0\cdot 2\beta + \beta\cdot 0 + (-\beta)\cdot 0 = 0\Rightarrow I_2 = 0.$$

Apply to $K'$

$\vec E' = (2\beta, 0, 0)$, $\vec B' = (B'_x, B'_y, \beta/c)$ (in SI; in natural units just $\beta$).

$I_2 = \vec E'\!\cdot\!\vec B' = 2\beta\cdot B'_x + 0 + 0 = 2\beta B'_x = 0\Rightarrow B'_x = 0$.

$I_1 = |\vec E'|^2 - c^2|\vec B'|^2 = 4\beta^2 - c^2(0 + B'_y{}^2 + \beta^2/c^2) = 3\beta^2 - c^2 B'_y{}^2$. Set $= -2\beta^2$: $c^2 B'_y{}^2 = 5\beta^2$.

$$\boxed{\;B'_x = 0,\quad B'_y = \pm\sqrt 5\,\beta/c\;\text{(in SI)},\;\pm\sqrt 5\,\beta\;\text{(natural units)}.\;}$$

Class identification

$I_1 < 0$: magnetic-like Lorentz class. Same structure as Problem 1.125 — demonstrates the systematic power of using the two field invariants to constrain the boost-transformation outputs without solving the full $\Lambda^\mu{}_\nu$ system explicitly.

$\vec E = (\alpha,0,0)$, $\vec B = (\alpha/c,0,2\alpha/c)$ in $A$. In $B$: $\vec E' = (E'_x,\alpha,0)$, $\vec B' = (\alpha/c,B'_y,\alpha/c)$. Find $E'_x,B'_y$. Then boost to $C$ along $+\hat x$ at velocity $v$.

Invariants in $A$

$\vec E = (\alpha, 0, 0)$, $\vec B = (\alpha/c, 0, 2\alpha/c)$. Use natural units where convenient.

$$|\vec E|^2 = \alpha^2,\quad c^2|\vec B|^2 = c^2(\alpha^2/c^2 + 4\alpha^2/c^2) = 5\alpha^2.$$

$$I_1 = \alpha^2 - 5\alpha^2 = -4\alpha^2.$$

$$\vec E\!\cdot\!\vec B = \alpha\cdot\alpha/c + 0 + 0 = \alpha^2/c\Rightarrow I_2 = \alpha^2/c.$$

Determine $E'_x, B'_y$ in frame $B$

$\vec E' = (E'_x, \alpha, 0)$, $\vec B' = (\alpha/c, B'_y, \alpha/c)$.

$I_1$: $E'_x{}^2 + \alpha^2 + 0 - c^2(\alpha^2/c^2 + B'_y{}^2 + \alpha^2/c^2) = E'_x{}^2 + \alpha^2 - 2\alpha^2 - c^2 B'_y{}^2 = E'_x{}^2 - \alpha^2 - c^2 B'_y{}^2$. Set $= -4\alpha^2$:

$$E'_x{}^2 - c^2 B'_y{}^2 = -3\alpha^2.$$

$I_2$: $E'_x\cdot\alpha/c + \alpha\cdot B'_y + 0 = (\alpha E'_x + \alpha c B'_y)/c = \alpha^2/c$, so

$$E'_x + c B'_y = \alpha\;\Longleftrightarrow\;E'_x = \alpha - c B'_y.$$

Substitute: $(\alpha - cB'_y)^2 - c^2 B'_y{}^2 = -3\alpha^2 \Rightarrow \alpha^2 - 2\alpha c B'_y + c^2 B'_y{}^2 - c^2 B'_y{}^2 = -3\alpha^2 \Rightarrow -2\alpha c B'_y = -4\alpha^2$. Therefore

$$\boxed{\;B'_y = 2\alpha/c,\quad E'_x = -\alpha.\;}$$

Boost $B\to C$ along $+\hat x$

Use field-transformation laws (parallel components unchanged; perpendicular components mix):

$$E_x^{''} = E'_x = -\alpha,\quad B_x^{''} = B'_x = \alpha/c.$$

$$E_y^{''} = \gamma(E'_y - v B'_z) = \gamma(\alpha - v\alpha/c) = \gamma\alpha(1 - \beta).$$

$$E_z^{''} = \gamma(E'_z + v B'_y) = \gamma(0 + v\cdot 2\alpha/c) = 2\gamma\alpha\beta.$$

$$B_y^{''} = \gamma(B'_y + v E'_z/c^2) = \gamma(2\alpha/c + 0) = 2\gamma\alpha/c.$$

$$B_z^{''} = \gamma(B'_z - v E'_y/c^2) = \gamma(\alpha/c - v\alpha/c^2) = \gamma\alpha(1-\beta)/c.$$

Summarising:

$$\vec E'' = (-\alpha,\;\gamma\alpha(1-\beta),\;2\gamma\alpha\beta),\quad \vec B'' = (\alpha/c,\;2\gamma\alpha/c,\;\gamma\alpha(1-\beta)/c).$$

Cross-check

$|\vec E''|^2 - c^2|\vec B''|^2 = \alpha^2 + \gamma^2\alpha^2(1-\beta)^2 + 4\gamma^2\alpha^2\beta^2 - \alpha^2 - 4\gamma^2\alpha^2 - \gamma^2\alpha^2(1-\beta)^2$. Simplify: $4\gamma^2\alpha^2\beta^2 - 4\gamma^2\alpha^2 = -4\gamma^2\alpha^2(1-\beta^2) = -4\alpha^2$ — matches $I_1$. ✓

A muon descends vertically from 10 km altitude through $B = 10\,\mu$T (south-to-north). Energy 2 GeV, negative charge. Compute the lateral deviation.

Force direction

Velocity $\vec v = -v\hat z$ (downward); field $\vec B = B\hat y$ (south-to-north, taking north as $+\hat y$); charge $-e$.

$$\vec F = q\vec v\times\vec B = (-e)(-v\hat z)\times B\hat y = +evB(\hat z\times\hat y) = -evB\hat x.$$

So the force is along $-\hat x$ — westward (taking east as $+\hat x$).

Cyclotron radius

From Problem 1.123, $R = p/(eB)$. For a 2 GeV muon:

$$R = \frac{2\,\text{GeV}/c}{e\cdot 10^{-5}\,\text{T}} = \frac{2\,\text{GeV}}{e\cdot 10^{-5}\,\text{T}\cdot c}.$$

Numerical shortcut: $R[\text{m}] = p[\text{GeV}/c]/(0.3\,B[\text{T}]) = 2/(0.3\cdot 10^{-5}) \approx 6.7\times 10^5$ m $= 667$ km.

Sagitta over a 10 km drop

For a circular arc of radius $R$ subtended by chord $L = \Delta z$, the sagitta (lateral offset at midpoint, or full offset at endpoint for a short arc with one endpoint tangent to the chord):

$$\Delta x \approx L^2/(2R) = (10\,\text{km})^2/(2\cdot 667\,\text{km}) = 100/1334\,\text{km} \approx 75\,\text{m}.$$

$$\boxed{\;|\Delta x| \approx 75\text{ m westward.}\;}$$

Muon survival check

2 GeV muon: $\gamma = E/m_\mu c^2 \approx 2000/105.66 \approx 18.9$. Mean decay length: $\beta\gamma c\tau_\mu \approx \gamma c\tau_\mu \approx 18.9\cdot 659\,\text{m} \approx 12.5$ km. Exceeds the 10 km descent — most muons survive to the ground.

Practical relevance

Cosmic-ray and atmospheric-neutrino detectors must correct for this geomagnetic deflection when reconstructing the parent particle direction. The 75 m offset at 10 km distance corresponds to ~0.4 degrees deflection — relevant for sub-GeV experiments and at all energies for precision pointing (Super-Kamiokande, IceCube).

Observer in $S$ has $\vec E, \vec B$. Boost along $+\hat x_1$ at velocity $u$ to $S'$ where $\vec B' = 0$. Express $\vec E'$ in terms of $\vec E$ and $u$.

Constraint from $\vec B' = 0$

The field-transformation law for $\vec B$ along a boost at $\vec u$:

$$\vec B' = \gamma\left(\vec B - \frac{\vec u\times\vec E}{c^2}\right)_\perp + \vec B_\parallel.$$

$\vec B' = 0$ forces both pieces to vanish: $\vec B_\parallel = 0$ and $\vec B_\perp = \vec u\times\vec E/c^2$.

Since $\vec u\times\vec E$ has no component along $\vec u$, $\vec B = \vec u\times\vec E/c^2$ is automatically perpendicular to $\vec u$, consistent. This is the electric-like Lorentz class (Problem 1.121 in reverse): if a frame exists where $\vec B$ vanishes, the original field must satisfy $\vec E\!\cdot\!\vec B = 0$ and $|\vec E|^2 > c^2|\vec B|^2$.

Transform $\vec E$

Parallel and perpendicular components transform as:

$$\vec E'_\parallel = \vec E_\parallel,\quad \vec E'_\perp = \gamma(\vec E_\perp + \vec u\times\vec B).$$

Substitute $\vec B = \vec u\times\vec E/c^2$:

$$\vec u\times\vec B = \vec u\times(\vec u\times\vec E)/c^2 = [\vec u(\vec u\!\cdot\!\vec E) - \vec E\,u^2]/c^2 = -\vec E_\perp u^2/c^2$$

(using $\vec u(\vec u\!\cdot\!\vec E) = \vec u\cdot u\,E_\parallel = u^2\vec E_\parallel/u\cdot u/u\cdot u$... cleaner: $\vec u\!\cdot\!\vec E = u E_\parallel$ where $E_\parallel = \vec E\!\cdot\!\hat u$, so $\vec u(\vec u\!\cdot\!\vec E) = u E_\parallel\vec u = u^2 E_\parallel\hat u = u^2\vec E_\parallel$). So $\vec u\times\vec B = u^2(\vec E_\parallel - \vec E)/c^2 = -u^2\vec E_\perp/c^2$.

Therefore $\vec E'_\perp = \gamma(\vec E_\perp - u^2\vec E_\perp/c^2) = \gamma\vec E_\perp(1 - u^2/c^2) = \vec E_\perp/\gamma$.

$$\boxed{\;\vec E'_\parallel = \vec E_\parallel,\quad \vec E'_\perp = \vec E_\perp/\gamma.\;}$$

Cross-check invariants

$\vec E\!\cdot\!\vec B = \vec E\!\cdot\!(\vec u\times\vec E)/c^2 = 0$ (since $\vec u\times\vec E\perp\vec E$). ✓

In $S$: $\vec E = 0$, $\vec B$ constant. Boost along $+\hat x$ at $v$. Find $\vec E',\vec B'$; verify invariants.

Field transformations with $\vec E = 0$

Apply the standard transformation laws under boost at $v\hat x$:

$$\vec E'_\parallel = \vec E_\parallel = 0,\quad \vec E'_\perp = \gamma(\vec E_\perp + \vec v\times\vec B)_\perp = \gamma(\vec v\times\vec B).$$

(Since $\vec v\times\vec B$ has no $\hat x$ component, it is purely perpendicular.)

$$\vec B'_\parallel = B_x\hat x,\quad \vec B'_\perp = \gamma(\vec B_\perp - 0) = \gamma\vec B_\perp.$$

$$\boxed{\;\vec E' = \gamma\vec v\times\vec B,\quad \vec B' = \vec B_\parallel + \gamma\vec B_\perp.\;}$$

Physical interpretation

The induced $\vec E'$ is the relativistic generalisation of the motional EMF. A wire moving through a magnetic field at $\vec v$ experiences an effective electric force per unit charge $\vec v\times\vec B$ on the comoving charge carriers — the fundamental mechanism of an electric generator. At low speeds $\gamma\approx 1$ and we recover the textbook motional-EMF expression.

Invariant check

$\vec B = (B_x, B_y, B_z)$. $\vec v\times\vec B = v\hat x\times(B_x\hat x + B_y\hat y + B_z\hat z) = v(B_y\hat z - B_z\hat y)\cdot(-1)$—careful with sign: $\hat x\times\hat y = \hat z$, $\hat x\times\hat z = -\hat y$, so $\vec v\times\vec B = v B_y\hat z - v B_z\hat y$.

$$|\vec E'|^2 = \gamma^2 v^2(B_y^2 + B_z^2).$$

$$|\vec B'|^2 = B_x^2 + \gamma^2(B_y^2 + B_z^2).$$

$$I_1 = |\vec E'|^2 - c^2|\vec B'|^2 = \gamma^2 v^2(B_y^2 + B_z^2) - c^2 B_x^2 - c^2\gamma^2(B_y^2 + B_z^2).$$

$= \gamma^2(B_y^2 + B_z^2)(v^2 - c^2) - c^2 B_x^2 = -c^2(B_y^2 + B_z^2) - c^2 B_x^2 = -c^2|\vec B|^2$.

Matches the original $I_1$ (since $\vec E = 0$). ✓

$\vec E'\!\cdot\!\vec B' = (\gamma v B_y\hat z\cdot\gamma B_z) + (-\gamma v B_z\hat y\cdot\gamma B_y) + 0 = \gamma^2 v B_y B_z - \gamma^2 v B_z B_y = 0$. ✓

Electron in a linac of length $L=3$ km (SLAC) with potential $U$. From rest at $t=0$. Find (a) trajectory $x(t)$, (b) traversal time, (c) energy $E(t)$.

Setup

Linear accelerator: potential $U$ across length $L$, giving uniform electric field $\mathcal E = U/L$ along $\hat x$. Force on the electron: $f = e\mathcal E = eU/L$, constant in the lab frame. Electron starts from rest at $t = 0$, $x = 0$.

Lab-frame motion: $p(t)$ and $E(t)$

Newton's relativistic law: $dp/dt = f$, so

$$p(t) = ft.$$

Work-energy: $E(t) = mc^2 + fx(t)$. Combine with mass-shell $E^2 = (pc)^2 + (mc^2)^2$:

$$(mc^2 + fx)^2 = (fct)^2 + (mc^2)^2.$$

(a) Trajectory $x(t)$

Expand and solve: $2mc^2 fx + f^2 x^2 = (fct)^2$. Treat as quadratic in $x$:

$$f x = -mc^2 + \sqrt{(mc^2)^2 + (fct)^2} = mc^2\left[\sqrt{1 + (fct/(mc^2))^2} - 1\right].$$

$$\boxed{\;x(t) = \frac{mc^2}{f}\left[\sqrt{1 + \left(\frac{fct}{mc^2}\right)^{\!2}} - 1\right].\;}$$

This is the standard hyperbolic worldline of constant-proper-force motion (Problem 1.118), in coordinate form.

(b) Traversal time

Set $x = L$ and solve for $t = T$:

$$\sqrt{1 + (fcT/(mc^2))^2} = 1 + fL/(mc^2),\quad\Rightarrow\quad (fcT/(mc^2))^2 = (fL/(mc^2))(2 + fL/(mc^2)).$$

Substitute $fL = eU$:

$$T^2 = \frac{(eU/(mc^2))(2 + eU/(mc^2))(mc^2)^2}{f^2 c^2} = \frac{L^2(2mc^2 + eU)}{eU c^2}.$$

$$\boxed{\;T = \frac{L}{c}\sqrt{1 + \frac{2mc^2}{eU}}.\;}$$

(c) Energy along the way

$$E(t) = \sqrt{(pc)^2 + (mc^2)^2} = \sqrt{(fct)^2 + (mc^2)^2} = \sqrt{(eUct/L)^2 + (mc^2)^2}.$$

$$\boxed{\;E(t) = \sqrt{(mc^2)^2 + (eUct/L)^2}.\;}$$

SLAC numerical context

SLAC linac: $L = 3$ km, $eU\sim 50$ GeV. Then $2mc^2/(eU) \sim 2(0.51\,\text{MeV})/(50\,\text{GeV}) \approx 2\times 10^{-5}$. So $T\approx (L/c)\sqrt{1 + 2\times 10^{-5}}\approx L/c \approx 10$ μs — the electron is ultra-relativistic almost immediately, the rest-mass correction to the transit time is $\sim 10^{-5}\cdot L/c \sim 0.1$ ns — below current synchronisation accuracy.