Special Relativity ā worked problems
Electron in a homogeneous magnetic field ā relativistic cyclotron
$\vec B = B\hat z$, no electric field. Electron has $\vec u = u\hat x$ at $t=0$. Find the trajectory.
The Lorentz force does no work, so $\gamma$ and $|\vec v|$ are constants. Equations $\gamma m_0\dot v_x = -eB v_y$, $\gamma m_0\dot v_y = eB v_x$ give $\omega_c = eB/(\gamma m_0)$ — the relativistic cyclotron frequency.
Trajectory: $\vec r(t) = R(\sin\omega_c t,\,1-\cos\omega_c t,\,0)$ with $R = u/\omega_c = \gamma m_0 u/(eB) = p/(eB)$. Circle of radius $R$, counterclockwise viewed from $+\hat z$.
Sanity. $R = p/(eB)$ — the universal bending-radius formula used in magnetic spectrometers and synchrotron design. Newtonian limit: $\omega_c\to eB/m_0$.
Field invariants: solving for $B'_y, B'_z$
In $K$: $\vec E = (cB,0,0)$, $\vec B = (B,0,0)$. In $K'$: $\vec E' = (0,2cB,cB)$ and $B'_x = 0$. Find $B'_y$, $B'_z$.
Invariants in $K$: $I_1 = |\vec E|^2 - c^2|\vec B|^2 = 0$ (parallel, $|\vec E| = c|\vec B|$), $I_2 = \vec E\!\cdot\!\vec B = cB^2$.
In $K'$: equating gives $(B'_y)^2 + (B'_z)^2 = 5B^2$ and $2B'_y + B'_z = B$. Quadratic: $$\boxed{\;B'_y = \frac{2(1\pm\sqrt 6)}{5}B,\quad B'_z = \frac{1\mp 4\sqrt 6}{5}B.\;}$$ The $\pm$ pair are two distinct points on the same Lorentz orbit.
Field invariants: solving for $B'_x, B'_z$
$\vec E = (\alpha,-\alpha,0)$, $\vec B = (0,0,2\alpha/c)$. In another frame: $\vec E' = (0,0,2\alpha)$, $\vec B' = (B'_x,\alpha/c,B'_z)$. Find $B'_x,B'_z$.
Invariants in $S$: $I_1 = 2\alpha^2 - 4\alpha^2 = -2\alpha^2$ (magnetic-like), $I_2 = 0$. In $S'$: $I_2$ gives $B'_z = 0$; $I_1$ gives $(B'_x)^2 = 5\alpha^2/c^2$. $$\boxed{\;B'_x = \pm\sqrt 5\,\alpha/c,\quad B'_z = 0.\;}$$
Field transformation via invariants: $B'_x, B'_y$
$\vec E = (0,\beta,-\beta)$, $\vec B = (2\beta,0,0)$. In $K'$: $\vec E' = (2\beta,0,0)$, $\vec B' = (B'_x,B'_y,\beta/c)$. Find $B'_x,B'_y$.
In natural units: $|\vec E|^2 = 2\beta^2$, $|\vec B|^2 = 4\beta^2$, $\vec E\!\cdot\!\vec B = 0$. In $K'$: $I_2$ gives $B'_x = 0$; $I_1$ gives $(B'_y)^2 = 5\beta^2$. $$\boxed{\;B'_x = 0,\quad B'_y = \pm\sqrt 5\,\beta\;(/c\text{ in SI}).\;}$$
Three-frame field chain: invariants then boost
$\vec E = (\alpha,0,0)$, $\vec B = (\alpha/c,0,2\alpha/c)$ 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 $v$.
$I_1 = -4\alpha^2$, $I_2 = \alpha^2/c$ in $A$. In $B$: $(E'_x)^2 - c^2(B'_y)^2 = -3\alpha^2$ and $E'_x = \alpha - cB'_y$. Solving: $$\boxed{\;B'_y = 2\alpha/c,\quad E'_x = -\alpha.\;}$$
Boost $B\to C$: $$\vec E'' = (-\alpha,\;\gamma\alpha(1-\beta),\;2\gamma\alpha\beta),\quad \vec B'' = (\alpha/c,\;2\gamma\alpha/c,\;\gamma\alpha(1-\beta)/c).$$ Cross-check with $I_1 = -4\alpha^2$ confirms the boost.
Muon deflection by Earth's magnetic field
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.
$\vec F = q\vec v\times\vec B = -evB\hat x$ — westward. Radius: $R = p/(eB) = 2$ GeV/(3000 V/m) $\approx 667$ km. Sagitta: $$\boxed{\;|\Delta x| = (\Delta z)^2/(2R) = 100/(2\cdot 667)\,\text{km} \approx 75\,\text{m westward.}\;}$$
Muon survival check: $\gamma\approx 19$, $\gamma c\tau_\mu \approx 12.6$ km — exceeds the 10 km descent. Cosmic-ray detectors must correct for this geomagnetic deflection when reconstructing the parent particle direction.
Frame in which $\vec B' = 0$: find $\vec E'$
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 $u$.
$\vec B' = 0$ forces $\vec B = \vec u\times\vec E/c^2$ in $S$ (electric-like Lorentz class). Then $$\boxed{\;\vec E'_\parallel = \vec E_\parallel,\quad \vec E'_\perp = \vec E_\perp/\gamma.\;}$$ Cross-check: $\vec E\!\cdot\!\vec B = 0$ and $|\vec E|^2 - c^2|\vec B|^2 = E_x^2 + (E_y^2+E_z^2)/\gamma^2 = |\vec E'|^2 \ge 0$ — electric-like, as expected.
Boosting a pure $\vec B$-field configuration
In $S$: $\vec E = 0$, $\vec B$ constant. Boost along $+\hat x$ at $v$. Find $\vec E',\vec B'$; verify invariants.
$$\boxed{\;\vec E' = \gamma\vec v\times\vec B,\quad \vec B' = \vec B_\parallel + \gamma\vec B_\perp.\;}$$ Induced $\vec E'$ is the relativistic motional EMF, generalising $\vec v\times\vec B$.
Invariant check: $|\vec E'|^2 - c^2|\vec B'|^2 = \gamma^2 v^2(B_y^2+B_z^2) - c^2 B_x^2 - c^2\gamma^2(B_y^2+B_z^2) = -c^2|\vec B|^2$ using $\gamma^2(v^2-c^2)=-c^2$. ā. $\vec E'\!\cdot\!\vec B' = 0$ ā.
Electron in a linear accelerator (hyperbolic motion)
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 $E(t)$.
Force $f = eU/L$. Work-energy: $E = mc^2 + fx$. With $E^2=(pc)^2+(mc^2)^2$ and $p=ft$:
(a) $\boxed{\;x(t) = (mc^2/f)\left[\sqrt{1+(fct/mc^2)^2} - 1\right].\;}$
(b) $\boxed{\;T = (L/c)\sqrt{1 + 2mc^2/(eU)}.\;}$
(c) $\boxed{\;E(t) = \sqrt{(mc^2)^2 + (eUct/L)^2}.\;}$
At SLAC ($eU\sim 50$ GeV), $2mc^2/eU \sim 2\times 10^{-5}$, so $T \approx L/c$ — the electron is ultra-relativistic almost immediately.
Field of a uniformly moving charge from boosted potentials
Charge $q$ moves at $v\hat x$. Rest frame: $\phi = q/(4\pi|x'|)$, $\vec A = 0$. (a) Find lab-frame $\vec E,\vec B$. (b) Cross-check via Lorentz invariants.
Boosting: $\phi_\text{lab} = \gamma q/(4\pi R)$, $\vec A = (\vec v/c^2)\phi$, with $R = \sqrt{\gamma^2(x-vt)^2 + y^2 + z^2}$. Differentiating: $$\boxed{\;\vec E = \frac{q(1-\beta^2)\hat r}{4\pi r^2 (1-\beta^2\sin^2\theta)^{3/2}},\quad \vec B = \vec v\times\vec E/c^2,\;}$$ the Liénard–Wiechert field for uniform motion — pancakes transversely as $\gamma\to\infty$.
Invariants check. $\vec E\!\cdot\!\vec B \propto \vec E\!\cdot\!(\vec v\times\vec E) = 0$. $|\vec E|^2 - c^2|\vec B|^2 = |\vec E|^2(1-\beta^2\sin^2\theta) = q^2/(16\pi^2 R^4)$ — matches the rest-frame value of $q^2/(16\pi^2|x'|^4)$.