Special Relativity — worked problems

Strategy

$B$ at rest in the lab after the decay means $A$'s 3-momentum was entirely transferred to $C$. Use 4-momentum conservation $P_A = P_B + P_C$ together with mass-shell relations.

Momentum and energy

$\vec p_B = 0 \Rightarrow \vec p_C = \vec p_A$ (3-momentum conservation). Energy conservation:

$$E_A = E_B + E_C = m_B + E_C \;\Longrightarrow\; E_C = E_A - m_B.$$

$C$ is on its mass shell: $E_C^2 = |\vec p_C|^2 + m_C^2 = |\vec p_A|^2 + m_C^2 = E_A^2 - m_A^2 + m_C^2$.

Solve for $E_A$

$(E_A - m_B)^2 = E_A^2 - m_A^2 + m_C^2$, so

$$-2 E_A m_B + m_B^2 = -m_A^2 + m_C^2 \;\Longrightarrow\; E_A = \frac{m_A^2 + m_B^2 - m_C^2}{2 m_B}.$$

Velocity of $A$

$\gamma_A = E_A/m_A$, $\beta_A = \sqrt{1 - 1/\gamma_A^2} = \sqrt{E_A^2 - m_A^2}/E_A = |\vec p_A|/E_A$. Use $E_A^2 - m_A^2 = $ standard Källén form:

$$E_A^2 - m_A^2 = \frac{(m_A^2 + m_B^2 - m_C^2)^2}{4m_B^2} - m_A^2 = \frac{[(m_A+m_B)^2 - m_C^2][(m_A-m_B)^2 - m_C^2]}{4 m_B^2}.$$

Therefore

$$\boxed{\;v_A = \frac{\sqrt{[(m_A + m_B)^2 - m_C^2][(m_A - m_B)^2 - m_C^2]}}{m_A^2 + m_B^2 - m_C^2}.\;}$$

Physical condition

For $B$ to land at rest, $A$'s lab velocity must precisely cancel $B$'s CM-frame backward kick. The expression is real iff $m_A + m_B > m_C$ and $|m_A - m_B| > m_C$ — the first is the usual decay threshold; the second restricts how unevenly the masses can be split for this particular boost configuration.

Limit $m_C\to 0$: $v_A\to (m_A^2 - m_B^2)/(m_A^2 + m_B^2)$ — the boost needed to bring $B$ to rest when the third daughter is massless (e.g., $A\to B + \gamma$).

4-momentum conservation

Particle 1: $P_1^\mu = (\gamma_1 m_1, \gamma_1 m_1 v_1, 0, 0)$. Particle 2 (at rest): $P_2^\mu = (m_2, 0, 0, 0)$. After fully inelastic collision, the composite particle has 4-momentum $P^\mu = P_1^\mu + P_2^\mu = (\gamma_1 m_1 + m_2, \gamma_1 m_1 v_1, 0, 0)$.

Invariant mass

$M^2 = P\!\cdot\!P$:

$$M^2 = (\gamma_1 m_1 + m_2)^2 - (\gamma_1 m_1 v_1)^2.$$

Expand: $\gamma_1^2 m_1^2 + 2\gamma_1 m_1 m_2 + m_2^2 - \gamma_1^2 m_1^2 v_1^2 = \gamma_1^2 m_1^2(1 - v_1^2) + 2\gamma_1 m_1 m_2 + m_2^2$. Use $\gamma_1^2(1-v_1^2) = 1$:

$$\boxed{\;M = \sqrt{m_1^2 + m_2^2 + 2\gamma_1 m_1 m_2}.\;}$$

Composite velocity

$$v = p^x/E = \frac{\gamma_1 m_1 v_1}{\gamma_1 m_1 + m_2}.$$

$$\boxed{\;v = \frac{\gamma_1 m_1 v_1}{\gamma_1 m_1 + m_2}.\;}$$

Limits

• Newtonian ($\gamma_1\to 1$): $M\to\sqrt{m_1^2 + 2m_1 m_2 + m_2^2} = m_1 + m_2$, and $v \to m_1 v_1/(m_1 + m_2)$ — classical inelastic collision.
• Ultra-relativistic ($\gamma_1\gg 1$): $M\sim\sqrt{2\gamma_1 m_1 m_2}$ — the famous $\sqrt{\gamma}$ scaling of fixed-target experiments.

Collider vs. fixed-target

To produce a heavy resonance of mass $M$ in fixed-target ($p_2 = 0$), the beam energy scales as $E_1\sim M^2/(2m_2)$ — quadratically. In a head-on collider ($p_1 = -p_2$), $M = 2 E_\text{beam}$ — linear. This is why the LHC uses colliders: at fixed-target you'd need TeV beams to reach $\sqrt s\sim 100$ GeV; at the LHC, two 6.5 TeV beams reach 13 TeV directly. The Higgs discovery (~125 GeV) was kinematically inaccessible to any realistic fixed-target machine.

(a) Invariant mass from two 4-momenta

$P^\mu = P_1^\mu + P_2^\mu$ with $P_i^\mu = \gamma_i m_i(1, \vec u_i)$. Compute $P\!\cdot\!P$:

$$m^2 = (\gamma_1 m_1 + \gamma_2 m_2)^2 - (\gamma_1 m_1\vec u_1 + \gamma_2 m_2\vec u_2)^2.$$

Expand: $\gamma_1^2 m_1^2(1 - u_1^2) + \gamma_2^2 m_2^2(1 - u_2^2) + 2\gamma_1\gamma_2 m_1 m_2(1 - \vec u_1\!\cdot\!\vec u_2)$.

Using $\gamma_i^2(1 - u_i^2) = 1$:

$$\boxed{\;m^2 = m_1^2 + m_2^2 + 2 m_1 m_2 \gamma_1\gamma_2(1 - \vec u_1\!\cdot\!\vec u_2).\;}$$

(b) Rewrite via relative velocity

From Problem 1.3 (or 1.44): if $v$ is the magnitude of the relative velocity of particle 1 in particle 2's rest frame,

$$\gamma(v) = \gamma_1\gamma_2(1 - \vec u_1\!\cdot\!\vec u_2).$$

Substituting:

$$\boxed{\;m^2 = m_1^2 + m_2^2 + 2 m_1 m_2\gamma(v).\;}$$

Invariant mass depends only on relative velocity — the inner-product structure of 4-momenta in disguise.

(c) Compare fixed-target vs. CM lab energies at same $v$

Setup A (fixed-target): $u_1 = 0$, $u_2 = v$. Total lab energy: $E_A = m_1 + \gamma(v) m_2$.

Setup B (CM): the same relative velocity is split, with each particle moving at $v/(1+\gamma(v))\cdot 1/\gamma$ in some boost — the precise expressions don't matter. Total lab energy in CM frame: $E_B = m$.

Both setups have the same $m$ (invariant). Therefore

$$\Delta E = E_A - E_B = m_1 + \gamma(v) m_2 - m = (\gamma_\text{new} - 1)m > 0,$$

where $\gamma_\text{new}$ is the Lorentz factor of the composite resonance in the lab frame for setup A. The "wasted" kinetic energy in fixed-target experiments goes into lab motion of the resonance, contributing nothing to the invariant mass available to produce new particles.

Numerical comparison

• SLAC fixed-target: 50 GeV electrons on stationary protons (mass $\sim 1$ GeV) gives $\sqrt s \approx \sqrt{2\cdot 50\cdot 1} = 10$ GeV.
• LHC head-on: two 7 TeV beams ⇒ $\sqrt s = 14$ TeV directly — ~1000$\times$ the fixed-target equivalent at the same per-particle beam energy.

Pion-rest-frame kinematics

Two-body decay $\pi\to\mu + \nu$ in the pion rest frame: $|\vec p_\mu^*| = |\vec p_\nu^*| \equiv p^*$. With $m_\nu = 0$:

$$E_\mu^* + p^* = m_\pi,\quad (E_\mu^*)^2 = p^{*2} + m_\mu^2.$$

Solve:

$$E_\mu^* = \frac{m_\pi^2 + m_\mu^2}{2 m_\pi},\quad p^* = \frac{m_\pi^2 - m_\mu^2}{2 m_\pi}.$$

Boost to lab and impose right-angle condition

Lab and rest frame related by boost at $v_\pi$ along $+\hat x$. Lab muon 3-momentum:

$$p_\mu^x = \gamma_\pi(p^*\cos\theta^* + v_\pi E_\mu^*),\quad p_\mu^y = p^*\sin\theta^*.$$

The muon emerges perpendicular to $\hat x$ means $p_\mu^x = 0$:

$$\cos\theta^* = -\frac{v_\pi E_\mu^*}{p^*}.$$

This requires backward emission in the pion rest frame — $\cos\theta^* < 0$ — consistent with the directional boost.

Lab energy

$E_\mu = \gamma_\pi(E_\mu^* + v_\pi p^*\cos\theta^*) = \gamma_\pi(E_\mu^* - v_\pi^2 E_\mu^*) = E_\mu^*/\gamma_\pi$.

Substitute $\gamma_\pi = E_\pi/m_\pi$:

$$\boxed{\;E_\mu = \frac{m_\pi^2 + m_\mu^2}{2 E_\pi}.\;}$$

Sanity

• Threshold $E_\pi = m_\pi$ (pion at rest): $E_\mu = (m_\pi^2 + m_\mu^2)/(2m_\pi) = E_\mu^*$. ✓ — pion-rest-frame value, decay in any direction including transverse.
• Ultra-relativistic pion ($E_\pi\gg m_\pi$): $E_\mu\to 0$ — orthogonal muons become impossibly soft, kinematically suppressed by the boost. Most decay muons emerge close to the pion direction in the lab.

This kinematic suppression is the reason the muon beam from a high-energy pion source is sharply forward-collimated — key to neutrino-beam design (NuMI, T2K, DUNE).

Pion-rest-frame kinematics

$\pi^-\to e^- + \bar\nu_e$ at rest. By 4-momentum conservation, electron and antineutrino emerge back-to-back with momenta $\pm p^*$ along some axis (say $\hat x$).

Two-body decay momentum: $p^* = \sqrt{[m_\pi^2 - (m + m_\nu)^2][m_\pi^2 - (m - m_\nu)^2]}/(2m_\pi)$.

Energies: $E_e^* = \sqrt{p^{*2} + m^2}$, $E_\nu^* = \sqrt{p^{*2} + m_\nu^2}$, with $E_e^* + E_\nu^* = m_\pi$. Velocities: $v_e^* = p^*/E_e^*$, $v_\nu^* = -p^*/E_\nu^*$.

Antineutrino velocity in electron rest frame

Boost from pion rest frame to electron rest frame along $\hat x$ at velocity $v_e^*$:

$$v_\nu^{(e)} = \frac{v_\nu^* - v_e^*}{1 - v_\nu^* v_e^*} = \frac{-p^*/E_\nu^* - p^*/E_e^*}{1 - (-p^*/E_\nu^*)(p^*/E_e^*)} = \frac{-p^*(E_e^* + E_\nu^*)/(E_e^* E_\nu^*)}{1 + p^{*2}/(E_e^* E_\nu^*)} = \frac{-p^* m_\pi}{E_e^* E_\nu^* + p^{*2}}.$$

So $|v_\nu^{(e)}| = m_\pi p^*/(E_e^* E_\nu^* + p^{*2})$.

Massless-neutrino limit

$m_\nu\to 0$: $E_\nu^* = p^*$, so denominator $= p^*(E_e^* + p^*)$. From $E_e^* + E_\nu^* = m_\pi$ and $E_\nu^* = p^*$: $E_e^* + p^* = m_\pi$. Numerator: $m_\pi p^*$. Therefore

$$|v_\nu^{(e)}| = \frac{m_\pi p^*}{p^* m_\pi} = 1 = c.$$

$$\boxed{\;|v_\nu^{(e)}|\to 1 = c.\;}$$

Comment

This is a cleaner restatement of "massless particles travel at $c$ in every frame", made manifest by the algebraic identity $E_e^* + p^* = m_\pi$. The formula automatically produces the speed-of-light limit when $m_\nu = 0$, without invoking any boost-invariance principle separately — it's encoded in the conservation laws.

For massive neutrinos (cosmological $m_\nu \lesssim 0.1$ eV), $|v_\nu^{(e)}|$ differs from $c$ by ~$(m_\nu/p^*)^2 \sim 10^{-17}$ for a few-MeV decay neutrino — far below any direct kinematic measurement, but accessible to oscillation experiments through the energy-dependent phase difference.

(a) Two-body decay kinematics

$\pi^+\to\mu^+ + \nu_\mu$ at rest. Treat $m_{\nu_\mu} = 0$. 4-momentum conservation: equal and opposite 3-momenta $\pm p$, with

$$m_\pi c^2 = E_\mu + E_\nu = \sqrt{(m_\mu c^2)^2 + (pc)^2} + pc.$$

Isolate the square root and square:

$$\sqrt{(m_\mu c^2)^2 + (pc)^2} = m_\pi c^2 - pc \Rightarrow (m_\mu c^2)^2 + (pc)^2 = (m_\pi c^2)^2 - 2 m_\pi c^2\cdot pc + (pc)^2.$$

Solving:

$$\boxed{\;p_{\nu_\mu} = \frac{(m_\pi^2 - m_\mu^2)c}{2 m_\pi}.\;}$$

Antimuon kinetic energy:

$$T_{\mu^+} = E_\mu - m_\mu c^2 = m_\pi c^2 - p c - m_\mu c^2 = \frac{(m_\pi - m_\mu)^2 c^2}{2 m_\pi}.$$

$$\boxed{\;T_{\mu^+} = \frac{(m_\pi - m_\mu)^2 c^2}{2 m_\pi}.\;}$$

Numerical values

$m_{\pi^+} c^2 = 139.57$ MeV, $m_\mu c^2 = 105.66$ MeV:

• $p_\nu c = (139.57^2 - 105.66^2)/(2\cdot 139.57)\,\text{MeV} \approx 29.8$ MeV.
• $T_{\mu^+} = (33.91)^2/(2\cdot 139.57)\,\text{MeV} \approx 4.12$ MeV.

The antimuon's $T/m_\mu c^2 \approx 0.04$ — non-relativistic in the pion frame. Almost all the kinetic energy released in pion decay goes to the neutrino.

(b) Antimuon mean flight distance

Antimuon momentum: $p_\mu = p_\nu = 29.8$ MeV/$c$, so $\beta_\mu\gamma_\mu = p_\mu/(m_\mu c) = 29.8/105.66 \approx 0.282$. With muon mean lifetime $\tau_\mu = 2.197\;\mu$s, $c\tau_\mu = 659$ m (in muon rest frame). Lab decay length (in the pion rest frame, which is where the antimuon is born):

$$\boxed{\;\langle d\rangle = \beta_\mu\gamma_\mu\,c\tau_\mu \approx 0.282\cdot 659\,\text{m} \approx 186\,\text{m}.\;}$$

Atmospheric-muon context

Cosmic-ray pions decay rapidly (proper lifetime $2.6\times 10^{-8}$ s); the resulting muons can travel ~186 m in the pion rest frame before themselves decaying. In the Earth rest frame, the atmospheric pion is highly boosted, so its progeny muons can traverse the troposphere — this is exactly the chain that produces the atmospheric muon flux measured by Super-Kamiokande.

Work-energy theorem

A constant electric force does work $qEr$ on the particle as it traverses distance $r$ from rest. Total relativistic energy:

$$\mathcal E(r) = m c^2 + qEr.$$

Use the mass-shell relation

$\mathcal E^2 = (pc)^2 + (mc^2)^2$, so

$$(pc)^2 = \mathcal E^2 - (mc^2)^2 = (mc^2 + qEr)^2 - (mc^2)^2.$$

Velocity from $v = pc^2/\mathcal E$

$$\frac{v^2}{c^2} = \frac{(pc)^2}{\mathcal E^2} = 1 - \left(\frac{mc^2}{mc^2 + qEr}\right)^{\!2}.$$

$$\boxed{\;v(r) = c\sqrt{1 - \left(\frac{mc^2}{mc^2 + qEr}\right)^{\!2}}.\;}$$

Limits

• Newtonian ($qEr\ll mc^2$): expand $(1 - x)^2 \approx 1 - 2x$ for small $x = qEr/(mc^2)$:
  $$v^2 \approx c^2\cdot 2qEr/(mc^2) = 2qEr/m \Rightarrow v\approx\sqrt{2qEr/m}.$$
  Familiar non-relativistic work-energy: $\tfrac{1}{2}mv^2 = qEr$. ✓
• Ultra-relativistic ($qEr\gg mc^2$): $v\to c$ asymptotically. Light-speed barrier preserved.

Hyperbolic motion

The worldline traced out is a hyperbola in $(ct, x)$ spacetime, asymptoting to the light cone. The particle has constant proper acceleration $\alpha = qE/m$ — same hyperbolic motion that defines Rindler observers (Problem 1.49, 2.28). Equivalence principle: a charged particle in a uniform electric field is locally indistinguishable from a particle at rest in a uniform gravitational field, as long as you ignore radiation reaction (which complicates the case of a single accelerating charge).

(a) Transform $F^{\mu\nu}$

Wire along $\hat z$ carrying current $I$ in $+\hat z$. At perpendicular distance $\rho = \sqrt{x^2 + y^2}$:

$$\vec B(\rho) = \frac{\mu_0 I}{2\pi\rho}\,\hat\varphi.$$

Pick a point on the $+\hat x$ axis, where $\hat\varphi = +\hat y$: $\vec B = (\mu_0 I/2\pi\rho)\hat y$, $\vec E = 0$ (wire neutral).

Frame $K'$ moves at $v\hat z$ relative to $K$. Standard transformations:

$$E'_\parallel = E_\parallel,\quad E'_\perp = \gamma(\vec E + \vec v\times\vec B)_\perp,\quad B'_\parallel = B_\parallel,\quad B'_\perp = \gamma(\vec B - \vec v\times\vec E/c^2)_\perp.$$

With $\vec v = v\hat z$, $\vec B = (\mu_0 I/2\pi\rho)\hat y$, $\vec E = 0$:

$$\vec v\times\vec B = v\hat z\times(\mu_0 I/2\pi\rho)\hat y = -(v\mu_0 I/2\pi\rho)\hat x.$$

So

$$\boxed{\;\vec E' = -\frac{\gamma v\mu_0 I}{2\pi\rho}\,\hat x,\quad \vec B' = \gamma\,\frac{\mu_0 I}{2\pi\rho}\,\hat y.\;}$$

(b) Transform the 4-current $J^\mu$

In $K$: $J^\mu = (c\rho_q, \vec J) = (0, 0, 0, j_z)$ — pure current, no charge density (neutral wire).

Boost to $K'$ moving at $v\hat z$:

$$\rho'_q c = \gamma(c\rho_q - v j_z/c) = -\gamma v j_z/c,\quad j'_z = \gamma(j_z - v\cdot 0) = \gamma j_z.$$

Linear charge density per unit length: $\lambda' = \rho'_q \cdot A = -\gamma v I/c^2$ (negative if $I > 0$), and current $I' = \gamma I$.

Cross-check: Coulomb + Ampère recover the same fields

Line charge $\lambda'$ produces $E'_\text{Coul} = \lambda'/(2\pi\varepsilon_0\rho) = -\gamma v I/(2\pi\varepsilon_0 c^2\rho)$. Using $1/(\varepsilon_0 c^2) = \mu_0$: $E' = -\gamma v\mu_0 I/(2\pi\rho)$ — same as in (a). ✓

Current $I'$ produces $B' = \mu_0 I'/(2\pi\rho) = \gamma\mu_0 I/(2\pi\rho)$ — matches. ✓

Physical interpretation (Purcell)

Boost along a neutral wire with drifting electrons: in $K'$ the positive ion lattice and the electron stream length-contract by different amounts (their lab-frame velocities differ, so their boosted velocities differ, and so do their $\gamma$ factors). Charge neutrality is observer-dependent! This is Purcell's classic demonstration that magnetism is just relativistic electrostatics — the same field tensor $F^{\mu\nu}$ appears as different mixtures of $\vec E$ and $\vec B$ in different frames.

Setup

Lorenz gauge: $\partial_\mu A^\mu = 0$. In vacuum, the inhomogeneous Maxwell equation $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu = 0$ becomes $\Box A^\nu = 0$. Plane-wave ansatz $A^\mu = \varepsilon^\mu e^{ik\cdot x}$ substitutes:

$$\Box A^\nu = -k^2 A^\nu = 0 \;\Longrightarrow\; k^2 = 0\quad\text{(null dispersion)}.$$

Lorenz gauge condition: $\partial_\mu A^\mu = ik_\mu\varepsilon^\mu e^{ik\cdot x} = 0 \Rightarrow k_\mu\varepsilon^\mu = 0$ (polarisation orthogonal to wavevector).

Field tensor

$$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu = i(k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)\,e^{ik\cdot x}.$$

$\vec B\perp\vec k$

$B_i = -\tfrac{1}{2}\epsilon_{ijk}F^{jk}$. Substitute: $\vec B = -i\vec k\times\vec\varepsilon\,e^{ik\cdot x}$. Then

$$\vec B\!\cdot\!\vec k = -i\vec k\!\cdot\!(\vec k\times\vec\varepsilon)\,e^{ik\cdot x} = 0$$

by the standard triple-product identity — any cross product is perpendicular to both factors.

$\vec E\perp\vec k$

$E_i = F^{i0} = i(k^i\varepsilon^0 - k^0\varepsilon^i)e^{ik\cdot x}$, so

$$\vec E = i(\vec k\varepsilon^0 - \omega\vec\varepsilon/c)\,e^{ik\cdot x}\quad\text{(restoring }c\text{)}.$$

$$\vec E\!\cdot\!\vec k = i(|\vec k|^2\varepsilon^0 - (\omega/c)\vec k\!\cdot\!\vec\varepsilon)\,e^{ik\cdot x}.$$

Use Lorenz: $k_\mu\varepsilon^\mu = (\omega/c)\varepsilon^0 - \vec k\!\cdot\!\vec\varepsilon = 0 \Rightarrow \vec k\!\cdot\!\vec\varepsilon = (\omega/c)\varepsilon^0$. Substitute:

$$\vec E\!\cdot\!\vec k = i\varepsilon^0[|\vec k|^2 - (\omega/c)^2]\,e^{ik\cdot x} = i\varepsilon^0\cdot 0\cdot e^{ik\cdot x} = 0$$

using the null-dispersion condition $\omega = c|\vec k|$.

Conclusion

Both transversality relations follow from the two constraints $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$, themselves a consequence of Maxwell's equations + Lorenz gauge. This is why EM plane waves are transverse: only 2 of the 4 components of $\varepsilon^\mu$ are physical (the two polarisations), reflecting the 2 physical degrees of freedom of a massless spin-1 field.

Compute $F_{\mu\nu}F^{\mu\nu}$

From Problem 1.120, the field tensor of a plane wave (in Lorenz gauge) is $F^{\mu\nu} = i(k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)e^{ik\cdot x}$, with $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$. Contract:

$$F_{\mu\nu}F^{\mu\nu} = -(k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)(k_\mu\varepsilon_\nu - k_\nu\varepsilon_\mu)\,e^{2ik\cdot x}.$$

Expanding: $-(2 k^2\varepsilon^2 - 2(k\!\cdot\!\varepsilon)^2)e^{2ik\cdot x} = 0$ — both terms vanish by the constraints.

Compute $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$

Pull a factor of two out from the antisymmetric pair: $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} \propto \epsilon\,k\varepsilon\,k\varepsilon$. Two of the four arguments are $k$, the totally antisymmetric tensor with two repeated indices vanishes. So this invariant is identically zero.

Translate to fields

The two invariants in terms of fields are

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

Vanishing of both:

$$\boxed{\;|\vec E| = c|\vec B|\quad\text{and}\quad \vec E\perp\vec B.\;}$$

Null Lorentz class

Both invariants vanish — the EM field of a free plane wave belongs to the null Lorentz class. Consequences:

• No frame trivialises it to pure $\vec E$ (would require $I_1 < 0$) or pure $\vec B$ (would require $I_1 > 0$).
• The field cannot be removed by any boost (in any frame, both $\vec E$ and $\vec B$ are non-zero, orthogonal, equal magnitude).

Free EM waves enjoy the larger 15-parameter conformal symmetry (SO(4,2)) of massless particles, not just Poincaré (10 parameters). The two invariants $I_1 = F^2$ and $I_2 = F\tilde F$ also serve as conformal scalars (weight zero) only because they vanish — conformal symmetry generically breaks under field rescaling, but vanishing invariants are scale-invariant trivially.