Electromagnetism was relativistic all along: the field tensor

Idea. There is no idea. This is the drill that makes every later index manipulation automatic, and the only way to get it wrong is to hurry.

1. The arrays. With \(E_y/c = 2.00138\times10^{-3}\ \mathrm T\) and \(B_z = 1.00\times10^{-3}\ \mathrm T\), in units of \(10^{-3}\) T,

\[F_{\mu\nu} = \begin{pmatrix}0&0&-2.00138&0\\0&0&1.00&0\\2.00138&-1.00&0&0\\0&0&0&0\end{pmatrix},\qquad F^{\mu\nu} = \begin{pmatrix}0&0&2.00138&0\\0&0&1.00&0\\-2.00138&-1.00&0&0\\0&0&0&0\end{pmatrix}.\]

Raising the first index only, \(F^\mu{}_\nu = \eta^{\mu\alpha}F_{\alpha\nu}\), which negates row 0:

\[F^{\mu}{}_{\nu} = \begin{pmatrix}0&0&2.00138&0\\0&0&1.00&0\\2.00138&-1.00&0&0\\0&0&0&0\end{pmatrix}\times10^{-3}\ \mathrm T.\]

Note that \(F^\mu{}_\nu\) is neither symmetric nor antisymmetric: mixed components of an antisymmetric tensor have no symmetry, because the two slots are of different kinds.

2. The trace. \(F^\mu{}_\mu = 0 + 0 + 0 + 0 = 0\). In general \(F^\mu{}_\mu = \eta^{\mu\nu}F_{\nu\mu}\) is the contraction of a symmetric array with an antisymmetric one, which vanishes identically: relabel \(\mu\leftrightarrow\nu\) and the sum equals minus itself.

3. The invariants.

\[\mathcal I_1 = B^2 - \frac{E^2}{c^2} = 1.00\times10^{-6} - 4.00554\times10^{-6} = -3.00554\times10^{-6}\ \mathrm{T^2},\]

and \(\mathcal I_2 = \vec E\cdot\vec B = 0\), since \(\vec E\parallel\hat y\) and \(\vec B\parallel\hat z\). Summing the products directly: the only nonzero terms of \(F_{\mu\nu}F^{\mu\nu}\) are \(F_{02}F^{02} + F_{20}F^{20} = -2(2.0013846)^2\times10^{-6} = -8.01108\times10^{-6}\) and \(F_{12}F^{12} + F_{21}F^{21} = 2(1.00)(1.00)\times10^{-6} = 2.00\times10^{-6}\); the sum is \(-6.01108\times10^{-6}\), and half of it is \(-3.00554\times10^{-6}\ \mathrm{T^2}\) ✓.

4. The quadratic form vanishes. Of the sixteen terms, the four diagonal ones carry \(F_{\mu\mu}\), which is zero because \(F_{\mu\mu} = -F_{\mu\mu}\). The remaining twelve pair up: for each \(\mu\lt\nu\) the terms \(F_{\mu\nu}u^\mu u^\nu\) and \(F_{\nu\mu}u^\nu u^\mu\) carry the same product of components and opposite coefficients, so each pair sums to zero. Concretely, for this field only two pairs are nonzero at all: \(F_{02}u^0u^2 = -2.00138\times10^{-3}\gamma^2cv_y\) against \(F_{20}u^2u^0 = +2.00138\times10^{-3}\gamma^2cv_y\), and \(F_{12}u^1u^2\) against \(F_{21}u^2u^1\). Six pairs, six zeros.

5. The four-force per unit charge. At \(v = 1.00\times10^{8}\ \mathrm{m\,s^{-1}}\), \(\beta = 0.333564\) and \(\gamma = 1.06075\), so \(u^\mu = (3.18005\times10^{8},\ 1.06075\times10^{8},\ 0,\ 0)\ \mathrm{m\,s^{-1}}\). With \(F^\mu{}_\nu\) above,

\[F^0{}_\nu u^\nu = 0,\qquad F^1{}_\nu u^\nu = 0,\qquad F^2{}_\nu u^\nu = F^2{}_0u^0 + F^2{}_1u^1 = (2.00138\times10^{-3})(3.18005\times10^{8}) - (1.00\times10^{-3})(1.06075\times10^{8}),\]

which is \(6.36451\times10^{5} - 1.06075\times10^{5} = 5.30376\times10^{5}\), in \(\mathrm{T\,m\,s^{-1}} = \mathrm{N\,C^{-1}}\). So the four-force per unit charge is \((0,\,0,\,5.30376\times10^{5},\,0)\), purely in \(\hat y\). Its Minkowski product with \(u\) is \(-f^0u^0 + f^iu^i = f^2u^2\), and \(u^2 = \gamma v_y = 0\) for this particle ✓.

Check. Dimensions: every entry of \(F\) is in tesla, and \(\mathrm{T\cdot m\,s^{-1}} = \mathrm{V\,m^{-1}}\) ✓, so part 5 is a force per unit charge as claimed. Limit: the three-force is \(F^2{}_\nu u^\nu/\gamma = 5.30376\times10^{5}/1.06075 = 5.00\times10^{5}\ \mathrm{N\,C^{-1}}\), and directly \(E_y - vB_z = 6.00\times10^{5} - (1.00\times10^{8})(1.00\times10^{-3}) = 5.00\times10^{5}\) ✓. Special case: at \(v = c^2B/E = 1.49793\times10^{8}\ \mathrm{m\,s^{-1}}\) the force would vanish — that is the drift speed, and P10 makes it the point.

Idea. Mass conservation is a scalar condition, one equation per worldline; but there is a four-parameter family of worldlines through each event, and four parameters is enough to kill ten functions.

1. Let \(p = mu\) with \(m\) possibly varying. Then \(\eta(p,p) = -m^2c^2\), and differentiating,

\[\frac{\mathrm d}{\mathrm d\tau}\big(-m^2c^2\big) = -2mc^2\frac{\mathrm dm}{\mathrm d\tau} = 2\,\eta\Big(\frac{\mathrm dp}{\mathrm d\tau}, p\Big) = 2\,\eta\big(qF^\cdot{}_\nu u^\nu,\ mu\big) = 2qm\,F_{\mu\nu}u^\mu u^\nu .\]

Dividing by \(-2mc^2\) and noting that only the symmetric part survives contraction with \(u^\mu u^\nu\),

\[\frac{\mathrm dm}{\mathrm d\tau} = -\frac{q}{c^2}F_{(\mu\nu)}u^\mu u^\nu .\]

2. Write \(S = F_{(\mu\nu)}\). The hypothesis is \(S(u,u) = 0\) for every \(u\) with \(\eta(u,u) = -c^2\) and \(u^0>0\). Let \(\mathcal T\) be the set of future-pointing timelike vectors; it is open, being the intersection of the preimages of the open sets \((-\infty,0)\) under \(w\mapsto\eta(w,w)\) and \((0,\infty)\) under \(w\mapsto w^0\), both continuous. Every \(w\in\mathcal T\) is \(\lambda u\) with \(\lambda = \sqrt{-\eta(w,w)}/c > 0\) and \(u\) unit-timelike, and \(S(w,w) = \lambda^2 S(u,u) = 0\). So the polynomial \(P(w) = S_{\mu\nu}w^\mu w^\nu\) in the four variables \(w^0,\dots,w^3\) vanishes on the nonempty open set \(\mathcal T\).

This is where openness is used. Fix \(w_0\in\mathcal T\) and any \(z\in V\); for \(|s|\) small, \(w_0 + sz\in\mathcal T\), so \(s\mapsto P(w_0+sz)\) is a quadratic polynomial in \(s\) vanishing on an interval, hence identically zero; in particular its \(s^2\) coefficient, \(P(z)\), is zero. So \(P\equiv0\) on all of \(V\).

Polarisation then gives, for any \(w,z\),

\[S(w,z) = \tfrac12\big[P(w+z) - P(w) - P(z)\big] = 0,\]

using symmetry of \(S\) to write \(P(w+z) = P(w) + 2S(w,z) + P(z)\). Hence \(S\equiv0\), i.e. \(F_{\mu\nu} = -F_{\nu\mu}\). \(\square\)

3. Take \(S_{01} = S_{10} = 1\) with every other component zero — symmetric and nonzero — and \(u = (c,0,0,0)\), the four-velocity of a particle at rest. Then \(S(u,u) = 2S_{01}u^0u^1 = 0\) because \(u^1 = 0\), so that one worldline says nothing at all about \(S\). A second example, with \(u\) in motion: \(S = \mathrm{diag}(1,\,-c^2/v_0^2,\,0,\,0)\) and \(u = \gamma_0(c, v_0, 0, 0)\) give \(S(u,u) = \gamma_0^2c^2 - \gamma_0^2c^2 = 0\). One worldline is never enough: one quadratic condition cannot kill ten independent functions, and only the whole open cone of directions can.

4. Sixteen functions, minus the ten independent components of the symmetric part, leaves six. Those six are \(F_{0i}\) (three, named \(-E_i/c\)) and \(F_{ij}\) with \(i\lt j\) (three, named \(B_k\)). The count \(\binom42 = 6\) is the whole reason electromagnetism has two three-vector fields and not one, or three.

Check. Dimensions in part 1: the contraction carries two factors of \(u\), so \([qF_{(\mu\nu)}u^\mu u^\nu/c^2] = \mathrm{C\,T\,(m\,s^{-1})^2}/(\mathrm{m^2s^{-2}}) = \mathrm{C\,T} = \mathrm{A\,s\cdot kg\,A^{-1}s^{-2}} = \mathrm{kg\,s^{-1}}\) ✓, a mass per second of proper time. Special case: if \(F\) is antisymmetric the right-hand side is zero and mass is constant ✓, which is the deck's four-force constraint. Limit: at \(v\ll c\) part 1 reads \(\dot m \approx -qF_{(00)}\), so even a purely "electrostatic" symmetric part would heat a charge at rest — visibly absurd, which is the physical form of the theorem.

Idea. A boost acts as the identity on the \(23\) block, so the only work is in the two mixed blocks, and the algebra of part 2 is one application of \(\gamma^2(1-\beta^2) = 1\) per pair.

1. The six rules.

• \(F'^{01} = \Lambda^0{}_\alpha\Lambda^1{}_\beta F^{\alpha\beta} = \Lambda^0{}_0\Lambda^1{}_1F^{01} + \Lambda^0{}_1\Lambda^1{}_0F^{10} = (\gamma^2 - \gamma^2\beta^2)F^{01} = F^{01}\). Hence \(E'_x = E_x\).
• \(F'^{02} = \gamma F^{02} - \gamma\beta F^{12} = \gamma\frac{E_y}{c} - \gamma\beta B_z\), so \(E'_y = \gamma(E_y - vB_z)\).
• \(F'^{03} = \gamma F^{03} - \gamma\beta F^{13} = \gamma\frac{E_z}{c} + \gamma\beta B_y\), so \(E'_z = \gamma(E_z + vB_y)\).
• \(F'^{12} = -\gamma\beta F^{02} + \gamma F^{12} = -\gamma\beta\frac{E_y}{c} + \gamma B_z\), so \(B'_z = \gamma(B_z - vE_y/c^2)\).
• \(F'^{13} = -\gamma\beta F^{03} + \gamma F^{13} = -\gamma\beta\frac{E_z}{c} - \gamma B_y\), and \(F'^{13} = -B'_y\), so \(B'_y = \gamma(B_y + vE_z/c^2)\).
• \(F'^{23} = F^{23}\), so \(B'_x = B_x\).

2. The first invariant.

\[E'^2 = E_x^2 + \gamma^2\big[(E_y - vB_z)^2 + (E_z + vB_y)^2\big],\qquad B'^2 = B_x^2 + \gamma^2\Big[\big(B_y + \tfrac{v}{c^2}E_z\big)^2 + \big(B_z - \tfrac{v}{c^2}E_y\big)^2\Big].\]

Expand the \(y\)-and-\(z\) pieces of \(B'^2 - E'^2/c^2\). The cross terms are \(2\gamma^2\big[\frac{v}{c^2}(B_yE_z - B_zE_y)\big]\) from \(B'^2\) and \(-\frac{\gamma^2}{c^2}\big[-2vE_yB_z + 2vE_zB_y\big] = -2\gamma^2\frac{v}{c^2}(E_zB_y - E_yB_z)\) from \(-E'^2/c^2\); they are equal and opposite, so they cancel. What is left is

\[\gamma^2\Big[B_y^2 + \tfrac{v^2}{c^4}E_z^2 + B_z^2 + \tfrac{v^2}{c^4}E_y^2 - \tfrac{E_y^2}{c^2} - \tfrac{v^2B_z^2}{c^2} - \tfrac{E_z^2}{c^2} - \tfrac{v^2B_y^2}{c^2}\Big] = \gamma^2(1-\beta^2)\Big[B_y^2 + B_z^2 - \tfrac{E_y^2 + E_z^2}{c^2}\Big],\]

and \(\gamma^2(1-\beta^2) = 1\). Adding the untouched \(B_x^2 - E_x^2/c^2\) gives \(\mathcal I_1' = \mathcal I_1\) ✓.

3. The second invariant. \(\vec E{}'\cdot\vec B{}' = E_xB_x + E'_yB'_y + E'_zB'_z\), and

\[E'_yB'_y = \gamma^2(E_y - vB_z)\big(B_y + \tfrac{v}{c^2}E_z\big),\qquad E'_zB'_z = \gamma^2(E_z + vB_y)\big(B_z - \tfrac{v}{c^2}E_y\big).\]

Expanding, the terms \(\gamma^2\frac{v}{c^2}E_yE_z\) and \(-\gamma^2\frac{v}{c^2}E_zE_y\) cancel, as do \(-\gamma^2vB_zB_y\) and \(+\gamma^2vB_yB_z\); what remains is \(\gamma^2(1-\beta^2)(E_yB_y + E_zB_z) = E_yB_y + E_zB_z\) ✓.

4. First order. \(\gamma = 1 + O(\beta^2)\), so to first order \(E'_y = E_y - vB_z\), \(E'_z = E_z + vB_y\), i.e. \(\vec E{}' = \vec E + \vec v\times\vec B\) for the boost along \(\hat x\); and likewise \(\vec B{}' = \vec B - \vec v\times\vec E/c^2\). Unit 25's Galilean field transformation kept the electric rule and dropped the \(\vec v\times\vec E/c^2\) term, because a consistent Galilean electromagnetism has to choose: keeping both makes the transformation fail to be a group action at order \(\beta^2\). The relativistic rules are the unique repair, and they are exactly what the \(\gamma\)s restore.

Check. Dimensions: \([vB] = \mathrm{V\,m^{-1}}\) ✓, \([vE/c^2] = \mathrm T\) ✓. Limit \(v\to0\): all six rules become the identity ✓. Number: for the running example's wire, \(E_y = 0\) and \(B_z = 10.0\ \mu\mathrm T\) give \(E'_y = -\gamma vB_z = -30.0015\ \mathrm{V\,m^{-1}}\) at \(v = 3.00\times10^{6}\ \mathrm{m\,s^{-1}}\), the deck's number ✓.

Another way

Part 2 and part 3 together are one line if you are allowed the index argument: \(\mathcal I_1\) and \(\mathcal I_2\) are full contractions, hence scalars, hence frame-independent. The brute-force route is worth doing once precisely so that you believe the index argument afterwards.

Idea. The two invariants are a complete classifier of the orbit of \(F\) under the Lorentz group, so the verdict is decided before any boost is constructed. Constructing the boost is then a one-line equation.

Case 1. \(\mathcal I_2 = \vec E\cdot\vec B = 0\) and \(\mathcal I_1 = 1.00\times10^{-6} - 4.00554\times10^{-6} = -3.006\times10^{-6}\ \mathrm{T^2} \lt 0\): purely electric somewhere. Boost along \(\hat x\) with \(v = c^2B_z/E_y = (8.98755\times10^{16})(1.00\times10^{-3})/(6.00\times10^{5}) = 1.498\times10^{8}\ \mathrm{m\,s^{-1}}\), i.e. \(\beta = cB/E = 0.4997\), \(\gamma = 1.154\). The surviving field is \(E'_y = E_y/\gamma = 6.00\times10^{5}/1.154434 = 5.20\times10^{5}\ \mathrm{V\,m^{-1}}\).

Case 2. \(\mathcal I_2 = 0\) and \(\mathcal I_1 = 4.00\times10^{-6} - 1.00139\times10^{-6} = +2.999\times10^{-6}\ \mathrm{T^2} \gt 0\): purely magnetic somewhere. Kill \(E'_y = \gamma(E_y - vB_z)\) with \(v = E_y/B_z = 1.500\times10^{8}\ \mathrm{m\,s^{-1}}\), \(\beta = E/(cB) = 0.5003\), \(\gamma = 1.155\). Surviving: \(B'_z = B_z/\gamma = 2.00\times10^{-3}/1.154967 = 1.732\times10^{-3}\ \mathrm T\).

Case 3. \(\mathcal I_2 = \vec E\cdot\vec B = (1.00\times10^{5})(1.00\times10^{-3}) = 1.00\times10^{2}\ \mathrm{V\,m^{-1}T} \ne 0\): neither. Since \(\mathcal I_2\) is invariant and equals \(\vec E\cdot\vec B\) in every frame, a frame with \(\vec E{}' = 0\) or \(\vec B{}' = 0\) would force \(\mathcal I_2 = 0\) there and hence everywhere. (A frame does exist in which \(\vec E{}'\parallel\vec B{}'\); finding it is a rotation followed by a boost, and is not asked for here.)

Case 4. \(\mathcal I_1 = (1.00/c)^2 - (1.00)^2/c^2 = 0\) and \(\mathcal I_2 = 0\), with \(F\ne0\): neither, and for the stronger reason that both invariants vanish. This is a null field: \(E = cB\) and \(\vec E\perp\vec B\) in every frame, no matter how you boost.

The general rule. Suppose \(\vec E\cdot\vec B = 0\) and \(E \lt cB\). Put \(\vec v = \vec E\times\vec B/|\vec B|^2\). Because \(\vec E\perp\vec B\) we have \(|\vec E\times\vec B| = EB\), so \(|\vec v| = E/B\), and \(|\vec v|/c = E/(cB) \lt 1\) ✓: it is a legitimate boost. The vector \(\vec v\) is perpendicular to both fields, so both are transverse to the boost, and the transverse rule gives

\[\vec E{}'_\perp = \gamma\big(\vec E + \vec v\times\vec B\big) = \gamma\Big(\vec E + \frac{(\vec E\times\vec B)\times\vec B}{|\vec B|^2}\Big) = \gamma\Big(\vec E + \frac{(\vec E\cdot\vec B)\vec B - |\vec B|^2\vec E}{|\vec B|^2}\Big) = 0,\]

using \(\vec E\cdot\vec B = 0\) and the triple product \((\vec a\times\vec b)\times\vec b = (\vec a\cdot\vec b)\vec b - |\vec b|^2\vec a\) (unit 08). So the field is purely magnetic there, and the mirror rule for \(E \gt cB\) is \(\vec v = c^2\,\vec E\times\vec B/E^2\), of magnitude \(c^2B/E \lt c\). \(\square\)

No rest frame for light. If a frame existed in which the wave were at rest, its fields there would be static, and in particular one could ask for a frame where \(\vec E{}' = 0\); but both invariants vanish, so \(\vec E{}' = 0\) would force \(\vec B{}' = 0\) too (from \(\mathcal I_1 = 0\)), hence \(F = 0\) in every frame, contradicting \(F\ne0\). One line, and it settles a question Einstein said he first asked at sixteen.

Check. Case 1 invariant in the new frame: \(-E'^2/c^2 = -(5.19735\times10^{5})^2/c^2 = -3.006\times10^{-6}\ \mathrm{T^2}\) ✓, the original value. Case 2: \(B'^2 = (1.73165\times10^{-3})^2 = 2.999\times10^{-6}\ \mathrm{T^2}\) ✓. Limits: in case 1, as \(cB\to E\) the required \(\beta\to1\) and the surviving field \(E/\gamma\to0\) — the null case is the boundary, approached but never reached ✓. Dimensions: \(c^2B/E\) has units \(\mathrm{m^2s^{-2}\,T/(V\,m^{-1})} = \mathrm{m\,s^{-1}}\) ✓.

Idea. The same bookkeeping as worked example 3, but the boost is now to the electrons' frame instead of the test charge's, and the two line densities move oppositely rather than one being at rest — so the leading term is quadratic, not linear.

1. In the electrons' rest frame: the lattice, of proper density \(\lambda\), moves at speed \(u\), so its density is \(\gamma_u\lambda\). The electrons, of proper density \(-\lambda/\gamma_u\) (their lab density \(-\lambda\) was already contracted by \(\gamma_u\)), are at rest, so their density is \(-\lambda/\gamma_u\). Adding,

\[\lambda'' = \gamma_u\lambda - \frac{\lambda}{\gamma_u} = \lambda\gamma_u\Big(1 - \frac1{\gamma_u^2}\Big) = \lambda\gamma_u\frac{u^2}{c^2} \approx \lambda\frac{u^2}{c^2} \gt 0 .\]

It is positive: in the electrons' frame the wire is positively charged, so a co-moving negative test charge is attracted — as it must be, since in the lab a charge drifting with the electrons is a parallel current and parallel currents attract.

2. With \(\lambda = 1.3602\times10^{4}\ \mathrm{C\,m^{-1}}\) and \(u = 7.3516\times10^{-5}\ \mathrm{m\,s^{-1}}\),

\[\lambda'' = \frac{(1.3602\times10^{4})(7.3516\times10^{-5})^2}{8.98755\times10^{16}} = 8.180\times10^{-22}\ \mathrm{C\,m^{-1}},\]

which is \(8.180\times10^{-22}/1.602177\times10^{-19} = 5.11\times10^{-3}\) elementary charges per metre: one extra proton's worth of charge per 196 metres of wire.

3. \(E''_y = \dfrac{\lambda''}{2\pi\varepsilon_0 r} = \dfrac{8.180\times10^{-22}}{1.11265\times10^{-12}} = 7.352\times10^{-10}\ \mathrm{V\,m^{-1}}\), pointing away from the wire. So the force per unit charge is \(7.352\times10^{-10}\ \mathrm{N\,C^{-1}}\).

4. In the lab, \(B = \mu_0I/2\pi r = 1.00\times10^{-5}\) T, and the charge moves at speed \(u\), so the magnetic force per unit charge is

\[uB = (7.3516\times10^{-5})(1.00\times10^{-5}) = 7.352\times10^{-10}\ \mathrm{N\,C^{-1}} \checkmark\]

to four figures. (The two differ by \(\gamma_u = 1 + 3\times10^{-26}\), which no arithmetic here resolves.)

5. Because in the electrons' frame both line densities are moving relative to the lab — one forwards, one backwards — so the first-order pieces of the two \(\gamma\) factors cancel and the residue is \(O(u^2/c^2)\). In the test charge's frame the imbalance was \(\gamma_v\lambda uv/c^2\): one factor of the electrons' small drift and one factor of the test charge's much larger speed, which is why that case is measurable and this one is not.

Check. Dimensions: \(\lambda u^2/c^2\) is \(\mathrm{C\,m^{-1}}\) ✓. Limit \(u\to0\): \(\lambda''\to0\) and the wire is neutral in every frame — correct, since with no current there is no boost that does anything ✓. Consistency: the ratio of this effect to worked example 3's is \(u/(\gamma_v v) = 2.45\times10^{-11}\), and indeed \(8.180\times10^{-22}/3.3381\times10^{-11} = 2.45\times10^{-11}\) ✓.

Idea. \(J\) is a vector, so the only invariant question you can ask about it is the sign of \(\eta(J,J)\); and that sign is exactly the answer to "can I find a frame where this wire is uncharged, or one where the current stops?".

1. \(u^\mu = \gamma(c,\vec v)\), so \(J^0 = \gamma\rho_0c\), and since \(J^0 = c\rho\) by definition, \(\rho = \gamma\rho_0\) ✓. The spatial part is \(\vec J = \gamma\rho_0\vec v = \rho\vec v\) ✓. Then \(\eta(J,J) = \rho_0^2\,\eta(u,u) = -\rho_0^2c^2\).

2. (\(\Rightarrow\)) If a frame has \(\rho' = 0\) then \(J'^0 = 0\), so \(\eta(J,J) = |\vec J{}'|^2 \gt 0\) — spacelike (and nonzero, since \(J\ne0\)). If a frame has \(\vec J{}' = 0\) then \(\eta(J,J) = -(J'^0)^2 \lt 0\) — timelike.

(\(\Leftarrow\)) Suppose \(J\) is spacelike. Rotate so \(\vec J\parallel\hat x\), giving \(J^\mu = (c\rho, J, 0, 0)\) with \(J^2 \gt c^2\rho^2\), so \(|c\rho/J| \lt 1\). Boost along \(\hat x\) with \(\beta = c\rho/J\): then \(J'^0 = \gamma(c\rho - \beta J) = \gamma(c\rho - c\rho) = 0\) ✓, and \(|\beta|\lt1\) makes it a legitimate boost. Suppose instead \(J\) is timelike; then \(|J/c\rho| \lt 1\) and the boost \(\beta = J/(c\rho)\) gives \(J'^1 = \gamma(J - \beta c\rho) = 0\) ✓. \(\square\)

3. \(J^\mu = (0,\ 1.00\times10^{6},\ 0,\ 0)\ \mathrm{A\,m^{-2}}\), and \(\eta(J,J) = +1.00\times10^{12} \gt 0\): spacelike. Consistent with part 2: the lab is a frame in which \(\rho = 0\), and no frame exists in which the current stops — you cannot outrun a current by running alongside it.

4. With \(\beta = v/c = 1.000692\times10^{-2}\) and \(\gamma = 1.0000500730\),

\[J'^0 = \gamma\big(J^0 - \beta J^1\big) = -(1.0000500730)(1.000692\times10^{-2})(1.00\times10^{6}) = -1.00074\times10^{4}\ \mathrm{A\,m^{-2}},\] \[\rho' = \frac{J'^0}{c} = -3.3381\times10^{-5}\ \mathrm{C\,m^{-3}} .\]

The cross-section is transverse to the boost and so unchanged, \(A = 1.00\times10^{-6}\ \mathrm{m^2}\), giving \(\lambda' = \rho'A = -3.3381\times10^{-11}\ \mathrm{C\,m^{-1}}\) — exactly worked example 3's number ✓. Two entirely different routes: there, contracting two line densities; here, boosting one four-vector.

5. Take \(\partial_\nu\) of both sides: \(\mu_0\partial_\nu J^\nu = \partial_\nu\partial_\mu F^{\nu\mu}\). The two facts: mixed partial derivatives commute for \(C^2\) fields (Clairaut, unit 03), so \(\partial_\nu\partial_\mu\) is symmetric in \(\mu\nu\); and \(F^{\nu\mu}\) is antisymmetric. The contraction of a symmetric with an antisymmetric pair vanishes (unit 16). Hence \(\partial_\nu J^\nu = 0\). \(\square\)

Check. Dimensions: \([J^\mu] = \mathrm{A\,m^{-2}}\) throughout, since \(c\rho\) has those units ✓. Limit: at \(v\to0\) the boost does nothing and \(\rho' \to 0\) ✓. Sign: \(\rho'\lt0\) means the wire is negatively charged in the charge's frame, attracting the positive test charge — which is the direction the lab's magnetic force pointed ✓.

Idea. Two of the three terms in the expanded divergence have nothing to do with the sources, and the Bianchi identity — which is the statement that \(F\) comes from a potential — is exactly what makes them cancel. Conservation of field energy is therefore a consequence of the existence of \(A\), which is a pleasing thing to know.

1. Symmetry. \(F^{\mu\alpha}F^\nu{}_\alpha = \eta_{\alpha\beta}F^{\mu\alpha}F^{\nu\beta}\), and swapping \(\mu\leftrightarrow\nu\) relabels the dummy pair \(\alpha\leftrightarrow\beta\) and uses the symmetry of \(\eta\); \(\eta^{\mu\nu}\) is symmetric. Hence \(T\) is symmetric.

Tracelessness. \(T^\mu{}_\mu = \mu_0^{-1}\big(F^{\mu\alpha}F_{\mu\alpha} - \tfrac14\eta^\mu{}_\mu F_{\alpha\beta}F^{\alpha\beta}\big)\), and \(\eta^\mu{}_\mu = \delta^\mu{}_\mu = 4\), so the bracket is \(F^{\mu\alpha}F_{\mu\alpha} - F_{\alpha\beta}F^{\alpha\beta} = 0\).

2. The divergence. Using \(\eta^{\mu\nu}\partial_\nu = \partial^\mu\),

\[\mu_0\,\partial_\nu T^{\mu\nu} = \big(\partial_\nu F^{\nu}{}_{\alpha}\big)F^{\mu\alpha} + F^{\nu\alpha}\,\partial_\nu F^{\mu}{}_{\alpha} - \tfrac12 F^{\alpha\beta}\,\partial^\mu F_{\alpha\beta} .\]

First term. Maxwell reads \(\partial_\nu F^{\beta\nu} = \mu_0J^\beta\), so \(\partial_\nu F^{\nu\beta} = -\mu_0J^\beta\) and, lowering, \(\partial_\nu F^\nu{}_\alpha = -\mu_0J_\alpha\). The first term is therefore \(-\mu_0F^{\mu\alpha}J_\alpha = -\mu_0F^\mu{}_\nu J^\nu\).

The lemma. Lower \(\mu\) and set \(X_\mu = F^{\nu\alpha}\partial_\nu F_{\mu\alpha}\). The Bianchi identity \(\partial_\nu F_{\mu\alpha} + \partial_\mu F_{\alpha\nu} + \partial_\alpha F_{\nu\mu} = 0\) gives

\[X_\mu = -F^{\nu\alpha}\partial_\mu F_{\alpha\nu} - F^{\nu\alpha}\partial_\alpha F_{\nu\mu} = +F^{\nu\alpha}\partial_\mu F_{\nu\alpha} - F^{\nu\alpha}\partial_\alpha F_{\nu\mu},\]

the first step using \(F_{\alpha\nu} = -F_{\nu\alpha}\). In the remaining term relabel the dummies \(\nu\leftrightarrow\alpha\) and use antisymmetry twice:

\[-F^{\nu\alpha}\partial_\alpha F_{\nu\mu} \;=\; -F^{\alpha\nu}\partial_\nu F_{\alpha\mu} \;=\; +F^{\nu\alpha}\partial_\nu F_{\alpha\mu} \;=\; -F^{\nu\alpha}\partial_\nu F_{\mu\alpha} \;=\; -X_\mu .\]

Hence \(2X_\mu = F^{\nu\alpha}\partial_\mu F_{\nu\alpha}\), i.e. \(X_\mu = \tfrac12F^{\alpha\beta}\partial_\mu F_{\alpha\beta}\) \(\square\) — which is precisely the third term with the opposite sign. The second and third terms cancel, and

\[\partial_\nu T^{\mu\nu} = -F^{\mu}{}_{\nu}J^{\nu}. \qquad\square\]

3. In vacuum \(J^\nu = 0\), so \(\partial_\nu T^{\mu\nu} = 0\): four local conservation laws, energy and the three momenta, in exactly unit 22's form.

4. In \(n\) dimensions \(\eta^\mu{}_\mu = n\), so the trace of the second term is \(-\tfrac14 n F_{\alpha\beta}F^{\alpha\beta}\) and the trace vanishes only if the coefficient is \(1/n\). With the coefficient fixed at \(\tfrac14\) by the definition, tracelessness is a four-dimensional accident. Since unit 22's perfect fluid has \(T^\mu{}_\mu = -\rho c^2 + 3p\), \(T^\mu{}_\mu = 0\) gives the radiation equation of state \(p = \rho c^2/3\), the one unit 39 uses for the early universe.

Check. Dimensions: \([T] = [F^2/\mu_0] = \mathrm{T^2\,A^2\,N^{-1}}\); with \(\mathrm T = \mathrm{N\,A^{-1}m^{-1}}\) that is \(\mathrm{N\,m^{-2}} = \mathrm{J\,m^{-3}}\) ✓, an energy density. Both sides of part 2 are then \(\mathrm{J\,m^{-4}} = \mathrm{N\,m^{-3}}\), and \([FJ] = \mathrm{T\,A\,m^{-2}} = \mathrm{N\,m^{-3}}\) ✓. Special case: for a static field with no current both sides vanish, and the statement reduces to the electrostatic Maxwell stress being divergence-free — unit 24's result ✓.

Idea. A null field has so little structure left that its stress-energy tensor is forced to be a single outer product of a null vector with itself, and everything else follows by inspection.

1. \(\vec E\cdot\vec B = 0\) since \(\hat y\perp\hat z\); and \(B^2 - E^2/c^2 = E^2/c^2 - E^2/c^2 = 0\). Both vanish, as they must for light.

2. Write \(E = E_0\cos(k(x-ct))\). Energy density: \(u = \tfrac{\varepsilon_0}{2}E^2 + \tfrac{1}{2\mu_0}\tfrac{E^2}{c^2} = \varepsilon_0E^2\), using \(1/(\mu_0c^2) = \varepsilon_0\). Poynting: \(\vec S = \tfrac1{\mu_0}\vec E\times\vec B = \tfrac{E^2}{\mu_0c}\,\hat y\times\hat z = \varepsilon_0cE^2\,\hat x = uc\,\hat x\). Stress: \(\sigma_{xx} = \varepsilon_0(E_x^2 - \tfrac12E^2) + \tfrac1{\mu_0}(B_x^2 - \tfrac12B^2) = -\tfrac{\varepsilon_0}{2}E^2 - \tfrac{1}{2\mu_0}B^2 = -u\), so \(T^{xx} = -\sigma_{xx} = u\); \(\sigma_{yy} = \varepsilon_0(E^2 - \tfrac12E^2) - \tfrac{1}{2\mu_0}B^2 = \tfrac{\varepsilon_0}{2}E^2 - \tfrac{\varepsilon_0}{2}E^2 = 0\), and likewise \(\sigma_{zz} = 0\); all off-diagonal spatial components vanish. Hence

\[T^{\mu\nu} = u\begin{pmatrix}1&1&0&0\\1&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix} = u\,n^\mu n^\nu, \qquad n^\mu = (1,1,0,0),\quad \eta_{\mu\nu}n^\mu n^\nu = 0 .\]

3. \(u\) depends on \(x\) and \(t\) only through \(\xi = x - ct\), so \(\partial_0 u = \tfrac1c\partial_t u = -u'(\xi)\) and \(\partial_1u = u'(\xi)\). Then

\[\partial_\nu T^{\mu\nu} = n^\mu n^\nu\partial_\nu u = n^\mu\big(\partial_0u + \partial_1u\big) = n^\mu\big(-u' + u'\big) = 0 \checkmark\]

— and the mechanism is that \(n^\nu\partial_\nu\) is the derivative along the null direction the wave travels in, and \(u\) is constant along it.

4. \(T^{x\nu}\) says that \(x\)-momentum crosses a surface of normal \(\hat x\) at rate \(T^{xx} = u\) per unit area. An absorber takes all of it, so the pressure is \(p_{\text{abs}} = \langle u\rangle = \langle S\rangle/c\). A mirror sends it back, doubling the momentum transfer: \(p_{\text{mir}} = 2\langle S\rangle/c\).

5. With \(\langle S\rangle = 1361\ \mathrm{W\,m^{-2}}\),

\[p_{\text{abs}} = \frac{1361}{2.99792458\times10^{8}} = 4.540\times10^{-6}\ \mathrm{Pa},\qquad p_{\text{mir}} = 9.080\times10^{-6}\ \mathrm{Pa}.\]

The amplitude: \(\langle S\rangle = \tfrac12\varepsilon_0cE_0^2\) (the time average of \(\cos^2\) is \(\tfrac12\)), so \(E_0 = \sqrt{2\langle S\rangle/\varepsilon_0c} = 1.013\times10^{3}\ \mathrm{V\,m^{-1}}\). Force on \(1.00\ \mathrm{km^2} = 1.00\times10^{6}\ \mathrm{m^2}\) of absorber: \(4.540\times10^{-6}\times10^{6} = 4.540\) N.

Check. Dimensions: \([S/c] = \mathrm{W\,m^{-2}}/(\mathrm{m\,s^{-1}}) = \mathrm{J\,m^{-3}} = \mathrm{Pa}\) ✓. Trace: \(T^\mu{}_\mu = u\,\eta_{\mu\nu}n^\mu n^\nu = 0\) ✓, as the general theorem demands. Limit: for a wave at normal incidence the pressure equals the energy density, which is the \(p = \rho c^2/3\) equation of state not applying — that one is for isotropic radiation, where the factor \(\tfrac13\) comes from averaging \(\cos^2\theta\) over directions. A sanity number: \(4.54\) N on a square kilometre is about the weight of a small apple, which is why solar sails need to be very large and very light.

Idea. Radiated power is an invariant, and there is only one invariant quadratic in the acceleration; that pins the formula, and the rest-frame limit pins the constant.

1. \(\dot\gamma = \gamma^3(\vec v\cdot\vec a)/c^2\), as in rung 3. Then

\[a^\mu = \frac{\mathrm{d}u^\mu}{\mathrm{d}\tau} = \gamma\frac{\mathrm d}{\mathrm dt}\big(\gamma c,\ \gamma\vec v\big) = \gamma\big(\dot\gamma c,\ \dot\gamma\vec v + \gamma\vec a\big) = \Big(\gamma^4\frac{\vec v\cdot\vec a}{c},\ \gamma^2\vec a + \gamma^4\frac{(\vec v\cdot\vec a)\vec v}{c^2}\Big).\]

2. \(a\cdot a = -(a^0)^2 + |\vec a_{\text{4-sp}}|^2\), i.e.

\[a\cdot a = -\gamma^8\frac{(\vec v\cdot\vec a)^2}{c^2} + \gamma^4a^2 + 2\gamma^6\frac{(\vec v\cdot\vec a)^2}{c^2} + \gamma^8\frac{(\vec v\cdot\vec a)^2v^2}{c^4}.\]

The first and last combine: \(\gamma^8\frac{(\vec v\cdot\vec a)^2}{c^2}\big(\frac{v^2}{c^2} - 1\big) = -\gamma^6\frac{(\vec v\cdot\vec a)^2}{c^2}\). Adding the middle term \(2\gamma^6(\vec v\cdot\vec a)^2/c^2\) leaves \(+\gamma^6(\vec v\cdot\vec a)^2/c^2\), so

\[a\cdot a = \gamma^4|\vec a|^2 + \gamma^6\frac{(\vec v\cdot\vec a)^2}{c^2} \;\gt\;0 .\]

For the second form, substitute \(|\vec v\times\vec a|^2 = v^2a^2 - (\vec v\cdot\vec a)^2\):

\[\gamma^6\Big(a^2 - \frac{v^2a^2}{c^2} + \frac{(\vec v\cdot\vec a)^2}{c^2}\Big) = \gamma^6\Big(\frac{a^2}{\gamma^2} + \frac{(\vec v\cdot\vec a)^2}{c^2}\Big) = \gamma^4a^2 + \gamma^6\frac{(\vec v\cdot\vec a)^2}{c^2} \checkmark\]

Positivity also follows from unit 27: \(a\) is \(\eta\)-orthogonal to the timelike \(u\), hence spacelike or zero.

3. In the instantaneous rest frame the dipole pattern is \(\propto\sin^2\theta\) about \(\vec a\) (unit 24, cited), which is even under \(\hat n\to-\hat n\), so the radiated momentum cancels and the radiated four-momentum in proper time \(\mathrm{d}\tau\) is \(\mathrm{d}P^\mu = (\mathrm{d}E_0/c,\ \vec 0)\). In any other frame \(\mathrm{d}E = \gamma\,\mathrm{d}E_0\) and \(\mathrm{d}t = \gamma\,\mathrm{d}\tau\), so \(\mathrm{d}E/\mathrm{d}t = \mathrm{d}E_0/\mathrm{d}\tau\): \(P\) is a scalar.

The assumption is the next step: that \(P\) is built from \(u\) and \(a\) alone and is quadratic in \(a\). Granted it, the only scalar available is \(a^\mu a_\mu\) (since \(\eta(u,a) = 0\) and \(\eta(u,u) = -c^2\) are constants), so \(P = K\,a^\mu a_\mu\), and the rest frame, where \(a^\mu a_\mu = |\vec a|^2\) and Larmor gives \(P = \mu_0q^2|\vec a|^2/6\pi c\), fixes \(K = \mu_0q^2/6\pi c\). The assumption excludes a dependence on \(\mathrm{d}a/\mathrm{d}\tau\), which is not obviously illegitimate; the formula is derived without it in Jackson §14.2–14.3, from the Liénard–Wiechert fields.

4. \(\vec a\parallel\vec v\): \((\vec v\cdot\vec a)^2 = v^2a^2\), so \(a\cdot a = \gamma^4a^2(1 + \gamma^2v^2/c^2) = \gamma^6a^2\), giving \(P\propto\gamma^6\). \(\vec a\perp\vec v\): \(\vec v\cdot\vec a = 0\), so \(a\cdot a = \gamma^4a^2\) and \(P\propto\gamma^4\). At a given force the comparison reverses, because a transverse force produces \(a = F/\gamma m\) and a longitudinal one \(a = F/\gamma^3m\) — which is unit 27's case against relativistic mass, reappearing.

5. Circular: \(a = v^2/\rho\), \(a\cdot a = \gamma^4v^4/\rho^2\), so

\[P = \frac{\mu_0q^2\gamma^4v^4}{6\pi c\rho^2},\qquad \Delta E = P\cdot\frac{2\pi\rho}{v} = \frac{\mu_0q^2\gamma^4v^3}{3c\rho} \;\xrightarrow{\ v\to c\ }\; \frac{\mu_0q^2c^2\gamma^4}{3\rho} = \frac{q^2\gamma^4}{3\varepsilon_0\rho},\]

using \(\mu_0c^2 = 1/\varepsilon_0\).

LEP. \(\gamma = 104.5\times10^{9}/0.510999\times10^{6} = 2.0450\times10^{5}\), so

\[\Delta E = \frac{(1.602177\times10^{-19})^2(2.0450\times10^{5})^4}{3(8.8541878\times10^{-12})(3096)} = 5.459\times10^{-10}\ \mathrm J = 3.41\ \mathrm{GeV\ per\ turn}.\]

LHC. \(\gamma = 6.8\times10^{12}\,\mathrm{eV}\cdot e/(m_pc^2) = 7247\), and the same formula with \(\rho = 2804\) m gives \(9.51\times10^{-16}\ \mathrm J = 5.94\ \mathrm{keV}\) per turn.

Comment. A factor \(5.7\times10^{5}\). LEP threw away three per cent of its beam energy every turn, which is what stopped it at about 104 GeV per beam and is why the same tunnel now holds a proton machine. The LHC's 5.94 keV per turn is negligible as a beam loss but is a real heat load on a pipe held at 1.9 K.

Check. Dimensions of \(\Delta E\): \([q^2/\varepsilon_0\rho] = \mathrm{C^2/(F\,m^{-1}\cdot m)} = \mathrm{C^2/F} = \mathrm J\) ✓. Limit: \(\gamma\to1\) gives \(\Delta E = q^2/3\varepsilon_0\rho\), the non-relativistic circular-orbit loss, and at \(\rho\to\infty\) it vanishes ✓ — straight-line motion at constant speed does not radiate. Cross-check: the published LHC figure at 7 TeV is 6.7 keV per turn; scaling ours by \((7/6.8)^4 = 1.123\) gives 6.67 keV ✓.

Idea. A hard problem in the lab (coupled linear ODEs with a driving term) becomes an easy one (uniform circular motion) in a frame chosen by the invariants. This is what the classification is for.

1. \(\mathcal I_2 = \vec E\cdot\vec B = 0\). \(E/c = 3.336\times10^{-5}\) T, so \(\mathcal I_1 = (0.100)^2 - (3.336\times10^{-5})^2 = 1.000\times10^{-2}\ \mathrm{T^2} \gt 0\) — the electric field can be removed.

2. With \(\vec E\parallel\hat y\), \(\vec B\parallel\hat z\), the required boost is along \(\hat x\), and \(E'_y = \gamma(E_y - vB_z) = 0\) gives

\[v_d = \frac{E}{B} = \frac{1.00\times10^{4}}{0.100} = 1.00\times10^{5}\ \mathrm{m\,s^{-1}}\ \hat x, \qquad \beta_d = 3.336\times10^{-4},\quad \gamma_d - 1 = 5.56\times10^{-8}.\]

The surviving field is \(B'_z = \gamma_d(B_z - v_dE_y/c^2) = B_z/\gamma_d = 0.0999999944\) T, which is \(0.100\) T to every digit that matters.

3. In the drift frame the electron starts with velocity \(-\vec v_d\) (it was at rest in the lab), speed \(1.00\times10^{5}\ \mathrm{m\,s^{-1}}\), and there is no electric field, so its speed is constant and the motion is a circle. Unit 23's problem sheet did that circle: \(r = mv/qB\) and \(\omega = qB/\gamma m\), which is \(qvB = \gamma mv^2/r\) rearranged. With \(\gamma\approx1\),

\[\omega = \frac{eB'}{m_e} = \frac{(1.602177\times10^{-19})(0.100)}{9.1093837\times10^{-31}} = 1.759\times10^{10}\ \mathrm{rad\,s^{-1}},\qquad T = \frac{2\pi}{\omega} = 3.572\times10^{-10}\ \mathrm s,\] \[r = \frac{m_ev_d}{eB} = \frac{(9.1093837\times10^{-31})(1.00\times10^{5})}{(1.602177\times10^{-19})(0.100)} = 5.686\times10^{-6}\ \mathrm m .\]

4. The lab trajectory is that circle plus a uniform drift at \(\vec v_d\): a cycloid, with cusps where the circular velocity exactly cancels the drift — that is, at the moments when the electron is instantaneously at rest, which includes the start. The maximum lab speed is reached half a period later, when the two add:

\[v_{\max} = 2v_d = 2.00\times10^{5}\ \mathrm{m\,s^{-1}}.\]

5. In general the condition for killing \(\vec E\) is \(\vec E + \vec v\times\vec B = 0\), and when \(\vec E\perp\vec B\) it is solved by \(\vec v_d = \vec E\times\vec B/|\vec B|^2\): substituting,

\[\vec v_d\times\vec B = \frac{(\vec E\times\vec B)\times\vec B}{|\vec B|^2} = \frac{(\vec E\cdot\vec B)\vec B - |\vec B|^2\vec E}{|\vec B|^2} = -\vec E \checkmark\]

Neither \(q\) nor \(m\) appears, so electrons and protons drift together at the same velocity — which is why this drift carries no current, and why it is the workhorse of plasma physics.

Check. Dimensions: \([E/B] = \mathrm{V\,m^{-1}/T} = \mathrm{m\,s^{-1}}\) ✓; \([m v/qB] = \mathrm{kg\,m\,s^{-1}/(C\,T)} = \mathrm m\) ✓. Limit: \(E\to0\) gives \(v_d\to0\) and the motion is a circle about a fixed centre ✓; \(B\to0\) gives \(v_d\to\infty\), correctly signalling that the classification fails once \(E \gt cB\). Number: \(r\omega = (5.686\times10^{-6})(1.759\times10^{10}) = 1.00\times10^{5}\ \mathrm{m\,s^{-1}} = v_d\) ✓, as circular motion requires.

Idea. A determinant is exactly the factor by which a top-degree antisymmetric object scales under a linear map, so the answer to part 1 is a determinant before any calculation starts.

1. Let \(T_{\mu\nu\rho\sigma} := (\Lambda^{-1})^\alpha{}_\mu(\Lambda^{-1})^\beta{}_\nu(\Lambda^{-1})^\kappa{}_\rho (\Lambda^{-1})^\delta{}_\sigma\,[\alpha\beta\kappa\delta]\). Swapping two of \(\mu,\nu,\rho,\sigma\) swaps the corresponding two factors, which by the total antisymmetry of \([\ \cdot\ ]\) changes the sign; so \(T\) is totally antisymmetric.

The space of such objects is one-dimensional: any totally antisymmetric \(T\) vanishes whenever two indices agree, and its value on any ordering of \((0,1,2,3)\) is \(\pm T_{0123}\). So \(T\) is determined by the single number \(T_{0123}\), and \(T = T_{0123}\,\epsilon\).

Evaluate: \(T_{0123} = \sum_{\alpha\beta\kappa\delta}[\alpha\beta\kappa\delta] (\Lambda^{-1})^\alpha{}_0(\Lambda^{-1})^\beta{}_1(\Lambda^{-1})^\kappa{}_2(\Lambda^{-1})^\delta{}_3\), which is the Leibniz expansion of \(\det(\Lambda^{-1})\). Hence \(T = \det(\Lambda^{-1})\,\epsilon\). \(\square\)

2. A Lorentz transformation satisfies \(\Lambda^\top\eta\Lambda = \eta\); taking determinants, \((\det\Lambda)^2\det\eta = \det\eta\), so \(\det\Lambda = \pm1\) and \(\det(\Lambda^{-1}) = \det\Lambda\). For proper transformations (\(\det\Lambda = +1\)) the components are unchanged; for improper ones (a reflection, \(\det\Lambda = -1\)) they flip sign. Since every boost and rotation is connected to the identity, and the determinant is continuous and takes values in \(\{\pm1\}\), every boost and rotation is proper.

3. \(\epsilon^{\mu\nu\rho\sigma} = \eta^{\mu\alpha}\eta^{\nu\beta}\eta^{\rho\kappa}\eta^{\sigma\delta} \epsilon_{\alpha\beta\kappa\delta}\); exactly one of the four raised indices is \(0\) in any nonzero component, contributing one factor \(\eta^{00} = -1\), so \(\epsilon^{\mu\nu\rho\sigma} = -\epsilon_{\mu\nu\rho\sigma}\) numerically. There are \(4! = 24\) nonzero components, each with \(\epsilon^{\ldots}\epsilon_{\ldots} = (-1)(\pm1)^2 = -1\), so the full contraction is \(-24\) ✓. Fixing one index, say \(\mu = \alpha = 0\), leaves \(3! = 6\) assignments of the others, each contributing \(-1\); by symmetry the same holds for each value of \(\mu\), and the result vanishes for \(\mu\ne\alpha\) (there is then no way to fill the remaining three slots without repeating an index). Hence \(\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\nu\rho\sigma} = -6\delta^\mu{}_\alpha\) ✓. The minus signs are the single \(\eta^{00} = -1\); in Euclidean signature they are absent.

4. \(\mathcal I_2 = \tfrac c4F_{\mu\nu}\tilde F^{\mu\nu} = \tfrac c8\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\). Under a reflection, \(F\) transforms as a tensor and \(\epsilon\) picks up \(\det\Lambda = -1\), so \(\mathcal I_2\) changes sign: a pseudoscalar. Explicitly: take \(\vec E = E\hat z\), \(\vec B = B\hat z\), so \(\mathcal I_2 = EB\). Reflect in the plane \(z = 0\): \(\vec E\) is a polar vector, so \(E_z \to -E_z\); \(\vec B\) is axial, so \(B_z\to +B_z\). The mirror configuration has \(\mathcal I_2 = -EB\) ✓. No rotation or boost can carry one into the other.

Check. Consistency of part 3: contracting \(\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\nu\rho\sigma} = -6\delta^\mu{}_\alpha\) over \(\mu = \alpha\) gives \(-6\times4 = -24\) ✓, the first identity. Euclidean limit: the same counting in three Euclidean dimensions gives \(\varepsilon_{ijk}\varepsilon_{ijk} = 6\) and \(\varepsilon_{ijk}\varepsilon_{ljk} = 2\delta_{il}\) ✓, the identities unit 14 uses. Physical special case: for a purely electric or purely magnetic field \(\mathcal I_2 = 0\), which is reflection-symmetric, and indeed such a field has a mirror image in its own Lorentz orbit ✓.

Idea. A third-order equation of motion has one more constant of integration than initial data can supply, and whichever way you spend it you buy a pathology. That is the honest content of this problem.

1. \(\tau_0 = \mu_0q^2/6\pi mc\). For an electron,

\[\tau_0 = \frac{(1.256637\times10^{-6})(1.602177\times10^{-19})^2}{6\pi(9.109384\times10^{-31})(2.997925\times10^{8})} = 6.266\times10^{-24}\ \mathrm s .\]

For a proton the charge is the same and the mass is \(1836\) times larger, so \(\tau_0 = 3.413\times10^{-27}\) s.

2. With \(F_{\text{ext}} = 0\), \(\dot a = a/\tau_0\), so \(a(t) = a_0e^{t/\tau_0}\). Unless \(a_0 = 0\), a free electron accelerates without limit and without a source of energy: the runaway. A factor \(e^{100}\) takes \(100\tau_0 = 6.27\times10^{-22}\) s.

3. For \(t \ge 0\): the general solution is \(a = F_0/m + Ce^{t/\tau_0}\), and boundedness as \(t\to+\infty\) forces \(C = 0\), so \(a(t) = F_0/m\). For \(t \lt 0\): the equation is homogeneous, so \(a(t) = Ae^{t/\tau_0}\), which is automatically bounded as \(t\to-\infty\). Continuity of \(a\) at \(t = 0\) — required because \(\ddot v\) must be at worst a step, so \(\dot v\) is continuous — gives \(A = F_0/m\). Hence

\[a(t) = \frac{F_0}{m}\times\begin{cases}e^{t/\tau_0}, & t \lt 0,\\[2pt] 1, & t\ge0.\end{cases}\]

The electron is already accelerating before the force is applied, with an \(e\)-folding time \(\tau_0 = 6.27\times10^{-24}\) s: pre-acceleration. Causality is violated, by a very short interval, but violated.

4. \(r_e/c = 2.818\times10^{-15}/2.998\times10^{8} = 9.40\times10^{-24}\) s, so \(\tau_0 = \tfrac23 r_e/c\) exactly — the pathology sets in at the scale where the electron's own field energy reaches \(mc^2\). Meanwhile \(\hbar/m_ec^2 = 1.288\times10^{-21}\) s, which is \(206\) times larger. Quantum mechanics therefore takes over long before the classical theory misbehaves: classical electrodynamics is an effective theory, and its lower cutoff is comfortably above the scale of its own inconsistency.

5. Neither has been observed because both live at \(t\sim\tau_0\), a time scale two hundred times shorter than the one at which the classical description of an electron has already failed. What is observed is ordinary radiation reaction — synchrotron damping in storage rings, and the recoil on electrons in ultra-intense laser fields — for which the Landau–Lifshitz reduction of the equation (replace \(\ddot v\) by its value from the external force alone) is accurate and free of both diseases.

Check. Dimensions: \([\mu_0q^2/mc] = \mathrm{N\,A^{-2}C^2/(kg\,m\,s^{-1})} = \mathrm{N\,s^2/(kg\,m\,s^{-1})} = \mathrm s\) ✓. Limit: \(\tau_0\to0\) (a heavy or weakly charged particle) returns Newton's second law exactly, with no runaway and no pre-acceleration ✓ — the pathology is a singular perturbation in \(\tau_0\), of precisely the kind unit 21 studied, with the small parameter multiplying the highest derivative. Consistency: the ratio \(\tau_0(\mathrm p)/\tau_0(\mathrm e) = 3.413\times10^{-27}/6.266\times10^{-24} = 1/1836\) ✓, the mass ratio.

Another way

Part 3 can be done in one step by writing the equation as \(a(t) = \frac1{m\tau_0}\int_t^\infty e^{-(s-t)/\tau_0}F_{\text{ext}}(s)\,\mathrm{d}s\), which is the unique bounded solution. The integral runs over the future, which displays the acausality without solving anything.