Radiation from a Moving Charge

These notes closely follow the original lecture derivation, with only minor additions for clarity, and use the conventions fixed earlier in the course.

1 Maxwell equations in covariant form

Starting from Maxwell’s equations, \[ \partial_\mu F^{\mu\nu}=-\mu_0J^\nu, \] with \(F^{0i}=E^i/c\), \(F^{ij}=\epsilon^{ijk}B_k\), and \(J^\mu=(c\rho,\mathbf J)\) as fixed in the conventions, one recovers the usual Maxwell equations component by component exactly as in Maxwell Equations in Covariant Form.

Introducing the four-potential, \[ F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu, \] the \(0i\) component reads \(F_{0i}=\partial_0A_i-\partial_iA_0=\frac1c\partial_tA_i-\partial_iA_0\); comparing with \(F_{0i}=-E_i/c\) (the lowered version of \(F^{0i}=E^i/c\)) gives \[ \frac{E_i}{c} = \partial_iA_0 - \frac{1}{c}\frac{\partial A_i}{\partial t}, \] and, from the spatial components, \[ \mathbf B=\nabla\times\mathbf A. \] In the electrostatic limit, \[ \frac{\partial A_i}{\partial t}=0, \qquad E_i=\partial_i(cA_0)=-\partial_i\phi, \] using \(cA_0=-\phi\) from \(A^\mu=(\phi/c,\mathbf A)\).

2 Wave equation for the four-potential

Substituting \(F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu\) into Maxwell’s equations and using that partial derivatives commute, \[ \partial_\mu F^{\mu\nu}=\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)=-\mu_0J^\nu, \] that is, \[ \Box A^\nu-\partial^\nu(\partial_\mu A^\mu) = -\mu_0J^\nu. \] Using the Lorenz gauge, \[ \partial_\mu A^\mu=0, \] the equations become \[ \boxed{\Box A^\nu=-\mu_0J^\nu.} \]

3 Retarded Green function

The Green function satisfies \[ \Box G(x-x')=\delta^{(4)}(x-x'). \] The retarded solution is \[ G_R(x-x') = \frac{\Theta(x^0-x'^0) \delta(x^0-x'^0-|\mathbf x-\mathbf x'|)} {4\pi|\mathbf x-\mathbf x'|}. \]

4 Current of a point charge

For a point charge, \[ J^\mu(x) = cq \int u^\mu(\tau) \delta^{(4)}(x-y(\tau)) \,\mathrm{d}\tau, \] where

• \(y^\mu(\tau)\) is the worldline;
• \(u^\mu=\mathrm{d}y^\mu/(c\,\mathrm{d}\tau)\) is the four-velocity, normalized as in the conventions.

5 Four-potential

Combining the Green function and the current, \[ A^\mu(x) = -\mu_0 \int G(x-z) J^\mu(z)\, \mathrm{d}^4z, \] Substituting the point-charge current and its own delta function collapses the \(z\)-integral onto the worldline, leaving a single integral over \(\tau\): \[ A^\mu(x) = \frac{\mu_0qc^2}{4\pi} \int \frac{ \Theta(x^0-y^0(\tau)) \delta(x^0-y^0(\tau)-|\mathbf x-\mathbf y(\tau)|) }{ |\mathbf x-\mathbf y(\tau)| } u^\mu(\tau)\, \mathrm{d}\tau. \] Only values satisfying \[ x^0-y^0(\tau)=|\mathbf x-\mathbf y(\tau)| \] contribute to the integral.

6 Covariant form

Using \[ (x-y)^2 = -(x^0-y^0)^2 + |\mathbf x-\mathbf y|^2, \] and \[ \delta(f(x)) = \sum_i \frac{\delta(x-x_i)} {|f'(x_i)|}, \] one finds \[ \Theta(x^0-y^0) \delta((x-y)^2) = \frac{ \delta(x^0-y^0-|\mathbf x-\mathbf y|) }{ 2|\mathbf x-\mathbf y| }. \] Therefore, \[ \boxed{ A^\mu(x) = \frac{\mu_0qc^2}{2\pi} \int \Theta(x^0-y^0(\tau)) \delta((x-y)^2) u^\mu(\tau) \,\mathrm{d}\tau. } \]

7 Retarded time

The retarded proper time is defined implicitly by \[ (x-y(\tau_r))^2=0, \] with \[ x^0>y^0(\tau_r). \] Equivalently, \[ t-t_r = \frac{ |\mathbf x-\mathbf y(t_r)| }{c}. \] Write this as \(f(t_r)=t-t_r-|\mathbf x-\mathbf y(t_r)|/c=0\). Since \(\mathrm{d}|\mathbf x-\mathbf y(t_r)|/\mathrm{d}t_r=-\hat{\mathbf n}\cdot\mathbf v\), with \(\hat{\mathbf n}\) pointing from the charge to the field point, differentiating with respect to \(t_r\) gives \[ \dot f = \frac{\hat{\mathbf n}\cdot\mathbf v}{c}-1<0, \] which never vanishes, showing that the solution is unique.

Example (uniform velocity). For a charge moving with constant velocity \(\mathbf v_0\), \(\mathbf y(t_r)=\mathbf y_0+\mathbf v_0t_r\). Writing \(\Delta\mathbf x_0=\mathbf x-\mathbf y_0\), the defining condition \(c(t-t_r)=|\mathbf x-\mathbf y(t_r)|\) squares to \[ c^2(t-t_r)^2=|\Delta\mathbf x_0-\mathbf v_0t_r|^2 =|\Delta\mathbf x_0|^2-2\Delta\mathbf x_0\cdot\mathbf v_0\,t_r+v_0^2t_r^2. \] Collecting powers of \(t_r\) and dividing by \(c^2\) gives a quadratic equation, \[ \left(1-\frac{v_0^2}{c^2}\right)t_r^2 -2\left(t-\frac{\Delta\mathbf x_0\cdot\mathbf v_0}{c^2}\right)t_r +\left(t^2-\frac{|\Delta\mathbf x_0|^2}{c^2}\right)=0, \] with two roots — an advanced and a retarded one. Only the smaller root is causal (\(t_r<t\)), so \[ \boxed{ t_r = \frac{ \left(t-\dfrac{\Delta\mathbf x_0\cdot\mathbf v_0}{c^2}\right) -\dfrac1c\sqrt{\left(ct-\dfrac{\Delta\mathbf x_0\cdot\mathbf v_0}{c}\right)^2-\left(1-\dfrac{v_0^2}{c^2}\right)\left(c^2t^2-|\Delta\mathbf x_0|^2\right)} } {1-v_0^2/c^2}. } \]

8 Liénard–Wiechert potentials

It remains to evaluate the \(\tau\)-integral in the boxed covariant potential using the same distributional identity as in the Covariant form section, now applied to \(g(\tau)=(x-y(\tau))^2\), which has a single zero at \(\tau=\tau_r\): \[ \delta((x-y(\tau))^2)=\frac{\delta(\tau-\tau_r)}{|g'(\tau_r)|}, \qquad g'(\tau)=-2c\,u(\tau)\cdot(x-y(\tau)), \] where \(g'\) follows from \(\mathrm{d}y^\mu/\mathrm{d}\tau=c\,u^\mu\). At \(\tau=\tau_r\), with \(\Delta x^\mu=x^\mu-y^\mu(\tau_r)\) having \(\Delta x^0=|\mathbf x-\mathbf y|\) and spatial part \(|\mathbf x-\mathbf y|\hat{\mathbf n}\), \[ u\cdot\Delta x=-\gamma|\mathbf x-\mathbf y|\left(1-\hat{\mathbf n}\cdot\mathbf v/c\right), \] so that \(|g'(\tau_r)|=2c\gamma|\mathbf x-\mathbf y|(1-\hat{\mathbf n}\cdot\mathbf v/c)\). The \(\tau\)-integral then just picks out \(u^\mu(\tau_r)/|g'(\tau_r)|\), and the factors of \(\gamma\) cancel against those hidden in \(u^\mu=\gamma(1,\mathbf v/c)\), giving \[ \boxed{ A^\mu = \frac{\mu_0qc}{4\pi} \left. \frac{(1,\mathbf v/c)} {|\mathbf x-\mathbf y| \left(1-\hat{\mathbf n}\cdot\mathbf v/c\right)} \right|_{t_r}. } \] Hence \[ V = \frac{q}{4\pi\varepsilon_0} \left. \frac{1} {|\mathbf x-\mathbf y| (1-\hat{\mathbf n}\cdot\mathbf v/c)} \right|_{t_r}, \] and \[ \mathbf A = \frac{\mu_0qc}{4\pi} \left. \frac{\mathbf v} {|\mathbf x-\mathbf y| (1-\hat{\mathbf n}\cdot\mathbf v/c)} \right|_{t_r}. \]

9 Derivatives of the potential

The field \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\) requires differentiating the covariant potential \[ A^\nu = \frac{\mu_0qc^2}{2\pi} \int \Theta(x^0-y^0(\tau)) \delta((x-y(\tau))^2) u^\nu(\tau)\,\mathrm{d}\tau \] with respect to \(x^\mu\). Only \(\Theta\) and \(\delta\) depend on \(x\) inside the integral (through \(x-y(\tau)\)), so \[ \partial_\mu A^\nu = \frac{\mu_0qc^2}{2\pi} \int \Big[ \big(\partial_\mu\Theta(x^0-y^0)\big)\delta((x-y)^2) + \Theta(x^0-y^0)\,\partial_\mu\big[\delta((x-y)^2)\big] \Big] u^\nu\,\mathrm{d}\tau. \]

Discarding the self-field term. The first term has \(\partial_\mu\Theta(x^0-y^0)\propto\delta^0_\mu\,\delta(x^0-y^0)\), which multiplies \(\delta((x-y)^2)\). A point where both \(x^0=y^0\) and \((x-y)^2=-(x^0-y^0)^2+|\mathbf x-\mathbf y|^2=0\) forces \(\mathbf x=\mathbf y\) as well, i.e. \(x=y\): this term is supported only at the charge’s own location and gives an infinite self-field contribution there. It is dropped, keeping only the field at points away from the source: \[ \partial_\mu A^\nu = \frac{\mu_0qc^2}{2\pi} \int \Theta(x^0-y^0)\,\partial_\mu\big[\delta((x-y)^2)\big]\,u^\nu\,\mathrm{d}\tau. \]

Trading the \(x\)-derivative for a \(\tau\)-derivative. By the chain rule, \[ \partial_\mu\big[\delta((x-y)^2)\big] = \delta'((x-y)^2)\,\partial_\mu(x-y)^2 = 2\Delta x_\mu\,\delta'((x-y)^2), \qquad \Delta x^\mu\equiv x^\mu-y^\mu(\tau). \] Differentiating instead along the worldline, using \(\mathrm{d}y^\mu/\mathrm{d}\tau=c\,u^\mu\), \[ \frac{\mathrm{d}}{\mathrm{d}\tau}\delta((x-y)^2) = \delta'((x-y)^2)\,\frac{\mathrm{d}}{\mathrm{d}\tau}(x-y)^2 = \delta'((x-y)^2)\,\big(-2c\,u\cdot\Delta x\big), \] so that \[ \delta'((x-y)^2)=-\frac{1}{2c\,u\cdot\Delta x}\frac{\mathrm{d}}{\mathrm{d}\tau}\delta((x-y)^2). \] Substituting, \[ \partial_\mu A^\nu = -\frac{\mu_0qc}{2\pi} \int \Theta(x^0-y^0) \left[\frac{\mathrm{d}}{\mathrm{d}\tau}\delta((x-y)^2)\right] \frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\,\mathrm{d}\tau. \]

Integration by parts. Moving the \(\tau\)-derivative off \(\delta\) and onto the rest of the integrand costs a sign; the boundary term and the piece hitting \(\Theta\) are, again, self-field contributions at \(x=y\) and are dropped: \[ \partial_\mu A^\nu = \frac{\mu_0qc}{2\pi} \int \Theta(x^0-y^0)\,\delta((x-y)^2)\, \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\right) \mathrm{d}\tau. \] The \(\tau\)-integral is now done exactly as in the Liénard–Wiechert derivation, using \(\delta((x-y)^2)=\delta(\tau-\tau_r)/(2c|u\cdot\Delta x|)\): \[ \partial_\mu A^\nu = \frac{\mu_0q}{4\pi} \frac{1}{|u\cdot\Delta x|} \left. \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\right) \right|_{\tau=\tau_r}. \]

Differentiating term by term. Writing the four-acceleration as \(a^\mu=\mathrm{d}u^\mu/\mathrm{d}\tau\), the product rule gives \[ \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\right) = \frac{\dot{(\Delta x_\mu)}\,u^\nu+\Delta x_\mu\,a^\nu}{u\cdot\Delta x} - \frac{\Delta x_\mu\,u^\nu}{(u\cdot\Delta x)^2}\,\frac{\mathrm{d}}{\mathrm{d}\tau}(u\cdot\Delta x). \] Since \(x\) is fixed, \(\dot{(\Delta x_\mu)}=-\mathrm{d}y_\mu/\mathrm{d}\tau=-c\,u_\mu\), and \[ \frac{\mathrm{d}}{\mathrm{d}\tau}(u\cdot\Delta x) = a\cdot\Delta x+u\cdot\dot{(\Delta x)} = a\cdot\Delta x-c\,u\cdot u = a\cdot\Delta x+c, \] using \(u\cdot u=-1\). Substituting both, \[ \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\right) = \frac{-c\,u_\mu u^\nu+\Delta x_\mu\,a^\nu}{u\cdot\Delta x} - \frac{\Delta x_\mu\,u^\nu}{(u\cdot\Delta x)^2}\big(\Delta x\cdot a+c\big), \] so that \[ \partial_\mu A^\nu = \frac{\mu_0q}{4\pi} \frac{1}{(u\cdot\Delta x)^2} \left[ -c\,u_\mu u^\nu+\Delta x_\mu\,a^\nu - \frac{\Delta x_\mu\,u^\nu}{u\cdot\Delta x}\big(\Delta x\cdot a+c\big) \right]_{\tau=\tau_r}. \] The term \(-c\,u_\mu u^\nu\) is symmetric under \(\mu\leftrightarrow\nu\) (raising the index on \(\mu\)), so it cancels in the antisymmetric combination \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\) and can be dropped when only the field, not the potential’s derivative itself, is wanted: \[ \boxed{ \partial_\mu A_\nu = \frac{\mu_0q}{4\pi} \frac{1}{(u\cdot\Delta x)^2} \left[ \Delta x_\mu\,a_\nu - \frac{\Delta x_\mu\,u_\nu}{u\cdot\Delta x}\big(\Delta x\cdot a+c\big) \right]_{\tau=\tau_r} +\;(\text{symmetric in }\mu\nu). } \]

Near field versus radiation field. Both \(u\cdot\Delta x\) and \(\Delta x_\mu\) scale with the distance \(R=|\mathbf x-\mathbf y|\) to the source: \(u\cdot\Delta x\sim R\) (Liénard–Wiechert section) and \(\Delta x_\mu\sim R\), while \(u^\nu\) and \(a^\nu\), evaluated at the fixed retarded time \(\tau_r\), do not depend on \(R\). In the boxed expression, the term \(\Delta x_\mu a_\nu/(u\cdot\Delta x)^2\) scales as \(R/R^2=1/R\): it carries the four-acceleration explicitly, and is the radiation field. In the second term, the factor \(\Delta x\cdot a\) inside the parenthesis itself grows as \(R\), so it contributes another \(R/(u\cdot\Delta x)^3\sim1/R\) piece to the radiation field, while the leftover \(c\) — the only acceleration-independent piece left in the whole expression — contributes \(\Delta x_\mu u_\nu c/(u\cdot\Delta x)^3\sim1/R^2\). This is exactly the statement in the original notes: terms carrying the acceleration fall off as \(1/R\); whatever survives without it falls off as \(1/R^2\) and is the near (velocity) field.

10 The radiation field

The boxed formula above is for \(\partial_\mu A_\nu\), not yet the field. Since the symmetric piece cancels in the antisymmetric combination, the full field tensor is obtained simply by antisymmetrizing the whole bracket (dropping nothing this time — this is exact, not just the radiative part): \[ F_{\mu\nu} = \frac{\mu_0q}{4\pi(u\cdot\Delta x)^2} \left[ \Delta x_\mu a_\nu-\Delta x_\nu a_\mu - \frac{\Delta x\cdot a+c}{u\cdot\Delta x} \big(\Delta x_\mu u_\nu-\Delta x_\nu u_\mu\big) \right]_{\tau=\tau_r}. \] Splitting off the radiative (\(1/R\)) part identified above — the terms carrying \(a^\mu\) — from the near-field (\(1/R^2\)) part left over, \[ F_{\mu\nu} = \underbrace{ \frac{\mu_0q}{4\pi(u\cdot\Delta x)^2} \left[ \Delta x_\mu a_\nu-\Delta x_\nu a_\mu - \frac{\Delta x\cdot a}{u\cdot\Delta x} \big(\Delta x_\mu u_\nu-\Delta x_\nu u_\mu\big) \right] }_{F_{\mu\nu}^{\rm rad}} \;-\; \underbrace{ \frac{\mu_0qc}{4\pi(u\cdot\Delta x)^3} \big(\Delta x_\mu u_\nu-\Delta x_\nu u_\mu\big) }_{F_{\mu\nu}^{\rm near}}. \]

Why the radiation field is transverse. Contract \(F_{\mu\nu}^{\rm rad}\) with \(\Delta x^\nu\). Every term picks up either \(\Delta x\cdot\Delta x\) or a combination that cancels exactly: \[ F_{\mu\nu}^{\rm rad}\Delta x^\nu = \frac{\mu_0q}{4\pi(u\cdot\Delta x)^2} \Big[ \Delta x_\mu(a\cdot\Delta x)-a_\mu(\Delta x\cdot\Delta x) -\frac{\Delta x\cdot a}{u\cdot\Delta x}\big(\Delta x_\mu(u\cdot\Delta x)-u_\mu(\Delta x\cdot\Delta x)\big) \Big]. \] Since \(\Delta x=x-y(\tau_r)\) is null, \(\Delta x\cdot\Delta x=0\) (the defining condition of \(\tau_r\)), so this collapses to \[ F_{\mu\nu}^{\rm rad}\Delta x^\nu = \frac{\mu_0q}{4\pi(u\cdot\Delta x)^2} \Big[\Delta x_\mu(a\cdot\Delta x)-\Delta x_\mu(a\cdot\Delta x)\Big] = 0. \] The same is not true of \(F_{\mu\nu}^{\rm near}\) (there is nothing to cancel the leftover \(u_\mu(\Delta x\cdot\Delta x)\) term against): only the radiation field is null in this sense, not the full field.

In 3-vector form. With \(\Delta x^0=R\equiv|\mathbf x-\mathbf y|\), \(\Delta x^i=R\hat n^i\), and \(F^{0i}=E^i/c\), \(F^{ij}=\epsilon^{ijk}B_k\), the statement \(F_{\mu\nu}^{\rm rad}\Delta x^\nu=0\) reads, for \(\mu=0\), \[ F_{0i}^{\rm rad}\Delta x^i=0 \;\Longrightarrow\; \hat{\mathbf n}\cdot\mathbf E_{\rm rad}=0, \] and for \(\mu=j\), using \(F_{j0}=E_j/c\) and \(F_{ji}=\epsilon_{jik}B_k\), \[ F_{j0}^{\rm rad}\Delta x^0+F_{ji}^{\rm rad}\Delta x^i=0 \;\Longrightarrow\; \frac{E_j}{c}=-\epsilon_{jik}\hat n^iB_k \;\Longrightarrow\; \boxed{ \mathbf B_{\rm rad}=\frac1c\,\hat{\mathbf n}\times\mathbf E_{\rm rad}, \qquad \hat{\mathbf n}\cdot\mathbf E_{\rm rad}=0. } \] This is exact, for a source moving with any velocity: the radiation field looks locally like a plane wave propagating along \(\hat{\mathbf n}\), which is why \(|\mathbf E_{\rm rad}|=c|\mathbf B_{\rm rad}|\) is used below. The explicit radiation field itself follows from \(E_i^{\rm rad}=-cF_{0i}^{\rm rad}\), \[ \mathbf E_{\rm rad}\cdot\hat{\mathbf e}_i = -\frac{\mu_0qc}{4\pi(u\cdot\Delta x)^2} \left[ \Delta x_0a_i-\Delta x_ia_0 -\frac{\Delta x\cdot a}{u\cdot\Delta x}\big(\Delta x_0u_i-\Delta x_iu_0\big) \right]_{\tau=\tau_r}, \] valid for any velocity of the source. Specializing to a source instantaneously at rest, \(u^\mu=(1,\mathbf 0)\), reduces this to the non-relativistic dipole field used in Thomson Scattering.

11 Radiation zone

Far from the source only the radiation field survives, with \(B=E/c\) as just shown. Substituting this into the energy density and Poynting vector from Energy of the Electromagnetic Field, \[ u=\frac1{2\mu_0}\left(\frac{E^2}{c^2}+B^2\right), \qquad \mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B, \] gives \[ \boxed{ u=\varepsilon_0E^2, \qquad \mathbf S=\frac{E^2}{\mu_0c}\,\hat{\mathbf n}=cu\,\hat{\mathbf n}. } \] Energy flows radially outward at the speed of light, as expected for radiation. (Here \(u\) is the field energy density, unrelated to the four-velocity \(u^\mu\) of the previous sections; both are standard uses of the same letter and are always distinguished by the presence of the index.)

12 Covariant radiation tensor

Write \(F_{\mu\nu}^{\rm rad}=C(\Delta x_\mu b_\nu-\Delta x_\nu b_\mu)\), with \(C\equiv\mu_0q/[4\pi(u\cdot\Delta x)^2]\) and \(b_\nu\equiv a_\nu-\lambda u_\nu\), \(\lambda\equiv(\Delta x\cdot a)/(u\cdot\Delta x)\), matching the bracket in the formula above term by term. Two facts make what follows easy: \[ \Delta x\cdot b=\Delta x\cdot a-\lambda(\Delta x\cdot u)=\Delta x\cdot a-\Delta x\cdot a=0, \qquad \Delta x\cdot\Delta x=0. \] The first is just \(\lambda\)’s definition; the second is the null condition used above. Together they make the field itself null, \[ F^{\alpha\beta}_{\rm rad}F_{\alpha\beta}^{\rm rad} = 2C^2\big[(\Delta x\cdot\Delta x)(b\cdot b)-(\Delta x\cdot b)^2\big] = 0, \] consistent with \(|\mathbf E_{\rm rad}|=c|\mathbf B_{\rm rad}|\) found above (a plane wave has \(F^{\alpha\beta}F_{\alpha\beta}=2(B^2-E^2/c^2)=0\) exactly when \(E=cB\)). So the second term of \(T^{\mu\nu}=\frac1{\mu_0}(F^{\mu\alpha}F^\nu{}_\alpha-\frac14\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta})\) drops out for the radiation piece, leaving \[ \mu_0T^{\mu\nu}_{\rm rad} = F^{\mu\alpha}_{\rm rad}F^{\nu}{}_{\alpha,\rm rad} = C^2(\Delta x^\mu b^\alpha-\Delta x^\alpha b^\mu)(\Delta x^\nu b_\alpha-\Delta x_\alpha b^\nu). \] Expanding the product, every term with a lone \(\Delta x\cdot b\) or \(\Delta x\cdot\Delta x\) vanishes, leaving only \[ \mu_0T^{\mu\nu}_{\rm rad}=C^2(b\cdot b)\,\Delta x^\mu\Delta x^\nu. \] Finally, expand \(b\cdot b=(a-\lambda u)\cdot(a-\lambda u)=a\cdot a-2\lambda(a\cdot u)+\lambda^2(u\cdot u)\). The middle term drops since \(a\cdot u=0\), and \(u\cdot u=-1\) turns the last term negative, so \[ b\cdot b=a\cdot a-\lambda^2=\frac{(a\cdot a)(\Delta x\cdot u)^2-(\Delta x\cdot a)^2}{(\Delta x\cdot u)^2}. \] Substituting \(C^2=\mu_0^2q^2/[16\pi^2(u\cdot\Delta x)^4]\), \[ \boxed{ T^{\mu\nu}_{\rm rad} = \frac{\mu_0q^2}{16\pi^2(\Delta x\cdot u)^6} \Big[(a\cdot a)(\Delta x\cdot u)^2-(\Delta x\cdot a)^2\Big] \Delta x^\mu\Delta x^\nu. } \] Important observations:

• proportional to \(\Delta x^\mu\Delta x^\nu\);
• null, since \(\Delta x^2=0\) on the light cone;
• directed along outgoing light rays;
• survives at arbitrarily large distances, since the \(1/|\Delta\mathbf x|^2\) falloff exactly matches the near-field terms falling off faster.

This is the field responsible for energy loss by the source.

13 Relativistic Larmor formula

“Power radiated” is ambiguous for an accelerated source unless a frame is specified: a general observer sees the flux Doppler-shifted by the retarded-time relation between source and observation time. The unambiguous, covariant quantity is the energy radiated per unit proper time of the source, computed as the flux of \(T^{\mu\nu}_{\rm rad}\) through a sphere surrounding the charge in its own instantaneous rest frame, where this subtlety is absent because the source is momentarily not moving. Since this power, so defined, and \(a^\mu a_\mu\) are both Lorentz scalars, a relation between them established in one frame (the simplest one, the rest frame) holds in every frame.

Step 1: compute the flux in the rest frame. There, \(u^\mu=(1,\mathbf 0)\), \(\Delta x\cdot u=-R\), and \(u_\mu a^\mu=0\) forces \(a^\mu=(0,\dot{\mathbf v}/c)\) for the ordinary acceleration \(\dot{\mathbf v}=\mathrm{d}\mathbf v/\mathrm{d}t\), so \[ a\cdot a=\frac{\dot v^2}{c^2}, \qquad \Delta x\cdot a=R\,\frac{\hat{\mathbf n}\cdot\dot{\mathbf v}}{c}, \] exactly as used for the Thomson-scattering radiation field. Substituting into the boxed \(T^{\mu\nu}_{\rm rad}\) above, with \(\chi\) the angle between \(\hat{\mathbf n}\) and \(\dot{\mathbf v}\) (kept distinct from the Heaviside \(\Theta\) used throughout this page), \[ (a\cdot a)(\Delta x\cdot u)^2-(\Delta x\cdot a)^2 = \frac{R^2}{c^2}\Big[\dot v^2-(\hat{\mathbf n}\cdot\dot{\mathbf v})^2\Big] = \frac{R^2\dot v^2}{c^2}\sin^2\chi, \] so that, using \(T^{0i}=S^i/c\) and \((\Delta x\cdot u)^6=R^6\), \[ \mathbf S\cdot\hat{\mathbf n}=cT^{0i}\hat n^i = c\cdot\frac{\mu_0q^2}{16\pi^2R^6}\cdot\frac{R^2\dot v^2\sin^2\chi}{c^2}\cdot R^2 = \frac{\mu_0q^2\dot v^2}{16\pi^2cR^2}\sin^2\chi. \]

Step 2: integrate over the sphere. The power radiated per solid angle is \[ \frac{\mathrm{d}P}{\mathrm{d}\Omega}=R^2\,\mathbf S\cdot\hat{\mathbf n}=\frac{\mu_0q^2\dot v^2}{16\pi^2c}\sin^2\chi. \] Using \(\int\sin^2\chi\,\mathrm{d}\Omega=2\pi\int_0^\pi\sin^3\chi\,\mathrm{d}\chi=2\pi\cdot\frac43=\frac{8\pi}{3}\), \[ P=\int\frac{\mathrm{d}P}{\mathrm{d}\Omega}\,\mathrm{d}\Omega = \frac{\mu_0q^2\dot v^2}{16\pi^2c}\cdot\frac{8\pi}{3} = \frac{\mu_0q^2\dot v^2}{6\pi c}, \] the familiar non-relativistic Larmor formula.

Step 3: covariantize. In this same rest frame, \(a^\mu a_\mu=\dot v^2/c^2\), so the result of Step 2 is exactly \[ P=\frac{\mu_0q^2c}{6\pi}\,a^\mu a_\mu. \] Both sides of this equation are Lorentz scalars (\(P\) by the proper-time construction above, \(a^\mu a_\mu\) trivially), and they agree in one frame; therefore they agree in every frame. The relation is thus valid for a source moving with arbitrary velocity, not just at the instant it is at rest: \[ \boxed{ P = \frac{\mu_0q^2c}{6\pi} a^\mu a_\mu. } \]

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