Past exam of the symmetry and relativity course of the University of Oxford / 2021 / 2

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

For an ultrarelativistic particle, relativistic beaming concentrates the synchrotron radiation within an angle of order $1/\gamma$ around its instantaneous velocity. A fixed viewing direction in the orbital plane receives an intense signal only when the tangent sweeps through that direction. One electron, or one sufficiently short bunch, therefore produces one pulse per revolution, separated by $\mathcal{T}\simeq 10.48\mu \mathrm{s}$. Several bunches produce the corresponding additional pulses per revolution.

The time during which the velocity points within that cone is of order $r/(\gamma c)$, but this is the emission interval. If $t_a$ is the arrival time at a distant detector,Near alignment, $1-\beta \cos \theta \simeq ({\gamma }^{-2}+{\theta }^2)/2=O({\gamma }^{-2})$. Thus synchrotron pulse compression reduces the observed duration by a further factor of order ${\gamma }^2$:A more explicit geometric estimate chooses the cone edges $\theta =\pm 1/\gamma$. With $\omega =u/r$, $x=r\sin \theta$ along the line of sight, the difference in arrival-time compression for a moving source isThe order-one coefficient depends on how pulse width is defined; the robust scale is $r/(c{\gamma }^3)$. The usual central-cone estimate $r/(2c{\gamma }^3)$ gives $4.1\times 10^{-18}\mathrm{s}$. These estimates agree on the attosecond scale and must not be confused with $r/(\gamma c)\simeq 2.84\times 10^{-10}\mathrm{s}$.^[ref]

Bunch qualification. The source says only that the bunches are short, without their longitudinal length $L_b$. Different electrons pass the same emitting tangent at times separated by up to about $L_b/u$, with essentially the same path length from that tangent to the detector. The bunch-length broadening of synchrotron radiation means that the measured bunch envelope is the single-electron pulse convolved with the bunch's passage-time profile; its scale is generally at least of order $\max \{L_b/u,r/(c{\gamma }^3)\}$, with additional detector bandwidth effects. For example, a $1\mathrm{mm}$ bunch, very short compared with this ring, gives an approximately $3.3\mathrm{ps}$ envelope, much longer than the single-electron duration. No unique numerical bunch-envelope width follows from the supplied data. The pulse argument also presumes that the accepted $+x$ direction is tangent to the orbit; a direction far outside the orbital plane receives no comparable intense forward pulse.