Past exam of the astrophysics course of the University of Oxford / 2022 / 4

An ultrarelativistic electron has kinetic energy much larger than its rest energy: $(\gamma -1)m_e{c}^2\gg m_e{c}^2$, equivalently $\gamma \gg 1$. Its speed obeys $v/c=\sqrt{1-{\gamma }^{-2}}\simeq 1-1/(2{\gamma }^2)$ and its total energy is approximately $pc$. At $\gamma =10^6$, $E=\gamma m_e{c}^2\simeq 511\mathrm{GeV}$.

Diffusive shock acceleration is viable in principle at $511\mathrm{GeV}$: repeated scattering off magnetic irregularities on opposite sides of a strong nonrelativistic shock wave can reach such energies. Actual viability requires a sufficiently large and long-lived accelerator with an acceleration time shorter than its escape and loss times; $\gamma$ alone does not establish those conditions.

Take positive velocity toward the upstream material, and let the front move at $+v_s$ in the upstream rest frame. For a strong adiabatic shock wave in a gas with adiabatic index $5/3$, the Rankine-Hugoniot conditions give ${\rho }_d/{\rho }_u=4$ and the shock-frame inflow speeds satisfy $u_1=v_s$, $u_2=v_s/4$. The signed velocities in the three frames areThus the two scattering media converge at $V=u_1-u_2=3v_s/4$. For a crossing particle with incident energy $E$ in its local plasma rest frame, a Lorentz transformation to the approaching other plasma giveswhere $\mu >0$ is the incident direction cosine measured toward the approaching medium and $pc\simeq E$. For an almost isotropic crossing population, flux weighting gives $⟨\mu ⟩={\int }_0^1{\mu }^{2}d\mu /{\int }_0^1\mu d\mu =2/3$. Scattering randomizes directions without changing energy in that plasma frame, and the reverse crossing has the same mean gain. ThereforeThese are first-order mean gains. They are not an increase of energy in one fixed frame caused by an instantaneous boundary crossing. Literally choosing the electron's instantaneous rest frame gives incident $E=m_e{c}^2$, and the other gas approaches at nearly $c$, so the expansion in small $V/c$ is invalid. This concrete frame check explains why the printed description of $V$ must instead refer to the upstream/downstream scattering frames.

The magnetic Lorentz force is perpendicular to velocity, so it does no work and $\gamma$ remains constant over an orbit when radiative losses are neglected. For the transverse momentum, $p_{\perp }=\gamma m_e{v}_{\perp }$ and $|{\dot{\mathbf{p}}}_{\perp }|={\omega }_g{p}_{\perp }=eB{v}_{\perp }$. HenceSince $10\mu \mathrm{G}=10^{-9}\mathrm{T}$,This is the gyromotion frequency, not the much higher characteristic synchrotron radiation frequency.

Relativistic beaming concentrates the emission into a forward cone about the instantaneous velocity. The aberration of light maps a rest-frame ray at ${\theta }^{\prime }=\pi /2$ to $\cos \theta =\beta$, so the characteristic cone half-angle is $\theta \simeq {\gamma }^{-1}$. Thus its conventional full width is $2/\gamma =2\times 10^{-6}\mathrm{rad}$.

The isotropic-emission assumption has no beam nulls. For example, its received bolometric angular power is proportional to $D^4$, where the Doppler factor is $D=[\gamma (1-\beta \cos \theta ){]}^{-1}>0$. In particular it is nonzero at $\theta =1/\gamma$, disproving a literal null there. A true dipole pattern for an accelerated electron is not isotropic in its rest frame and needs a specified polarization and viewing plane. The width just quoted is a characteristic beaming angle, not a null-to-null width under the stated premise.

For a circular orbit seen in its orbital plane, let the velocity direction sweep past the observer through $-1/\gamma <\psi <1/\gamma$. The emission interval is $\Delta t_{\mathrm{em}}\simeq 2/(\gamma {\omega }_g)=1.14\times 10^{-2}\mathrm{s}$. The observed interval must also include light-travel compression:Its ratio to the gyromotion period is $\Delta t_{\mathrm{obs}}/T_g\simeq 2/(3\pi {\gamma }^3)=2.1\times 10^{-19}$. Since the boundary of a smooth beam is conventional, only the scale ${\omega }_g^{-1}{\gamma }^{-3}$ is invariant under reasonable pulse-width definitions. For helical motion the viewing geometry and pitch angle are additional inputs; the numerical estimate assumes transverse motion and a sightline intercepted by the cone.^[ref]

One explanation is an upper electron energy imposed by the accelerator: finite age, finite confinement size, escape, or a balance of acceleration and losses limits $\gamma$ to ${\gamma }_{\max }$. Since the synchrotron critical frequency scales as ${\nu }_c\propto B{\gamma }^2\sin \alpha$, the absence of higher-energy electrons produces a corresponding high-frequency cutoff, broadened by the single-particle spectrum.

The other is synchrotron cooling, with inverse Compton scattering often contributing: $\dot{\gamma }=-b{\gamma }^2$ gives $t_{\mathrm{cool}}=1/(b\gamma )$. High-energy electrons disappear fastest, so a population of age $t$ develops a cooling scale ${\gamma }_b\sim 1/(bt)$ and ${\nu }_b\propto B/(b^2{t}^2)$. A burst population can develop a strong cutoff. With continuous power-law injection, the simplest steady cooling model instead steepens the electron energy index by one and the synchrotron radiation spectral index by $1/2$, giving a break rather than necessarily an exponential cutoff. The detailed spectral shape therefore distinguishes finite acceleration from aging and injection history.