Relativistic Particles & Radiation

Step 1: Covariant form of radiated power

The radiated power must be a Lorentz scalar. The unique scalar quadratic in the 4-acceleration $a^\mu = du^\mu/d\tau$ is:

$$P = -\frac{\mu_0 q^2 c}{6\pi}a_\mu a^\mu$$

Step 2: Express $a_\mu a^\mu$ in terms of lab quantities

The 4-acceleration has components $a^\mu = \gamma^2(\gamma^2\mathbf{v}\cdot\mathbf{a}/c, \gamma^2(\mathbf{v}\cdot\mathbf{a})\mathbf{v}/c^2 + \mathbf{a})$. Computing the invariant:

$$-a_\mu a^\mu = \gamma^4\left[a^2 + \gamma^2\frac{(\mathbf{v}\cdot\mathbf{a})^2}{c^2}\right] = \gamma^6\left[a^2 - \frac{|\mathbf{v}\times\mathbf{a}|^2}{c^2}\right]$$

Step 3: Relativistic Larmor formula

$$\boxed{P = \frac{\mu_0 q^2 c}{6\pi}\gamma^6\left(a^2 - \frac{|\mathbf{v}\times\mathbf{a}|^2}{c^2}\right)}$$

Step 4: Acceleration parallel to velocity

When $\mathbf{a} \parallel \mathbf{v}$: $P = \mu_0 q^2 c\gamma^6 a^2/(6\pi)$. This applies to linear accelerators. The $\gamma^6$ factor makes relativistic linear acceleration extremely radiatively costly.

Step 5: Acceleration perpendicular to velocity

When $\mathbf{a} \perp \mathbf{v}$: $P = \mu_0 q^2 c\gamma^4 a^2/(6\pi)$. This applies to synchrotrons and storage rings. The power scales as $\gamma^4$, which is why electron synchrotrons are limited to ~100 GeV while proton synchrotrons reach TeV energies.

Step 1: Opening angle of the radiation cone

Relativistic beaming concentrates the radiation into a forward cone of half-angle $\Delta\theta \sim 1/\gamma$.

Step 2: Duration of the pulse

An observer sees radiation only while the beam sweeps past, covering an arc $\Delta\phi \sim 2/\gamma$. The retarded time interval is $\Delta t_{\text{ret}} = 2R/(c\gamma)$.

Step 3: Time compression by the Doppler effect

Due to the relativistic approach, the observed pulse duration is compressed: $\Delta t_{\text{obs}} = \Delta t_{\text{ret}}(1-\beta) \approx \Delta t_{\text{ret}}/(2\gamma^2)$.

Step 4: Characteristic frequency

$$\omega_c \sim \frac{1}{\Delta t_{\text{obs}}} \sim \frac{2\gamma^2}{2R/(c\gamma)} = \frac{\gamma^3 c}{R}$$

The exact result includes a factor of $3/2$: $\omega_c = \frac{3}{2}\gamma^3(c/R)$.

Step 5: Spectrum

The spectral distribution involves the Airy integral: $dP/d\omega \propto (\omega/\omega_c)\int_{\omega/\omega_c}^\infty K_{5/3}(x)\,dx$ where $K_{5/3}$ is a modified Bessel function. It rises as $\omega^{1/3}$ for $\omega \ll \omega_c$ and falls exponentially for $\omega \gg \omega_c$.

Step 1: Aberration formula

If a photon is emitted at angle $\theta'$ in the rest frame, in the lab frame: $\cos\theta = (\cos\theta' + \beta)/(1+\beta\cos\theta')$.

Step 2: Maximum emission angle

In the rest frame, maximum emission is at $\theta' = \pi/2$. In the lab: $\cos\theta_{\max} = \beta$, i.e., $\sin\theta_{\max} = 1/\gamma$. For $\gamma \gg 1$: $\theta_{\max} \approx 1/\gamma$.

Step 3: Solid angle compression

The isotropic rest-frame radiation ($4\pi$ sr) is compressed into a forward cone of solid angle $\Delta\Omega \sim \pi/\gamma^2$. The intensity is enhanced by a factor of $\sim\gamma^2$.

Step 4: Doppler boosting of power

The emitted power transforms as $dP/d\Omega = \mathcal{D}^4(dP'/d\Omega')$ where $\mathcal{D} = [\gamma(1-\beta\cos\theta)]^{-1}$ is the Doppler factor. For $\theta \approx 0$: $\mathcal{D} \approx 2\gamma$.

Step 5: Observational consequences

A synchrotron source moving toward the observer has apparent luminosity boosted by $\sim\gamma^4$, while one moving away is suppressed. This explains the one-sidedness of relativistic jets in AGN and the extreme brightness of blazars (jets aimed at Earth).

Step 1: Circular orbit condition

At radius $R_0$: $p = eBR_0$ (relativistically exact). For the orbit to remain at $R_0$, both $p$ and $B$ must change together.

Step 2: Faraday's law gives the accelerating field

The induced electric field at $R_0$: $E \cdot 2\pi R_0 = -d\Phi/dt$ where $\Phi = \int_0^{R_0}B(r)2\pi r\,dr$.

Step 3: Rate of momentum change

$dp/dt = eE = \frac{e}{2\pi R_0}\frac{d\Phi}{dt}$. From the orbit condition: $dp/dt = eR_0\,dB(R_0)/dt$.

Step 4: Equate and integrate

$eR_0\,dB(R_0)/dt = \frac{e}{2\pi R_0}\frac{d\Phi}{dt}$. Integrating: $\Phi = 2\pi R_0^2 B(R_0)$.

Step 5: The betatron condition

$$\boxed{B(R_0) = \frac{\langle B\rangle}{2} = \frac{\Phi}{2\pi R_0^2}}$$

The field at the orbit must equal half the average field within the orbit. This is achieved by shaping the pole pieces. The first betatron (Kerst, 1940) accelerated electrons to 2.3 MeV; modern betatrons reach 300 MeV.

Step 1: Power for circular motion

For perpendicular acceleration in a circular orbit: $a = v^2/R \approx c^2/R$ and $P = \mu_0 q^2 c\gamma^4 a^2/(6\pi)$.

Step 2: Substitute $a = c^2/R$

$$P = \frac{\mu_0 q^2 c^5\gamma^4}{6\pi R^2} = \frac{2r_0 c}{3}\frac{E^4}{(mc^2)^4 R^2}$$

where $r_0 = q^2/(4\pi\epsilon_0 mc^2)$ is the classical particle radius.

Step 3: Revolution period

$T = 2\pi R/v \approx 2\pi R/c$ for ultrarelativistic particles.

Step 4: Energy loss per revolution

$$\Delta E = PT = \frac{4}{3}\frac{r_0}{(mc^2)^3}\frac{E^4}{R}$$

Step 5: Mass dependence

$$\boxed{\Delta E \propto \frac{E^4}{m^4 R}}$$

Since $m_p/m_e \approx 1836$, protons lose $(1836)^4 \approx 10^{13}$ times less energy than electrons at the same energy. This is why the largest circular electron collider (LEP) reached 209 GeV while the LHC reaches 14 TeV with protons.