** Gyro Radiation**

Larmor's equation is nonrelativistic,
so we first consider gyro
radiation by a particle with $v \ll c$. The magnetic
force $\vec{F}$ on
a particle with charge $e$ moving with velocity $v \ll c$ is given by Ampere's
law:

$$\bbox[border:3px blue solid,7pt]{\vec{F} = { e (\vec{v} \times \vec{B}) \over c}}\rlap{\quad \rm {(5A1)}}$$

The magnetic
force is perpendicular to the particle velocity: $\vec{F} \cdot \vec{v}
= 0$. No power is transferred to the charged particle and its kinetic
energy $mv^2 /2$ remains constant. Also, the component of velocity
$v_\parallel$ parallel to the magnetic field doesn't change. Since both
$\vert v \vert$ and $v_\parallel$ are constant, the magnitude of the
component of
velocity $\vert v_\bot \vert$ perpendicular to the magnetic field must
also be constant. In a constant magnetic field, the particle moves
along the magnetic field line on a uniform helical path with constant
linear and angular speeds. In the inertial frame moving with velocity
$v_\parallel$, the particle orbits in a circle perpendicular to the
magnetic field with the angular velocity $\omega$ needed to balance the
centripetal and magnetic forces.

$$m\vert \dot{v} \vert = m \omega^2 R
= \vert \vec{F} \vert ~,$$

where $R$ is the radius of the circle.

$$m
\omega^2 R = { e \over c} \vert \vec{v} \times \vec{B} \vert = {e \over
c} \omega R B$$ $$ \omega = { e B \over m c}$$

All nonrelativistic
electrons in the same magnetic field orbit at the same frequency and
hence emit electromagnetic radiation at the same frequency. We can
define the
gyro frequency as

$$\bbox[border:3px blue solid,7pt]{\omega_{\rm G} \equiv { e B
\over m c}}\rlap{\quad \rm {(5A2)}}$$

This equation defines $\omega_{\rm G}$ regardless
of $v/c$. The gyro frequency equals the actual orbital frequency only
if $v \ll c$. The electron
gyro frequency is

$$\omega_{\rm G} = { e B \over m_{\rm e} c } = { 4.8 \times 10^{-10} {\rm ~statcoul} \times B \over 9.1 \times 10^{-28} {\rm ~g} \times 3 \times 10^{10} {\rm ~cm~s}^{-1} } \approx 17.6 \times 10^6 {\rm ~rad~s}^{-1} \times B{\rm ~(Gauss)}$$

Let $\nu_{\rm G} \equiv \omega_{\rm
G} / (2 \pi)$ so the electron
gyro frequency in MHz is:

$$\bbox[border:3px blue solid,7pt]{{\nu_{\rm G} \over {\rm MHz}} = 2.8
\biggl( { B \over {\rm Gauss}
}\biggr)}\rlap{\quad \rm {(5A3)}}$$

Example: What is the electron gyro frequency in the interstellar magnetic field of our Galaxy, $B \approx 5\,\mu$G?

$$\nu_{\rm G} = 2.8 {\rm ~MHz} \times
5 \times 10^{-6} {\rm ~Gauss}
= 14 {\rm ~Hz}$$

Since Larmor's formula states that
the radiation field varies with
the same frequency as the acceleration, this interstellar gyro
frequency is far too
low to yield observable radio emission. Gyro radiation is observable
only in very strong magnetic fields. An extreme astrophysical example
is the magnetic field of a neutron star, $B \sim 10^{12}$ Gauss.
For example, the binary X-ray source Hercules X-1 exhibits an X-ray
absorption line at photon energy $E \approx 34$ keV.

Gyro-resonance absorption line near 34 keV (Gruber et al. 2001, ApJ, 562, 499).

This spectral feature is thought to
be a gyro resonance absorption.
Therefore the frequency of this absorption line directly measures the
magnetic field strength near Her X-1. The observed photon energy
corresponds to the frequency

$$\nu = {E \over h} \approx { 34 \times
10^3 {\rm ~eV} \times 1.60 \times 10^{-12} {\rm erg~eV}^{-1} \over 6.63
\times 10^{-27} {\rm erg~s} } \approx 8.2 \times 10^{18} {\rm ~Hz}$$

Equating this frequency to the gyro frequency yields the magnetic field
near the neutron star:

$$B = { 2 \pi \nu_{\rm G}\, m_{\rm e}\, c \over
e}$$

$$B \approx { 2 \pi \times 8.2 \times 10^{18} {\rm ~Hz} \times 9.1
\times 10^{-28} {\rm ~g} \times 3 \times 10^{10} {\rm ~cm~s}^{-1} \over
4.8 \times 10^{-10} {\rm ~statcoul} } \approx 2.9 \times 10^{12} {\rm
~Gauss}$$