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Cyclotron resonant scattering

A1  ?? Obtain the equations of motioin for the electron in the form
\[\begin{cases}\ddot{X} = \dots\\ \ddot{Y} = \dots\end{cases}\]

Ответ: The Lorentz force is acting on the electron:
\[\begin{cases}
\ddot{X} = \frac{eB}{m} \dot{Y} \\
\ddot{Y} = -\frac{eB}{m} \dot{X}
\end{cases}\]
A2  ?? Integrate the second equation with respect to time, and using substitution, obtain the equation of harmonic oscillations
\[ \ddot{X} + \omega_0^2 X = C.\]Express $\omega_0$ in terms of $m$, $e$ and $B$.

After integrating the second equation we have:
\[ \dot{Y} = -\frac{eB}{m} (X-X_0).\]Thus, with substitution $X \to X + X_0$:
\[\ddot{X}+\left( \frac{eB}{m} \right)^2 X = 0\]

Ответ: \[\omega_0 = \frac{eB}{m}\]
A3  ?? Similarly to question A1, obtain the equation of motion for $x(t)$ and $y(t)$.

Ответ: \[
\begin{cases}
\ddot{X} + \ddot{x} = \omega_0 (\dot{Y} + \dot{y}) - \frac{eE}{m} e^{i \omega t} \\
\ddot{Y} + \ddot{y} = -\omega_0 (\dot{X} + \dot{x})
\end{cases}\]
A4  ?? Similarly to question A2, eliminate $y$ and obtain the equation of driven oscillations for $x(t)$.

Ответ: \[\ddot{x}+ \omega_0^2x = -\frac{eE}{m}e^{i\omega t}\]
A5  ?? What is the frequency of EM wave $\omega$ when the resonance takes place?

Let's substitute the solution $x=A e^{i \omega t}$ correspoing to driven oscillations
\[ A \left( (i \omega)^2 + \omega_0^2 \right) e^{i \omega t}= - \frac{eE}{m} e^{i \omega t}\]There is resonance when $\omega=\omega_0$.

A6  ?? From the quantum point of view explain the presence of several cyclotron resonant scattering features. Obtain the value of the mean magnetic field $B$ on the surface of the V0332+53 pulsar.

If a system can absorb an EM field with a frequency of $\omega$, then it can absorb photons with an energy of $\hbar \omega$. Then, the system can also absorb a photon with energy $n \hbar \omega$ due to the absorption of multiples portions of energy each equal to $\hbar \omega$.

There are two cyclotron resonant scattering features (e.g. 272 rev) on the graph: $E_\mathrm{cycl}=26~\text{keV}$ and $2E_\mathrm{cycl}=50~\text{keV}$. So, the magnetic field
\[ B = \frac{m E_\mathrm{cycl}}{\hbar e}=2 \cdot 10^{8}~\text{T}\]