Radiation Lecture 04

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<latex> \documentclass[11pt]{article} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\wz{\omega_0} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\inv#1{\frac1{#1}} \def\hf{\frac12} \def\bfieldTemplate:\vec B \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\ef{\vec E} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}


\subsection*{ Estimating Cross-Section}

The absorption coefficient, written in terms of Einstein constants is: $$\alpha_\nu={n_1\bot\phi(\nu)\over4\pi}h\nu=n_1\sigot$$ Thus, the cross-section of an atom for absorption of a photon is: $$\sigot={\bot\phi(\nu)h\nu\over4\pi}$$ To estimate $\bot$, we use the fact that, ignoring g's, $\bot\sim\bto$, and ${\ato\over\bto}={2h\nu^3\over c^2}$. Then using the approximation that that $\phi(\nu)\sim\inv{\Delta\nu}$, we get: $$\sigot\sim{\ato\over\left({2h\nu^3\over c^2}\right)} \inv{\Delta\nu}$$ $$\boxed{\sigot\sim{\lambda^2\over8\pi}{\ato\over\Delta\nu}}$$ In a single atom, $\Delta\nu\sim\ato$, so $\sigot\sim{\lambda^2\over8\pi}$.

\subsection*{ Order of Magnitude Interaction of Radiation and Matter}

\subsection*{Energy Levels} \begin{itemize} \item Electronic Transitions:\par We'll start with the Bohr atom. We begin by quantizing angular momentum: $$m_ev_ea_0=n\hbar$$ If we balance the force required to keep the $e^-$ in a circular orbit with the electric force: $${m_ev_e^2\over a_0}={Ze^2\over a_0^2}$$ $$a_0={\hbar^2\over m_ee^2Z}\approx0.52{\AA\over Z}$$ A {\it Rydberg} is the energy required to ionize an H atom from the ground state. It is $\sim13.6eV$. We can estimate it by integrating the electric force from $r=a_0$ to $r=\infty$, but in reality, there is another factor of 2: $$\boxed{Rydberg={Ze^2\over2a_0}={Z^2e^4m_e\over2\hbar^2}=13.6\cdot Z^2eV}$$ \item Fine Structure:\par Fine Structure comes from the interaction of the magnetic moment of the $e^-$ with the a $\bfield$ caused by the Lorentz-transformed Coulomb field of the proton (generated by the $e^-$'s motion). The energy of a dipole interaction is $E=\vec\mu\cdot\bfield$, so we'd expect that: $$\Delta E\sim\mu B$$ To estimate B: $B\sim{Ze\over a_0^2}{v\over c}$, and ${v\over c}\sim {e^2\over\hbar c}=\alpha=\inv{137}$ ($\alpha$ is the fine structure constant): $$B\sim{Ze^3\over\hbar ca_0^2}$$ Estimating $\mu$: The $\bfield$ of a dipole goes as $B_{di}\sim{\mu\over r^3}$, so $\mu\sim B_e\eval{r_e}r_e^3$, where $r_e$ is the classical $e^-$ radius. Estimating that the rest mass energy of the $e^-$ should be about the electrostatic potential energy left in the $e^-$, $m_ec^2\sim{e\over r_e}$: $$r_e\sim{e^2\over m_ec^2}$$ We can estimate $B_e$ by reverting to Maxwell's equation: $$\dcb={4\pi J\over c}$$ $${B_e\over2\pi r_e}={4\pi\over c}{I\over4\pi r_e^2}$$ Again, we estimate that $I\sim{e\over t_{spin}}$, and because the electron spin is quantized, $\hbar=m_er_e^2{2\pi\over t_{spin}}$. After some algebra, we get that: $$\mu_e={e\hbar\over2m_ec}$$ \centerline{(Bohr magneton for $e^-$)} $$\mu_p={Ze\hbar\over2m_pc}$$ \centerline{(Bohr magneton for nucleus)} Getting back to the energy: $$\begin{aligned}\Delta E&\sim{e\hbar\over2m_ec}{Ze\over a_0^2}\alpha\\ &\sim Z^4\alpha^2\cdot Ryd\\ \end{aligned}$$

\item Hyperfine Structure:\par Instead of using the Bohr magneton, we use the intrinsic magnetic moment (spin) of the nucleus: $$B\eval{p}\sim{\mu_p\over a_0^3}\sim B_{fine}{m_e\over m_p}$$ Thus: $$\Delta E\sim\Delta E_{fine}{m_e\over m_p}$$

Note that Fine and Hyperfine are magnetic dipole transitions. $e^-$ level transitions are {\it electric} dipole transitions. Magnetic dipole transitions are generally weaker.

\item Vibrational Transitions in Molecules:\par Our general technique with vibrational transitions is to model them as harmonic oscillators. Thus, they should have the characteristic harmonic energy series: $$E_n=(n+\hf)\hbar\wz$$ For a harmonic oscillator, $\wz=\sqrt{k\over m}$. We estimate that since the force for a spring is $k\cdot x$, and that force should be about the Coulomb force on $e^-$'s. If we say that atoms stretch with respect to each other about a Bohr radius: $$ka_0\sim{e^2\over a_0^2}$$ $$\Delta E\eval{vib\atop trans}\sim Ryd\cdot\sqrt{m_e\over A\cdot m_p}$$ where A is the atomic mass \# of our atoms.

\item Rotational Transitions in Molecules:\par The thing to remember is that angular momentum comes in units of $\hbar$. \end{itemize}

\subsection*{ Damped Simple Harmonic Oscillator}

We can model $e^-$ and their atoms as a lattice of springs connecting $e^-$ to fixed points. Say that we send a plane wave at these springs that looks like: \def\eikrwt{e^{i(\vec k\vec r-wt)}} $$\ef=E_0\eikrwt$$ We'll say that the displacement of $e^-\ll{2\pi\over k}$. The equation of motion for a single electron is then: $$\ddot x+\gamma\dot x+\wz^2x={eE_0\eikrwt\over m_e}$$ where $\gamma$ is our dampening factor. Note that $\bfield$ is absent here; we're neglecting it because it is small for reasonable energies. It turns out this equation has the steady-state solution: $$x={-e\over m_e}{\left((\wz^2-\omega^2)+i\omega\gamma\right)\over \left((\wz^2-\omega^2)^2+(\omega\gamma)^2\right)}E_0\eikrwt$$ The limiting cases of this equation explain many phenomena. For example: $\gamma=0$ (loss-less propagation).\par In this case, our solution looks like: $$x={-e\over m_e}{E_0\eikrwt\over(\wz^2-\omega^2)}$$ From this we can use the dispersion relation to relate $\omega$ (frequency) and $k$ (phase).

\subsection*{ The Dispersion Relation}

Let's solve Maxwell's equations. Recall that $\vec J$ is the current density: $$\begin{aligned}\vec J&=n\cdot e\cdot\dot x\\ &={ne(-i\omega)(-eE_0\eikrwt)\over m_e(\wz^2-\omega^2)}=\sigma\ef\\ \end{aligned}$$ where $\sigma\equiv conductivity={-ne^2i\omega\over m_e(\wz^2-\omega^2)}$, and n is the \# density of electrons. So on to Maxwell's equations: $$\begin{aligned}\dce&=-{1\over c}{\partial\bfield\over\partial t}\\ ikE&={i\omega B\over c}\\ \end{aligned}$$ $$\begin{aligned}\dcb&={4\pi\vec J\over c}+{1\over c}{\partial\ef\over\partial t}\\ -ikB&=\left({4\pi\sigma\over c}-{i\omega\over c}\right)E\\ \end{aligned}$$ Combining these two equations we get: $$\boxed{\left({ck\over\omega}\right)^2=1+{4\pi ne^2\over m_e(\wz^2-\omega^2)}}$$ This is our dispersion relation. The plasma frequency is defined as: $$\omega_p^2\equiv {4\pi ne^2\over m_e}$$ and the index of refraction is: $$\eta\equiv{ck\over\omega}$$ Rewritten in these terms, our dispersion relation in plasma is: $$\boxed{\eta^2=1-{\omega_p^2\over\omega^2}}$$


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