Examples of V-E Day in the following topics:
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- Let $v dE$ be the density of photons having energy in the range $dE$.
- $\displaystyle \frac{v dE}{E} = \frac{v' dE'}{E'} = \text{ Lorentz Invariant} $
- $\displaystyle \frac{d E_f}{d t} = \frac{d E_f'}{d t'} = c \sigma_T \int E_f' v' d E'$
- $\displaystyle \frac{d E_f}{d t} = c \sigma_T \int E'^2 \frac{v' dE'}{E'} = c \sigma_T \int E'^2 \frac{v dE}{E}$
- $\displaystyle \frac{d E}{d t} = -c\sigma_T \int E v dE = -\sigma_T c U_\text{ph}$
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- $\displaystyle d P \propto e^{-E/kT} d^3 {\bf v} = \exp\left(\frac{-mv^2}{2 k T} \right ) d^2 {\bf v}$
- $\displaystyle \frac{dW(T,\omega)}{d\omega dV dt} = \frac{\int_{v_\mbox{min}}^\infty \frac{dW (v,\omega)}{d\omega d V dt} v^2 e^{-\beta m v^2/2} d v} {\int_0^\infty v^2 e^{-\beta m v^2/2} d v}$
- If we look at the emission for a particular velocity, the emisision rate diverges as $v \rightarrow 0$, but the phase space vanishes faster; however, it is stll reasonable to cut off the integral at some minimum velocity.
- $\displaystyle \epsilon_\nu^{ff} \equiv \frac{d W}{dV dt d\nu} \\ = \frac{2^5 \pi e^6}{3 m c^3} \left ( \frac{2\pi}{3km} \right )^{1/2} T^{-1/2} Z^2 n_e n_i e^{-h\nu/kT} {\bar g}_{ff} \\ = 6.8 \times 10^{-38} Z^2 n_e n_i T^{-1/2} e^{-h\nu/kT} {\bar g}_{ff}$
- $\displaystyle \epsilon^{ff} = \frac{2^5 \pi e^6}{3 h m c^3} \left ( \frac{2\pi k T}{3m} \right )^{1/2} Z^2 n_e n_i {\bar g}_{B} \\ = 1.4 \times 10^{-27} Z^2 n_e n_i T^{1/2} {\bar g}_{B}$
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- $\displaystyle \frac{d N_0^+(v) }{N_0} = \frac{g_e g_0^+}{g_0} \exp \left [ -\frac{E_I + \frac{1}{2} m_e v^2}{kT} \right ]$
- $\displaystyle \frac{d N_0^+}{N_0} = \frac{8 \pi m_e^3}{h^3} \frac{g_0^+}{N_e g_0} \exp \left [ -\frac{E_I + \frac{1}{2} m_e v^2}{kT} \right ] v^2 d v$
- $\displaystyle \frac{N_0^+ N_e}{N_0} = \left ( \frac{2\pi m_e kT}{h^2} \right)^{3/2} \frac{2g_0^+}{g_0} e^{-E_I/kT}$
- $\displaystyle \frac{N^+ N_e}{N} = \left ( \frac{2\pi m_e kT}{h^2} \right)^{3/2} \frac{2 U^+(T)}{U(T)} e^{-E_I/kT}.$
- $\displaystyle \frac{N_{j+1} N_e}{N_j} = \left ( \frac{2\pi m_e kT}{h^2} \right)^{3/2} \frac{2 U_{j+1}(T)}{U_j(T)} e^{-E_{I,j,j+1}/kT}.$
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- $\displaystyle E_x = q(x - v t)(1-\beta^2)\frac{\gamma^3}{\left[y^2+\gamma^2(x-vt)^2\right]^{3/2}} \\ = \frac{q\gamma (x-vt)}{r^3} \\ E_y = \frac{q\gamma y}{r^3} \\ E_z = \frac{q\gamma z}{r^3}$
- $\displaystyle E_x = -\frac{qv \gamma t}{(\gamma^2 v^2 t^2 + b^2)^{3/2}} B_x =0 \\ E_y = \frac{q \gamma b}{(\gamma^2 v^2 t^2 + b^2)^{3/2}} B_y =0 \\ E_z = 0 B_z =\beta E_y$
- $\displaystyle {\hat E}_x(\omega) = \frac{1}{2\pi} \int E_x(t) e^{i\omega t}dt = -\frac{q \gamma v}{2\pi} \int_{-\infty}^\infty t \left ( \gamma^2 v^2 t^2 + b^2 \right )^{-3/2} e^{i\omega t} dt \\ {\hat E}_y(\omega) = \frac{1}{2\pi} \int E_y(t) e^{i\omega t}dt = \frac{q \gamma b}{2\pi} \int_{-\infty}^\infty \left ( \gamma^2 v^2 t^2 + b^2 \right )^{-3/2} e^{i\omega t} dt.$
- $\\ {\hat E}_y(\omega) = \frac{q}{\pi b v} \left [ \frac{\omega b}{\gamma v} K_1 \left ( \frac{\omega b}{\gamma v} \right ) \right ].$
- $\displaystyle {\hat E}(\omega) \approx \frac{1}{2\pi} \int E_y(t) e^{i\omega t}dt = \frac{q \gamma b}{2\pi} \int_{-\infty}^\infty \left ( \gamma^2 v^2 t^2 + b^2 \right )^{-3/2} dt = \frac{q }{v b\pi}$
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- In practice viscosity and thermal conduction save the day, so although the wave may get really steep, a discontinuity doesn't actually form.
- $\displaystyle \left ( \frac{1}{2} v_1^2 + w_1 \right ) \rho_1 v_1 = \left ( \frac{1}{2} v_2^2 + w_2 \right ) \rho_2 v_2 .$
- $\displaystyle v_1-v_2=j\left ( V_1 - V_2 \right )~\textrm{so}~\left (v_1-v_2\right)^2 = \left ( p_2 - p_1 \right ) \left (V_1 - V_2\right ) = -\Delta p \Delta V.$
- $\displaystyle p_2 = p_1 \frac{\frac{\gamma}{\gamma-1} V_1 - \frac{1}{2} \left ( V_1 + V_2 \right ) }{\frac{\gamma}{\gamma-1} V_2 - \frac{1}{2} \left ( V_1 + V_2 \right ) }.$
- $\displaystyle v_1^2 = j^2 V_1^2 = -\frac{\Delta p}{\Delta V} V_1^2~\textrm{and}~ v_2^2 = -\frac{\Delta p}{\Delta V} V_2^2.$
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- $\displaystyle -\omega^2 \hat{\bf d}(\omega) = -\frac{e}{2\pi} \int_{-\infty}^\infty \dot{\bf v} e^{i\omega t} dt.$
- $\displaystyle \Delta v = \int_{-\infty}^\infty \frac{b}{R} \frac{1}{m} \frac{Ze^2}{R^2} = \frac{Z e^2}{m} \int_{-\infty}^\infty \frac{b}{\left ( b^2 + v^2 t^2 \right )^{3/2}} dt = \frac{2 Z e^2}{m b v}$
- $\displaystyle \frac{d W}{d\omega dV dt} = n_e n_i 2\pi v \int_{b_\mbox{min}}^{b_\mbox{max}} \frac{8 Z^2 e^6}{3\pi c^3 m^2 v^2 b^2} b d b \\ = \frac{16 e^6}{3 c^3 m^2 v} n_e n_i Z^2 \int_{b_\mbox{min}}^{b_\mbox{max}} \frac{db}{b}\\ = \frac{16 e^6}{3 c^3 m^2 v} n_e n_i Z^2 \ln \left (\frac{b_\mbox{max}}{b_\mbox{min}} \right )$
- $\displaystyle v \sim \Delta v(b_\mbox{min}^{(1)}) = \frac{2 Z e^2}{m b_\mbox{min}^{(1)} v} \rightarrow b_\mbox{min}^{(1)} \sim \frac{2 Z e^2}{m v^2}.$
- $\displaystyle \frac{d W}{d\omega dV dt} = \frac{16 \pi e^6}{3 \sqrt{3} c^3 m^2 v} n_e n_i Z^2 g_{ff}(v,\omega)$
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- The magnitude of the centripetal force is $\frac{m_ev^2}{r_n}$, while the Coulomb force is $\frac{Zk_e e^2}{r^2}$.
- $\displaystyle E= \frac{1}{2} m_e v^2 - \frac{Z k_e e^2}{r} = - \frac{Z k_e e^2}{2r}$
- Now, here comes the Quantum rule: As we saw in the previous module, the angular momentum $L = m_e r v$ is an integer multiple of $\hbar$:
- $\displaystyle E = -\frac{Zk_e e^2}{2r_n } = - \frac{ Z^2(k_e e^2)^2 m_e }{2\hbar^2 n^2} \approx \frac{-13.6Z^2}{n^2}\mathrm{eV}$
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- $\displaystyle {\bf d}_{if} = -\frac{e\hbar}{m \omega_{if}} \frac{1}{\sqrt{V}} \left ( \frac{Z^3}{\pi a_0^3} \right )^{1/2}\int e^{-Zr/a_0} \nabla_{\bf r} e^{i {\bf q}\cdot {\bf r}} d^3 x \\ = -\frac{i e\hbar {\bf q} }{m \omega_{if}} \frac{1}{\sqrt{V}} \left ( \frac{Z^3}{\pi a_0^3} \right )^{1/2}\int e^{-Zr/a_0} e^{i {\bf q}\cdot {\bf r}} d^3 x \\ = -\frac{ie \hbar {\bf q}}{m \omega_{if}} \frac{1}{\sqrt{V}} \left ( \frac{Z^3}{\pi a_0^3} \right )^{1/2}2\pi \int_0^\infty r^2 dr \int_{-1}^1 d \mu e^{-i q r \mu} e^{-Zr/a_0} \\ = -\frac{i e \hbar {\bf q}}{m \omega_{if}} \frac{1}{\sqrt{V}} \left ( \frac{Z^3}{\pi a_0^3} \right )^{1/2}\frac{4\pi}{q} \int_0^\infty r dr e^{-Zr/a_0} \sin qr \\ = -\frac{i e\hbar {\bf q}}{m \omega_{if}} \frac{1}{\sqrt{V}} \left ( \frac{Z^3}{\pi a_0^3} \right )^{1/2} \frac{8\pi a_0^3 Z}{\left (Z^2 + q^2 a_0^2\right)^2}$
- $\displaystyle |{\bf d}_{if} |^2 = \frac{256\pi}{V} \left ( \frac{Z}{a_0} \right )^5 \left ( \frac{Z^2}{a_0^2} + q^2 \right )^{-6} e^2 q^2 \approx \frac{256\pi e^2}{V} \left ( \frac{Z}{a_0} \right )^5 q^{-10}.$
- where $f(v)$ is the Maxwellian velocity distribution, $N_e$ is the electron density and $N_+$ is the ion density.
- $\displaystyle \frac{\sigma_{bf}}{\sigma_{fb}} = \frac{N_+ N_e}{N_n} e^{h\nu/kT} \frac{f(v) c^2 h}{8\pi m \nu^2}$
- where we have used $h\nu = \frac{1}{2} m v^2 + E_I$ to eliminate $d\nu$ and $dv$.
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- $\displaystyle I'(E',\mu') = F_0 \left (\frac{E'}{E}\right)^2 \delta (E-E_0)\\ \displaystyle = F_0 \left (\frac{E'}{E_0}\right)^2 \delta (\gamma E' (1+\beta\mu') -E_0) \\ \displaystyle = \frac{F_0}{\gamma\beta E'} \left (\frac{E'}{E_0}\right)^2 \delta \left (\mu' - \frac{E_0-\gamma E'}{\gamma\beta E'} \right )$
- where we have assumed that $E_f'=E'$.
- $\displaystyle j'(E_f') = \frac{N' \sigma_T E_f' F_0}{2 E_0^2 \gamma \beta}~\text{ if }~ \frac{E_0}{\gamma (1+\beta)} < E_f' < \frac{E_0}{\gamma(1-\beta)}$
- $\displaystyle j(E_f,\mu_f) = \frac{E_f}{E_f'} j'(E_f') \\ \displaystyle = \frac{N \sigma_T E_f F_0}{2 E_0^2 \gamma^2 \beta} \\ \displaystyle ~~\text{ if }~ \frac{E_0}{\gamma (1+\beta)(1-\beta \mu_f)} < E_f < \frac{E_0}{\gamma(1-\beta)(1-\beta \mu_f)} \nonumber$
- $\displaystyle \frac{d E}{dV dt dE_f} = 4 \pi E_f j(E_f) \\ \displaystyle = \frac{3}{4} c \sigma_T C \int d E \left ( \frac{E_f}{E} \right ) v(E) \int_{\gamma_1}^{\gamma_2} d \gamma \gamma^{-p-2} f \left ( \frac{E_f}{4\gamma^2 E} \right ) \\ \displaystyle = 3\sigma_T c C 2^{p-2} E_f^{-(p-1)/2} \times \\ \displaystyle \nonumber ~~~\int d E E^{(p-1)/2} v(E) \int_{x_1}^{x_2} x^{(p-1)/2} f(x) dx$
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- $I = s_0 = |\epsilon_1 \cdot {\bf E}|^2 + |\epsilon_2 \cdot {\bf E}|^2 = a_1^2 + a_2^2 \\ Q = s_1 = |\epsilon_1 \cdot {\bf E}|^2 - |\epsilon_2 \cdot {\bf E}|^2 = a_1^2 - a_2^2 \\ U = s_2 = 2 \Re \left [ (\epsilon_1 \cdot {\bf E})^* (\epsilon_2 \cdot {\bf E}) \right ] = 2 a_1 a_2 \cos \left (\delta_2 - \delta_1 \right ) \\ V = s_3 = 2 \Im \left [ (\epsilon_1 \cdot {\bf E})^* (\epsilon_2 \cdot {\bf E}) \right ] = 2 a_1 a_2 \sin \left (\delta_2 - \delta_1 \right )$
- $I = s_0 = |\epsilon_+ \cdot {\bf E}|^2 + |\epsilon_- \cdot {\bf E}|^2 = a_+^2 + a_-^2 \\ Q = s_1 = 2 \Re \left [ (\epsilon_+ \cdot {\bf E})^* (\epsilon_- \cdot {\bf E}) \right ] = 2 a_+ a_- \cos \left (\delta_- - \delta_+ \right ) \\ U = s_2 = 2 \Im \left [ (\epsilon_+ \cdot {\bf E})^* (\epsilon_- \cdot {\bf E}) \right ] = 2 a_+ a_- \sin \left (\delta_- - \delta_+ \right ) \\ V = s_3 = |\epsilon_+ \cdot {\bf E}|^2 - |\epsilon_- \cdot {\bf E}|^2 = a_+^2 - a_-^2$