e

Examples of e in the following topics:

• Inverse Compton Spectra - Single Scattering

  • $\displaystyle I'(E',\mu') = F_0 \left (\frac{E'}{E}\right)^2 \delta (E-E_0)\\ \displaystyle = F_0 \left (\frac{E'}{E_0}\right)^2 \delta (\gamma E' (1+\beta\mu') -E_0) \\ \displaystyle = \frac{F_0}{\gamma\beta E'} \left (\frac{E'}{E_0}\right)^2 \delta \left (\mu' - \frac{E_0-\gamma E'}{\gamma\beta E'} \right )$
  • where we have assumed that $E_f'=E'$.
  • $\displaystyle j'(E_f') = \frac{N' \sigma_T E_f' F_0}{2 E_0^2 \gamma \beta}~\text{ if }~ \frac{E_0}{\gamma (1+\beta)} < E_f' < \frac{E_0}{\gamma(1-\beta)}$
  • $\displaystyle j(E_f,\mu_f) = \frac{E_f}{E_f'} j'(E_f') \\ \displaystyle = \frac{N \sigma_T E_f F_0}{2 E_0^2 \gamma^2 \beta} \\ \displaystyle ~~\text{ if }~ \frac{E_0}{\gamma (1+\beta)(1-\beta \mu_f)} < E_f < \frac{E_0}{\gamma(1-\beta)(1-\beta \mu_f)} \nonumber$
  • $\displaystyle \frac{d E}{dV dt dE_f} = 4 \pi E_f j(E_f) \\ \displaystyle = \frac{3}{4} c \sigma_T C \int d E \left ( \frac{E_f}{E} \right ) v(E) \int_{\gamma_1}^{\gamma_2} d \gamma \gamma^{-p-2} f \left ( \frac{E_f}{4\gamma^2 E} \right ) \\ \displaystyle = 3\sigma_T c C 2^{p-2} E_f^{-(p-1)/2} \times \\ \displaystyle \nonumber ~~~\int d E E^{(p-1)/2} v(E) \int_{x_1}^{x_2} x^{(p-1)/2} f(x) dx$

• A.2 Parseval's Theorem

  • $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = 2\pi \int_{-\infty}^{\infty} |{\hat E}(\omega)|^2 d \omega.$
  • $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = \int_{-\infty}^{\infty} d t \int_{-\infty}^{\infty} {\hat E}(\omega') e^{-i\omega' t} d \omega'.
  • \int_{-\infty}^{\infty} {\hat E}^*(\omega) e^{i\omega t} d \omega \\ \displaystyle = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} d t d\omega' d \omega {\hat E}(\omega') {\hat E}^*(\omega) e^{-i\omega' t} e^{i\omega t} $
  • The integral over time is simply Fourier transform of $2\pi e^{-i\omega' t}$ which we know,
  • $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = 2\pi \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} d\omega' d \omega {\hat E}(\omega') {\hat E}^*(\omega) \delta (\omega -\omega') \\ \displaystyle = 2 \pi \int_{-\infty}^{\infty} d \omega {\hat E}(\omega) {\hat E}^*(\omega) =2 \pi \int_{-\infty}^{\infty} |{\hat E}(\omega)|^2 d \omega$

• The Natural Exponential Function: Differentiation and Integration

  • First, we determine the derivative of $e^{x}$ using the definition of the derivative:
  • Since $e^{x}$ does not depend on $h$, it is constant as $h$ goes to $0$.
  • Therefore $\ln(e^x) = x$ and $e^{\ln x} = x$.
  • Let's consider the example of $\int e^{x}dx$.
  • Since $e^{x} = (e^{x})'$ we can integrate both sides to get:

• Inverse Compton Power - Single Scattering

  • $\displaystyle \frac{v dE}{E} = \frac{v' dE'}{E'} = \text{ Lorentz Invariant} $
  • $\displaystyle \frac{d E_f}{d t} = \frac{d E_f'}{d t'} = c \sigma_T \int E_f' v' d E'$
  • $\gamma^2 - 1 \gg E/(m c^2)$, so $E_f'=E'$, so we have
  • $\displaystyle \frac{d E_f}{d t} = c \sigma_T \int E'^2 \frac{v' dE'}{E'} = c \sigma_T \int E'^2 \frac{v dE}{E}$
  • $\displaystyle \frac{d E}{d t} = -c\sigma_T \int E v dE = -\sigma_T c U_\text{ph}$

• Repeated Scattering

  • $\displaystyle E_f = \frac{E_i}{1+ \frac{E_i}{mc^2} \left ( 1 - \cos\theta \right )}$
  • $\displaystyle E_f \approx E_i \left [ 1 - \frac{E_i}{mc^2} \left ( 1 - \cos\theta \right ) \right ] = E_f \left [ 1 - \frac{E_i}{mc^2} \right ]$
  • $\displaystyle \langle E_f - E_i \rangle = -\frac{\langle E_i^2 \rangle}{mc^2} + \frac{\alpha k T}{mc^2} \langle E_i \rangle = 0.\\ \displaystyle = -\frac{12 (kT)^2}{mc^2} + \frac{3 \alpha (kT)^2}{mc^2} = 0$
  • $\displaystyle \frac{E_f - E_i}{E_i} = \frac{1}{mc^2}\left ( 4 k T - E_i \right )$
  • $\displaystyle \frac{E_f - E_i}{E_f} \approx \frac{4}{3} \left [ 12 \left ( \frac{kT}{mc^2} \right )^2 \right ] = 16 \left ( \frac{kT}{mc^2} \right )^2.$

• Repeated Scattering with Low Optical Depth

  • $\displaystyle A \equiv \frac{E_f}{E_i} \sim \frac{4}{3} \langle \gamma^2 \rangle = 16 \left ( \frac{kT}{mc^2} \right )^2.$
  • The probability that a photon will scatter as it passes through a medium is simply $\tau_{es}$ if the optical depth is low, and the probability that it will undergo $k$ scatterings $p_k \sim \tau_{es}^k$ and its energy after $k$ scatterings is $E_k=A^k E_i$, so we have
  • $\displaystyle I(E_k) = I(E_i) \exp \left ( \frac{\ln\tau_{es} \ln\frac{E_k}{E_i}}{\ln A} \right ) = I(E_i) \left ( \frac{E_k}{E_i} \right )^{-\alpha}$
  • $\displaystyle P = \int_{E_i}^{A^{1/2}mc^2} I(E_k) dE_k = I(E_i) E_i \left [ \int_1^{A^{1/2} mc^2/E_i} x^{-\alpha} dx \right ].$

• Waves

  • ${\bf \nabla} \times \left ({\bf \nabla} \times {\bf E} \right ) = {\bf \nabla} \left ( {\bf \nabla} \cdot {\bf E} \right ) - {\bf \nabla}^2 {\bf E}.$
  • $\displaystyle {\bf \nabla}^2 {\bf E} -\frac{1}{c^2} \frac{\delta ^2 {\bf E}}{\delta t^2} = 0$
  • $\displaystyle {\bf E}= \Re \left ({\bf \hat a}_1 E_0 e^{i({\bf k \cdot r}-\omega t)} \right ) ~~\text{ and }~~~ {\bf B}= \Re \left ( {\bf \hat a}_2 B_0 e^{i({\bf k \cdot r}-\omega t)} \right )$
  • d t \left ( \Re E_0 \Re B_0 \cos^2 \omega t + \Im E_0 \Im B_0 \sin^2 \omega t \right ) \\ \displaystyle = \frac{c{\bf \hat k}}{8\pi} \left ( \Re E_0 \Re B_0 + \Im E_0 \Im B_0 \right ) = \frac{c{\bf \hat k}}{8\pi} \Re E_0^* B_0 \\ \displaystyle = \frac{c{\bf \hat k}}{8\pi} |E_0|^2= \frac{c{\bf \hat k}}{8\pi} |B_0|^2$
  • $\displaystyle \left < U \right > = \frac{1}{16\pi} \Re \left ( E_0 E_0^* + B_0 B_0^* \right ) = \frac{1}{8\pi} |E_0|^2 = \frac{1}{8\pi} |B_0|^2$

• The Spectrum

  • $\displaystyle {\hat E} (\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} E(t) e^{i\omega t} d t.$
  • $\displaystyle E (t) = \int_{-\infty}^{\infty} {\hat E}(\omega) e^{-i\omega t} d \omega.$
  • Because $E(t)$ is real find that ${\hat E}(-\omega)={\hat E}^*(\omega)$ so we don't have to keep track of the negative frequencies.
  • $\displaystyle \int_{-\infty}^{\infty} |E|^2(t) dt = 2\pi \int_{-\infty}^{\infty} |{\hat E}(\omega)|^2 d \omega.$
  • $\displaystyle \int_{-\infty}^{\infty} E^2(t) dt = 4\pi \int_0^{\infty} |{\hat E}(\omega)|^2 d \omega.$

• Energy of a Bohr Orbit

  • 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 $\hbar$:
  • $\displaystyle E = -\frac{Zk_e e^2}{2r_n } = - \frac{ Z^2(k_e e^2)^2 m_e }{2\hbar^2 n^2} \approx \frac{-13.6Z^2}{n^2}\mathrm{eV}$

• Second (or E) Conjugation