# 高能量天体物理学 High Energy Astrophysics PHYS201501

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$$H=\frac{p^{2}}{2 m_{e}}-V(r)$$
Making the usual substitution: $\boldsymbol{p}=-i \hbar \nabla$ we get the relevant form of (114):
$$\left(-\frac{\hbar^{2}}{2 m_{e}} \nabla^{2}-V(r)\right) \psi(\boldsymbol{r})=E \psi(\boldsymbol{r})$$
It is convenient to use atomic units where the natural unit of length is the Bohr radius: $a_{0} \equiv \hbar^{2} / m e^{2}=0.52910^{-8} \mathrm{~cm}$, and the natural unit of energy is twice the Rydberg constant: $e^{2} / a_{0} \equiv 2 R y=27.2 \mathrm{eV}=4.3610^{-11}$ erg. In these units, $e=\hbar=m=1$.
Equation (119) then takes the form:
$$\left(\frac{1}{2} \nabla^{2}+E+V(r)\right) \psi(\boldsymbol{r})=0$$

## PHYS201501COURSE NOTES ：

$$\frac{1}{2} m v^{2}=\frac{Z e^{2}}{2 r}$$
For the ground-state:
$$r \simeq \frac{a_{0}}{Z},$$
and thus:
$$v \simeq\left(\frac{Z^{2} e^{2}}{m a_{0}}\right)^{1 / 2}=(Z \alpha) c$$

# 高能天体物理学|PH40113 High energy astrophysics代写

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The rigorous theory of Cherenkov radiation (more correctly called Cherenkov-Vavilov radiation after its co-discoverers) was developed by Frank and Tamm from which the basic formulae are derived [6].
From this theory the emission formula is derived:
$$\begin{array}{cc} \mathrm{d} E / \mathrm{d} t=\left(e^{2} / c^{2}\right) & \int \sin ^{2} \theta \omega \mathrm{d} \omega . \ \text { I } & \text { II } \quad \text { III } \end{array}$$
The basic components of this equation were expounded by Jelley [11]. The mechanism is quite different from other, more familiar, radiation mechanisms such as synchrotron radiation or bremsstrahlung. The three factors in the Cherenkov formula for radiation yield can be understood by analogy with the single elementary classical dipole.

## PH40113 COURSE NOTES ：

$$\delta u=s(\varphi) \int_{0}^{\varphi} c(\psi) q(\psi) \mathrm{d} \psi-c(\varphi) \int_{0}^{\varphi} s(\psi) q(\psi) \mathrm{d} \psi$$
where $s(\varphi)$ and $c(\varphi)$ are two independent solutions of the homogenous equation $f_{, \rho \varphi}+(1-3 \zeta \cos \varphi) f=0$. These are the so-called Mathieu functions:
$$s(\varphi)=\sin \varphi+O(\zeta) ; \quad c(\varphi)=\cos \varphi+O(\zeta)$$
The O-terms are actually series in $\sin n \varphi$ or $\cos n \varphi$ of order $\zeta$ or smaller, whose explicit expression we fortunately don’t need. The whole solution is now