Tex練習：compton電子

$$\displaystyle \frac{E'}{E}=\displaystyle \frac{1}{1+\alpha (1-cos\theta )}$$
$$\alpha =\displaystyle \frac{E}{m_{0}c^{2 } }$$
クラインー仁科
$$\displaystyle \frac{d\sigma }{d\mit\Omega }=\displaystyle \frac{r_{0 } }{2}\left( \displaystyle \frac{E'}{E} \right) ^{2}\left( \displaystyle \frac{E'}{E}+\displaystyle \frac{E}{E'}-sin^{2}\theta \right) $$
$$E_{e}=E-E'=E\displaystyle \frac{\alpha (1-cos\theta )}{1+\alpha (1-cos\theta )}$$
$$\displaystyle \frac{d\sigma }{dE_{e } }=\displaystyle \frac{d\sigma }{d\mit\Omega }\displaystyle \frac{d\mit\Omega }{d\theta }\displaystyle \frac{d\theta }{dE_{e } }$$
$$ d\mit\Omega =2\pi sin\theta d\theta ,　　　　 \displaystyle \frac{d\mit\Omega }{d\theta }=2\pi sin\theta $$\

$$\displaystyle \frac{dE_{e } }{d\theta }=E\displaystyle \frac{\alpha sin\theta }{\left[ 1+\alpha \left( 1-cos\theta \right) \right] ^{2 } }$$

$$\displaystyle \frac{d\sigma }{dE_{e } }=\displaystyle \frac{d\sigma }{d\mit\Omega }2\pi sin\theta \displaystyle \frac{\left[ 1+\alpha \left( 1-cos\theta \right) \right] ^{2 } }{E\alpha sin\theta }=\displaystyle \frac{\pi r_{0 } }{E\alpha }\left( 1+cos^{2}\theta \right) \left\{ 1+\displaystyle \frac{\alpha ^{2}\left( 1-cos\theta \right) ^{2 } }{\left( 1+cos^{2}\theta \right) \left[ 1+\alpha \left( 1-cos\theta \right) \right] } \right\} $$