# 当相对论文明遇到牛顿力学文明 精选

$\delta m \ V + (m- \delta m)\ dv =0$

$\delta m \ V + m\ dv =0$

$dv = - V \delta m/m$

$\Delta v = -V \ln (m_i/m_f)$

$m_i = m_f + M = m_f \ e^{-\Delta v/V} \\ M = m_f \ (e^{-\Delta v/V} -1)$

$M_1 = m_f \ (e^{U/V} -1)\\ M_2 = (m_f +M_1) \ (e^{U/V} -1) \\ M = M_1 + M_2 = m_f (e^{U/V}-1) + m_f [1+ (e^{U/V}-1)] (e^{U/V}-1)\\ M=m_f\ (e^{2U/V}-1)$

$\gamma_0 \ \delta m (v+V)/\sqrt{1-v^2} + ( m -\gamma_0\ \delta m) (v+dv)/\sqrt{1-(v+dv)^2} = m v /\sqrt{1-v^2}$

$\gamma_0 \ \delta m (v+V)/\sqrt{1-v^2} + ( m - \gamma_0 \ \delta m) (v+dv) [1+v \ dv /(1-v^2)]/\sqrt{1-v^2} = m v /\sqrt{1-v^2}$

$\gamma_0 \ \delta m (v+V)+ ( m -\gamma_0 \ \delta m) (v+dv) [1+v \ dv /(1-v^2)] = m v$

$( m - \gamma_0 \ \delta m) (v+dv) [1+v \ dv /(1-v^2)] \\ = (mv +m\ dv - v \ \gamma_0 \ \delta m) [1+v \ dv /(1-v^2)]\\ =mv + m \ dv - v \ \gamma_0 \ \delta m + mv^2 \ dv /(1-v^2)$

$\gamma_0 \ \delta m\ (v+V) + m \ dv - v \ \gamma_0 \delta m + mv^2 \ dv /(1-v^2) =0 \\ \gamma_0 \ V \ \delta m+ m \ dv /(1-v^2) =0$

$d m/m = \frac{dv}{V (1-v^2)}$

$V \ dm/m = \frac{dv}{ 1-v^2 } = \frac{1}{2}(\frac{dv}{1-v} + \frac{dv}{1+v})$

$V \ln m = \frac{1}{2}(\ln (1+v) - \ln (1-v))= \frac{1}{2} \ln\frac{1+v}{1-v} +C$

$V \ln m/m_i = \frac{1}{2}(\ln (1+v) - \ln (1-v))= \frac{1}{2} \ln\frac{1+v}{1-v} - \frac{1}{2} \ln\frac{1+v_i}{1-v_i}$

$\frac{m_f}{m_f+M} =\left[\frac{(1-v_i)\ (1+v)}{(1+v_i) \ (1-v)}\right]^{\frac{1}{2\ V}}$

$M = m_f \left[ (\frac{1-U}{1+U})^{\frac{1}{2 V}} -1\right]$

$M = m_f \left[ (\frac{1-U}{1+U})^{\frac{1}{V}} -1\right]$

$R_{n/r}= \left[ (\frac{1+U}{1-U})^{\frac{1}{V}} -1\right] /(e^{2U/V}-1)$

https://blog.sciencenet.cn/blog-684007-1304964.html

## 全部精选博文导读

GMT+8, 2022-5-23 17:05