相対論 (1) - 定数、換算式 -¶

定数¶

$E &= m c^2 \\ &= \gamma m_0c^2 \\ &= \frac{m_0c^2}{\sqrt{1 - \beta^2}} \\$

$E &= E_0 + E_k \\ &= m_0 c^2 + E_k$

$E = \sqrt{p^2 c^2 + m_0^2 c^4}$

$E_k &= m_0c^2 (\gamma - 1) \\ &= m_0c^2 \left( \frac{1}{\sqrt{1 - \beta^2}} - 1 \right)$

$E_k &= E - E_0 \\ &= E - m_0 c^2$

$p = m_0c \gamma \beta$

$p = \frac{m_0c \beta}{\sqrt{1 - \beta^2}}$

$p = m_0c \sqrt{\gamma^2 - 1}$

$pc = E \beta$

$pc = \sqrt{E^2 - m_0^2 c^4}$

$pc = \sqrt{K^2 + 2Km_0c^2}$

$\beta = \frac{\sqrt{K^2 + 2Km_0c^2}}{K + m_0c^2} = \frac{\sqrt{\left( \frac{K}{m_0c^2} \right)^2 + 2 \frac{K}{m_0c^2}}}{\frac{K}{m_0c^2} + 1}$

$\beta = \sqrt{1 - \left( \frac{m_0c^2}{E} \right)^2 }$

$\beta = \sqrt{1 - \frac{1}{\gamma^2}} = \frac{\sqrt{\gamma^2 - 1}}{\gamma}$

$\beta = \frac{\gamma\beta}{\sqrt{1 + (\gamma\beta)^2}}$

$\gamma = \frac{K}{m_0c^2} + 1$

$\gamma = \frac{E}{m_0c^2}$

$\gamma = \sqrt{1 + (\gamma\beta)^2}$

$\gamma\beta = \sqrt{\gamma^2 - 1}$

$\gamma\beta = \frac{\beta}{\sqrt{1 - \beta^2}}$