# Momentum NOT as Spatial Energy, 1.2

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\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

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\displaystyle{\begin{aligned} E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ E_r &= m c^2 \\ \\ E &= E_{k} + E_{r} \\ \end{aligned}}

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$\displaystyle { p = \left(p^{0},p^{1},p^{2},p^{3}\right) = \left({E \over c},p_{x},p_{y},p_{z}\right) = \left({E \over c},\bf p \right) }$

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$\displaystyle { p = \left({E \over c},p_{x},p_{y},p_{z}\right)}$

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\displaystyle{\begin{aligned} E &= E_{k} + E_{r} \\ \\ E_r &= m c^2 \\ E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ \\ E_k &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\ &= {\frac {1}{2}}mv^{2}+{\frac {3}{8}}m{\frac {v^{4}}{c^{2}}} + \dotsb \\ \end{aligned}}

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\displaystyle{ \begin{aligned} p \cdot p &= -(p^0)^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 \\ &= -(\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\ \\ p \cdot p &= -m^{2}c^{2} \\ \\ -m^{2}c^{2} &= -(\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\ -m^{2}c^{2} &= -(\frac{E}{c})^2 + |\mathbf{p}|^2 \\ \end{aligned} }

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\displaystyle{\begin{aligned} E &= E_{k} + E_{r} \\ \end{aligned}}

\displaystyle{\begin{aligned} E &= E_{k} + E_{r} \\ \end{aligned}}

$\displaystyle { p = \left(p^{0},p^{1},p^{2},p^{3}\right) = \left({E \over c},p_{x},p_{y},p_{z}\right) = \left({E \over c},\bf p \right) }$

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\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

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\displaystyle{\begin{aligned} E_k &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\ &= {\frac {1}{2}}mv^{2}+{\frac {3}{8}}m{\frac {v^{4}}{c^{2}}} + \dotsb \\ \\ E_r &= m c^2 \\ E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ \\ E &= E_{k} + E_{r} \\ \end{aligned}}

— Me@2022-01-20 11:33:48 AM

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