# Momentum as Spatial Energy

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In special relativity, four-momentum is the generalization of the classical three-dimensional momentum to four-dimensional spacetime. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in spacetime. The contravariant four-momentum of a particle with relativistic energy $E$ and three-momentum $p = (p_x, p_y, p_z) = \gamma m \bf v$, where $\bf v$ is the particle’s three-velocity and $\gamma$ the Lorentz factor, is

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

The quantity $m \bf v$ of above is ordinary non-relativistic momentum of the particle and $m$ its rest mass. The four-momentum is useful in relativistic calculations because it is a Lorentz covariant vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.

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$\displaystyle{ p\cdot p=\eta _{\mu \nu }p^{\mu }p^{\nu }=p_{\nu }p^{\nu }=-{E^{2} \over c^{2}}+|\mathbf {p} |^{2}=-m^{2}c^{2}}$

— Wikipedia on Four-momentum

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1. Newtonian momentum 牛頓動量

$\displaystyle{m \bf v}$

2. Three-momentum 相對論三維動量

$\displaystyle{\mathbf{p} = \gamma m \bf v}$, $\displaystyle{\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$

3. Four-momentum 時空動量（相對論四維動量）

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

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\displaystyle{ \begin{aligned} \| p \| &= \sqrt{p \cdot p} \\ \\ p \cdot p &= -(p^0)^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 \\ \end{aligned}}

\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 \\ \end{aligned} }

\displaystyle{ \begin{aligned} \| p \| &= \sqrt{p \cdot p} &= \sqrt{- \left(\frac{E}{c}\right)^2 + (p_x)^2 + (p_y)^2 + (p_z)^2} \\ \end{aligned} }

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

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\displaystyle{ \begin{aligned} p \cdot p &= -m^{2}c^{2} \\ \end{aligned} }

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

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

\displaystyle{ \begin{aligned} \frac{E}{c} &= \dotsb + \dotsb \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ &= 0 + m^2 c^2 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= m^2 c^2 \\ \\ E &= m c^2 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} E_r &= m c^2 \\ \end{aligned}}

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

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

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

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\displaystyle{ \begin{aligned} E_k &= \sqrt{p^2 c^2 + m^2 c^4} - m c^2 \\ \end{aligned}}

1. Four-momentum 時空動量

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

2. Total energy 總能量

$\displaystyle{E = p^0 c}$

3. Three-momentum (3-space momentum) 相對論三次元空間動量

$\displaystyle{\mathbf{p} = \left(p^{1},p^{2},p^{3}\right) = \left( p_x, p_y, p_z \right)}$

$\displaystyle{\mathbf{p} = \gamma m \mathbf{v} = \frac{m \mathbf{v}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$

1. 總能量 $\displaystyle{E}$ 會因為，時間的均勻性而守恆。「時間均勻性」的意思是，物理定律不會隨時間改變。

2. 空間動量 $\displaystyle{\mathbf{p}}$ 會因為，空間的均勻性而守恆。「空間均勻性」的意思是，宇宙間任何兩個不同地方，物理定律必為相同。

3. 結果，因為時間分量 $\displaystyle{\frac{E}{c}}$ 和空間分量 $\displaystyle{\mathbf{p}}$ 都守恆，時空動量 $\displaystyle{p=\left({E \over c},\mathbf{p}\right)}$ 作為整體，亦會守恆。

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\displaystyle{\begin{aligned} E_{\text{k}} &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\ \\ E_{\text{k}} &\approx {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} \end{aligned}}

— Me@2022-01-16 05:39:11 PM

— Me@2022-01-17 01:17:51 PM

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I would flip it and say momentum [is] energy in the space-like direction. When at rest (four velocity $u = (c, {\bf 0})$), you are moving entirely in the time direction. Four momentum is:

$\displaystyle{p_{\mu} = mu_{\mu} = (mc^2, {\bf 0})}$

For a observer boosted to velocity $\displaystyle{ \bf v }$:

$\displaystyle{p_{\mu} = mu_{\mu} =\gamma(mc^2, m{\bf v}c)=(mc^2+T, {\bf p}c)}$

[Y]our rest mass (energy) now appears as momentum (and the timelike term has kinetic energy added to it).

That last equality may make it clear that when in motion ($c=1$):

$\displaystyle{m \rightarrow m+T}$

$\displaystyle{{\bf 0}\rightarrow {\bf 0} + {\bf p}}$

which could be interpreted as “Kinetic energy is momentum in the time-like direction”.

— answered Apr 25 ’18 at 1:13

— JEB

— Physics Stack Exchange

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