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[ "article:topic", "Lorentz transformations", "authorname:nwalet", "license:ccbyncsa", "showtoc:no" ]
Physics LibreTexts

10.1: Lorentz Transformations of Energy and Momentum

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  • As you may know, like we can combine position and time in one four-vector \(x=({\vec{x}}, ct)\), we can also combine energy and momentum in a single four-vector, \(p=({\vec{p}}, E/c)\). From the Lorentz transformation property of time and position, for a change of velocity along the \(x\)-axis from a coordinate system at rest to one that is moving with velocity \({\vec{v}} = (v_x,0,0)\) we have

    \[ \begin{align} x' &= \gamma(v) (x-v/c t),\\[5pt] t'&=\gamma (t-xv x/c^2),\end{align}\]

    we can derive that energy and momentum behave in the same way,

    \[\begin{align} p'_x &= \gamma(v) (p_x - E v/c^2) \\[5pt] &= m u_x \gamma(|u|), \nonumber\\[5pt] E' &= \gamma(v) (E - v p_x) \\[5pt] &= \gamma(|u|) m_0 c^2. \label{eqLorentzE} \end{align}\]

    To understand the context of these equations remember the definition of \(\gamma\)

    \[\gamma(v) = \dfrac{1}{\sqrt{1-\beta^2}}\]



    In Equation \ref{eqLorentzE}, we have also re-expressed the momentum energy in terms of a velocity \({\vec{u}}\). This is measured relative to the rest system of a particle, the system where the three-momentum \({\vec{p}}=0\).

    Now all these exercises would be interesting mathematics but rather futile if there was no further information. We know however that the full four-momentum is conserved, i.e., if we have two particles coming into a collision and two coming out, the sum of four-momenta before and after is equal,

    \[\begin{aligned} E^{\mathrm{in}}_1+E^{\mathrm{in}}_2 &= E^{\mathrm{out}}_1+E^{\mathrm{out}}_2, \nonumber\\ {\vec{p}}^{\mathrm{in}}_1+{\vec{p}}^{\mathrm{in}}_2 &= {\vec{p}}^{\mathrm{out}}_1+{\vec{p}}^{\mathrm{out}}_2. \end{aligned}\]