# 10.1: Lorentz Transformations of Energy and Momentum

- Page ID
- 15052

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

\[x' = \gamma(v) (x-v/c t),\quad t'=\gamma (t-xv x/c^2),\] 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 \(\gamma\)

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

and

\[\beta=\frac{v}{c}.\]

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,

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