$$\require{cancel}$$

# 10.1: Lorentz Transformations of Energy and Momentum

[ "article:topic", "Lorentz transformations", "authorname:nwalet" ]

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}