Lorentz transformations of energy and momentum

10.1 Lorentz transformations of energy and momentum

As you may know, like we can combine position and time in one four-vector $x = (x, c t)$ , we can also combine energy and momentum in a single four-vector, $p = (p, E ∕ c)$ . 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 $v = (v_{x}, 0, 0)$ we have

x^{'} = γ (v) (x - v ∕ c t), t^{'} = γ (t - x v x ∕ c^{2}),

(10.1)

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

\begin{array}{rcl} p_{x}^{'} & = & γ (v) (p_{x} - E v ∕ c^{2}) = m u_{x} γ (| u |), \\ E^{'} & = & γ (v) (E - v p_{x}) = γ (| u |) m_{0} c^{2} . & (10.2) \end{array}

To understand the context of these equations remember the deﬁnition of $γ$

γ (v) = 1 ∕ \sqrt{1 - β^{2}}, β = \frac{v}{c} .

(10.3)

In Eq. (10.2) we have also re-expressed the momentum energy in terms of a velocity $u$ . This is measured relative to the rest system of a particle, the system where the three-momentum $p = 0$ .

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{array}{rcl} E_{1}^{i n} + E_{2}^{i n} & = & E_{1}^{o u t} + E_{2}^{o u t}, \\ p_{1}^{i n} + p_{2}^{i n} & = & p_{1}^{o u t} + p_{2}^{o u t} . & (10.4) \end{array}

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