Transformation of Energy and Momentum

Consider two inertial reference frames, $S$ and $S'$. Let $S'$ move with velocity ${\bf v} = v\,{\bf e}_x$, and be in a standard configuration, with respect to $S$. Let ${\bf p}$ and $E$ be some particle's momentum and energy, respectively, in $S$. Likewise, let ${\bf p}'$ and $E'$ be the particle's momentum and energy, respectively, in $S'$. We have seen that the transformation of spacetime coordinates, (3.110)–(3.113), implies that the spacetime interval, $(c\,{\mit\Delta} t)^2-\vert{\mit\Delta}{\bf r}\vert^2$, takes the same value in all inertial frames of reference. Given that $(E/c)^2-\vert{\bf p}\vert^2$ also takes the same value in all inertial frames of reference, it seems reasonable to assume, by analogy, that the components of ${\bf p}$ and $E$ in our two inertial reference frames are related as follows:

$$p_x' =\gamma\left(p' -\frac{v\,E}{c^2}\right),$$ (3.180)

$$p_y' = p_y,$$ (3.181)

$$p_z' = p_z,$$ (3.182)

$$E' = \gamma\,(E- v\,p_x).$$ (3.183)

We can easily test out the previous transformation rule. Suppose that the particle is at rest in $S$. It follows that $E=m_0\,c^2$ and $p_x=p_y=p_z=0$. Hence, Equations (3.180)–(3.183) yield

$$p_x' = -\gamma\,m_0\,v = - m\,v,$$ (3.184)

$$p_y' =0,$$ (3.185)

$$p_z' =0,$$ (3.186)

$$E' = \gamma\,m_0\,c^2 = m\,c^2$$ (3.187)

in $S'$. In other words, in the frame $S'$, in which the particle moves with velocity $-{\bf v}$, we have ${\bf p}'= -m\,{\bf v}$ and $E'=m\,c^2$. Of course, these are the correct results. (See Sections 3.3.2 and 3.3.4.)