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# 15.27: Energy and Momentum

A moving particle has energy arising from its momentum and also from its rest mass, and we need to find an expression relating energy to rest mass and momentum. It is fairly easy and it goes like this:

$$E^{2}=m^{2}c^{4}=c^{2}(m^{2}c^{2}-m^{2}u^{2}+m^{2}u^{2})=c^{2}[m^{2}(c^{2}-u^{2})+p^{2}]$$

$$=c^{2}\left(\frac{m_{0}^{2}(c^{2}-u^{2})}{1-\frac{u^{2}}{c^{2}}}+p^{2}\right)=c^{2}(m_{0}^{2}c^{2}+p^{2})$$

Thus we obtain for the energy in terms of rest mass and momentum

$E^{2}=(m_{0}c^{2})^{2}+(pc)^{2}. \label{15.27.1}$

If the speed (and hence momentum) is zero, the energy is merely $$m_{0}c^{2}$$. If the rest mass is zero (as, for example, a photon) and the energy is not zero, then $$E\ =\ pc\ =\ muc$$. But also $$E=mc^{2}$$, so that, if the rest mass of a particle is zero and the energy is not, the particle must be moving at the speed of light. This could be regarded as the reason why photons, which have zero rest mass, travel at the speed of photons. If neutrinos have zero rest mass, they, too, will travel at the speed of light; if they are not massless, they won’t.

In addition to Equation $$\ref{15.27.1}$$, which relates the energy to the magnitude of the momentum, it will be of interest to see how the components of momentum transform between reference frames. As usual, we are considering frame $$\Sigma'$$ to be moving with respect to $$\Sigma$$ at a speed $$v$$ with respect to $$\Sigma$$. There is no difficulty with the $$y$$- and $$z$$- components. We have merely $$p'_{y'}=p_{y}$$ and $$p'_{z'}=p_{z}$$. However:

$$p_{x}=mu_{x}=\frac{m_{0}u_{x}}{\left(1-\frac{u_{x}^{2}}{c^{2}}\right)^{\frac{1}{2}}}$$ and $$p'_{x'}=m'u'_{x'}=\frac{m_{0}u'_{x'}}{\left(1-\frac{u_{x'}^{2}}{c^{2}}\right)^{\frac{1}{2}}}$$.

Also $$u'_{x'}=\frac{u_{x}-v}{\left(1-\frac{u_{x}^{2}}{c^{2}}\right)}$$, from which $$\left(1-\frac{u'^{2}_{x'}}{c^{2}}\right)^{\frac{1}{2}}=\frac{\left(1-\frac{u_{x}^{2}}{c^{2}}\right)^{\frac{1}{2}}\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}}{1-\frac{u_{x}v}{c^{2}}}$$.

After a little algebra, we obtain

$$p_{x}=\frac{m_{0}(u_{x}-v)}{\left(1-\frac{u^{2}}{c^{2}}\right)^{\frac{1}{2}}\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}}.$$

And this is

$p'_{x'}=\frac{p_{x}-\frac{vE}{c^{2}}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}}=\gamma\left(p_{x}-\frac{vE}{c^{2}}\right) \label{15.27.2}$

The inverse is found in the usual way:

$p_{x}=\gamma\left(p'_{x'}+\frac{vE'}{c^{2}}\right) \label{15.27.3}$

If we eliminate $$p'_{x'}$$ from Equations $$\ref{15.27.2}$$ and $$\ref{15.27.3}$$, we’ll find $$E'$$ in terms of $$E$$ and $$p_{x}$$:

$E'=\gamma(E-vp_{x}). \label{15.27.4}$

Thus the transformations between energy and the three spatial components of momentum is similar to the transformation between time and the three space coordinates, and are described by a similar 4-vector:

$\begin{pmatrix} p'_{x'} \\ p'_{y'} \\ p'_{z'} \\ \frac{iE'}{c}\end{pmatrix}\ =\ \begin{pmatrix} \gamma&0&0&i\beta\gamma \\\ 0&1&0&0\\0&0&1&0\\-i\beta\gamma&0&0&\gamma \end{pmatrix}\begin{pmatrix} p_{x} \\ p_{y} \\ p_{z} \\ \frac{iE}{c}\end{pmatrix}. \label{15.27.5}$

The reader should multiply this out to verify that it does reproduce Equations $$\ref{15.27.3}$$ and $$\ref{15.27.4}$$.