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\markright{{\sf  Relativistic energy-momentum}}
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\begin{document}


\section{11.5\quad Relativistic Energy-Momentum.}

\subsection{11.5.1}

\quad \thinspace \textbf{a. }We will not play the game of pretending that
relativity is new to us, making some ``reasonable'' assumptions and showing
by ``simple'' arguments that Einstein's two postulates dictate that 
\[
\mathbf{p}=m\gamma \mathbf{v\quad \quad \quad \quad \quad \quad \quad \quad
\quad \quad }\gamma =\frac 1{\sqrt{1-v^2/c^2}} 
\]
\begin{equation}
E=m\gamma c^2.\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad
\quad \quad \quad \quad \quad \quad  \label{1}
\end{equation}

\smallskip\ 

\textbf{b}. Rather, we note that, if there are to be relativistically
conserved quantities corresponding to the components of momentum, they must
transform like the spatial components of a $4$-vector. Hence there must be a 
$4$-vector that is conserved in collisions and decay processes. Next we try
out the most obvious choice.

\smallskip\ 

\textbf{c.} We note that $dx_\alpha /ds$, with $ds^2=c^2dt^2-d\mathbf{x}^2$
is a $4$-vector and that $mcdx_\alpha /ds$ is too, if $m$ is an invariant
(scalar). For small $\mathbf{v,}$ the spatial components of $mcdx_\alpha /ds$
give the classical momentum, if $m$ is identified with the mass. Hence it is
logical to suppose that $mcdx_\alpha /ds$ is conserved.

\smallskip\ 

\textbf{d. }The zeroth component, $mcdx_0/ds=mc\gamma $ (note $%
ds^2=c^2dt^2/\gamma ^2)$ will now be identified with $E/c,$ because for
small velocities $mc^2\gamma =mc^2/\sqrt{1-v^2/c^2}\simeq mc^2+(1/2)mv^2.$

\smallskip

\textbf{e.} It is important to note that the fourth conserved relativistic
quantity does not simply reduce to the kinetic energy $(1/2)mv^2,$ but
contains in addition the ``rest energy'' $mc^2.$ \textsl{Mass and energy are
not separately conserved.} In summary 
\begin{eqnarray}
p_\alpha &=&mc\frac{dx_\alpha }{ds}=m\gamma v_\alpha =\frac{mv_\alpha }{%
\sqrt{1-v^2/c^2}}\quad \quad \quad \alpha =1,2,3  \label{2} \\
p_0 &=&\quad \quad \,\,\frac Ec=m\gamma c=\frac{mc}{\sqrt{1-v^2/c^2}}\,. 
\nonumber
\end{eqnarray}

\newpage

\subsection{11.5.2}

\quad \quad \textbf{a. }The ``length'' of the vector $p_{\alpha }$ is simply 
$mc$ (from $\mathsf{p}\cdot \mathsf{p}=m^{2}c^{2}\mathsf{dx}\cdot \mathsf{dx}%
/ds^{2})$. This is consistent with the assumption that $m$ is invariant.
Only in the rest frame (where $d\mathbf{x/}dt=0)$ does $m$ coincide with the
``mass'' defined as the ratio of force to acceleration (we shall see this
later). For this reason, $m$ is sometimes called the ``rest mass'', but we
shall simply call it ``mass''.

\smallskip\ 

\textbf{b.} From 
\[
\mathsf{p}\cdot \mathsf{p}=p_{0}^{2}-\mathbf{p}^{2}=(E/c)^{2}-\mathbf{p}^{2}
\]
we obtain 
\begin{equation}
E=\sqrt{c^{2}\mathbf{p}^{2}+m^{2}c^{4}.}  \label{3}
\end{equation}

\smallskip\ 

\textbf{c. }The relations (\ref{1})(\ref{2})(\ref{3}) are verified daily (at
great expense)\ in the operation of particle accelerators.

\smallskip\ 

\textbf{d.} The derivation of the relation (\ref{3}) between ``space'' and
``time'' components has made use of the invariant product $\mathsf{p}\cdot 
\mathsf{p}.$ We could also have derived (\ref{3}) by eliminating $v^{2}$
from $\mathbf{p}^{2}=m^{2}\gamma ^{2}v^{2},$ $\gamma
^{2}=(1-v^{2}/c^{2})^{-1},$ $E^{2}=m^{2}\gamma ^{2}c^{4},$ but the algebra
would be more involved (especially if we don't know what relation we are
looking for). Generally, the use of $4$-vectors and invariant ($4$%
-dimensional) dot products is very advantageous in deriving relations
between conserved quantities and in relating collision parameters in
different frames.

\smallskip\ 

\textbf{e. }The typical examples involve transformations between the center
of momentum system (usually called the center of mass system, because that's
what it is in the newtonian limit) and the laboratory system. The treatment
given in the rest of this section follows sections 12.2, 12.3 and 12.4 of
Jackson's first edition. In the new edition this material is relegated to
problems 11.16 through 11.23.

\smallskip \ 

\textbf{f. }To save writing, we often set $c=1,$ so that the symbols $p,E,m,v
$ stand for $cp,E,mc^{2},v/c.$

\newpage 

\subsection{11.5.3}

\smallskip\ 

\quad \textbf{a.} As a first example, consider a simple decay of the type $%
\pi \rightarrow \mu +\nu $ or $K^{\pm }\rightarrow \pi ^{\pm }+\pi ^{0}.$ In
general: $M\rightarrow m_{1}+m_{2}.$

\smallskip\ 

\textbf{b. }The conservation laws are $\mathsf{P}=\mathsf{p}_{1}+\mathsf{p}%
_{2}.$ By definition of mass, $\mathsf{P}\cdot \mathsf{P}=M^{2},$ $\mathsf{p}%
_{1}\cdot \mathsf{p}_{1}=m_{1}^{2},$ $\mathsf{p}_{2}\cdot \mathsf{p}%
_{2}=m_{2}^{2}.$

\smallskip

\textbf{c. }In the frame where $M$ is at rest, $\mathbf{p}_{2}=-$ $\mathbf{p}%
_{1}$ and $E_{1}+E_{2}=M.$ Clearly then the decay is completely determined
by kinematics alone. Since $\left| \mathbf{p}_{2}\right| =$ $\left| \mathbf{p%
}_{1}\right| ,$ we solve $\sqrt{\left| \mathbf{p}_{1}\right| ^{2}+m_{1}^{2}}+%
\sqrt{\left| \mathbf{p}_{1}\right| ^{2}+m_{2}^{2}}=M$ and proceed by
squaring (twice) to eliminating the square roots. A more elegant way is to
use 4-vector dot products as follows.

\smallskip

\textbf{d. }To find $E_{1},$ write $\mathsf{p}_{2}=\mathsf{P}-\mathsf{p}_{1}$
and square to obtain $m_{2}^{2}=M^{2}+m_{1}^{2}-2\mathsf{P}\cdot \mathsf{p}%
_{1};$ then use $\mathsf{P}\cdot \mathsf{p}_{1}=ME_{1},$ since $\mathsf{P}%
=(M,\mathbf{0).}$ Hence 
\begin{equation}
E_{1}=\frac{M^{2}+m_{1}^{2}-m_{2}^{2}}{2M}.  \label{4}
\end{equation}
Then $\left| \mathbf{p}_{1}\right| ^{2}=E_{1}^{2}-m_{1}^{2}$ and the kinetic
energy is 
\[
T_{1}=E_{1}-m_{1}=\frac{\left( M-m_{1}\right) ^{2}-m_{2}^{2}}{2M}
\]
and similarly for $E_{2}$ and $T_{2}.$

\smallskip

\textbf{e.} The formula for $T_{1}$ can be rewritten in many ways,
introducing the \textsl{mass defect} $\Delta =M-m_{1}-m_{2}.$ If we
substitute $m_{2}=\left( M-m_{1}\right) -\Delta $ we find 
\begin{equation}
T_{1}=\Delta \left( 1-\frac{m_{1}}{M}-\frac{\Delta }{2M}\right)   \label{4'}
\end{equation}
and similarly for $T_{2}.$

\smallskip

\textbf{f.} If the mass $M$ decays into any number of masses $m_{i},$ then a
formula of type (\ref{4}) gives the \textsl{maximum} kinetic energy $T_{i}$
for the $i$-th particle (Problem 11.18).

\newpage

\subsection{11.5.4}

\quad \quad \textbf{a.} As a second example, consider a two-particle
collision $1+2\rightarrow 3+4.$ In the center of momentum (CM) frame, $%
\mathbf{p}_{1}^{\prime }+\mathbf{p}_{2}^{\prime }=0$ and $\mathbf{p}%
_{3}^{\prime }+\mathbf{p}_{4}^{\prime }=0.$ The first condition defines the
CM frame $K^{\prime },$ the second follows from momentum conservation.
(space part of \textsf{p}$_{1}^{\prime }+$\textsf{p}$_{2}^{\prime }=$\textsf{%
p}$_{3}^{\prime }+$\textsf{p}$_{4}^{\prime }).$

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\smallskip\ 

\textbf{b. }The relation between $E_{1}^{\prime }$ and $E_{2}^{\prime }$ is
easily obtained by comparing the two expressions 
\begin{eqnarray*}
\left( \mathsf{p}_{1}^{\prime }+\mathsf{p}_{2}^{\prime }\right) \cdot \left( 
\mathsf{p}_{1}^{\prime }-\mathsf{p}_{2}^{\prime }\right) &=&\left(
E_{1}^{\prime }+E_{2}^{\prime }\right) \left( E_{1}^{\prime }-E_{2}^{\prime
}\right) \\
\left( \mathsf{p}_{1}^{\prime }+\mathsf{p}_{2}^{\prime }\right) \cdot \left( 
\mathsf{p}_{1}^{\prime }-\mathsf{p}_{2}^{\prime }\right) &=&\mathsf{p}%
_{1}^{\prime }\cdot \mathsf{p}_{1}^{\prime }-\mathsf{p}_{2}^{\prime }\cdot 
\mathsf{p}_{2}^{\prime }=m_{1}^{2}-m_{2}^{2}.
\end{eqnarray*}
Letting $E^{\prime }=E_{1}^{\prime }+E_{2}^{\prime }$ be the total energy
(also equal to $E_{3}^{\prime }+E_{4}^{\prime })$, we find $E_{1}^{\prime
}-E_{2}^{\prime }=\left( m_{1}^{2}-m_{2}^{2}\right) /E^{\prime },$ hence 
\[
E_{1}^{\prime }=\frac{E^{\prime }+m_{1}^{2}-m_{2}^{2}}{2E^{\prime }}\quad
\quad \quad E_{2}^{\prime }=\frac{E^{\prime }+m_{2}^{2}-m_{1}^{2}}{%
2E^{\prime }} 
\]
and similarly 
\begin{equation}
E_{3}^{\prime }=\frac{E^{\prime }+m_{3}^{2}-m_{4}^{2}}{2E^{\prime }}\quad
\quad \quad E_{4}^{\prime }=\frac{E^{\prime }+m_{4}^{2}-m_{3}^{2}}{%
2E^{\prime }}  \label{5}
\end{equation}

\smallskip \ 

\textbf{c.} These formulae show how the total energy $E^{\prime }$ is
apportioned among the particles. The close resemblance to (\ref{4}) of 3e
should be noted. The values of $\mathbf{p}_{i}^{\prime \,2}$ $(i=1,2,3,4)$
are found from $E_{i}^{\prime \,2}=m_{i}^{2}+\mathbf{p}_{i}^{\prime \,2}.$

\smallskip \ 

\textbf{d.} $E^{\prime }$ must be larger than $m_{1}+$ $m_{2}$ and larger
than $m_{3}+$ $m_{4}.$ It is easy to see that, with these restrictions, $%
E_{i}^{\prime }$ is an increasing function of $E^{\prime }$ in (\ref{5}). If 
$m_{3}+$ $m_{4}>m_{1}+$ $m_{2},$ a total kinetic energy greater than the
threshold $T^{\prime }=m_{3}+$ $m_{4}-\left( m_{1}+m_{2}\right) $ must be
available for the reaction to proceed.

\newpage

\subsection{11.5.5}

\quad \quad \textbf{a. }The other important aspect of two body kinematics is
the explicit relation between the CM frame and the laboratory frame $K$
where $2$ is initially at rest. The energy of particle $1,$ $E_1,$ is the
usual parameter that is directly accesible to experimental determination.

\smallskip\ 

\textbf{b.}The relation between $\ E_{1}$ and $E^{\prime }$ is found by
evaluating the scalar product $\left( \mathsf{p}_{1}+\mathsf{p}_{2}\right)
\cdot \left( \mathsf{p}_{1}+\mathsf{p}_{2}\right) =\left( \mathsf{p}%
_{1}^{\prime }+\mathsf{p}_{2}^{\prime }\right) \cdot \left( \mathsf{p}%
_{1}^{\prime }+\mathsf{p}_{2}^{\prime }\right) .$ With $\mathbf{p}_{2}=0$
and $\mathbf{p}_{1}^{\prime }+\mathbf{p}_{2}^{\prime }=0$ this gives $\left(
E_{1}+m_{2}\right) ^{2}-\mathbf{p}_{1}^{2}=E^{\prime \,2},$ and using $%
E_{1}^{2}=\mathbf{p}_{1}^{2}+m_{1}^{2},$ 
\begin{equation}
E^{\prime \,2}=m_{1}^{2}+m_{2}^{2}+2\ E_{1}m_{2}.  \label{6}
\end{equation}

\smallskip\ 

\textbf{c.} The value of $\mathbf{p}_{1}^{\prime }$ can be found by squaring
both sides of $2E^{\prime }E_{1}^{\prime }=E^{\prime
\,2}+m_{1}^{2}-m_{2}^{2} $ (see (\ref{5})) to obtain $4E^{\prime \,2}\mathbf{%
p}_{1}^{\prime 2}=(E^{\prime \,2}+m_{1}^{2}-m_{2}^{2})^{2}-4E^{\prime
\,2}m_{1}^{2}.$ Inserting (\ref{6}) on the r.h.s. we find simply 
\begin{equation}
\mathbf{p}_{1}^{\prime }=\frac{m_{2}\mathbf{p}_{1}}{E^{\prime }}  \label{7}
\end{equation}

\smallskip\ 

\textbf{d. }The other important parameter is the CM velocity $\mathbf{v}%
_{CM}.$ One could \textsl{define}, quite generally, 
\[
\mathbf{v}_{CM}=\frac{\mathbf{p}_{1}+\mathbf{p}_{2}}{E_{1}+E_{2}}, 
\]
which in our case reduces to 
\begin{equation}
\mathbf{v}_{CM}=\frac{\mathbf{p}_{1}}{E_{1}+m_{2}}.  \label{com}
\end{equation}
A Lorentz transformation to the frame $K^{\prime }$, moving with velocity $%
\mathbf{v}_{CM},$ will then show that $\mathbf{p}_{1}^{\prime }+\mathbf{p}%
_{2}^{\prime }=0.$

One can also follow the inverse route: define the CM system by $\mathbf{p}%
_{1}^{\prime }+\mathbf{p}_{2}^{\prime }=0,$compute $\mathbf{p}_{1}^{\prime }$
and $\mathbf{p}_{2}^{\prime }$ by invariant products (as we just did), and
then obtain (\ref{com}) as follows. By a Lorentz transformation (with $%
\mathbf{p}_{2}=0$), we obtain $\mathbf{p}_{2}^{\prime }=-\gamma _{CM}m_{2}%
\mathbf{v}_{CM}$ and $E_{2}^{\prime }=\gamma _{CM}m_{2},$ hence $\mathbf{p}%
_{2}^{\prime }=-\mathbf{v}_{CM}E_{2}^{\prime }$ and $\mathbf{p}_{1}^{\prime
}=\mathbf{v}_{CM}E_{2}^{\prime }=m_{2}\mathbf{p}_{1}/$ $E^{\prime }$ (using (%
\ref{7})). To get $\mathbf{v}_{CM}=\mathbf{p}_{1}/\left( E_{1}+m_{2}\right) $
we need to show that $E_{2}^{\prime }E^{\prime }=m_{2}\left(
E_{1}+m_{2}\right) .$ This follows easily from \textsf{p}$_{2}^{\prime
}\cdot \left( \mathsf{p}_{1}^{\prime }+\mathsf{p}_{2}^{\prime }\right) =%
\mathsf{p}_{2}\cdot \left( \mathsf{p}_{1}+\mathsf{p}_{2}\right) .$

\smallskip\ 

\textbf{e.} There is, clearly, an almost endless variety of formulas
involving $p,E,v,\gamma $ and the scattering angles. See Jackson, 1-st
edition, 12.2 to 12.4.

\end{document}
