It is natural to extend~\Eq{(12.2)} by adding a fourth component, thus \[ m\, \frac{dx}{ds},\quad m\, \frac{dy}{ds},\quad m\, \frac{dz}{ds},\quad m\, \frac{dt}{ds}. \Tag{(12.5)} \] \PageSep{31} By \Eq{(12.3)} the fourth component is equal to~$M$. Thus the momenta and mass (relative mass) form together a symmetrical expression, the momenta being space-components, and the mass the time-component. We shall see later that the expression~\Eq{(12.5)} constitutes a vector, and the laws of conservation of momentum and mass assert the conservation of this vector. The following is an analytical proof of the law of variation of mass with velocity directly from the principle of conservation of mass and momentum. Let $M_{1}$,~$M_{1}'$ be the mass of a body as measured by $S$ and $S'$ respectively, $v_{1}$,~$v_{1}'$ being its velocity in the $x$-direction. Writing \[ \beta_{1} = (1 - v_{1}^{2}/c^{2})^{-\frac{1}{2}},\quad \beta_{1}' = (1 - v_{1}'^{2}/c^{2})^{-\frac{1}{2}},\quad \beta = (1 - u^{2}/c^{2})^{-\frac{1}{2}}, \] we can easily verify from~\Eq{(6.2)} that \[ \beta_{1}v_{1} = \beta\beta_{1}'(v_{1}' - u). \Tag{(12.6)} \] Let a number of such particles be moving in a straight line subject to the conservation of mass and momentum as measured by~$S'$, viz.\ \[ \sum M_{1}' \quad\text{and}\quad \sum M_{1}' v_{1}' \quad\text{are conserved.} \] Since $\beta$ and $u$ are constants it follows that \[ \sum M_{1}' \beta (v_{1}' - u)\quad\text{is conserved.} \] Therefore by \Eq{(12.6)} \[ \sum M_{1}' \beta_{1} v_{1}/\beta_{1}'\quad\text{is conserved.} \Tag{(12.71)} \] But since momentum must also be conserved for the observer~$S$ \[ \sum M_{1} v_{1}\quad\text{is conserved.} \Tag{(12.72)} \] The results \Eq{(12.71)} and \Eq{(12.72)} will agree if \[ M_{1}/\beta_{1} = M_{1}'/\beta_{1}', \] and it is easy to see that there can be no other general solution. Hence for different values of~$v_{1}$, $M_{1}$~is proportional to~$\beta_{1}$, or \[ M = m(1 - v^{2}/c^{2})^{-\frac{1}{2}}, \] where $m$~is a constant for the body.