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it is found that k is different from unity we shall be forced to the conclusion that either our conception of mass in the classical mechanics or our conception of charge in the classical electrical theory is in need of essential modification. Again, if k = I and if we assume, as is natural, that $t\left(e_{v}\right)=l\left(e_{v}\right)$ then we have an experimental disproof of the theory of relativity. In fact we have such a disproof unless $k=1/\left(1-\beta ^{2}\right)$, provided of course that we assume $t\left(e_{v}\right)=l\left(e_{v}\right)$.

From these remarks it is obvious that, whatever may be the result of the experiments, they will certainly lead to important conclusions of a fundamental nature; that is, we have here a crucial experiment, one that cannot fail to lead somewhither. It is to be hoped that some laboratory worker will soon perform the requisite experiments; the writer, who is a mathematician, can only regret that he cannot conveniently carry out the work himself.

A variation of the experiment of Bucherer would seem to be sufficient for the purpose here. Bucherer's results were obtained by subjecting the moving electron to a magnetic field and also to an electric field each at right angles to the line of motion. A variation of the direction of these fields relative to the line of motion of the electron would probably afford a means of making the necessary measurements for the experimental proof of the relations requisite for use in the preceding discussion.

§ 7. Mass and Energy.

If, as is frequently done, we employ for the definition of the kinetic energy E the relation (compare §1)

$E=\int _{0}^{v}Mdv=\int _{0}^{v}mvdv$

it is clear that for the mass m we should take the longitudinal mass $l\left(m_{v}\right)$. Then let $m_{0}$ denote the mass of the body at rest, $E_{0}$ its energy when at rest (that is, the energy due to its internal activity), and $E_{v}$ its energy when moving at the velocity v. Then clearly $E=E_{v}-E_{0}$, so that in view of theorem II. we have

$E=E_{v}-E_{0}=\int _{0}^{v}{\frac {m_{0}vdv}{\left(1-\beta ^{2}\right)^{\frac {3}{2}}}}$

whence, on integration, we have

$E=E_{v}-E_{0}=m_{0}c^{2}\left({\frac {1}{\sqrt {1-\beta ^{2}}}}-1\right)$

(1)

Hence for the kinetic energy of a moving body we have

$E=m_{0}c^{2}\left({\frac {1}{2}}\beta ^{2}+{\frac {3}{8}}\beta ^{4}+\dots \right)$