Page:Thomson1881.djvu/10

 wave-length was the length of the tube, and so, by Stokes's law, could not produce a luminous phosphorescence.

It may be useful to form a rough estimate of the electromotive force which could be produced by a moving particle.

By equation (1) we see, if the particle be moving parallel to the axis of x with velocity p, that the greatest value of F at a point distant R from the centre of the particle is

$$\mu p\left(\frac{1}{R}-\frac{a^{2}}{5R^{3}}\right)$$.

Now the greatest value of e, as before, is $$K\times3\times10^{12}\times a^{2}$$,

$$\mu K=\frac{1}{9\times10^{20}}$$;

hence the greatest value of F at the surface of the particle

$$=\frac{3\times4\times10^{12}pa}{5\times9\times10^{20}}$$.

Now during the collision let us represent p by p0 cos kr, where $$\frac{2\pi}{k}$$ is less than the period of vibration of green light; R must be therefore at least $$3\times10^{15}$$; for a particle of air a is of the order 10-7. Substituting, we get

$$\frac{dF}{dt}=-\frac{4\times10^{5}}{15\times10^{20}}Rp_{0}\sin Rt$$;

or the maximum value of $$\frac{dF}{dt}$$ is

$$\frac{4\times3\times10^{20}}{15\times10^{20}}p_{0}=\frac{4}{5}p_{0}$$

Now at present we know nothing about p0; but it must be very much greater than the mean velocity of the air-molecules, which is about $$5\times 10^{4}$$; if we substitute this value for it, we get the maximum value of $$\tfrac{dF}{dt}$$ or the maximum electromotive force to be about $$4\times 10^{4}$$, or about $$\tfrac{1}{2500}$$ of a volt per centimetre. Now, for sunlight the maximum electromotive force is about 6 volts per centimetre (Maxwell's 'Electricity and Magnetism,' § 793); and when we consider the immense number of particles which must be striking the glass at each instant, we have no difficulty in conceiving that the magnitude of the electromotive force due to the moving particle may be sufficient to cause phosphorescence. To show the rapidity with