These are also the values of the electric and magnetic forces produced at points outside, by Lorentz's electron, or by any charged system in which, when at rest, the charge is distributed with spherical symmetry and which, when in motion, suffers the Lorentz-FitzGerald contraction. E is the force exerted by the moving charge e, upon a unit charge which is at rest at the point P. If the unit charge at P is in motion with the velocity u, then the force exerted upon it, which we may call (E), is

 $(E) = E + u \times H$ (3)

where u × H represents the vector product. Thus the force on a charge at rest at the point, P, is in the direction of r, but this is riot true in general if it is in motion.

Let us consider first the special case when the two charges have the same velocity, u=v. Let the two components of E parallel and perpendicular to v be E1 and E2, respectively. The force v × H will be parallel to E2 and in the opposite direction and its magnitude will be vH. So that the corresponding components of (E) are

 $(E)_1 =E cos \theta$ (4)

and

 $(E)_2 = E sin \theta vH$ (5)

or since

$\mathbf{H}=\frac{v}{V^{2}}\mathbf{E}\ \sin\theta$,

 $(E)_2 = sin \theta (1-\beta ^2 )$ (5)

These are the components of the actual force on the moving charge at P; if it is of opposite sign to the charge e, the force will have the direction given in fig. 2.

When θ = 0, (E)2 = 0 and

$(E)_{1}=E=V^{2}\frac{e(1-\beta^{2})}{r^{2}}$

which is (1-β²)) times the value of the electrostatic force when the charges are at rest: this corresponds to the gravitational case of p. 501 when the force was in the direction of motion.

When $\theta=\frac{\pi}{2}, (E)1 =0$, and

$(E)_{2}=E(1-\beta^{2})=V^{2}\frac{e}{r^{2}}\sqrt{1-\beta^{2}}$

which also agrees with the corresponding case for gravitation. If we apply this electromagnetic law of force to gravitation we are at first sight confronted with the difficulty that the magnitude of the force varies not only with the distance but