Page:Motion of Electrification through a Dielectric.djvu/4

poles. The vector-potentials of these currents are also radial, and their tensors are $\scriptstyle {{\frac {1}{2}}qu}$ and $\scriptstyle {-{\frac {1}{2}}qu}$. We have now merely to find their resultant when the linear element is indefinitely shortened, add on to the former qu/r, and multiply by μ₀, to obtain the complete circuital vector-potential of qu, viz.:—

$\scriptstyle {\mathbf {A} =\mu _{0}q\left({\frac {u}{r}}-{\frac {1}{2}}u\nabla {\frac {dr}{ds}}\right),\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(5)}}}$

where r is the distance from q to the point P when A is reckoned, and the differentiation is to s, the axis of the convection-current. Both it and the space-variation are taken at P. The tensor of u is u. Though different and simpler in form (apart from the use of vectors) this vector-potential is, I believe, really the same as the one used by J. J. Thomson. From it we at once find, by the method described in § 4, the mutual energy of a pair of point-charges q₁ and q₂, moving at velocities u₁ and u₂, to be

$\scriptstyle {M=\mu _{0}q_{1}q_{2}\left({\frac {\mathbf {u} _{1}\mathbf {u} _{2}}{r}}-{\frac {1}{2}}u_{1}u_{2}{\frac {d^{2}r}{ds_{1}ds_{2}}}\right),\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(6)}}}$

when at distance r apart. Both axial differentiations are to be effected at one end of the line r.

As an alternative form, let ε be the angle between u₁ and u₂, and let the differentiation to s₁ be at ds₁ that to s₂ at ds₂, as in the German investigations relating to current-elements; then^[1]

$\scriptstyle {M=\mu _{0}q_{1}q_{2}u_{1}u_{2}\left({\frac {\cos \epsilon }{r}}+{\frac {1}{2}}{\frac {d^{2}r}{ds_{1}ds_{2}}}\right).\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(7)}}}$

Another form, to render its meaning plainer. Let λ₁, μ₁, v₁ and λ₂, μ₂, v₂ be the direction-cosines of the elements referred to rectangular axes, with the x-axis, to which λ₁ and λ₂ refer, chosen as the line joining the elements. Then^[2]

$\scriptstyle {M={\frac {\mu _{0}q_{1}q_{2}u_{1}u_{2}}{2r}}\left(2\lambda _{1}\lambda _{2}+\mu _{1}\mu _{2}+v_{1}v_{2}\right).\quad \cdots \cdots \cdots \cdots \cdots \cdots {\text{(8)}}}$

J. J. Thomson's estimate is^[3]

$\scriptstyle {M={\frac {1}{3}}\mu _{0}q_{1}q_{2}u_{1}u_{2}{\frac {\cos \epsilon }{r}}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(9)}}}$

Comparing this with (8) we see that there is a notable difference.

7. The mutual energy being different, the forces on the charges, as derived by J. J. Thomson by the use of Lagrange's equations, will be different. When the speeds are constant, we shall have simply the before-described vector product (4) for the "electromagnetic force;" or

$\scriptstyle {\mathbf {F} _{1}=\mu _{0}q_{1}\mathbf {V} u_{1}\mathbf {H} _{2},\quad \quad \mathbf {F} _{2}=\mu _{0}q_{2}\mathbf {V} u_{2}\mathbf {H} _{1},\quad \cdots \cdots \cdots \cdots \cdots {\text{(10)}}}$

if F₁ is the electromagnetic force on the first and F₂ that on the second element, whilst H₁ and H₂ are the magnetic forces. Similar changes are needed in the other parts of the complete mechanical forces.

1. The Electrician, Dec. 28, 1888, p. 230. [p. 501, vol. II.].
2. The Electrician, Jan 24, 1885, p. 221. [vol.I., p. 446].
3. "Applications of Dynamics to Physics and Chemistry," chap. iv.; and Phil. Mag., April 1881.