# A Dynamical Theory of the Electromagnetic Field/Part IV

PART IV.— MECHANICAL ACTIONS IN THE FIELD.

Mechanical Force on a Moveable Conductor.

(76) We have shown (§§ 34 & 35) that the work done by the electromagnetic forces in aiding the motion of a conductor is equal to the product of the current in the conductor multiplied by the increment of the electromagnetic momentum due to the motion.

Let a short straight conductor of length $a$ move parallel to itself in the direction of $x$ , with its extremities on two parallel conductors. Then the increment of the electromagnetic momentum due to the motion of $a$ will be

$a\left({\frac {dF}{dx}}{\frac {dx}{ds}}+{\frac {dG}{dx}}{\frac {dy}{ds}}+{\frac {dH}{dx}}{\frac {dz}{ds}}\right)\delta x$ That due to the lengthening of the circuit by increasing the length of the parallel conductors will be

$-a\left({\frac {dF}{dx}}{\frac {dx}{ds}}+{\frac {dG}{dy}}{\frac {dy}{ds}}+{\frac {dH}{dz}}{\frac {dz}{ds}}\right)\delta x$ The total increment is

$a\delta x\left\{{\frac {dy}{ds}}\left({\frac {dG}{dx}}-{\frac {dF}{dy}}\right)-{\frac {dz}{ds}}\left({\frac {dF}{dz}}-{\frac {dH}{dx}}\right)\right\}$ which is by the equations of Magnetic Force (B), p. 482,

$a\delta x\left({\frac {dy}{ds}}\mu \gamma -{\frac {dz}{ds}}\mu \beta \right)$ Let X be the force acting along the direction of $x$ per unit of length of the conductor, then the work done is $Xa\delta x$ .

Let C be the current in the conductor, and let $p',q',r'$ be its components, then

$Xa\delta =Ca\delta xx\left({\frac {dy}{ds}}\mu \gamma -{\frac {dz}{ds}}\mu \beta \right),$ $\left.{\begin{array}{ccc}\mathrm {or} &&X=\mu \gamma q'-\mu \beta r'.\\\\\mathrm {Similarly,} &&Y=\mu \alpha r'-\mu \gamma p',\\\\&&Z=\mu \beta p'-\mu \alpha q'.\end{array}}\right\}$ (J)

These are the equations which determine the mechanical force acting on a conductor carrying a current. The force is perpendicular to the current and to the lines of force, and is measured by the area of the parallelogram formed by lines parallel to the current and lines of force, and proportional to their intensities.

Mechanical Force on a Magnet.

(77) In any part of the field not traversed by electric currents the distribution of magnetic intensity may be represented by the differential coefficients of a function which may be called the magnetic potential. When there are no currents in the field, this quantity has a single value for each point. When there are currents, the potential has a series of values at each point, but its differential coefficients have only one value, namely,

${\frac {d\varphi }{dx}}=\alpha ,\ {\frac {d\varphi }{dy}}=\beta ,\ {\frac {d\varphi }{dz}}=\gamma .$ Substituting these values of $\alpha ,\beta ,\gamma$ in the expression (equation 38) for the intrinsic energy of the field, and integrating by parts, it becomes

$-\sum \left\{\varphi {\frac {1}{8\pi }}\left({\frac {d\mu \alpha }{dx}}+{\frac {d\mu \beta }{dy}}+{\frac {d\mu \gamma }{dz}}\right)\right\}dV$ The expression

 $\sum \left({\frac {d\mu \alpha }{dx}}+{\frac {d\mu \beta }{dy}}+{\frac {d\mu \gamma }{dz}}\right)dV=\sum mdV$ (39)

indicates the number of lines of magnetic force which have their origin within the space V. Now a magnetic pole is known to us only as the origin or termination of lines of magnetic force, and a unit pole is one which has $4\pi$ lines belonging to it, since it produces unit of magnetic intensity at unit of distance over a sphere whose surface is $4\pi$ .

Hence if $m$ is the amount of free positive magnetism in unit of volume, the above expression may be written $4\pi m$ , and the expression for the energy of the field becomes

 $E=-\sum \left({\frac {1}{2}}\varphi m\right)dV$ (40)

If there are two magnetic poles $m_{1}$ and $m_{2}$ producing potentials $\varphi _{1}$ and $\varphi _{2}$ in the field, then if ma is moved a distance $dx$ , and is urged in that direction by a force $X$ , then the work done is $Xdx$ , and the decrease of energy in the field is

$d\left({\frac {1}{2}}\left(\varphi _{1}+\varphi _{2}\right)\left(m_{1}+m_{2}\right)\right)$ and these must be equal by the principle of Conservation of Energy.

Since the distribution $\varphi _{1}$ is determined by $m_{1}$ , and $\varphi _{2}$ by $m_{2}$ , the quantities $\varphi _{1}m_{1}$ and $\varphi _{2}m_{2}$ will remain constant.

It can be shown also, as Green has proved (Essay, p. 10), that

$m_{1}\varphi _{2}=m_{2}\varphi _{1}$ so that we get

$Xdx=d\left(m_{2}\varphi _{1}\right)$ $\left.{\begin{array}{l}X=m_{2}{\frac {d\varphi _{1}}{dx}}=m_{2}\alpha _{1},\\\\\mathrm {where} \ \alpha _{1}\ \mathrm {represents\ the\ magnetic} \ \\\mathrm {intensity\ due\ to} \ m_{1}.\ {\mathsf {Similarly}}\\\\Y=m_{2}\beta _{1},\\\\Z=m_{2}\gamma _{1}.\end{array}}\right\}$ (K)

So that a magnetic pole is urged in the direction of the lines of magnetic force with a force equal to the product of the strength of the pole and the magnetic intensity.

(78) If a single magnetic pole, that is one pole of a very long magnet, be placed in the field, the only solution of $\varphi$ is

 $\varphi _{1}=-{\frac {m_{1}}{\mu }}{\frac {1}{r}}$ (41)

where $m_{1}$ is the strength of the pole and $r$ the distance from it.

The repulsion between two poles of strength $m_{1}$ and $m_{2}$ is

 $m_{2}{\frac {d\varphi _{1}}{dr}}={\frac {m_{1}m_{2}}{\mu r^{2}}}$ (42)

In air or any medium in which $\mu =1$ this is simply ${\tfrac {m_{1}m_{2}}{r^{2}}}$ , but in other media the force acting between two given magnetic poles is inversely proportional to the coefficient of magnetic induction for the medium. This may be explained by the magnetization of the medium induced by the action of the poles.

Mechanical Force on an Electrified Body.

(79) If there is no motion or change of strength of currents or magnets in the field, the electromotive force is entirely due to variation of electric potential, and we shall have (§ 65)

$P=-{\frac {d\Psi }{dx}},\ Q=-{\frac {d\Psi }{dy}},\ R=-{\frac {d\Psi }{dz}}$ Integrating by parts the expression (I) for the energy due to electric displacement, and remembering that P, Q, R vanish at an infinite distance, it becomes

${\frac {1}{2}}\sum \left\{\Psi \left({\frac {df}{dx}}+{\frac {dg}{dy}}+{\frac {dh}{dz}}\right)\right\}dV$ or by the equation of Free Electricity (G), p. 485,

$-{\frac {1}{2}}\sum (\Psi e)dV$ By the same demonstration as was used in the case of the mechanical action on a magnet, it may be shown that the mechanical force on a small body containing a quantity $e_{2}$ of free electricity placed in a field whose potential arising from other electrified bodies is $\Psi _{1}$ , has for components
 $\left.{\begin{array}{c}X=e_{2}{\frac {d\Psi _{1}}{dx}}=-P_{1}e_{2},\\\\Y=e_{2}{\frac {d\Psi _{1}}{dy}}=-Q_{1}e_{2},\\\\X=e_{2}{\frac {d\Psi _{1}}{dz}}=-R_{1}e_{2},\end{array}}\right\}$ (D)

So that an electrified body is urged in the direction of the electromotive force with a force equal to the product of the quantity of free electricity and the electromotive force.

If the electrification of the field arises from the presence of a small electrified body containing ex of free electricity, the only solution of $\Psi _{1}$ is

 $\Psi _{1}={\frac {k}{4\pi }}{\frac {e_{1}}{r}}$ (43)

where $r$ is the distance from the electrified body.

The repulsion between two electrified bodies $e_{2}$ , $e_{2}$ is therefore

 $e_{2}{\frac {d\Psi _{1}}{dr}}={\frac {k}{4\pi }}{\frac {e_{1}e_{2}}{r^{2}}}$ (44)

Measurement of Electrical Phenomena by Electrostatic Effects.

(80) The quantities with which we have had to do have been hitherto expressed in terms of the Electromagnetic System of measurement, which is founded on the mechanical action between currents. The electrostatic system of measurement is founded on the mechanical action between electrified bodies, and is independent of, and incompatible with, the electromagnetic system; so that the units of the different kinds of quantity have different values according to the system we adopt, and to pass from the one system to the other, a reduction of all the quantities is required.

According to the electrostatic system, the repulsion between two small bodies charged with quantities $\eta _{1},\eta _{2}$ of electricity is

${\frac {\eta _{1}\eta _{2}}{r^{2}}}$ where $r$ is the distance between them.

Let the relation of the two systems be such that one electromagnetic unit of electricity contains $v$ electrostatic units; then $\eta _{1}=ve_{1}$ and $\eta _{2}=ve_{2}$ , and this repulsion becomes

 $v^{2}{\frac {e_{1}e_{2}}{r^{2}}}={\frac {k}{4\pi }}{\frac {e_{1}e_{2}}{r^{2}}}$ by equation (44), (45) (45)

whence $k$ , the coefficient of "electric elasticity" in the medium in which the experiments are made, i. e. common air, is related to $v$ , the number of electrostatic units in one electromagnetic unit, by the equation

 $k=4\pi v^{2}$ (46)
The quantity $v$ may be determined by experiment in several ways. According to the experiments of MM. Weber and Kohlrausch,

$v$ =310,740,000 metres per second.

(81) It appears from this investigation, that if we assume that the medium which constitutes the electromagnetic field is, when dielectric, capable of receiving in every part of it an electric polarization, in which the opposite sides of every element into which we may conceive the medium divided are oppositely electrified, and if we also assume that this polarization or electric displacement is proportional to the electromotive force which produces or maintains it, then we can show that electrified bodies in a dielectric medium will act on one another with forces obeying the same laws as are established by experiment.

The energy, by the expenditure of which electrical attractions and repulsions are produced, we suppose to be stored up in the dielectric medium which surrounds the electrified bodies, and not on the surface of those bodies themselves, which on our theory are merely the bounding surfaces of the air or other dielectric in which the true springs of action are to be sought.

Note on the Attraction of Gravitation.

(82) After tracing to the action of the surrounding medium both the magnetic and the electric attractions and repulsions, and finding them to depend on the inverse square of the distance, we are naturally led to inquire whether the attraction of gravitation, which follows the same law of the distance, is not also traceable to the action of a surrounding medium.

Gravitation differs from magnetism and electricity in this; that the bodies concerned are all of the same kind, instead of being of opposite signs, like magnetic poles and electrified bodies, and that the force between these bodies is an attraction and not a repulsion, as is the case between like electric and magnetic bodies.

The lines of gravitating force near two dense bodies are exactly of the same form as the lines of magnetic force near two poles of the same name; but whereas the poles are repelled, the bodies are attracted. Let $E$ be the intrinsic energy of the field surrounding two gravitating bodies $M_{1}$ , $M_{2}$ , and let $E'$ be the intrinsic energy of the field surrounding two magnetic poles $m_{1}$ , $m_{2}$ equal in numerical value to $M_{1}$ , $M_{2}$ , and let X be the gravitating force acting during the displacement $\delta x$ , and $X'$ the magnetic force,

$X\delta x=\delta E,\ X'\delta x=\delta E';$ now $X$ and $X'$ are equal in numerical value, but of opposite signs; so that

$\delta E=-\delta E'\,$ or

${\begin{array}{ll}E&=C-E\\\\&=C-\sum {\frac {1}{8\pi }}\left(\alpha ^{2}+\beta ^{2}+\gamma ^{2}\right)dV\end{array}}$ where $\alpha ,\beta ,\gamma$ are the components of magnetic intensity. If R be the resultant gravitating force, and R' the resultant magnetic force at a corresponding part of the field,

$R=-R'\ \mathrm {and} \ \alpha ^{2}+\beta ^{2}+\gamma ^{2}=R^{2}=R'^{2}$ Hence

 $E=C-\sum {\frac {1}{8\pi }}R^{2}dV$ (47)

The intrinsic energy of the field of gravitation must therefore be less wherever there is a resultant gravitating force.

As energy is essentially positive, it is impossible for any part of space to have negative intrinsic energy. Hence those parts of space in which there is no resultant force, such as the points of equilibrium in the space between the different bodies of a system, and within the substance of each body, must have an intrinsic energy per unit of volume greater than

${\frac {1}{8\pi }}R^{2}$ where R is the greatest possible value of the intensity of gravitating force in any part of the universe.

The assumption, therefore, that gravitation arises from the action of the surrounding medium in the way pointed out, leads to the conclusion that every part of this medium possesses, when undisturbed, an enormous intrinsic energy, and that the presence of dense bodies influences the medium so as to diminish this energy wherever there is a resultant attraction.

As I am unable to understand in what way a medium can possess such properties, I cannot go any further in this direction in searching for the cause of gravitation.