The Actions Exerted By A Pair Of Oppositely Charged Ions

Let us next consider the actions exerted by a pair of oppositely charged ions, placed close to each other, and remaining so during their motion.

For convenience of mathematical treatment, we may even reason as if the two charges penetrated each other, so that, if they are equal, ϱ ′ = − ϱ {\displaystyle \varrho ‘=-\varrho } {\displaystyle \varrho ‘=-\varrho }.

On the other hand v ′ = v {\displaystyle {\mathfrak {v}}’={\mathfrak {v}}} {\displaystyle {\mathfrak {v}}’={\mathfrak {v}}} ; hence, by (I) and (II),

d ′ = − d {\displaystyle {\mathfrak {d}}’=-{\mathfrak {d}}} {\displaystyle {\mathfrak {d}}’=-{\mathfrak {d}}} and H ′ = − H {\displaystyle {\mathfrak {H}}’=-{\mathfrak {H}}} {\displaystyle {\mathfrak {H}}’=-{\mathfrak {H}}}.

Now let us put in the field, produced by the pair of ious, a similar pair with charges e and e ′ = − e {\displaystyle e’=-e} {\displaystyle e’=-e}, and moving with the common velocity v {\displaystyle {\mathfrak {v}}} {\displaystyle {\mathfrak {v}}}. Then, by (10) — (13),

k 2 = − β α k 1 , k 3 = − β α k 1 , k 4 = k 1 . {\displaystyle k_{2}=-{\frac {\beta }{\alpha }}k_{1},\ k_{3}=-{\frac {\beta }{\alpha }}k_{1},\ k_{4}=k_{1}.} {\displaystyle k_{2}=-{\frac {\beta }{\alpha }}k_{1},\ k_{3}=-{\frac {\beta }{\alpha }}k_{1},\ k_{4}=k_{1}.}

The total force on the positive particle will be

k 1 + k 2 = k 1 ( 1 − β α ) {\displaystyle k_{1}+k_{2}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle k_{1}+k_{2}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}

and that on the negative ion

k 3 + k 4 = k 1 ( 1 − β α ) {\displaystyle k_{3}+k_{4}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle k_{3}+k_{4}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}.

These forces being equal and having the same direction, there is no force tending to separate the two ions, as would be the case in an electric field. Nevertheless, the pair is acted on by a resultant force

2 k 1 ( 1 − β α ) {\displaystyle 2k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle 2k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}.

If now β be somewhat larger than α {\displaystyle \alpha } \alpha, the factor 2 ( 1 − β α ) {\displaystyle 2\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle 2\left(1-{\frac {\beta }{\alpha }}\right)} have a certain negative value –ε, and our result may be expressed as follows:

If we wish to determine the action between two ponderable bodies, we may first consider the forces existing between the positive ions in the one and the positive ions in the other. We then have to reverse the direction of these forces, and to multiply them by the factor ε. Of course, we are led in this way to Newton’s law of gravitation.

The assumption that all ponderable matter is composed of positive and negative ions is no essential part of the above theory. We might have confined ourselves to the supposition that the state of the aether which is the cause of gravitation is propagated in a similar way as that which exists in the electromagnetic field.

Instead of introducing two pairs of vectors ( d , H {\displaystyle {\mathfrak {d,\ H}}} {\displaystyle {\mathfrak {d,\ H}}}) and ( d ′ , H ′ {\displaystyle {\mathfrak {d’,H’}}} {\displaystyle {\mathfrak {d’,H’}}}), both of which come into play in the electromagnetic actions, as well as in the phenomenon of gravitation, we might have assumed one pair for the electromagnetic field and one for universal attraction.

For these latter vectors, say d , H {\displaystyle {\mathfrak {d,\ H}}} {\displaystyle {\mathfrak {d,\ H}}}, we should then have established the equations (I), ϱ {\displaystyle \varrho } \varrho being the density of ponderable matter, and for the force acting on unit mass, we should have put

− η { 4 π V 2 d + [ v . H ] } {\displaystyle -\eta \left{4\pi V^{2}{\mathfrak {d}}+{\mathfrak {\left[v.\ H\right]}}\right}} {\displaystyle -\eta \left{4\pi V^{2}{\mathfrak {d}}+{\mathfrak {\left[v.\ H\right]}}\right}},

where η {\displaystyle \eta } \eta is a certain positive coefficient.