Page:Lorentz Grav1900.djvu/6

 If Q were removed, these forces together would be 0, as has already been remarked. On the other hand, the force (8) taken by itself, would then likewise be 0. Indeed, its value is

or, by theorem $$\tfrac{n^{2}e^{2}b}{V^{2}}S_{\mathsf{x}}$$, if $$S_{\mathsf{x}}$$ be the flow of energy in a direction parallel to the axis of x. Now, it is clear that, in the absence of Q, any plane must be traversed in the two directions by equal amounts of energy.

In this way we come to the conclusion that the force (7), in so far as it depends on the part, ($$\mathfrak{d}_{1}$$), is 0, and from this it follows that the total value of (7) will vanish, because the part arising from the combination of ($$\mathfrak{d}_{1}$$) and ($$\mathfrak{d}_{2}$$), as well as that which is solely due to tho vibrations of Q, are 0. As to the first part, this may be shown by a reasoning similar to that used at the end of, the preceding §. For the second part, the proof is as follows.

The vibrations excited by Q in any point A of the surrounding aether are represented by expressions of the form

$\frac{1}{r}\vartheta\cos n\left(t-\frac{r}{V}+\varepsilon\right)$,

where $$\vartheta$$ depends on the direction of the line QA, and r denotes the length of this line. If, in differentiating such expressions, we wish to avoid in the denominator powers of r, higher than the first — and this is necessary, in order that (7) may remain free from powers higher than the second — $$\tfrac{1}{r}$$ and $$\vartheta$$ have to be treated as constants. Moreover, the factors $$\vartheta$$ are such, that the vibrations are perpendicular to the line QA. If, now, A coincides with P, and QA with the axis of x, in the expression for bx we shall have $$\vartheta=0$$, and since this factor is not to be differentiated, all terms in (7) will vanish.

Thus, the question reduces itself to (8) or (9). If, in this last expression, we take for $$\mathfrak{d}$$ and $$\mathfrak{H}$$ their real values, modified as they are by the motion of Q, we may again write for the force

$\frac{n^{2}e^{2}b}{V^{2}}S_{\mathsf{x}}$;|undefined

this time, however, we have to understand by $$S_{\mathsf{x}}$$ the flow of energy as it is in the actual case.