( 564 ) 
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 
rf ED (by Be =de Dil, vann Pe 
22} 
or, by Poyntine’s theorem 
= S, if Sx be the flow of energy 
in a direction parallel to the axis of «. 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 (d,), is 0, and from this it follows 
that the total value of (7) will vanish, because the part arising from — 
the combination of (dj) and (d,), as well as that which is solely due 
to the 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 pre- 
ceding §. 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 
1 9 ; r 
En cos n ( he aa e), 
where @ 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 7, higher than the 
first — and this is necessary, in order that (7) may remain free from 
1 
powers higher than the second — — and & have to be treated as 
ge 
constants. Moreover, the factors are such, that the vibrations are 
perpendicular to the line QA. If, now, A coincides with P, and 
QA with the axis of z, in the expression for d‚ we shall have 
9 = 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 > and Hltheir real values, modified as they 
are by the motion of Q, we may again write for the force 
ne eb 
2 x 5 
this time, however, we have to understand by Sx the flow of 
energy as it is jn the actual case. 
