In the case of two resonators the integration in (23) presents no difficulty; but when there are a larger number, it is preferable to calculate the emission of energy in the dispersed waves from the work which would have to be done to generate them at the resonators (in the absence of primary waves)—a method which entails no integration. We continue to suppose that all the resonators are similarly situated, so that it suffices to consider the work done at one of them—say the first. From (15)
1 — ikr . ^ e~ikR\ d-^r a R } ' dr ra'
The pressure is proportional to i-fy, and the part of it which is in the same phase as d^rjdr is proportional to
^ sin IcR
Accordingly the work done at each source is proportional to
Hence altogether by (19) the energy dispersed by n resonators is that carried by an area $ of primary wave-front, where
^ sin IcR
1 T" ^ TTi
TT I",* cos kR \' [1 - sin IcR]*' '"TcR ' + ' l + *
the constant factor being determined most simply by a comparison with the case of a single resonator, for which n = 1 and the S's vanish. We fall back on (24) by merely putting n = 2, and dropping the signs of summation, as there is then only one R.
If the tuning is such as to make the effect of the aggregate of resonators a maximum, the cosines in (29) are to be dropped, and we have
a- "v/7r,' ............................ (so)
As an example of (29), we may take 4 resonators at the angular points of a square whose side is b. There are then 3 jft's to be included in the summation, of which two are equal to b and one to b \f2, so that (28) becomes
kb kb A/2 j
(31)mined by (18), or (21).