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In the double integral containing the cosine, let us take first the integration with respect to 02j for which the limits are 61 + 8 and 2vr.    We have
- 61} ddz =  sin #! - sin 8 ;
and since the limits for Ol are 0 and 2?r  8, we get as the expectation of intensity
If 8" be neglected, this reduces to
If 8 = TT, we have 2 (1  4/7r2) ; and if 8  ZTT, we have 4, the only available situations being 0l = 0, #2 = 2?r, equivalent to phase identity.
This treatment might perhaps be extended to a greater value, or even to the general (integral) value, of n ; but I content myself with the simplifying supposition that 8 is very small.
In (40) the integration with respect to On supposes 6l , 0Z    0n-i already fixed. If 8  0, every term such as
?!... den
=     cos (0r - ea) d0ader -4-
d9a {sin (27r - ea) + sin 0f] 4- 4?r2 = 0,
and the expectation is n simply, as we have already seen.    In the next approximation the correction to n will be .of order 8, and we neglect S2.
In evaluating (40) there are \n (n -i 1) terms under the sign of summation, but these are all equal, since there is really nothing to distinguish one pair from another. If we put <r = 1, r = 2, we have to consider
0&...d0n+ Jl...d01d0a...d0n.......(43)
The integration with respect to 0n extends over the range from 0 to 2-Tr with avoidance of the neighbourhood of 6l, 02, ... 0n-i- For each of these there is usually a range 28 to be omitted, but this does not apply when any of them happen to be too near the ends of the range or too near one another. This complication, however, may be neglected in the present approximation. Then
fcos (0B - 0a) d0n = cos (0a - 0X). (27T - 28 (n - 1)}, be equated to