Dynamical Theory of Diffraction. 229 



On transformation, this can obviously be written 



so that 



,2 



f< 



B 2 



I£ now z = r cos 0, c?S = 27rr 2 sin d0, 

 the integral becomes 



27rr 2r/cos^v sin6> ^ 



2 sinhrv 



= -±7rr . 



Putting now r = ct y and remembering that dS = r 2 JH, 

 where d£l is the solid angle subtended by dS at the centre,, 



sinh c^v * ff a "^ , . Wo 

 = -r- 1 U d* +... + .. . hill. 



... "S^p «&/£+•■•+■••) mb 



c \7 4ttJ * y 



Writing now u=lct, &c, we have the symbolic solution 

 replaced by 



f = M f(* + ^ &c.) <zn + i_ J [* j/(* + kt, &c.)l rfn. 



9. Hence, to find the effect at 0(.f, y, z) at time t, due to 

 a disturbance/, F, we describe a sphere of radius ct, with 

 as centre, note the effects, at points over the sphere, and take 

 the sum as in the above expression. 



In other words, if n is the initial velocity, and f is the 

 initial displacement in the direction of x of any point on a 

 sphere of radius ct, having for origin, then the displacement 

 at at any time t, in the direction of x, can be briefly 

 written 



which is Poisson's result. (Rayleigh's ; Sound/ vol. ii.) 



