Radiation and Vibrations in Conical Pipes. 73 
eliminated. On differentiating equation (2) with respect to 
time, we obtain 
_ d*s _ dtu^ jPv L d*w 
dC 2 ~dx~dt + dydt + dzdt' ' * ' W 
Again, if we differentiate equation (6) to x t (7) to y, and 
(8) to z, and then add the results, we have 
jd 2 s dh_ d 2 s\_ dhi d 2 v d 2 w 
~ a \d^ 2 + df + dz 2 )~dtdx + a7dy + dfdz' ' ^ W) 
But since dhi/dx dt = d 2 u/dt dx, because the order of partial 
differentiation is indifferent, the right sides of (9) and (10) 
are identical. Thus, equating their left sides, we have 
2 s (> /d 2 s d 2 s d 2 i 
dt 2 ~ "{dx 2 ' df ' dzVI m / 1X n 
or d*s o c 
d 2 d 2 d 2 
where for brevity's sake y 2 denotes -r-, + -=-= + -—-,. 
J dx 2 dy 2 dz 1 
This is the general form of the differential equation appli- 
cable to small vibratory disturbances of a light compressible 
medium in space of three dimensions. 
Plane Waves. — As a check upon this result, let us reduce 
it to the case of plane waves in the plane of yz, and proceeding 
therefore in the direction x. Thus s is a function of t and x 
only and is independent of y and z. Hence S/' 2 s = d 2 s/dx i 
and (11) becomes 
d 2 s a d 2 s 
= a 2 - 
dt 2 dx 2 
(12) 
which is the well-known form of equation for plane waves. 
To examine if a has the right value here let the specific 
volume be U, then p = l/TJ and dp=— dJJ/J] 2 . 
••• a2 =|=Cwu) u=EU=E/ ' 5= ^'- (13) 
in which E denotes the volume elasticity and <y the ratio of 
the two specific heats. Thus, the ordinary value is seen to 
hold. 
Solution for Spherical Waves. — Let us now transform to 
polar coordinates defined by 
x = r sin 6 cos <£, y = r sin 6 sin <£, and z — r cos 0. (14. 
