178 
DR. T. J. I’A. BROMWICH ON THE 
As regards the transformation of (I'l) to (i’ll), it is sufficient to note that the 
gradient of P has the spherical polar components 
1 ^, 1 dT 
dr r do r sin 6 d<p 
and that (r, 0, O) corresponds to the Cartesian vector {x, y, z). 
To obtain the formulse corresponding to (l’2) we observe that the vector on the 
right is equal to the vector-product of the two vectors, 
(., .) and m-. 
dx dy dz 
and that these two are represented by 
(r, 0, O) and 
3Q 1 3Q 1 0Q 
dr r do r sin 0 d<p 
Thus the vector-product has the spherical polar components 
1 3Q 0Q 
sm 0 d(p 
Consequently equations (1'2) now become 
00 
0Hi _ . 
( 1 - 21 ) 
while (1*4) and (l'5) give 
(1-41) 
0 H 2 
dt 
1 0Q 
sin 0 d(j> 
— M 
0H-5 _ 0Q 
dt do 
ywK 0^P p. 0Q 3 / 
(1-51) 
/xK ^ 9 _1 
a / • .0Q\ , 1 0^Q 
sm 0^ H-^ 
dt^ 0'r\ dr/ r^sin0d0\ 00/ r^ sin‘^0 dtp' 
1 as , 1 a /. SQ\ , 1 as 
A consideration of these formulae suggests that further simplifications can be 
obtained by writing 
( 1 - 6 ) 
„ au r. mK 3'U 
which together satisfy equation (l‘4l); and then equation (l’5l) leads to the 
equation for U ;— 
1 a^u 
juKd^u d^u 1 a/, ^au, 
—r — sm 0 — M- 
df dr^ r^ sin 0 00 
0 / r^ siiP 0 d(p^ 
(1-7) 
