12 L. J. PETERS AND J. BARDEEN 
lI 
ky? = ko? = 2 = w/c? in the air 
cealaniaiey oe Ue { 1 0 = upw vile (17) 
k,? = k? = ppw? + ipow in the earth 
where ¢ is the velocity of light. 
Cylindrical coordinates (7, z, @) with the origin at the dipole will be used. 
The electric intensity E and the magnetic intensity H are: 
E, = H.= HH, =0 
E, = 0°I1/drdz 
1 a oll 
Bee <(r | (18) 
ry or or 
1k,2 oll 
A, = — — 
po Or 
We shall take the Hertzian function for the air to be IIo, and the Hertzian 
function for the earth to be II. The boundary conditions then are 
RoI = k*x 
} atz=0 (19) 
Allo/az = dx/dz 
and 
éil 
—=0Oatz=—h. (20) 
Oz 
The Hertzian function for a dipole in free space is 
le = gale sols e707)" 7 (dx) dd (21) 
oops o (A? — Bo?) 12 ° ; 
We, therefore, take for IIo 
2 x 
ral RTE? vt ae el LUT 2 (2 ky?) 1/2 
To f = aan fo) |e To(dr)dd s>0 (22) 
and for II 
as JS [gr(AjerO“ED™ + go A)er#O—¥I'] To(Ar) ad. (23) 
0 
The functions f, g:, and gs are chosen so as to satisfy the boundary conditions 
Eqs. (19) and (20). The solution for Iq is 
Fs if De aad et PE OAT IN = eae 
o -R2(A? — Ro*)2/? + Rg? — &*)1/? tanh (A? — #*)"/? 
The normal field for a homogeneous earth of conductivity a is 
Tl. = ihe 2h2Ae 707-2 Tr) dd ; (25) 
0 k?(d? = ky?) 1/2 -- Ro*(X? = R2)t/2 
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