40 Mr. 0. Heaviside on Maxwell's Electromagnetic 



To obtain the E due to Eq constant from e= — oo to 0, use the 

 first of (36) ; change H to E, Hq to Eq, and change the sign of 

 cr, not forgetting in the/^s. To obtain the corresponding H due 

 to Eg, use the second of (36) ; change E to H, Hq to Eo, and 

 /A to c. So 



E = p.e-.{l-../.(l-£)+l^/.(l-iy-...}, 



H=io.E,e-p.{l + .(^.(l-i)+(|-',.(l-i)V...}^ 



where the accent means that the sign of a is changed in the/'s. 

 From these, without going any further, we can obtain a 

 general idea of the growth of the waves to the right and left 

 of the origin, because the series are suitable for small values 

 of at. But, reserving a description till later, notice that E in 

 (36) and H in (37) must be true on both sides of the origin; on 

 expanding them in powers of z we consequently find that the 

 coefficients of the odd powers of z vanish, by the first of (28), 

 and what is left may be seen to be the expansion of 



E = i/.t;Ho6-p'Jo[°"(.^2-t;V^)^], . . (38) 



the complete solution for E due to Hq. Similarly, 



H = iwEoe-p'Jor^ (/^-i;^^)*] . . . (39) 



is the complete solution for H due to Eq. In both cases the 

 initial distribution was on the left side of the origin ; but, if 

 its sign be reversed, it may be put on the right side, without 

 altering these solutions. 



Similarly, by expanding the first of (36) and first of (37) 

 in powers of z we get rid of the even powers of z^ and produce 

 the solutions given by me in a previous paper*, which, how- 

 ever, it is needless to write out here, owing to the complexity. 



13. Arbitrary Initial States. — Knowing the solutions due 

 to the above distributions, we find those due to initial Yj^da at 

 the origin, or Hoc/a, by differentiation to z ; and for this we 

 do not need the firsts of (36) and (37) but only the seconds. 



* " Electromagnetic Waves," § 8 (Phil. Mag. Feb. 1888). 



