516 Mr. E. Edser on the Reflexion of Electromagnetic 



the energy brought up to unit area o£ the mirror is equal to 

 (2/S7rc 2 )E- l 2 (c-[-v)dt, and the energy carried away from the 

 same area is (2/87rc 2 )E 2 2 (c — v)dt. If p Y denotes the instan- 

 taneous pressure on the mirror, the work done per unit area 

 of the advancing mirror \& p x vdt. Thus 



Writing the expression within the brackets on the right of 

 this equation in the form 



{E 2 {c-v) .Ez-^ic + v) .EJ, 

 and using the relation 



«Y-/3S=4{( a -/3)( 7 + S) + («+/3)(y-S)}, 

 we obtain the equation 



PlV= 8^ {tE,(c ~ p)-El(c + , ' ):i(El+:B * ) 



+ P,(o-«)+B 1 (e+t>)](B 1 -B,)} 



= 8^[ E ^-")-Ei(^ + «)](E 1 + E 2 ), 



from (3). Then, substituting from (9), we obtain 



iV=-Sx(E,+E 2 ) .... (ii) 



=Cx|.^'(E 1 -E 2 ), 



from (3) 



p^Cx^^Ei-E,), . . . (12) 



and from (10), (l/2c)(E 1 -E 2 ) is equal to that part of the 

 magnetic force which is not derived from the current and is 

 independent of the motion of the mirror. This result will 

 be found to afford valuable guidance in the ensuing investi- 

 gation. 



We are now in a position to solve the problem of reflexion 

 from a moving mirror when the angle of incidence is not 

 equal to zero. Two cases arise. In the first, the electric 

 field of the incident waves is perpendicular to the plane of 

 incidence. In the second case, the electric field of th& 

 incident waves is parallel to the plane of incidence. 



