704 Sir J. Larmor on the Reflexion of Electromagnetic 



up to the first power of that quantity, the Maxwellian 

 expression for the electric force in a moving body (viz. the 

 intensity of force acting on electrons connected with the 

 body) is deducible from fundamental principles; and its ex- 

 pression thus obtained satisfies the Faraday circuital relation 

 expressing the electromotive effect round a circuit moving 

 with the medium *. Apply then this principle to a very 

 narrow circuit made up of two very near parallel lines, one 

 on each side of the reflecting interface, both travelling with 

 it, and connected at their ends : as the area enclosed here 

 tends to nothing, the electric force related to the moving 

 medium (as distinguished from the aethereal force, which is 

 what would act on an electron at rest) must be the same 

 along both lines. 



Again, the magnetic flux must be continuous, In our 

 present problem, on which there is no field of force trans- 

 mitted into the reflector, but only a shielding current sheet, 

 we have thus tangential electric force and normal magnetic 

 force continuous. These are, I think, the interfacial con- 

 ditions used by Mr. Edser. But as only one relation between 

 the intensities of the incident and reflected wave-trains is to 

 be determined from them, they ought to be identical ; and 

 in fact the one is involved in the other through the Faraday 

 circuital relation. As already stated, they hold good only 

 when (yjc) 2 is neglected ; if we go beyond that, the nature of 

 the linkage of the electron with the aether may have to come 

 into consideration and we get into unexplored regions. 



We consider now the problem of a simple harmonic wave- 

 train. If the aethereal (not electric) forces E and E' in the 

 incident and reflected waves are perpendicular to the plane 

 of reflexion, and given as by Mr. Edser's equations (13) and 

 (14) but with reversal of directions, say if they are 



E = a cos m (# cos 6 +y sin 6 — ct), 



W = a f cos m'(x cos 0' +y sin 6' + ct), 



then the magnetic forces are — E/c and E'/c, and are in the 

 plane of reflexion, and the vanishing of their total com- 

 ponent along the normal gives 



E . E' . 



— sin 6 — — sin 0' = 0, 



c c 



that is 



a sin = a' sin 6' (B) 



This equation, in conjunction with the kinematic relations (A), 

 solves the problem. 



* Of. my book 'vEther and Matter,' pp. 98, 113. 



