156 PROFESSOR W. THOMSON ON THE 



plane, obtained by multiplying their intensities by the cosine of those angles re- 

 spectively, will be each equal to 



i cos w sin w . 



Hence the absorptions of heat which they will produce at the surface of separa- 

 tion of the metals per unit of area per second will be, 



— j i cos o> sin co t 6 , and y i cos w sin oi t (p , 



respectively. According to the general principle of the superposition of thermo- 

 electric actions stated above, the sum of these is the rate of absorption of heat per 

 unit of surface, when the two systems of currents coexist. But the resultant of 

 these systems is simply the given longitudinal current in the bar, with no flow, 

 either out of it or into it, across any of its sides. Hence, a simple current of in- 

 tensity i, parallel to the sides of the bar, causes absorption of heat at the side 

 C D, amounting to 



j i cos oo sin oi t ((f) — 6) , 



per unit of area per second ; and the same demonstration shows that an equal 

 amount of evolution must be produced at the opposite side CD'. These effects 

 take place quite independently of the matter round the bar, since the metal 

 carrying electric currents which we supposed to exist at the sides of the bar in 

 the course of the demonstration, can exercise no influence on the phenomena. 



152. If I denotes the length of the bar, the area of each of the sides perpendi- 

 cular to the plane of the diagram will be I a; and therefore, the absorption over 

 the whole of the side C D, and the evolution over the whole of the other side C D', 

 per second will be 



j i I a cos &> sin w t (</> — 6) , 



or j y -j- cos o» sin oi t (<p — 6) . 



It is obvious, that there can be neither evolution nor absorption of heat at the 

 two other sides. 



153. An investigation, similar to that which has just been completed, shows 

 that if the actual current enter from a conductor of the standard metal at one 

 end of the bar, and leave it by a conductor of the same metal at its other end, 

 the absorption and evolution of heat at these ends respectively will amount to 



|7(l0 cos 2 co + t(f> sin 2 oS) 



per second. 



154. Let us now suppose the two sides C D, C D to be kept at uniform tem- 

 peratures, T, T", and the two ends to be kept with equal and similar distributions 

 of temperatures, whether a current is crossing them or not. Then if a current of 

 strength 7 be sent through the bar from left to right of the diagram, in a circuit 





