PHYSICS: E. H. HALL 
33 
of the |8 at I equal to that of the j8 at r? We can answer this question readily. 
Starting in a (fig. 3), close to the T 2 junction, going across the bridge and 
then down (8 to r, we have as the gain of negative potential 
P r -P a2 = — (EABF + FBCH) = — X EABCH (4) 
Ge Ge 
Starting again at the same point as before and going down a, then across 
the lower bridge to (3, we get 
Pi — P a2 = — (EADG + GDCH) = —XEADCH. (5) 
Ge Ge 
We have, then, by subtraction 
P r ~ Pi = ~X A BCD. (6) 
Ge 
If now we choose to connect / and r by means of some very large resistance, 
the current will flow from r to I through this resistance, while the conditions 
within a and /3 will remain almost unchanged. We shall have a thermo-elec- 
tric current passing clockwise around the circuit of Fig. 2, or Fig. 3, and the 
e.m.f. maintaining this current will be ~ X A B C D. The amount of elec- 
Ge 
trical work done by a gram of electrons, Ge units, in passing once around this 
circuit is A B C D, which is, of course, equal to the mechanical work of the 
pressure-volume cycle through which the gram of electrons passes. 
Under Hypothesis B. — In this case, as I have elsewhere shown, 1 we have, as 
the condition of electrical equilibrium in a detached bar with a temperature 
gradient, 
(7) 
\K 1 / 
% 
Hence 
fdl= l fndl 1 I dp _ 1_ P(dR + dT\ 
e I e ) n e 1 n 2e } n\ R T I 
But 
and so 
p,-p 2 =i- r vdp + -t- fp ivi ~ p 2 v\ ' 
Ge J pi 2&e \ / 
(8) 
