32 
PHYSICS: E. H. HALL 
Under hypothesis A. — In order to have electrical equilibrium within the bar, 
we must have 
dp = fn dl (2) 
where / is the electrostatic force acting on a single electron in the direction 
of increasing T, and dl is the element of length of the bar corresponding to 
the change of pressure dp. Hence the decrease of negative electrostatic 
potential from the cold to the hot end of the bar is 
ni ni r*i r*pz 
P.-:*-! »«I M.I [k.L \ vd p, ( 3 ) 
e I e I n e n Ge K 
•JO tJo *J 0 «-/ pi 
where e is the electron charge, v is the volume of 1 gram of the electrons, and 
G is the number of electrons therein. The value of the last integral is evi- 
dently equal to the area E A D G for the bar a. and to the area F B C H for 
the bar (3. 
Each of the lines A D and B C corresponds to a state of equilibrium for its 
metal- Let us now suppose the two metals to be joined together at the hot 
end while an additional piece of the metal /3, at temperature T h is joined to 
the Ti end of a, as in figure 2, leaving a gap between two pieces of j8, with 
temperature T 1 at both / and r. We may, if it seems desirable, instead of 
joining the two metals directly, insert at each junction a bar, or bridge, of 
alloy changing in composition from pure a at the left end to pure (3 at the 
right end, the whole length being at the temperature of the a and |3 adjacent. 
See Figure 3. Through each of the bridges we shall have a gradual change of 
m, from the valuable proper to a. to the value proper to (3. 
At the instant when the junctions indicated are effected there will probaoly 
Le a slight movement of electrons from one metal to the other, lowering a 
little the electric potential of one bar as a whole and raising that of the other ; 
but the gradient of mechanical pressure, and so the balancing gradient of elec- 
tric potential, will remain sensibly unaltered in each. Along each of the iso- 
thermal bridges of alloy the general law of equilibrium will be the same as 
that which we have seen to hold in each of the homogeneous but now isother- 
mal bars; that is, dp equals fndl. The volume-pressure changes through 
the bridge at T } will be those shown by the isothermal line A B of figure 1 ; 
the corresponding changes through the bridge at T\ will be those shown by 
the isothermal D C. The increase of negative electric potential through the 
bridge at T 2 from a to /3 will be, according to equation (3), represented by 
~ X {the area EABF). The corresponding increase at the T\ junction 
Ge 
will be i X (the area GDC H). 
Ge 
We still have equilibrium, the circuit being open at Ir. But is the potential 
