SEPTEMBER 27, 1918] 
Fig. 1. 
tum M=Co=nmro—2nma about this 
same axis. The ratio of the angular momen- 
tum to the magnetic moment is 
Cu 2m 
B e€ 
The vectors representing the angular mo- 
mentum and the magnetic moment are thus 
in the same or opposite directions according as 
e is positive or negative. 
Tf now the body of which this molecule is a 
part is set into rotation with angular velocity 
© about an axis A, the molecule, or the orbital 
ring, behaving like the wheel of a gyroscope, 
will strive, as it were, to take up a position 
with its axis of revolution coincident with that 
of the impressed rotation; but it will be pre- 
vented from turning so far by a torque 7 due 
“to the action of the rest of the body and 
brought into existence by the displacement. In 
a minute time kinetie equilibrium will be 
reached, and the axis of the orbit will then 
continuously trace out a cone making a con- 
stant angle @ with a line through its center 
parallel to the-axis of the impressed rotation. 
When this state has been reached, as is known 
from dynamics, and as can easily be estab- 
SCIENCE 
305 
lished by applying the second law of motion,? 
by Lagrange’s equations, or otherwise, 
T= sin 9-Ca-2 (145 2 cosa) 
«Ww 
Now imagine the body, instead of being ro- 
tated, to be placed in a uniform magnetic field 
whose intensity H is directed along the pre- 
vious axis of rotation, and consider a mole- 
cule whose magnetic axis, after displacement 
by the field, makes the angle @ with H. The 
wm 5 
mga eee | ae 
Fig. 2. 
molecule would keep on turning under the ac- 
tion of the field until its axis coincided with 
H, but is prevented from doing so by the 
torque 7” upon it due to the action of the rest 
of the body and brought into existence by the 
displacement. This torque is well known to be 
T’' = nH sin@ 
2The expression for ZT can be found readily 
from Fig. 2. Let A denote the moment of in- 
ertia of the ring about a diameter, and f§ the 
angle between the vector representing J, the total 
angular momentum of the ring, and the vector rep- 
resenting w. J can be resolved into two rectangu- 
lar components, one parallel to the axis of the im- 
pressed rotation, viz., J cos(@—f), which is con- 
stant, and one perpendicular to this axis, viz., J 
sin (9 — 8), which has the constant rate of change 
OJ sin (@— £8). By the second law of motion this 
is equal to the torque 7. Expanding this expres- 
sion for 7, substituting for J cos & and J sin £, 
the components of J parallel and perpendicular to 
the axis of the ring, their equals C(w + 0 cos @) 
and AQsin #, and noting that 4—=4 C, we ob- 
tain the relation sought. 
