188 HENEY A. EOWLAND 



Hence 



Z' - Xa+Ya' 



where a, /?, f ; a!, /3', -f ; and a", /5", /' are the direction cosines of the 

 new axes with reference to the old. 

 We then find 



E= - \fff{ X* (jfcy + JkJP + V) + Y* ( V 2 + V + V 2 ) + Z\k 



+ 2YZ 



The most simple and in many respects the most interesting cases 

 are when the crystal has only one optic or magnetic axis. In this 



CclSG $2 ' ' ~ wy 



Hence 



where , a! and a!' are the direction cosines of the magnetic axis with 

 respect to the coordinate axes. 



The first case to consider is that of a mass of crystal in a uniform 

 magnetic field. The magnetic forces which enter the equation are 

 those due to the magnetic action of the body as well as to the field in 

 which the body is placed. In the case of very weak magnetic or 

 diamagnetic bodies the forces are almost entirely those of the field alone. 

 Hence in the case under consideration we may put F = and Z = 0. 



Hence 



and if v is the volume of the body 



As this expression is the same at all points of the field there is no 

 force acting to translate the body from one part of the field to another. 

 The moment of the force tending to increase <p, where <p cos -1 , is 



j pi 



-.- = v X" 1 (k^Tc^ sin <p cos <p . 



By observing the moment of the force which acts on a crystal placed 

 in a uniform magnetic field we can thus find the value of k i k 2 or 

 the difference of the magnetic constant along the axis and at right 

 angles to it. The differences of the constants can also be found in the 

 case of crystals with three axes by a similar process. 



The next case which I shall consider is that of a bar hanging in a 



