Diamond ivith Theoretical Carbon Atoms. 263 



three corner atoms is zero in every tetrahedron, the axis of 

 the central atom as determined by the tetrahedron alone must 

 take the direction of the axis of one of the corner atoms. In 

 this case the axis of J takes the direction of that of 0. It might, 

 however, take the direction of the axis of anyone of the o^her 

 three corner atoms as far as this one tetrahedron is con- 

 cerned, and remain stable in each of four positions for small 

 displacements. A large displacement would be required 

 before it would leave one position and go over to another. 



Translational Forces. 



The consideration of the translational forces exerted upon 

 some selected atom is not apparently capable of such simple 

 treatment as was the case with the cubic rock-salt crystal. 

 In rock-salt every atom has a duplicate at the same distance 

 on opposite sides of the selected atom with axes parallel to 

 each other. The general formulae (23), (24), and (25) of 

 the former paper * show that the translational force upon 

 is equal and opposite for each pair of such atoms, and is 

 therefore zero for the whole crystal, irrespective of the 

 dimensions of an edge of the elementary cube. The con- 

 ditions for stability of the equilibrium determine the length 

 of the edge of this cube. 



In the diamond now under consideration we may eliminate 

 all atoms in planes numbered + 2 and —2, and +4 and —4. 

 An inspection shows that each atom in plane + 2 has a corre- 

 sponding atom in plane —2 at an equal distance from along 

 a diameter with axes parallel to each other. The total of 

 these two planes, therefore, gives zero translational force 

 on 0. For a similar reason atoms in plane +4 cancel those 

 in plane —4, and atoms in plane zero cancel each other. 



The forces due to the nearest atoms in planes +1, zero, 

 — 1, and —3 have been calculated by the formulae referred to, 

 giving the following results. The axis of z must by the 

 formula always lie in the direction of the axis of the atom, 0, 

 upon which the force is to be calculated. The axis of y 

 must by the formula lie in the direction of the vector kx /', 

 where k is a unit vector along the axis of atom 0, and k' a 

 unit vector along the axis of the second atom. The result is 

 that as the formula stands the directions of the x and y axes 

 differ when the force of different atoms upon is figured. 

 They must always lie in the plane of the hexagon, however, 

 7-12 inclusive, in fig. 3, the z axis being downward perpen- 

 dicular to this plane. The directions of the x and y axes will, 

 therefore, be specified for each atom calculated. It will be 



* Loc. cit. 



