1901—2.] Lord Kelvin on Molecular Dynamics of a Crystal. 223 
though in ordinary language any one of them would be called a 
strut if its force is push instead of pull on the atoms to which it is 
applied. Calling A 1} A 2 , A 3 . . . the atoms at one end of the row, 
suppose the tie between A x and A 2 to be removed, and A 1 allowed 
to take its position of equilibrium. A single equation gives the 
altered distance A 1 A 2 , which we shall denote by + Let 
an altered tie be placed between A 4 and A 2 to keep them at this 
altered distance during the operations which follow. Next remove 
the tie between A 2 and A 3 , and find by a single equation the altered 
distance a + j[X 0 . After that remove the tie between A s and A 4 and 
find, still by a single equation, the altered distance a + Y x 3 , and so on 
till we find 1 ^ 7 or 1 a; 8 or Y x h small enough to be negligible. Thus 
found, x x v jj&g, jajg, .... jXt give a first approximation to the devi- 
ations from equality of distance for complete equilibrium. Eepeat 
the process of removing the ties in order and replacing each one 
by the altered length as in the first set of approximations, and we 
find a second set 2 x v 2 x 2 , 2 x 3 ... . Go on similarly to a third, 
fourth, fifth, sixth .... approximation till we find no change by 
a repetition of the process. Thus, by a process essentially con- 
vergent if the equilibrium with which we started is stable, we find 
the deviations from equality of consecutive distances required for 
equilibrium when the system is left free in the neighbourhood of 
each end, and all through the row (except always the constraint to 
remain in a straight line). By this proceeding applied to the curve 
of fig. 7 and the case of equilibrium a =*680, the following suc- 
cessive approximations were found : — 
Xi 
^3 
a 4 
a? 6 
x 7 
1st Approximation . 
+ •018 
- *009 
+ •004 
- -002 
+ •001 
rH 
O 
O 
1 
■ooo 
2nd 
+ •026 
- *014 
+ •007 
- -003 
+ •002 
3rd 
+ •031 
- -018 
+ *009 
- -005 
+ •003 
4th ,, 
+ •034 
- -020 
+ -on 
- -006 
5th 
+ "036 
- -022 
+ •012 
- *007 
6th 
+ "03/’ 
- *023 
+ •013 
7th ,, 
+ •038 
- -024 
8th 
+ -039 
Thus our final solution, with a = ‘680, is 
x ± = +-039, x 2 = --024, a 3 = +-013, x 4 = -‘007, ||= + -003, 
x Q = - - 001, x 7 = -000. 
