204 Prof. Kelland's Reply to some Objections against the 



dV d?V d 3 V 

 the differential coefficients -7-?, -nm t^j & c « are zero; and 



df dj l d/ a 



since the force put in play on a particle by a displacement I 



dV 

 depends on the expansions of -r-p &c, and therefore of V 



in terms of 5, it is evident that the force is zero. The equili- 

 brium is consequently what is technically called neuter. 



The following investigation is copied from the paper above 

 referred to. The complete demonstration of the proposition 



d"V 

 that , fn is equal to zero, involves some little analysis ; and 



as it leads to a number of most important results, as, for in- 

 stance, that 2 m (x — /) n f{r) =— 2 m r 2 "/(r), I will 



reserve it to my next communication. 

 When V = Sw 



let V'=Sw 



1 



•(* -/- uf + (y - g - /3) 2 + (z - h - yf 



a, /3, y being the increments of/5 g 3 and h. 



Now if we put a{x —f) + /3 (3/ — g) + y (z — h) — 6, a 2 



+ /3 2 + y 2 = 8% and expand V, there results 



_., -,- / 1 2e-8 2 1.3 (2s-8 2 ) 2 \ 

 W = V + Zm(j —^~ + — r a + &c.) 



~ V + 2 ' W 1 2r^" + 2 ? 



We have obtained our reductions by introducing the results 

 of symmetry. Thus the coefficient of 8 2 is zero. By pro- 

 ceeding a step further, we get 



,1.3.5.7 16 1 4 , . \ 



