on the Physical Properties of Gases. 179 



To find the state of equilibrium of the gas, we must put 



dx ^dy __ ^_n 

 dt ^ dt ~~ dt~ 



Suppose .i?i, ?/i, zi one set of roots of the resulting equation. 

 This will give a stable state to the gas if, when ^i, ?/i, z^ receive 

 any small variations, the gas tends to return to its former 

 state. Suppose that x^, ?/i, ^i are disturbed to Xi + f , ?/i + r), 

 ^1 + ?; then, by Taylor's theorem, since /i(.i'i, ?/i, ^i) = 0, &c., 

 di^dJ^^^dU^^df, , 



dt 



dxi 



drj _ 

 dt' 



_df2 

 dxi 



d%_ 

 dt 



.df^ 



dxi 



diji 



dz. 



f+ . . ■ ., 



f+ • • • • ^ 



and these equations must be such that ^, 77, ^ never become 

 large, and ultimately vanish. To solve them we put 



f=a6^*, r] = l3e^^, ^=ye^^, 



and obtain, in the ordinary way, to find X, the equation 



= 0. 



dx ' 



dh dA 



dy dz 



df, 

 d^' 



df2 . df^ 

 dy ^' dz 



dh 

 dx' 



dh df, ^ 

 dy^ dz 



This is a cubic ; and we therefore get three values of \, say X^, 

 X2, ^3- Whence 



f = a^e^i' + a^e^^ + a^e^ 



V 



?= 



There vdll be two cases, according as all the roots are real or 

 two imaginary: — 



I. All real. Here the single condition that f &c. may not 

 increase is that X^, X2, X3 must all be negative. 



II, One real (X^) and two imaginary (X±//,^). In this case 



f = oiie^i^ + 0126^* cos (/j.t + ^3), v= "', ?=.... 



Here, then, X^ and X must both be negative. Both cases are 

 contained in the statement that the real part of the roots of the 

 cubic must be negative. 



The two cases belong to two totally different kinds of sta- 

 bility. In case I. the gas returns directly to its normal state, 

 but only arrives exactly at it after an infinite time. In 



N2 



