370 Lord Rayleigh on the Momentum and 



As in other applications of the virial theorem, the forces X 

 are divided into two groups, internal and external. The 

 latter reduces to the forces between the ends (pistons) and 

 the gas. If p 1 be the pressure on the pistons — it will be the 

 same on the average at both ends — the external virial is per 

 unit of area ±p x l simply. As regards the internal virial, I 

 do not remember to have seen its value stated, probably 

 because in hydrodynamics the mechanical properties of a 

 fluid are not usually traced to forces acting between the 

 particles. There can be no doubt, however, what the value 

 is. If we suppose that the whole mass of gas in (27) is at 

 rest, the left-hand member vanishes, so that the sum of the 

 internal and external virial must vanish. Under a uniform 

 pressure p , the internal virial is therefore %p Q l. In an actual 

 gas the virial for any part can depend "only on the local 

 density, so that whether the gas be in motion or not, the 

 value of the internal virial is 



-*j) 



(28). 



JO 



Hence (27) gives 



kinetic energy = \p^~ J pdx 



= UPi-Pq)1— l\ (p-p )dx . (29). 



If the gas be subject to Boyle's law, pressure is proportional 

 to density, and the last term in (29) disappears. The 

 additional pressure on the ends {pi— po) is thus equal to 

 twice the density of the kinetic energy. 



In general, 



p-Po = a Xp-Po)+U"(Po) • (p-po)\ 

 and Ci r* 



J o (i>-|>o)tftf = i/''(po)J {p-poYdw. 



If we introduce expressly the integration w 7 ith respect to t 

 already implied, (29) gives 



1 1 (Pi -Po) dt = po jj TJ 2 dx dt + if ' (> ) \$ ( P ~ Po ) Hx dt 



\ Po 



Po 



^ i JJ 



XJHx dt, 



regard being paid to (9). Equation (10) is thus derived very 

 simply from the virial. 



