Laws of Molecular Force, 249 



for a spherical mass. To cast it into the form of an integral, 

 take any molecule m amongst the number n in a spherical 

 vessel of radius R ; gather it to its centre as a true particle 

 and spread the remaining n— 1 in a uniform continuous mass 

 separated from m by a small spherical vacuum of radius a, so 

 chosen that the virial of m and the continuous mass is the 

 same as that of m and the n— 1 molecules. 



Suppose m at the point 0, and the centre of the vessel at 

 C, and let OC = c. Take 00 as axis of x and any two rec- 

 tangular axes through as axes of y and z. Let polar 

 coordinates r <f> be related to these in the usual manner. 

 Then the equation to the surface of the sphere is 



(x-e) 2 ±f+z 2 =W or ?' 2 -2crsin0cos<£ + c 2 -R 2 =O. 

 Let 7\ and r 2 be the two roots of this equation in r so that 

 r 1 r 2 =R 2 —c 2 . Then m 2 /r 3 can be replaced by 



mpr 2 sin 6 dd dcf> dr/r*, 

 and 



2m 2 /r 3 by (Yfm/> sin 6 dd d(f> drjr. 

 If we integrate with respect to r on one side of the plane ys from 

 a to i\ and on the other from a to r 2 and add the two results, 

 then we have to take 6 and (/> each between and ?r, thus : 



2m 2 /?< 3 = IT* mp sin ddO <ty Tpdr/r + | *dr/rl 



The two integrals in brackets give 



log rtfja* =\og (R 2 -c 2 )/a 2 . 

 Hence 



2m*/t* = 27rmp log (R 2 + c 2 )/a 2 . 

 To perform the second summation we can first add the values 

 of the last expression for all the molecules at distance c from 

 the centre of the vessel, and write the result in the form 



2irp^pcHclog (R 2 -c 2 )/a 2 , 

 and we then have 



\ . i223Am 2 /^ = 6AttV f R ~ a c 2 dc log (R 2 -c 2 )/a 2 . 



•n i • ^° 



Evaluating the integral, this becomes 



A7ry|g(R-a) 3 log— _(R_ a )3__ R2(R_ a ) +-^-log — 



in which, neglecting unity in comparison with the large 

 number R/a, we get 



A7ryR 3 {4 log 2R/a -1(3/3}. 



Phil, May, 8. 5. Vol. 35. No. 214. March 1893. 8 



