352 



so that 



sa"^/^(l + aL^^/2)cla 



p -TL p sa ' (i + aLdJ-,/ 



(3.17) Ab = 3 / sdt/a"^ = 3 / , ^ 



^ ^ ^l-a^-ka-^(^-l^- 



Ss^Sa^ 



Since s is given as the indefinite integral of the right 

 side of (3.11:3), Ab is really a double integral but we can 

 change it to a single integral by the follov;ing argument: 

 During the first part of the bubble motion, that is, 

 until the bubble reaches its maximum size, s increases 

 from zero to s, . During the second part of the motion, 

 from the bubble maximum to the bubble minimum, s increases 

 from S-, to s = 2s^. The momentum at any time during the 

 second half of the motion is i minus the momentimi devel- 

 oped as the bubble contracts from its given size to its 

 minimum size. Because of the symmetry of the bubble mo- 

 tion, this minus momentum is the same as the momentum de- 

 veloped when the bubble increases from its initial size 

 to the given size. We may then conclude that the dis- 

 tance moved during the bubble expansion just balances the 

 distance which would be produced by the minus momentum so 

 that the total distance moved is given by one term 



(3.18) 



Ab = 3s 



a 



a"^/^(l + aL^y2)da 

 ^ h-a^ - ka-^C^-l) - 3i2/2a5' 



(3.19) I^ 



= 3s(I^ + ^ L^J-lIq), 



^ p a-^/2 da 



/rrT3 . kVsTFir _^i2j^ ' 



1 



, ^ a--/2 da 



(3.20) lo = ' a aa 



"8 



a 

 This proves (1.27) 



^ \/l - a^ - ka-=5(*^"l) - 3i2/2a^' 



