Lord Rayleigh on Waves. 271 



takes to arrive at B'. Now if <r denotes the small breadth of 

 the tube A D at any point, and v the velocity, the total stream 



A 













c 



~~^^- 



^*<^ 











— — — _ 











~^~~~~---^^ 



j.B 













13 



Js> 







- 





B' 









a 











D' 



is <tv and is constant. Denoting it by K, we have 



The time t occupied by a particle in removing from A to B is 

 therefore 



= I — - = I -xr-= area AD-r-K. 



And if t f represents the time between A / and B', 



K being the same in both cases, since the total streams are by 

 supposition equaL Thus 



t : t' = area AD : area A'D', 



and it remains to prove that AD is greater than A'D'. 



If we draw equipotential lines in such a manner that the 

 small spaces cut off between them and AB, CD are squares, 

 then we know that the same series of equipotential lines will 

 divide the space between A'B', C'D', into small squares also. 

 Now if a line be divided into a given number of parts, the sum 

 of the squares of the parts will be a minimum when the parts 

 are all equal. Hence the space AD is greater than if the 

 squares described on the parts of AB were all equal, and there- 

 fore a fortiori greater than the space A'D', which consists of 



