288 



Mr B. F. Gwyther, On the solution of the [May 25, 



We may obtain another expression for this by comparing with (12), 

 thus 



2w 2 = n {a 2 - 4>fg + 6 2 - 4ge + c 2 - 4e/} 



dy dz, 



dxj \dx dy) j 



The first of these forms simplifies in the case we are specially 

 treating, for by (6) 



dx 



[dy 



+ 



^X = l( d l\ 



dz J 



dtj 



etc., 



and 



®2 = 1 



n = a . 



As we have been treating the unit of volume as being the unit mass 

 we may say that co 2 is equal to the kinetic energy per unit of 

 volume, and that actually and not on an average. The energy 

 is not oscillatory between the wholly kinetic and the wholly 

 potential as in the case of a pendulum, but when the kinetic 

 energy is zero then co 2 is zero. If it were not so we might expect 

 that there would be a luminous trace of the path of light owing to 

 the dying away of the oscillations, whereas this expression shews 

 that the light passes and leaves the medium at rest without trace 

 of its passage. 



We have also the formula 



a 2 j /# _ drA 2 /df _ d£\ 2 (dr, _ df \ 2 ) 

 2 \\dy dz) \dz dx) \dx dy) ) 



_ i 



— 2 



dt 



d V \ 2 , (dW 



+ y + [w\ =m » 



which I shall refer to later. 



Secondly, to evaluate &) 3 , write it as a determinant and border 

 it thus, using the relations (11), 



2a> =1 



1, 

 d% 



dx' 



dr} 



dx' 



dX 



dx ' 



0, 

 dx' 



dr) , di; 



(JjJb (JjJb 



o, 



o, 



dr] 

 dx 



, df; , d% d£ 



1 y-7/v» 2 sf/Y* /"//*» 



2k 



dx 



dr} 

 dx' 



dx 

 i d£ t dr} 



CLOG CLOG 



, d% d£ , d% , dr} 



(JjJb \AjJU \XitAj iX/JU 



2k. 



dx 



