478 
PROF. J. W. NICHOLSON AND DR. T. R. MERTON ON THE 
whence, by eliminating dyjdx and e kq , we derive 
2 I r e xq ' . 2 \ r q?e xqr — ( 2 I r q r e xq ’) = k ^ {'Z I r q r e xq 
\ 9 
Let X,. = I r e xqr , so that X l5 X 2 , X„ are all necessarily positive by physical 
considerations. Then the sign of ddyfdx 2 is that of 
n n f n \ 2 r — n s —n 
2 X r . 2 \q r 2 - 2 \q r ) = 2 2 X r X,(g-g s ) 2 . 
This is a sum of squares with positive coefficients, and is essentially positive for all 
values of x. Thus d 2 y/dx 2 is always positive. Even if some of the quantities q r were 
positive, this proof would be equally valid, and curves of equal density on the 
photograph would have d?y\dx 2 positive at all points. 
The curves obtained with any of Stark’s resolutions, if the components followed 
the simple exponential law, would all be included in this class. For they involve a 
possible central component l 0 e~ qx and pairs of components. A pair at points on the 
same side of both axes, if of separation 2 <x, produces an intensity 2 l cosh q<re~ qx at a 
point x, where 21 cosh q<r is constant. At a point x between the axes the intensity 
is Ie~ qa (e qx + e~ qx ), and the pair is equivalent to two central components, one increasing 
and the other decreasing. Any Stark resolution is therefore, under this law, 
equivalent to a set of components whose axes coincide, and the theorem follows. 
As already stated, this is the only possible exponential arrangement with the 
property in question, and this theorem in itself provides a very convincing argument 
in favour of the suggested theory of broadening, even when a close scrutiny is 
required to reveal the components. 
Let now a central component be superposed on a symmetrical doublet of separation 
2 o\ If the suffixes 1 and 2 refer to the component and the doublet, the curve of 
intensity I c is, between x — 0 and x — a, 
I^pytana _ q ie -?i* + 2l 2 e~ q2lT cosh xg 2 
and, between x = <x and x = oc } 
X c e pytana — I 1 e _ff|Z + 2 l 2 e _?2Z cosh o-q 2 . 
The two branches meet at x = <r, and the ratio of the two values of dy/dx at 
x = a is 
ji - sinh *q 2 . /jl + cosh aq 2 . e ~", 
which is obviously less than unity. It can be negative if I 2 is sufficiently large 
compared with I 1} and then the curve would have a sharp peak at x = <x. Otherwise. 
