430 PROCEEDINGS OF THE AMERICAN ACADEMY 



comparatively small number of observations, it seems difficult to decide 

 whether the apparently greater brilliancy towards the edge along the 

 equatorial diameter is real, or is due to errors of observation. Column 

 7 gives the mean of all the measurements taken, and column 8 the 

 probable error of this mean. 



If the sun had no atmosphere, its disk as seen from a distance would 

 appear uniformly bright, since the light emitted by one square metre 

 in any given direction is inversely as the cosine of the angle of emis- 

 sion, while, owing to foreshortening, its apparent area is proportional to 

 the cosine of the same angle. Let us next suppose it surrounded by a 

 homogeneous atmosphere not perfectly transparent. Evidently the 

 absorption will dejiend on the distance which the light has to j^ass 

 through it, and will be greatest at the edges, and least at the centre ; 

 or the disk will appear brightest at the centre and darkest at the ex- 

 terior, as is actually the case. To determine the law of this variation, 

 let the radius of the disk equal unity, x the apparent distance of any 

 point from the centre, /* the height of the atmosphere, h the brightness 

 of any portion of the disk were there no atmosphere, a the proportion 

 of light which would traverse a thickness of the atmosphere e(jual to 

 unity, or to the sun's radius. Call v also the distance the light from 

 the point x must traverse before emerging from tlie solar atmosphere, 

 and y the apparent brightness of the same point. It is readily proved 



that V = V (1 -j- liY — aj""^ — V 1 — ic^, and that y = hcC ; therefore, 



y 



^l„sja+hr-x^—sji- 



is the equation which gives the brightness of any point of the sun's 

 disk, assuming that its atmosphere is homogeneous. From any three 

 corresponding values of x and y we can compute a, b, and h. Assum- 

 ing from the above observations y = I for x ==: 0, y= .782 for 

 a; =.75, and y = .374 for x= 1, and taking logarithms, we deduce 

 the three equations of conditions : — / 



= log i -|- A log a ; 



— .10G8 = log i + (v/(l-f /O' — .5625 — .GG14) log a ; 



— .4271 = log b + (^2A + /r) log a. 



Subtracting these equations, we eliminate b ; and dividing one of the 

 resultant equations by the other, eliminates a. "\Ve thus deduce the 

 equation : — 



V/.4375 + 2h + h^ — .25 v/2A -\- /r — .7 oh — .6614 = 0. 



