136, 137, 138] 



LOGARITHMIC POTENTIAL. 



337 



138. Application to Logarithmic Potential. If in 56) 



we put U= 1, we obtain 



57 ) 



which is the divergence theorem in two 

 dimensions. If the function V is harmonic 

 everywhere within the contour, we have 



0- 



Applying this to the harmonic func- 

 tion logr, where P, the fixed pole from 

 which r is measured, is outside the contour, 



Cdlogr -, I l..r 4 /*cos(rw) 

 / 5^- ds = I - TT- ds = / - 



J dn J r dn J r 



If the pole P is within the contour, we draw a circle K of any 

 radius about the pole, and apply the theorem to the area outside of 

 this circle and within the contour, obtaining the sum of the integrals 

 around C and K equal to zero, or 



rf\\ 



59 ) 



cos (rri) 



These two results are Gauss's theorem for two dimensions. They 

 may of course be deduced geometrically in the same way as for three 

 dimensions, 118. We may now deduce Poisson's equation for the 

 logarithmic potential as in 123 for the Newtonian Potential. The 

 logarithmic potential due to a mass dm being dmlogr gives rise 

 to the flux of force 2ttdm outward through any closed contour 

 surrounding it, and a total mass m causes the flux 



= 2it / / 



pdxdy. 

 Put in terms of the potential this is 



60) Cj^ ds = - f f^Vdxdy = 2% 



C A A 



and since this is true for any area of the plane, we must have 



61) JV=-2itii. 



This is Poisson's equation for the logarithmic potential. 



25* 



