NinjaSlayer said:
Jackson electrodynamics 3rd. p244
I understood that
G=\frac{e^{ikR}}{R}
is a spetial solution for
( \nabla ^2 + k^2 )G =0 (R>0) .
Firstly
<br />
G=\frac{e^{ikR}}{R}<br />
is not strictly a solution of homogenous equation (i.e. equation without delta function
on RHS) because G is not well behaved for R=0 and because of it one has to consider all
derivatives near the point R=0 in distributional sense. It is true that for all points
R \neq 0 we have ( \nabla ^2 + k^2 )G =0 but it's not true for
R=0 as G blows up there.
NinjaSlayer said:
but,why G=\frac{e^{ikR}}/{R} satisfy
( \nabla ^2 + k^2 )G =-4\pi \delta (\mathbf{R}) ?
There is a nice trick which shows why it is the case. Let's integrate the expression
( \nabla ^2 + k^2 )G over small ball of radious r in the limit r goes to 0.
Notice that:
<br />
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G = <br />
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2)G<br />
Gauss theorem tells
us that we can change integration over ball K(0,r) for integration over sphere:
<br />
\int_{K(0,r)} d^3 x \nabla ^2 G = \int_{K(0,r)} d^3 x ~ \textrm{div} ~\textrm{grad} G<br />
= \int_{S(0,r)} dS \vec{n} ~\textrm{grad} G<br />
So we have:
<br />
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G = \lim_{r \rightarrow 0} <br />
\int_{S(0,r)} dS \left( \frac{(ik R -1)\exp(ikR)}{R^2} \right) = -4\pi<br />
Because integration of ( \nabla ^2 + k^2 )G gives -4\pi and
( \nabla ^2 + k^2 )G is zero everywher beside R=0 it has
to be equal to -4\pi \delta(\vec{R}).
How to normalize the Green function?
( \nabla ^2 + k^2 )\frac{e^{ikR}}{R}=...calculate...=0.
Because ( \nabla ^2 + k^2 )G =-4\pi \delta (\vec{R}) is not a homogenous
equation if you multiply G by a constant it won't be a solution to this eqaution.
