# Jackson p244,Green function for wave equation

#### NinjaSlayer

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) .

but,why G=\frac{e^{ikR}}/{R} satisfy
( \nabla ^2 + k^2 )G =-4\pi \delta (\mathbf{R}) ?

How to normalize the Green function?
( \nabla ^2 + k^2 )\frac{e^{ikR}}{R}=...calculate...=0.

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#### paweld

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
$$G=\frac{e^{ikR}}{R}$$
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.

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:
$$\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G = \lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2)G$$
Gauss theorem tells
us that we can change integration over ball K(0,r) for integration over sphere:
$$\int_{K(0,r)} d^3 x \nabla ^2 G = \int_{K(0,r)} d^3 x ~ \textrm{div} ~\textrm{grad} G = \int_{S(0,r)} dS \vec{n} ~\textrm{grad} G$$
So we have:
$$\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G = \lim_{r \rightarrow 0} \int_{S(0,r)} dS \left( \frac{(ik R -1)\exp(ikR)}{R^2} \right) = -4\pi$$
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.

#### NinjaSlayer

Oh...It's a very nice trick.

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