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jack476
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I'm working through the problems in the first chapter of Jackson and I'm still grappling with the interpretation of Green's functions.
I understand that if I have the Poisson equation ##\nabla^2\phi(x) = \frac{-\rho (x)}{\epsilon_0}## and the Green's function ##G(x, x^\prime)## then in general ##\phi(x) = \frac{\rho (x)}{\epsilon_0}\ast G(x, x^\prime) ##.
What's bothering me is that, if my understanding is correct, this would give ##\phi(x) = \frac{1}{\epsilon_0}\int \rho(x^\prime)G(x,x^\prime)dV^\prime##. But this is missing a factor of ##\frac{1}{4\pi}##, for instance, in the case of a point charge where Jackson says that we have ##G = \frac{1}{|x-x^\prime|}## then we have ##\phi(x) = \frac{1}{\epsilon_0}\int \frac{\rho(x^\prime)}{|x-x^{\prime}|}dV^\prime##, which is not correct because it's missing that factor.
The only way I can see it is that, properly speaking, for a linear differential operator ##L## a Green's function is a solution to ##LG(x, x^\prime) = \delta (x-x^\prime)##, but Jackson has defined Green's functions to be solutions to the differential equation ##\nabla^2 G(x,x^\prime) = -4\pi \delta(x-x^\prime)##, so that ##G## is not a Green's function for the Laplacian but rather for the operator ##\frac{-\nabla^2}{4\pi}##. If we represent Poisson's equation in terms of this operator by multiplying each side by ##\frac{-1}{4\pi}## then we get ##\frac{-\nabla^2}{4\pi}\phi(x) = \frac{\rho(x)}{4\pi \epsilon_0}## and now we can use the convolution to get the correct formula: ##\phi(x) = \frac{\rho (x)}{4\pi \epsilon_0}\ast G(x, x^\prime) =\frac{1}{4\pi \epsilon_0}\int \rho(x^\prime)G(x, x^\prime)dV^\prime##, but that just seems kind of inelegant.
Can someone please help me understand all this? That missing ##\frac{1}{4\pi}## has been ruining my entire day.
I understand that if I have the Poisson equation ##\nabla^2\phi(x) = \frac{-\rho (x)}{\epsilon_0}## and the Green's function ##G(x, x^\prime)## then in general ##\phi(x) = \frac{\rho (x)}{\epsilon_0}\ast G(x, x^\prime) ##.
What's bothering me is that, if my understanding is correct, this would give ##\phi(x) = \frac{1}{\epsilon_0}\int \rho(x^\prime)G(x,x^\prime)dV^\prime##. But this is missing a factor of ##\frac{1}{4\pi}##, for instance, in the case of a point charge where Jackson says that we have ##G = \frac{1}{|x-x^\prime|}## then we have ##\phi(x) = \frac{1}{\epsilon_0}\int \frac{\rho(x^\prime)}{|x-x^{\prime}|}dV^\prime##, which is not correct because it's missing that factor.
The only way I can see it is that, properly speaking, for a linear differential operator ##L## a Green's function is a solution to ##LG(x, x^\prime) = \delta (x-x^\prime)##, but Jackson has defined Green's functions to be solutions to the differential equation ##\nabla^2 G(x,x^\prime) = -4\pi \delta(x-x^\prime)##, so that ##G## is not a Green's function for the Laplacian but rather for the operator ##\frac{-\nabla^2}{4\pi}##. If we represent Poisson's equation in terms of this operator by multiplying each side by ##\frac{-1}{4\pi}## then we get ##\frac{-\nabla^2}{4\pi}\phi(x) = \frac{\rho(x)}{4\pi \epsilon_0}## and now we can use the convolution to get the correct formula: ##\phi(x) = \frac{\rho (x)}{4\pi \epsilon_0}\ast G(x, x^\prime) =\frac{1}{4\pi \epsilon_0}\int \rho(x^\prime)G(x, x^\prime)dV^\prime##, but that just seems kind of inelegant.
Can someone please help me understand all this? That missing ##\frac{1}{4\pi}## has been ruining my entire day.
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