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Jackson writes,

The function $1/|\mathbf{x} - \mathbf{x}'|$ is only one of a class of functions depending on the variables $\mathbf{x}$ and $\mathbf{x}'$, and called Green functions, which satisfy (1.31). In general,

$\nabla'^2 G(\mathbf{x}, \mathbf{x}') = -4\pi \delta(\mathbf{x} - \mathbf{x}')$

A bit further down, ...

$$ \Phi(\mathbf{x}) = \frac{1}{4\pi \epsilon_0} \int_V \rho(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') \, d^3x' + \frac{1}{4\pi}\oint_S \left[G(\mathbf{x}, \mathbf{x}') \frac{\partial \Phi}{\partial n'} - \Phi(\mathbf{x}') \frac{\partial G(\mathbf{x}, \mathbf{x}')}{\partial n'}\right] \, da' $$

Since the integration is performed over the primed coordinates, it would seem that $G$, as a function of $\mathbf{x}$, represents the field due to a point source at $\mathbf{x}'$. Shouldn't it then satisfy the equation $\nabla^2 G(\mathbf{x}, \mathbf{x}') = -4\pi \delta(\mathbf{x} - \mathbf{x}')$, that is, with the Laplacian taken over the unprimed coordinates? Taking the Laplacian over the primed coordinates would seem to depend on the variation in the field at a fixed point while the point source is moved, which does not seem meaningful.

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    $\begingroup$ What do you think is the relationship between $G(x,x')$ and $G(x',x)$? $\endgroup$ – Muphrid Dec 27 '13 at 5:45
  • $\begingroup$ If you're trying to point out that $\nabla^2 G(\mathbf{x}, \mathbf{x}') = -4\pi \delta(\mathbf{x} - \mathbf{x}')$ is also true by symmetry, yeah, I get that. But why would he write it in the form he gave? $\endgroup$ – Brian Bi Dec 27 '13 at 7:08
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I think that $\nabla^2 G(\mathbf{x}, \mathbf{x}')= -4\pi \delta(\mathbf{x} - \mathbf{x}')$, is the correct equation.

Generally, we may write $G(\mathbf{x}, \mathbf{x}')$ as $H(x^+, x^-)$, with $x^\pm = x\pm x'$. So we have, with $\partial_x = \partial_+ + \partial_- $ and $\partial_{x'} = \partial_+ - \partial_- $ :

$\nabla^2 G(\mathbf{x}, \mathbf{x}') = (\partial_+^2 + \partial_-^2 + 2 \partial_+ \partial_-) H(x^+, x^-)$

$\nabla'^2 G(\mathbf{x}, \mathbf{x}') = (\partial_+^2 + \partial_-^2 - 2 \partial_+ \partial_-) H(x^+, x^-)$

So the $2$ expressions are different. Now, in the special case, where $G(\mathbf{x}, \mathbf{x}')= H(x^+, x^-)$ could be written $G(\mathbf{x}, \mathbf{x}')=H(x^+, x^-) = H_+(x^+) + H_-(x^-)$, then we have $\partial_+ \partial_- H(x^+, x^-) = 0$, and the $2$ expressions are equivalent. For instance, this is the case, when $G(\mathbf{x}, \mathbf{x}')$ depends only of $x^-=x-x'$

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Logically speaking, you may have a point. This is because, you have made the distinction that the primed co-ordinates indicates the source point and the non-primed coordinates indicates the evaluation point. However, we know that the homogeneity and isotropy of space implies that the laws of physics are invariant w.r.t spacial translations and rotations. This implies that the Green's functions should not change in form if the source and evaluation point are swapped. In other words $G(x,x') = G(x',x)$. Hence, mathematically it does not matter whether the derivatives are taken w.r.t. the primed or non-primed coordinates.

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