How Does Dirac Delta Substitution Relate to Helmholtz's Decomposition Theorem?

[muzialis]

Hi All,

I found (Wikipedia page on Helmotz's decomposition theorem) the follwoing equality, which puzzles me:
$$\delta(x-y) = - (4 \pi)^{-1} \nabla^{2} \frac{1}{\vert x - y \vert}$$
I am not sure I understand, the r.h.s seems to me a proper function. The page mentions this a sa position, not an approximation, can maybe anuvboy help?

Many thanks

 

[Office_Shredder]

The right hand side is a function which has a singularity when x=y. For fixed y, if you "apply the divergence theorem" (which depending on the exact statement that you use may or may not actually apply) to the right hand side you integrating over a ball you will see that you get 1 if your sphere contains y, and 0 if your sphere does not contain y, which is the inspiration for this (in particular, the right hand side functions exactly as the delta function when it's under an integral sign).

 

[muzialis]

Office_shredder,

I clearly see your point. I am still puzzled by the notation, as the equality sign seems to me an abuse.

Thank you very much

Best Regards

 

[vanhees71]

The equation is correct, but it has to be understood properly, namely in the sense of distributions. The $\delta$ distribution is defined as a linear functional on an appropriate space of test functions (e.g., the $C^{\infty}$ functions with compact support or Schwartz's space of quickly falling $C^{\infty}$ functions) by
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{y} \delta^{(3)}(\vec{x}-\vec{y}) f(\vec{y})=f(\vec{x}).$$
In this sense you have to read the equation. To prove it, you must show that for any test function
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} f(\vec{y}) \left (-\Delta_{\vec{x}} \frac{1}{4 \pi |\vec{x}-\vec{y}|} \right )=f(\vec{x}),$$
which is not entirely trivial.

The intuitive physicist's way to understand it is that
$$G(\vec{x},\vec{y})=\frac{1}{4 \pi |\vec{x}-\vec{y}|}$$
is (read as function of $\vec{x}$) the electric potential of a unit-point charge located at $\vec{y}$, fulfilling the equation (valid in Heaviside-Lorentz units)
$$\Delta_{\vec{x}} G(\vec{x},\vec{y})=-\rho(\vec{x})=-\delta^{(3)}(\vec{x}-\vec{y}).$$
In other words, it's the Green's function of the differential operator $-\Delta$.

 

[muzialis]

Office_shredder,

it looks like I have understood less than what I thought...
Your explanation made perfect sense until I tried a direct computation.
For simplicity in 1D, I fixed y = 1.
Then, if I understood, the indicated function should behave as a delta-distributuion under the integral sign.
As $$\int_{-\infty}^{\infty} \delta \mathrm{d}x = 1$$ the same should be expected for the mentioned function, which in 1D behaves as $$\frac{1}{(x-y)^{4}}$$ and is hence not even integrable.
I thought this might be a problem related to the 1D setting.
I tried in 2D then, using the divergence theorem now
$$\int_{V} \nabla^{2} \frac{1}{\vert x -y \vert} \mathrm{d}V = ..$$
and then I got lost...

Could you please possibly expand on your explanation?
thanks a lot, much appreciated

 

[muzialis]

PS: I think I might be closer now...I just realized what was in front of me clearly, that the fraction is the Green's function for the operator nabla^2, am I right? Thanks a lot to all