Three dimensional dirac function

[johnkay]

Homework Statement

Show that if $$r = \sqrt{x^2 + y^2 + z^2}$$ then
$$\nabla^2 \left( \frac{1}{r} \right) = -4 \pi \delta^3(r)$$

Homework Equations

I've heard Green's theorem should help me... not quite certain how.

The Attempt at a Solution

I took the divergence of the left hand side and got

$$\nabla \left( \frac{1}{r} \right) = \frac{-1}{r^2}\hat{r}$$

I guess i should stick $$\nabla \cdot \frac{-1}{r^2}\hat{r}$$ into the divergence theorem/green's function but how will that show me I am dealing with a dirac delta function in three dimensions?

 

[mighty2000]

Thank you for your inquiry. You are correct in thinking that Green's theorem can help you solve this problem. Here is a step-by-step solution to help you understand how to approach it:

1. Begin by taking the Laplacian of the function 1/r:

\nabla^2 \left( \frac{1}{r} \right) = \nabla \cdot \nabla \left( \frac{1}{r} \right) = \frac{1}{r} \nabla^2(1) + \nabla \cdot \left( \frac{-1}{r^2} \nabla r \right)

2. Simplify the first term using the identity \nabla^2(1) = 0:

\nabla^2 \left( \frac{1}{r} \right) = \nabla \cdot \left( \frac{-1}{r^2} \nabla r \right) = \frac{-1}{r^2} \nabla^2(r) + \nabla \cdot \left( \frac{-1}{r^2} \nabla r \right)

3. Use the vector identity \nabla \cdot (f \nabla g) = f \nabla^2 g + \nabla f \cdot \nabla g to simplify the second term:

\nabla^2 \left( \frac{1}{r} \right) = \frac{-1}{r^2} \nabla^2(r) + \frac{-1}{r^2} \nabla \cdot (\nabla r) = \frac{-1}{r^2} \nabla^2(r)

4. Now, use the definition of the Laplacian in spherical coordinates \nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta}