Dirac delta, generalizations of vector calculus and sigh vagueness

[nonequilibrium]

Although I am an aspiring physicist, I cannot cope with the physicist's love for vagueness when it comes to yielding math. Exactness is simply not a luxury that can be ignored, certainly not in theoretical physics.

But okay, I realize the dirac delta function can be made exact by the use of, for example, distributions, although the measure-theoretic concept seems prettier to me, so maybe I should delve into that. Anyway, something I cannot make sense of, even with distributions:

$$\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) = 4 \pi \delta^3(\vec r)$$

(source: p70 from Griffiths' Electrodynamics)

I suppose it can't be anything else than the definition of the gradient of the function between parentheses in such a way that the good ol' divergence theorem

$$\int\int\int_V \vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) \mathrm dV = \int \int_{\partial V} \frac{\vec{r}}{r^3} \cdot \mathrm d \vec S$$

is once again true, because of course normally the theorem is simply not defined for the function (and domain) at hand, so we can view the definition as a generalization of the divergence theorem by use of distributions.

But one thing is bugging me: is it clear that the definition is well-defined? In the sense that: how do we know that there is not another situation where $$\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right)$$ can turn up as the limiting, normally non-defined case of a certain equality (just like the divergence theorem), in which case we no longer have a choice of definition and in which case we must substitute the dirac delta function as defined? Do we know that the same definition will extend also that equality (again: by use of distributions)?

Thank you! I'm learning to cope... (although I don't know if that's good or not)

 

[jostpuur]

It is important to understand, that if we are using ordinary integrals and derivatives, then following three equations are true (with some additional assumptions)

$$\int \big(\nabla f(x)\big) \cdot \frac{x - x'}{\|x - x'\|^3} d^3x = -4\pi f(x')\quad\quad\quad\quad (1)$$

$$\nabla \cdot \int f(x') \frac{x - x'}{\|x - x'\|^3} d^3x' = 4\pi f(x)\quad\quad\quad\quad (2)$$

$$\int f(x')\Big(\nabla\cdot \frac{x - x'}{\|x - x'\|^3}\Big) d^3x' = 0\quad\quad\quad\quad (3)$$

The equation (3) is the one that is the most poorly understood. It is a very simple thing, but even it cannot be understood if not clearly stated.

But okay, I realize the dirac delta function can be made exact by the use of, for example, distributions, although the measure-theoretic concept seems prettier to me, so maybe I should delve into that. Anyway, something I cannot make sense of, even with distributions:

$$\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) = 4 \pi \delta^3(\vec r)$$

(source: p70 from Griffiths' Electrodynamics)

The rigor definition of distributions and Dirac measure are not going to help you in understanding how to prove equations (1) and (2). The proofs can be carried out with ordinary calculus, and IMO a convergence result from the measure theory can be handy too.

 

[nonequilibrium]

How can it be proven? The left-hand side of the equation does not have any meaning, does it?

 

[jostpuur]

How can it be proven? The left-hand side of the equation does not have any meaning, does it?

You mean, how can this

$$\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) = 4 \pi \delta^3(\vec r)$$

be proven?

It should be interpreted as a notation for these equations:

$$\int \big(\nabla f(x)\big) \cdot \frac{x - x'}{\|x - x'\|^3} d^3x = -4\pi f(x')\quad\quad\quad\quad (1)$$

$$\nabla \cdot \int f(x') \frac{x - x'}{\|x - x'\|^3} d^3x' = 4\pi f(x)\quad\quad\quad\quad (2)$$

The heuristic idea is in doing integration by parts in (1) and changing the order of derivation and integration in (2).

Before you can prove an equation to be true, you must know what it means. Your equation means my equation (1) (and maybe (2) too). So you should focus in trying to prove the equations (1) and (2). I would recommend (1) first, and (2) second.

 

[nonequilibrium]

If I multiply both sides of my equation with f(x) and then integrate them both, I get the RHS of (2) but the LHS of (3), no?

 

[jostpuur]

If I multiply both sides of my equation with f(x) and then integrate them both, I get the RHS of (2) but the LHS of (3), no?

It depends what your equations left side means. If it is an ordinary derivative, then your equation is wrong. If it is a distributional derivative, then your equation is right but it needs lot of interpreting.

The derivatives in my equations are ordinary derivatives, not distributional derivatives. So if your equation has distributional derivative, you don't get the left side of my equation (3).

If you define a function like this:

$$\phi(x) = \left\{\begin{array}{l}<br /> +\infty,\quad x=x_0\\<br /> 0,\quad x\neq x_0\\<br /> \end{array}\right.$$

then

$$\int\limits_{\mathbb{R}^n} \phi(x) d^nx = 0$$

Ok? The infinite value at one point does not affect the integral.