# Helmholtz theorem: “There is no function that has zero divergence and zero curl everywhere and goes to zero at infinity”. How do I know this?

I am trying to proof Helmholtz’s theorem and I am currently using David J. Griffith’s book on electrodynamics to do this. Within the book’s Appendix B, (I have not finished the book, and am on the last sub chapter of the first chapter, the theory of vector fields) I have read along and understood until, “However, it so happens that there is no function with zero divergence and zero curl everywhere and goes to zero at infinity. (See sect 3.1.5)”.

Where sect 3.1.5 is “Boundary conditions and Uniqueness theorem”. Which gives the first uniqueness theorem as “The solution to Laplace’s equation in some volume V is uniquely determined if V is specified on boundary surface S”. While I understand individually what they mean, I don’t see how it proves the original quote. Any help is appreciated.

• Does this help: If $\text{curl} \mathbf u = 0$ then $\mathbf u = \text{grad} \phi$ for some scalar $\phi$ and from $\text{div} \mathbf u = 0$ you get $\text{div grad} \phi = \nabla^2 \phi =0$. May 5 at 21:58
• That makes a lot of sense yeah, thank you. And by uniqueness theorem that it applies to all functions I am guessing. Thank you May 5 at 22:11
• Though I have a question, does curl u = 0 enforces u = grad phi. Where phi is some scalar? May 5 at 22:16
• yes, $\text{curl} \mathbf = 0$ is equivalent to that $\oint \mathbf u \cdot d\ell =0$ for any loop from which you can uniquely define $\phi(\mathcal P) =\int_{\mathcal P_0}^{\mathcal P} \mathbf u \cdot d\ell$ for some fixed $\mathcal P_0$ and variable $\mathcal P$, and finally $\frac{d\phi}{d\ell}=\text{grad} \phi =\mathbf u$ May 5 at 22:28
• I'm not good at English, but I thought that Griffith's English in the first quote might be misleading. I think it should be "there is no non zero function with zero ..." instead of "there is no function with zero ..."? May 5 at 22:33

Helmholtz 3d/4d theorem on electrodynamics is a special case of a general theorem:

1: let $$A = \sum A_i dx^i$$ a 1-form with $$d A = \partial_k A_i dx^k\wedge dx^i =0$$ ($$R^3 : \nabla \times A=0$$) everywhere in R^n, then all path integrals from the origin 0 to any point x is independent of the path, integrals over closed paths vanish.

2: The path integral defines a potential $$\Phi$$ such that $$A = d \Phi =\sum \partial_k \Phi dx^k$$.

3: Now let $$R^n$$ be a space with an euclidean scalar product $$< dx^i, dx^k > \text{-->} \delta^{ik}$$

and let $$dV = dx^1 \wedge ... \wedge dx^n$$ be the antisymmetric volume n-form.

Then one can define the metric dual form of A by taking the scalar product over the wedge product, commuting all factors to the left and contract

$$< dV, \sum A_i dx^i > == \sum < dx^1 .. dx^n , A_i dx^i > = \sum (-1)^k A_k \partial_{dx^k}dV$$

where the partial derivative is short for "delete_that" in monomials.

So from the 1-form, conventionally seen in physics as a vector field by its component vector, we get a n-1-form, that forms a flow over surfaces. Its differential n-form is its production density. Again taking its dual, it is a scalar function, the sum over the second derivatives.

4:$$< dV, d < dV, A > > = \nabla . A = \Delta \Phi$$

The n-form derivative $$$$ is called the divergence of A by the genereral formula

$$\nabla . B = \lim_{dV -> 0} 1/dV \int_{\partial_x^k } B . dO^k$$,

that is the flow outward of the surface equals the volume integral over the production volume density. Taking the point limit, its the density. By this definition, B is not supposed to be differentiable; only the surface integrals must exist and have a point limit.

With these definitions the theorem reads:

Let A be a closed differential form in R^n everywhere: $$partial_i A_k == \partial_k A_i =0$$, then it is exact, ie. the derivative of a 2-times continuously differentiable function $$\Phi$$.

Any 2-times differentiable function $$\Phi$$ from $$R^n \to C$$ has $$\partial_{i,k}\Phi==\partial_{k,i}\Phi$$. $$\nabla \times A=0$$ is only a very weak smoothness condition to guarantee the existence of $$\Phi$$.

This is important, because in the physics of Maxwell, potential functions are distributions, generally, having their sources in its singular points, lines and surfaces.

So the question reduces to: if a n-1-form over surfaces is the dual of the derivative of a 2-differentiable potential, then it is the density of a potential with $$\Delta \Phi ==0$$.

5: The algebraic real kernel of $$\Delta$$ as a differential map are the linear functions.

Liouville: A harmonic bounded function is constant.

Hilbert spaces: The dimension of the space of square integrable complex functions with $$\Delta \psi=0$$ is zero.

But: In distributional spaces of continous functions eg, a surface of a jump of the derivative produces a delta-density of the Laplacian.

Standard example the capacitor with $$\Phi={{ 0,x<0}, {x, x<1}, {1, x>=1} }$$

5: On a flat torus, that is a unit cube in $$R^n$$ with opposite sides identified, all this does not apply because of toplogy: there are n linear independent constant vector fields along the n- axis.

This fact is a common case in electrodynamics: On a the superconducting torus you have two situations: The current as a a coil, field inside, or current around the axis, field outside. Both have rot A=0 div A=0, as well as rot j=0, div j=0.

1. The complex case is especially important, because of its intimate relation to complex analysis.
$$\partial_{(x+ I y)} f(x + I y) == f'(x+ I y)$$ if $$f$$ is independent of $$x-I y$$ $$\forall x,y \partial_{(x+ I y)} f(x + I y) = 0$$

Liouville: A complex differentiable bounded function is constant.

Proof: $$\partial_{x+ I y,{x-Iy}} f(x+I y) =\Delta f(x+I Y) = 0$$

Caveat:$$\Delta Exp[ k x + I k y] ==0$$ and so $$\Delta Exp[ k x + I k y] Exp[-(x^2+y^2)/s^2]/s$$

can be made arbitrarily small in a very large volume

• I suggest you use \langle and \rangle for angle brackets May 6 at 8:00