Examining the article on Wikipedia Helmholtz decomposition, compatible with the explanations of the book Introduction to Electrodynamics $4^{\mathrm{th}}$ edition David J. Griffiths §1.6 the theory of vector fields (subparagraph 1.6.1 The Helmholtz Theorem), Helmholtz's theorem, is also known as the fundamental theorem of vector calculus.

As an irrotational vector field has a scalar potential and a solenoidal vector field has a vector potential, the Helmholtz decomposition states that a vector field (satisfying appropriate smoothness and decay conditions) can be decomposed as the sum of the form ${\displaystyle -\operatorname {grad} \Phi +\operatorname {curl} \mathbf {A}}$ , where $\Phi$ is a scalar field, called scalar potential, and $A$ is a vector field, called a vector potential.

What is the reason or the motivation of the use of this theorem? What are its advantages?

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    $\begingroup$ I think it woud be fair to put into the post at least some equivalent of the work, you expect from the answerer. Please define here what CH equations are and what have you read about them. $\endgroup$ – Cryo Jul 16 '19 at 1:55
  • $\begingroup$ @Cryo I've edited many times. I hope that this version is clear. $\endgroup$ – Sebastiano Jul 18 '19 at 8:58
  • $\begingroup$ When I am more healthy, I will try to change my question as much as possible. Thank you very much. $\endgroup$ – Sebastiano Jul 18 '19 at 19:29
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    $\begingroup$ en.wikipedia.org/wiki/Helmholtz_decomposition is not a bad start. The second section shows that Helmholtz theorem allows to separate and arbitrary vector field (e.g. electric field $\mathbf{E}$) into two other fields that are determined by the boundary values of the initial field, as well as by its curl and divergence. So in case of electric field the boundary value is $\mathbf{E}\to 0\,@\,\infty$, the curl is zero, and divergence is proportional to charge density, hence $\mathbf{E}=-\boldsymbol{\nabla}V$, for scalar potential $V$ $\endgroup$ – Cryo Jul 19 '19 at 1:24
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    $\begingroup$ Where did you read about Clebsch in connection with the Helmholtz decomposition theorem? $\endgroup$ – DanielC Jul 19 '19 at 15:43

This is answers follows from the comments to the initial text. First of all I think there is one major problem with the question:

It is too general. Helmholtz Decomposition (HD) is a mathematical technique. There is no need to use it, in principle, but often it can make life easier. The best way to learn about the cases where HD is useful is to do specific examples, rather than ask general questions. For example, there are books by Griffiths and Jackson (Classical Electrodynamics). There are also more rigorous treatments by Morse & Feshbach for example. Going towards even more maths and less physics, the Helmholtz theorem is treated by Arfken & Weber (Mathematical Methods...).

Having said this, I can try to give my best personal view why HD is useful. Lets us take the statement of the theorem by Arfken & Weber (Ch3. Helmholtz's Theorem).

* A vector field is uniquely specified by giving its divergence and its curl within a simply connected region and its normal component on the boundary. *

So let us look at the electrostatic Maxwell's Equations.

$$\boldsymbol{\nabla}\cdot \mathbf{E}=\frac{\rho}{\epsilon}$$


Helmholtz theorem tells us that these equations are sufficient to fully and uniquely constrain the solution for electric field ($\mathbf{E}$) given boundary conditions. So any solution you can find for this problem is the only correct one.

Better still the Wikipedia link I gave (https://en.wikipedia.org/wiki/Helmholtz_decomposition) goes further and states that electric field can be written as:

$$\mathbf{E}=-\boldsymbol{\nabla}\phi + \boldsymbol{\nabla}\times\mathbf{K}$$

and then:

$$\phi\left(\mathbf{r}\right) = \frac{1}{4\pi}\int_V d^3 r' \frac{\boldsymbol{\nabla'}\cdot\mathbf{E}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|} - \frac{1}{4\pi}\oint_{\partial V} d^2 r'\frac{\mathbf{\hat{n}'}\cdot\mathbf{E}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}$$

Where $V$ is the simply connected region. Provided $\mathbf{E}$ vanishes at the surface of $V$ (i.e. at $\partial V$) faster than $1/r$ (here is your boundary condition), we get:

$$\phi\left(\mathbf{r}\right) = \frac{1}{4\pi}\int_V d^3 r' \frac{\rho\left(\mathbf{r}'\right)/\epsilon}{\left|\mathbf{r}-\mathbf{r}'\right|} $$

Similar treatment exists for $\mathbf{K}$, see link. So using Helmholtz decomposition you can find the electrostatic field.

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  • $\begingroup$ Finally :-) I am very happy with an answer and I thank you with all my heart for your patience and for the work that you have inserted. Thank you very much and I really hope that something will change in this site. Thank you very much. I'm editing the math part, hoping you won't get angry. $\endgroup$ – Sebastiano Jul 20 '19 at 11:32
  • $\begingroup$ Glad it helped :-) $\endgroup$ – Cryo Jul 20 '19 at 12:56
  • $\begingroup$ I too am happy and with all my heart I am happy to find people who are willing to understand me. If you need me on LaTeX...available :-) I was also happy to increase your score :-) $\endgroup$ – Sebastiano Jul 20 '19 at 19:15

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