So I am trying to write a report on the Helmholtz decomposition theorem on $\mathbb{R}^3$. The theorem states that under certain conditions, every vector field $\textbf{F}:U \subseteq \mathbb{R}^3 \to V \subseteq \mathbb{R}^3$ can be decomposed into a curl-free and a divergence-free component as: \begin{equation} \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} \end{equation} where $\phi$ and $\textbf{A}$ are the correspoding scalar and vector potentials of the field. So I tried to apply the theorem on the following vector fields: \begin{equation} \textbf{F}_1(x,y,z)=(-y,x,0) \quad \text{and} \quad \textbf{F}_2(x,y,z)=(x,-y,0) \end{equation} For the vector field $\textbf{F}_1$ I calculated that: \begin{equation} \nabla \times \textbf{F}_1=(0,0,2) \quad \text{and} \quad \nabla \cdot \textbf{F}_1=0 \end{equation} Therefore it is said that $\textbf{F}_1$ is a divergence-free field. In the same way I calculated that: \begin{equation} \nabla \times \textbf{F}_2=(0,0,0) \quad \text{and} \quad \nabla \cdot \textbf{F}_2=0 \end{equation} Therefore is said to be a curl-free and divergence-free field. Now, Helmholtz decomposition states that the sources of the field which is to be decomposed are defined as: \begin{equation} \phi=\nabla \cdot \textbf{F} \quad \text{και} \quad \textbf{A}=\nabla \times \textbf{F} \end{equation} which I find it to be terribly wrong since if I apply the divergence of a field decomposed by Helmholtz I would have: \begin{equation} \nabla \cdot \textbf{F}=-\nabla^2 \phi \end{equation} which is not $\phi$ but the Laplace equation of $\phi$! Is the notation wrong? I am really confused.

Moreover, if lets say I knew the curl and the divergence of $\textbf{F}_1$ in that particular example would it be possible to construct the whole field $\textbf{F}_1$? Helmholtz says that I can but when I try to do that I get (according to the wrong way): \begin{equation} \nabla \cdot \textbf{F}_1=0 \quad \text{and} \quad \nabla \cdot \textbf{F}_1=\phi \end{equation} Therefore $\phi=0$?? If that is true then $\textbf{F}_1=\nabla \times \textbf{A}=(0,0,0)$ which is obviously not $\textbf{F}_1$.

Finally, the Laplacian term which defines the harmonic field, appears into the second example and somehow the Laplacian is $\phi(x,y)=1/2(x^2-y^2)$. How is this done?

I believe that if I understand the application on the simple examples then I will be able to see the whole picture.

Thank you!


2 Answers 2


If ${\bf F} = - \nabla \phi + \nabla \times {\bf A}$, we get $\nabla \cdot {\bf F} = - \Delta \phi$. So we want to get $\phi$ by solving a Laplace equation. Then we get $\bf A$ by solving $\nabla \times {\bf A} = {\bf F} + \nabla \phi$.

In your first example, ${\bf F}_1$ is indeed divergence-free, so you can take any $\phi$ whose Laplacian is $0$. You could take $\phi = 0$, but if you prefer you can use any harmonic function. For example, let's take $\phi = x$ which makes $\nabla \phi = [1,0,0]$. So now you want $\nabla \times {\bf A} = [1-y,x,0]$. One possible solution is ${\bf A} = [xz, yz-z,0]$. There are many possible solutions: the decomposition is far from unique.

  • 1
    $\begingroup$ In order for it to be unique I have to implement boundary conditions, is that right? $\endgroup$
    – Bazinga
    Jun 16, 2016 at 22:37

First of all, the wrong thing that you think is wrong, is totally wrong! :) So, yes, you are right!

If you take the divergence and curl of the following equation

$$ \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} $$

then you will get, respectively

$$\boxed{ \begin{align} \nabla \cdot \textbf{F} &= - \nabla^2 \phi \\ \nabla \times \textbf{F} &= - \nabla^2 \textbf{A} \end{align} }\tag{1}$$

where use has been made of the identitiy

$$\nabla \times \nabla \times \textbf{A} = \nabla (\nabla \cdot \textbf{A})-\nabla^2 \textbf{A}$$

and the assumption

$$\nabla \cdot \textbf{A} =0 $$

which has been made for simplicity.

In conclusion, for a given vector field $\textbf{F}$, you should solve the Poisson equations mentioned in $(1)$ for $\phi$ and $\textbf{A}$.

  • $\begingroup$ So, the question now is: Why are the field sources defined the way they are as $\phi=\nabla \cdot \textbf{F}$, $\textbf{A}=\nabla \times \textbf{F}$? $\endgroup$
    – Bazinga
    Jun 16, 2016 at 22:23
  • 1
    $\begingroup$ Why do you think that they are defined in this way? :) They are not! :) $\endgroup$ Jun 16, 2016 at 22:29
  • $\begingroup$ Also, in order to keep track for the examples provided, for $\textbf{F}_1$ it is $\nabla \cdot \textbf{F}_1=0$ therefore I have to solve the Laplace equation $\nabla^2 \phi=0$ (or equivalently the homogenous Poisson) in order to find $\phi$. But that requires boundary conditions right? $\endgroup$
    – Bazinga
    Jun 16, 2016 at 22:29
  • $\begingroup$ @Mitscaype: Of course, it requires boundary conditions. :) $\endgroup$ Jun 16, 2016 at 22:30
  • $\begingroup$ So in order to decompose all it is needed is to apply the curl and div of the field. But if someone asks of me to build the field from those two components I have to solve the correspoding Poisson system. Right? $\endgroup$
    – Bazinga
    Jun 16, 2016 at 22:32

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