Prove or disprove the following proposition:

For any smooth plane vector field $\mathbf{H}=\left(H_x,H_y\right)$, there exist scalar potentials $\phi$, $\psi$ such that

$H_x=\frac{\partial \phi }{\partial x}+\frac{\partial \psi }{\partial y}$

$H_y=\frac{\partial \phi }{\partial y}-\frac{\partial \psi }{\partial x}$


2 Answers 2


Your proof is right (and I voted it up accordingly). But this is a result that's worth proving a few different ways, because the different ways lead to different insights. So I'll give some alternative proofs.

Proof 2 (yours is proof 1):

Taking linear combinations of the equations we're trying to solve, we get an equivalent pair of equations. $$ H_x+iH_y=\left({\partial\over\partial x}+i{\partial\over\partial y}\right)(\phi-i\psi) $$ $$ H_x-iH_y=\left({\partial\over\partial x}-i{\partial\over\partial y}\right)(\phi+i\psi). $$ If we define some shorthand, $$ \partial_\pm={\partial\over\partial x}\pm i{\partial\over\partial y}, $$ $$ f_{\pm}=\phi\pm i\psi, $$ $$ H_{\pm}=H_x\pm iH_y, $$ we can write these as $$ H_+=\partial_+f_-, $$ $$ H_-=\partial_-f_+. $$ Now we can think of these as two uncoupled equations for the two unknowns $f_\pm$. Once we've solved for them, we can get $\phi,\psi$ from them. So we have to prove that each of these two equations has a solution.

To solve for $f_-$, we start by finding a solution $g$ to $$ \nabla^2g=H_+. $$ (This is Poisson's equation, so it has a solution.) Then set $f_-=\partial_-g$. Because of the identity $$ \nabla^2=\partial_+\partial_-, $$ we have $$\partial_+f_-=\partial_+\partial_-g=\nabla^2g=H_+, $$ which is what we wanted.

A similar construction works for $f_-$. In fact, every step of the argument for $f_-$ is just the complex conjugate of the corresponding step for $f_+$ (as long as the $H$'s are real). That's the way it has to be for $\phi,\psi$ to end up real.

I have a personal reason I like this argument. I work a lot with maps of linear polarization, which are spin-2 fields rather than spin-1 (vector) fields. The equivalent of the Helmholtz theorem for spin-2 fields is called the $E$-$B$ decomposition (in the cosmology literature anyway). The above argument generalizes in a nice way to spin-2 (and presumably higher spins).

Proof 3:

Intuitively, it seems like we ought to be able to get the 2-D result from the 3-D Helmholtz theorem, and it turns out that we can. Here's one way to think about it.

Extend the vector field ${\bf H}$ to be a function of $(x,y,z)$: $$ {\bf H}_{3D}(x,y,z)={\bf H}(x,y). $$ For simplicity, imagine that $z$ extends only over a finite interval, say 0 to $2\pi$, and has periodic boundary conditions (i.e., points with $z=0$ are identified with those with $z=2\pi$). It's not necessary to do this, but it makes things cleaner.

Helmholtz's theorem says that there are functions $\phi,{\bf G}$ such that $$ {\bf H}_{3D}=\nabla\phi+\nabla\times{\bf G}. $$ If we knew that $\phi,{\bf G}$ were independent of $z$ and that ${\bf G}$ pointed in the $z$ direction, we'd be done. But we don't (yet) know that.

Here's the trick. Both $\phi$ and ${\bf G}$ are nice, smooth functions, so they can be expanded in Fourier series in $z$: $$ \phi(x,y,z)=\sum_{n=-\infty}^\infty \phi_n(x,y)e^{inz}, $$ and similarly for ${\bf G}$. Substituting these into the previous equation, we get $$ {\bf H}_{3D}=\sum_n\left(\nabla(\phi_ne^{inz})+\nabla\times({\bf G}_ne^{inz})\right). $$ Each term on the right has $z$-dependence of the form $e^{inz}$ (even after writing out the derivatives). But the left side is independent of $z$. By the uniqueness of Fourier series, it follows that all the terms with $n\ne 0$ on the right vanish! So $$ {\bf H}_{3D}=\nabla\phi_0+\nabla\times{\bf G}_0. $$ We know that $\phi_0,{\bf G}_0$ depend only on $(x,y)$, not $z$. Setting $\phi(x,y)=\phi_0(x,y)$ and $\psi(x,y)=G_z(x,y)$ gives the desired result.

This one could almost certainly be phrased more compactly in more formal mathematical language, probably involving symmetry groups. The basic idea is that the problem we're solving is invariant under translations in $z$, and the operations we're performing (in some sense) "commute" with these translations. That means it's possible to find a solution that respects that symmetry.

"Proof" 4:

The mathematicians won't like this one, but I think it's a nice way to think about it anyway.

Take 2-D fourier transforms of everything in sight: $$ {\bf H}(x,y)=\int \tilde{\bf H}(k_x,k_y)e^{i{\bf k}\cdot{\bf r}}d^2k, $$ and similarly for $\phi,\psi$. When you do, the equations you're trying to solve decouple -- that is, each value of ${\bf k}$ can be solved independently: $$ \tilde{\bf H}({\bf k})=i{\bf k}\tilde{\phi}({\bf k})+i\hat{\bf z}\times{\bf k}\tilde\psi({\bf k}). $$ (There may be a sign error in the above, but it's "morally" true.)

This equation can be solved algebraically for each ${\bf k}$. In fact, the solution has a nice physical meaning: decompose $\tilde{\bf H}$ into components parallel and perpendicular to ${\bf k}$. $\tilde\phi$ is the parallel component, and $\tilde\psi$ is the perpendicular component.

I like this one because it too generalizes to the spin-2 case nicely, and provides better intuition than anything else I can think of about the "meaning" of the $E$-$B$ decomposition for spin-2 fields.

The reason I say the mathematicians won't like it is because not everything has a Fourier transform, and the convergence of Fourier transforms isn't normal pointwise convergence. But for physics applications, it's generally a fine way to think about things.

  • $\begingroup$ Thanks for a very informative answer. I'm marking yours instead of mine as the answer. $\endgroup$
    – a06e
    May 28, 2011 at 22:32

We can certainly find a $\psi$ that solves the following partial differential equation

$\frac{\partial ^2\psi }{\partial x^2}+\frac{\partial ^2\psi }{\partial y^2}=\frac{\partial H_x}{\partial y}-\frac{\partial H_y}{\partial x}$

It then follows that

$\frac{\partial }{\partial y}\left(H_x-\frac{\partial \psi }{\partial y}\right)-\frac{\partial }{\partial x}\left(H_y+\frac{\partial \psi }{\partial x}\right)=0$

This shows that the vector field $(H_x-\frac{\partial \psi }{\partial y},H_y+\frac{\partial \psi }{\partial x})$ is curl-free/solenoidal. If the domain is {\bf simply connected} it follows that there's a scalar field $\phi$ such that

$H_x-\frac{\partial \psi }{\partial y}=\frac{\partial \phi }{\partial x}$

$H_y+\frac{\partial \psi }{\partial x}=\frac{\partial \phi }{\partial y}$


$H_x=\frac{\partial \phi }{\partial x}+\frac{\partial \psi }{\partial y}$

$H_y=\frac{\partial \phi }{\partial y}-\frac{\partial \psi }{\partial x}$.

If the domain isn't simply connected then "curl-free/solenoidal vector field" doesn't always imply "the vector field is conservative".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.