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In a physics paper I have encountered the following claim:

$\vec{\nabla}\times\vec{\nabla}\times(e^{i\vec{k}\cdot\vec{x}}\vec{\omega})=k^2(e^{i\vec{k}\cdot\vec{x}}\vec{\omega})$

where $\omega=\vec{\nabla}\times\vec{v}$ is a divergence-free vector field, which means that:

$\vec{\nabla}\times\vec{\nabla}\times\vec{\omega}=\vec{\nabla}(\vec{\nabla}\cdot\vec{\omega})-\nabla^2\vec{\omega}=-\nabla^2\vec{\omega}$

The proof of this equality must be really simple, but no matter what I attempt, I am unable to arrive at the desired result. I have tried to apply the following two properties:

  • $\vec{\nabla}\times\vec{\nabla}\times\vec{F}=\vec{\nabla}(\vec{\nabla}\cdot\vec{F})-\nabla^2\vec{F}$

  • $\nabla\cdot(\phi\vec{F})=\phi(\vec{\nabla}\cdot\vec{F})+\vec{F}\cdot(\vec{\nabla}\phi)$

I have tried to apply them in this order, and then in reverse order, but no luck either way. Is the statement in the paper really true? It looks very simple and makes me wonder if there is a very simple way to prove it that I'm missing. Any advice or tips on how to proceed would be really appreciated!

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    $\begingroup$ Is the curl taken with respect to $\vec x$? (Rather than $\vec k$?) $\endgroup$
    – Alex Ortiz
    Nov 22, 2023 at 17:44
  • $\begingroup$ Why not just write this out in components instead of continuing to use identities? None of the definitions are too bad. You also need to understand what it means to apply a Laplacian to a vector quantity. $\endgroup$
    – podiki
    Nov 22, 2023 at 17:46
  • $\begingroup$ @AlexOrtiz Yes, that's right, the derivatives in the $\nabla$ operator are with respect to $\vec{x}$. Sorry, I forgot to specify that. $\endgroup$ Nov 22, 2023 at 17:47
  • $\begingroup$ @podiki Since the authors of the paper stated that the equality holds because $\vec{\nabla}\times\vec{\nabla}\times\vec{\omega}=-\nabla^2\vec{\omega}$, I thought it should be easy to prove the equality by using vector identities. Working in components with the curl of the curl seems a bit daunting... $\endgroup$ Nov 22, 2023 at 17:49
  • $\begingroup$ I assume you can apply the product rule for the curl in succession: proofwiki.org/wiki/Product_Rule_for_Curl $\endgroup$
    – Alex Ortiz
    Nov 22, 2023 at 17:51

1 Answer 1

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Let $\omega = \langle \omega_1,\omega_2,\omega_3\rangle$. Using the first identity you have listed

$$\begin{align}\nabla \times \nabla \times (e^{ikx}\omega) = \nabla(\nabla \cdot (e^{ikx}\omega)) - \Delta(e^{ikx}\omega)\end{align}$$ Next, let's look at the just the first component. We calculate the first term using your second identity

$$\begin{align}(\nabla(\nabla \cdot (e^{ikx}\omega)))_1 &= (\nabla(e^{ikx}\nabla\cdot \omega+\omega\cdot \nabla(e^{ikx})))_1\\ &= (\nabla(\omega\cdot ik e^{ikx}))_1\\ &= e^{ikx}\left((\partial_{x_1}\omega)\cdot k -k_1 \omega\cdot k \right)\end{align}$$

Meanwhile, for the first component of the second term we find $$\begin{align}\Delta(e^{ikx}\omega)_1 = \Delta(e^{ikx}\omega_1) = e^{ikx}\left(\Delta\omega_1 + 2ik\cdot\nabla\omega_1-k^2\omega_1\right)\end{align}$$

Notice that this will only give $k^2\omega_1$, if

  1. $\omega$ is constant with respect to $x$
  2. $\omega\cdot k=0$.

My guess is that there is a conflation between $\omega(x)$ and its Fourier transform $\omega(k)$. Notice that if $\omega(x)$ is divergence free, then the modes of the Fourier transform satisfy $k\cdot\omega(k)=0$ as desired and would be $x$ independent.

Now suppose that $\omega(k)$ is such that $k\cdot\omega(k)=0$ then we can calculate

$$\begin{align}\nabla \times \nabla \times (e^{ikx}\omega(k)) &= \nabla(\nabla \cdot (e^{ikx}\omega(k))) - \Delta(e^{ikx}\omega(k))\\ &=i\nabla(\omega(k)\cdot ke^{ikx}) - \omega(k)\Delta(e^{ikx})\\ &= k^2e^{ikx}\omega(k)\end{align}$$

Edit: To address the comment below which requires some extra work.

  1. Yes the terminology and notation is confusing, but calling a function and its Fourier transform the same name is a common abuse of notation in physics literature.
  2. It is important because the fact that $\nabla \cdot \omega(x) = 0$ translates exactly into $k\cdot\omega(k) = 0$ where

$$\omega(k) = \int e^{-ikx}\omega(x)dx$$ 3. To observe this we calculate

$$\begin{align} k\cdot\omega(k)&= \int e^{-ikx}k\cdot\omega(x)dx\\ &=\int i\nabla(e^{-ikx})\cdot\omega(x)dx\\ &=-i\int e^{-ikx}\nabla\cdot\omega(x)dx = 0\end{align}$$

where I have used integration by parts and assumed that $\omega(x)$ vanishes at infinity so that the boundary terms do not contribute.

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  • $\begingroup$ Thank you for your answer! It makes sense, however, the way the paper phrases the assumption leaves me confused. They literally say that, since $\vec{\omega}$ is divergence free, $\vec{\nabla}\times\vec{\nabla}\times\vec{\omega}=-\nabla^2\vec{\omega}$, so that $(\vec{\nabla}\times)^{-2}(e^{i\vec{k}\cdot\vec{x}}\vec{\omega})=k^{-2}(e^{i\vec{k}\cdot\vec{x}}\vec{\omega})$. But what does the fact that $\vec{\nabla}\times\vec{\nabla}\times\vec{\omega}=-\nabla^2\vec{\omega}$ have to do with anything then? And why is $\vec{k}\cdot\vec{\omega}(\vec{k})=0$ if $\vec{\omega}$ is divergence-free? Thanks! $\endgroup$ Nov 23, 2023 at 9:57
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    $\begingroup$ I edited my original answer to address these additional questions. $\endgroup$
    – podiki
    Nov 23, 2023 at 14:34

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