# Why $\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf{u})\neq\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{u})-\nabla^2\mathbf{u}$?

It is well known that for some vector field $$\mathbf{u}$$ the following holds: $$\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf{u})=\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{u})-\nabla^2\mathbf{u}.$$

Let's consider the following vector field in cylindrical coordinates with the unit vectors $$\hat{\mathbf{r}}, \hat{\boldsymbol\phi}, \hat{\mathbf{z}}$$:

$$\mathbf{A}=0\hat{\mathbf{r}}+1\hat{\boldsymbol\phi}+0\hat{\mathbf{z}}.$$

For cylindrical coordinate frame we know that divergence, curl, and Laplacian are written as respectively:

$$\boldsymbol\nabla\cdot\mathbf{A} = \frac{1}{r}\frac{\partial(rA_r)}{\partial r} + \frac{1}{r}\frac{\partial A_\phi}{\partial \phi} + \frac{\partial A_z}{\partial z},$$ $$\boldsymbol\nabla\times\mathbf{A}=\left(\frac{1}{r}\frac{\partial A_z}{\partial\phi}-\frac{\partial A_\phi}{\partial z}\right)\hat{\mathbf{r}}+\left(\frac{\partial A_r}{\partial z}-\frac{\partial A_z}{\partial r}\right)\hat{\boldsymbol\phi}+\left(\frac{1}{r}\frac{\partial(rA_\phi)}{\partial r}-\frac{1}{r}\frac{\partial A_r}{\partial \phi}\right)\hat{\mathbf{z}},$$ $$\nabla^2 =\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2}{\partial\phi^2}+\frac{\partial^2}{\partial z^2}.$$

On the one hand, for the left side using these formulae we have:

$$\boldsymbol\nabla\times\hat{\boldsymbol{\phi}} = \frac{\hat{\mathbf{z}}}{r},$$ and

$$\boldsymbol\nabla\times\frac{\hat{\mathbf{z}}}{r}=\frac{\hat{\boldsymbol\phi}}{r^2}.$$

However, on the other hand, for the right side we shall have zeros, because all partials of $$1$$ are equal to zero. Where am I wrong?

• The Laplacian you wrote is only true for scalar fields. Aug 24, 2020 at 13:35
• Is the Laplacian for vector field and for scalar field different? I thought I could use the same form of Laplacian for each component.
– lR55
Aug 24, 2020 at 13:36
• In Cartesian coordinates one can just apply it to each direction. However in general it is defined using your general formula. Aug 24, 2020 at 13:38
• In cartesian coordinates, they look very similar: $\nabla^2 \mathbf{A} = \sum_{i=1}^n \frac{\partial^2\mathbf{A}}{\partial x_i^2} = \sum_{i,j=1}^n\frac{\partial^2A_j}{\partial x_i^2} \mathbf{e}_j$; but as you change to other coordinate systems, "weird things" happen. The "problem" is that the basis vectors also change from point to point, and you have to keep track of this during the differentiation (and for me this was all very confusing until I learnt some basic Riemannian geometry and covariant differentiation). See this for example Aug 24, 2020 at 13:39
• @peek-a-boo, I'm aware that basis vectors are changing during the differentiation for curvilinear orthogonal coordinate frames (never used non-orthogonal, though). So, when deriving Laplacian for scalar field, I saw that this weird things (for changing basis vectors) was accounted. So, I thought the form of Laplacian I wrote could be used for each component of a vector field.
– lR55
Aug 24, 2020 at 13:45

$$\nabla^2 =\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2}{\partial\phi^2}+\frac{\partial^2}{\partial z^2}$$ is supposed to be used for a scalar field only.

For the vector field, however, it is easy to check that Laplacian is given by:

$$\nabla^2 \mathbf{u}= \hat{\boldsymbol\rho}\left(\nabla^2 u_r-\frac{u_r}{r^2}-\frac{2}{r^2}\frac{\partial u_\phi}{\partial \phi}\right)+\hat{\boldsymbol\phi}\left(\nabla^2 u_\phi-\frac{u_\phi}{r^2}+\frac{2}{r^2}\frac{\partial u_r}{\partial \phi}\right)+\hat{\mathbf{z}}\nabla^2u_z.$$

For the vector field given, we have $$u_r=u_z =0, u_\phi = 1$$, and Laplacian will be equal to $$\nabla^2 \hat{\boldsymbol{\phi}}=-\frac{\hat{\boldsymbol{\phi}}}{r^2}.$$

So, it means that $$\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf{u})=\frac{\hat{\boldsymbol{\phi}}}{r^2}=\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{u})-\nabla^2\mathbf{u}=-(-\frac{\hat{\boldsymbol{\phi}}}{r^2})$$, so the identity holds.