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I am stucked with the following demonstration. The vector laplacian formula is: $Δa = ∇(∇a) - ∇×(∇×a)$ , where $a$ is a vector field. I have to demonstrate that the vector laplacian in cartesian coordinates is: $Δa = (∇∇ax)ux +(∇∇ay)uy +(∇∇az)uz$ where: $ux$,$uy$ and $uz$ are the unit vectors, and $∇∇$ stands for nabla's operator to square. Thanks in advance.

victor26567
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\begin{eqnarray} \nabla^2 a &=& \nabla (\nabla \cdot a) - \nabla \times (\nabla \times a) \\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- \varepsilon_{ijk} \varepsilon_{kbc} \partial_j \partial_b a_c\right] \\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- (\delta_{ib} \delta_{jc}-\delta_{ic}\delta_{jb}) \partial_j \partial_b a_c\right]\\ &=& \hat{\mathrm{e}}_i \left[ \partial_i \partial_j a_j- \partial_j \partial_i a_j+ \partial_j \partial_j a_i\right] \\ &=& \hat{\mathrm{e}}_i \left[\partial_j \partial_j a_i\right] \, . \end{eqnarray}

secavara
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  • I dont see the identity... – victor26567 Feb 20 '18 at 14:57
  • Ok, can you be more explicit about what you don't understand? – secavara Feb 20 '18 at 15:14
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    I suppose in your response, that êi stands for unit vector and i is the subindex? And another thing that I dont understand it is the derivates that you put, I mean, you have to have a derivative parcial, respect to x,y and z. And the last thing is the letters that you use (delta and epsilon), where they come from? – victor26567 Feb 20 '18 at 15:17
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    Oh ok. So here I am using the Einstein summation convention, see the wiki link. This means that when there are repeated indices there is a sum over such index. Here each index runs from 1 to 3 representing the Cartesian directions $x$, $y$ and $z$. The $\delta$ is a Kronecker delta and the $\varepsilon$ is the Levi-Civita symbol. These are useful in describing these vector identities (and tensors in general) in terms of their components. ... – secavara Feb 20 '18 at 15:22
  • It would be useful for you to familiarize with these techniques and, in my personal experience, they can be tricky to follow when you first encounter them, but once you get some practice you'll have a very useful tool in your toolbox. The $\hat{\mathrm{e}}_i$ are indeed Cartesian unit vectors. Take a look of the links, I would say it's unlikely I could teach you these in an answer or series of comments as they do require practice. – secavara Feb 20 '18 at 15:25
  • and can you please put it in the traditional way? haha. I didnt have idea about einstein notation, Im a student haha. If you could please put that ina more familiar way I would appreciate it a lot. – victor26567 Feb 20 '18 at 15:26
  • Let us continue this discussion in chat. – victor26567 Feb 20 '18 at 15:29
  • Sorry @victor26567 but I currently don't have the time to go through these topics. Maybe you can ask another question here or in the physics StackExchange regarding this or look for help in other references. I also learned about these as a student. – secavara Feb 20 '18 at 15:34
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I was also interested in this question so I am writing down what I came with. I've read on wiki that it can be seen as a particular case of the vector triple product formula (Lagrange's formula):

$\mathbf{u}\times (\mathbf{v}\times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\ \mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\ \mathbf{w}$

The demonstration (also on wiki) for first coordinate is:

$\begin{align} (\mathbf{u} \times (\mathbf{v} \times \mathbf{w}))_x &= \mathbf{u}_y(\mathbf{v}_x\mathbf{w}_y - \mathbf{v}_y\mathbf{w}_x) - \mathbf{u}_z(\mathbf{v}_z\mathbf{w}_x - \mathbf{v}_x\mathbf{w}_z) \\ &= \mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - \mathbf{w}_x(\mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z) \\ &= \mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - (\mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z)\mathbf{w}_x + (\mathbf{u}_x\mathbf{v}_x\mathbf{w}_x - \mathbf{u}_x\mathbf{v}_x\mathbf{w}_x) \\ &= \mathbf{v}_x(\mathbf{u}_x\mathbf{w}_x + \mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - (\mathbf{u}_x\mathbf{v}_x + \mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z)\mathbf{w}_x \\ &= \mathbf{v}_x (\mathbf{u}\cdot\mathbf{w}) - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}_x \end{align}$

(compared to initial demonstration, I've just made sure that the scalar is kept on the right when doing the substitution explained below to avoid any inconsistency)

now replacing $\mathbf{u}$ and $\mathbf{v}$ with $\nabla$ and $\mathbf{w}$ with $\mathbf{a}$, we get:

$ \begin{align} (\mathbf{\nabla} \times (\nabla \times \mathbf{a}))_x &= \left(\nabla (\nabla.\mathbf{a})\right)_x - (\nabla \cdot \nabla) a_x \\ &= \left(\nabla (\nabla.\mathbf{a})\right)_x - \Delta a_x \end{align} $

fdeloche
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