Im trying to symplify formulas like: $$\operatorname{div}(\operatorname{rot}\vec{F}),\qquad \operatorname{rot}(\operatorname{rot}\vec{F}) $$

or something more strange like:


To do this I want to use the tensor notation and by this I mean using the Einstein convention, Levi-Civita symbol, Kronecker delta and all that. The problem is that I don't understand the rules and what its allowed and what not.

As an example:

$$\operatorname{div}(f( r)\cdot \textbf r)= \partial_i(f( r)\cdot \textbf r)_i = \partial_i(f( r)\cdot x_i)= \partial _if(r)x_i+f(r)\partial_ix_i= f'(r)\frac{x_i}{r}x_i+3f(r)=rf'(r)+3f(r) $$

Any help on how to do this kind of problems or where I can find useful examples?

I'm working in flat space.


  • $\begingroup$ What would you consider "simpler"? If I told you that $\nabla \cdot (\nabla \times F) = \partial_a \epsilon^{abc} \partial_b [g_{cd} F^d]$, would you really regard that as "simpler"? $\endgroup$ – Muphrid Jun 21 '13 at 19:55
  • $\begingroup$ I might be able to help if you are working in flat space. If you're doing work in some kind of curved space, you need some help from someone with experience in differential geometry. If you are working in flat space, put that in your question and you might get some ready answers. $\endgroup$ – bob.sacamento Jun 21 '13 at 19:56
  • $\begingroup$ @Muphrid No, I would like to get that $\nabla \times (\nabla \times \vec F) = \nabla (\nabla \cdot \vec F) - \nabla ^2 \vec F$ using the tensor notation. $\endgroup$ – Msegade Jun 21 '13 at 20:08
  • $\begingroup$ @bob.sacramento Yes, I'm working in flat space. $\endgroup$ – Msegade Jun 21 '13 at 20:12
  • $\begingroup$ @Msegade If that's really what you mean to ask, you should say so. Is there anything else that you've not stated that you're interested in seeing? $\endgroup$ – Muphrid Jun 21 '13 at 20:19

The gradient, divergence, curl, and Laplacian can be written in the following way: $$\begin{eqnarray*} (\mathrm{grad}\, f)_i &=& \partial_i f \\ \mathrm{div}\,{\bf F} &=& \partial_i F_i \\ (\mathrm{rot}\,{\bf F})_i &=& \epsilon_{ijk} \partial_j F_k \\ \Delta &=& \partial^2 = \partial_i\partial_i. \end{eqnarray*}$$ The Levi-Civita symbol has some key properties. For example, it is totally antisymmetric so $\epsilon_{ijk}=-\epsilon_{jik} = \epsilon_{jki}$. In addition it has an important multiplicative property: $$\begin{eqnarray*} \epsilon_{ijk}\epsilon_{ilm} &=& \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}. \end{eqnarray*}$$ The Levi-Civita symbol and the Kronecker delta are just numbers, so they commute with derivatives. Below we give some sample calculations.

Example 1: $$\begin{eqnarray*} \mathrm{div}(\mathrm{rot}\,{\bf F}) &=& \partial_i (\mathrm{rot}\,{\bf F})_i \\ &=& \partial_i (\epsilon_{ijk} \partial_j F_k) \\ &=& \epsilon_{ijk} \partial_i \partial_j F_k \\ &=& \epsilon_{ijk} \partial_j \partial_i F_k \\ &=& -\epsilon_{jik} \partial_j \partial_i F_k \\ &=& -\epsilon_{ijk} \partial_i \partial_j F_k \\ &=& 0 \end{eqnarray*}$$

Example 2: $$\begin{eqnarray*} [\mathrm{rot}(\mathrm{rot}\,{\bf F})]_i &=& \epsilon_{ijk} \partial_j (\mathrm{rot}\,{\bf F})_k \\ &=& \epsilon_{ijk} \partial_j (\epsilon_{klm}\partial_l F_m) \\ &=& \epsilon_{ijk} \epsilon_{klm} \partial_j \partial_l F_m \\ &=& \epsilon_{kij} \epsilon_{klm} \partial_j \partial_l F_m \\ &=& (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \partial_j \partial_l F_m \\ &=& (\partial_m \partial_i - \delta_{im} \partial^2)F_m \\ &=& \partial_i(\mathrm{div}\,{\bf F}) - (\Delta{\bf F})_i \end{eqnarray*}$$ Therefore, $$\mathrm{rot}(\mathrm{rot}\,{\bf F}) = \mathrm{grad}(\mathrm{div}\,{\bf F}) - \Delta{\bf F}.$$

  • $\begingroup$ @Muphrid: Thanks, that looks better. $\endgroup$ – user26872 Jun 21 '13 at 21:16
  • $\begingroup$ @Msegade: Glad to help. $\endgroup$ – user26872 Jun 21 '13 at 21:38

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