# Tensor notation proof of Divergence of Curl of a vector field

Prove $$\nabla\cdot(\nabla\times \vec{F})=\vec{0}$$ using tensor notation.

Here is my shot at it:

$$\nabla\cdot(\nabla\times \vec{F})=\vec{0}$$ becomes $$\partial_{i}(\epsilon_{ijk}\partial_{j}F_{k})$$ Using the product rule.

$$\epsilon_{ijk}[F_{k}(\partial_{i}\partial_{j})+\partial_{j}(\partial_{i}F_{k})] = \epsilon_{ijk}F_{k}(\partial_{i}\partial_{j})+\epsilon_{ijk}\partial_{j}(\partial_{i}F_{k})$$ After permutation. $$\epsilon_{jki}F_{i}\partial_{j}\partial_{k}-\epsilon_{kji}\partial_{j}\partial_{k}F_{i}$$

So wouldn't this look like $$\vec{F}\cdot(\nabla\times \nabla)-\vec{F}\cdot(\nabla\times \nabla)=\vec{0}$$? I am pretty sure you are not allowed to cross the gradient operator with itself. I don't think this is right.

• There is no product, so you cannot use the product rule: $\partial_j F_k$ is not the product of $\partial_j$ and $F_k$. Mar 1, 2020 at 20:22

You've rewritten $$\partial_i(\partial_jF_k)$$ as $$F_k\partial_i\partial_j+\partial_j\partial_iF_k$$. That would work if $$\partial_j$$ were an ordinary quantity you just multiply by $$F_k$$, but of course it's not. Indeed, your strategy also requires acknowledging $$\partial_i$$ is instead a differential operator, obeying the famous product rule.

The correct treatment needs no product rule. As @DavideMorgante's answer noted, you can just use the same symmetric indices argument in the proof of $$A\cdot A\times F=0$$ for a "normal" (i.e. non-operator-valued) vector $$A$$, since $$\partial_i\partial_j=\partial_j\partial_i$$ is just as true as $$A_iA_j=A_jA_i$$.

• So I can treat this as a triple product with two of the same "vectors"? Mar 1, 2020 at 20:29
• @bsafaria You can if you follow the usual proof closely enough to know exactly which vectors it assumes commutes, to make sure they're only the leftmost two. For example, don't you dare assume $\nabla\cdot F\times\nabla$ works the same way. But given the memorization that needs, you may as well just do it all from scratch.
– J.G.
Mar 1, 2020 at 20:30
• @bsafaria Forget that two-term sum, it's wrong anyway. As I said, the product rule doesn't enter into it. To explain why $\epsilon_{ijk}\partial_i\partial_jF_k=0$, note it contracts the $i\leftrightarrow j$-symmetric $\partial_i\partial_jF_k$ with the $i\leftrightarrow j$-antisymmetric $\epsilon_{ijk}$, which vanishes if $i=j$. Meanwhile, each choice of $i,\,j,\,k$ with $i\ne j$ pairs with the $i\leftrightarrow j$ equivalent, e.g.$$\epsilon_{123}\partial_1\partial_2F_3+\epsilon_{213}\partial_2\partial_1F_3=\epsilon_{123}\partial_1\partial_2F_3-\epsilon_{123}\partial_1\partial_2F_3=0.$$
– J.G.
Mar 1, 2020 at 20:47
• @bsafaria In other words,$$2\epsilon_{ijk}\partial_i\partial_jF_k=\epsilon_{ijk}\partial_i\partial_jF_k+\epsilon_{jik}\partial_j\partial_iF_k=\epsilon_{ijk}\partial_i\partial_jF_k-\epsilon_{ijk}\partial_i\partial_jF_k=0.$$
– J.G.
Mar 1, 2020 at 22:06
• @TedBlack The former, surely. The latter's final term would be $(\nabla\times F)\cdot\nabla G$. In particular, the terms sum to $\nabla\times(G\nabla F)$.
– J.G.
Oct 16 at 12:19

The most simple way is by noticing that $$\partial_i\partial_j$$ is completely symmetric under the exchange of the two indices while $$\epsilon_{ijk}$$ is completely anti-symmetric. Now you use the fact that

The contraction of a symmetric quantity with an antisymmetric one is always zero

You can easily see this by computing by hand the product $$\epsilon_{ijk}\partial_i\partial_j$$

At this point it's clear that $$\partial_i(\epsilon_{ijk}\partial_jF_k) = (\epsilon_{ijk}\partial_i\partial_j)F_k = 0$$

• I see, so I can transform it into where I dot the gradients to get $\nabla^2F$-$\nabla^2F$? Mar 1, 2020 at 19:53
• @bsafaria what do you mean? Mar 1, 2020 at 19:55
• If I'm understanding this right. I can just arrange it into $-\epsilon_{ikj}F_{j}\partial_{i}\partial_{k}+\epsilon_{jki}\partial_{i}\partial_{k}F_{j}$ which would mean the what I said above since you are now doting the gradient with itself. Mar 1, 2020 at 20:02
• @bsafaria Don't fall into the trap that most students do, which is to translate index notation back into something vectorial at every step. The thing about index notation is that while you are going through the procedure, you will end up with intermediaries that cannot be written in standard vector or matrix notation. Not every step is translatable. The point of learning index notation is not to tie it back to an old context, but to be comfortable with manipulations in the new context. Mar 1, 2020 at 20:02
• Nit-pick: $\partial_i\partial_j$ isn't a tensor, and neither is $\epsilon_{ijk}$. Mar 1, 2020 at 20:24

Here's a solution using matrix notation, instead of index notation.

In three dimensions, each vector is associated with a skew-symmetric matrix, which makes the cross product equivalent to matrix multiplication, i.e. \eqalign{ A &= \left[\begin{array}{r} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{array}\right] \\ Af &=a\times f } This suggests that the curl operation is \eqalign{ \nabla\times f &= \left[\begin{array}{r} 0 & -\partial_3 & \partial_2 \\ \partial_3 & 0 & -\partial_1 \\ -\partial_2 & \partial_1 & 0 \end{array}\right]\cdot\pmatrix{f_1\\f_2\\f_3} \\ &= \pmatrix{ \partial_2f_3-\partial_3f_2 \\ \partial_3f_1-\partial_1f_3 \\ \partial_1f_2-\partial_2f_1 \\ } } And the operation of interest becomes \eqalign{ \nabla\cdot\nabla\times f &= \pmatrix{\partial_1&\partial_2&\partial_3} \cdot \pmatrix{ \partial_2f_3-\partial_3f_2 \\ \partial_3f_1-\partial_1f_3 \\ \partial_1f_2-\partial_2f_1 \\ } \\ &= \partial_1(\partial_2f_3-\partial_3f_2) + \partial_2(\partial_3f_1-\partial_1f_3) + \partial_3(\partial_1f_2-\partial_2f_1) \\ &= \partial_1\partial_2(f_3-f_3) + \partial_2\partial_3(f_1-f_1) + \partial_3\partial_1(f_2-f_2) \\ &= 0 \\ } which is zero for any $$f$$ vector.

This answer uses the rules of tensor calculus with both upper and lower indices.

Let us define the divergence of a tensor field $$V^i$$ by using the covariant derivative $$\nabla_jV^i$$; where the curl is given by $$V^i = \epsilon^{ijk}\nabla_jU_k$$, and $$\epsilon^{ijk}$$ is the Levi-Cività symbol:

$$\nabla_iV^i = \nabla_i(\epsilon^{ijk}\nabla_jU_k) = \epsilon^{ijk}\nabla_i(\nabla_j U_k) = \epsilon^{ijk}(\nabla_i\nabla_jU_k + \nabla_j\nabla_iU_k) = \epsilon^{ijk}\nabla_i\nabla_jU_k + \epsilon^{ijk}\nabla_j\nabla_iU_k.$$

Now, since the last term does not contain live indices, we can let index $$i$$ become $$j$$ and vice-versa to give:

$$\nabla_iV^i = \epsilon^{ijk}\nabla_i\nabla_jU_k + \epsilon^{jik}\nabla_i\nabla_jU_k = \epsilon^{ijk}\nabla_i\nabla_jU_k - \epsilon^{ijk}\nabla_i\nabla_jU_k = 0,$$

by a single permutation in the last Levi-Cività symbol.

While other answers are correct, allow me to add a detailed calculation. We can write the divergence of a curl of $$\vec {F}$$ as:

$$\nabla\cdot(\nabla\times \vec{F}) = \partial_i (\epsilon_{ijk} \partial_j F_k)$$

We would have used the product rule on terms inside the bracket if they simply were a cross-product of two vectors. But as we have a differential operator, we don't need to use the product rule. We get:

$$\nabla\cdot(\nabla\times \vec{F}) = \epsilon_{ijk} \partial_i \partial_j F_k$$

Differential operators do commute, so $$\partial_i \partial_j$$ will be a symmetric quantity, we can write it as: $$\partial_i \partial_j = \frac{1}{2}\big(\partial_i \partial_j + \partial_j \partial_i \big)$$. Hence we get:

$$\Rightarrow \nabla\cdot(\nabla\times \vec{F}) = \epsilon_{ijk} \frac{1}{2}\big(\partial_i \partial_j + \partial_j \partial_i \big) F_k$$ $$= \frac{1}{2}\big(\epsilon_{ijk}\partial_i \partial_j F_k + \epsilon_{ijk}\partial_j \partial_i F_k \big) = \frac{1}{2}\big(\epsilon_{ijk}\partial_i \partial_j F_k - \epsilon_{jik}\partial_j \partial_i F_k \big)$$ $$= \frac{1}{2}\big(\nabla \cdot (\nabla \times \vec F) - \nabla \cdot (\nabla \times \vec F \big) \big) = 0$$

Where in the second line we used the anti-symmetric property of Levi-Civita tensor: $$\epsilon_{ijk} = -\epsilon_{jik}$$.