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I am looking for a vector expression for the curl of a composite vector-valued function. In other words $$ \nabla\times\mathbf{A}(\mathbf{B}) = ? $$ In indicial notation, this can be written as $$ \left(\nabla\times\mathbf{A}(\mathbf{B})\right)_i = -\epsilon_{ijk}\left(\frac{\partial A_k}{\partial B_n}\right)\frac{\partial B_n}{\partial x_j} $$ where $\epsilon_{ijk}$ is the Levi-Civita symbol and Einstein summation is used. so far, the simplest expression we could write was
$$ \left(\nabla\times\mathbf{A}(\mathbf{B})\right)_i = \epsilon_{ijk}\left[\left(\frac{\partial\mathbf{A}}{\partial\mathbf{B}}\right)\cdot\left(\nabla\otimes\mathbf{B}\right)^T\right]_{jk} $$ where $\cdot$ is used to denote the usual matrix product and $\otimes$ the Kronecker outer product. Could this be written all in vector notation?

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  • $\begingroup$ That is to say, you want $\nabla \times (A \circ B)$? The curl of the composition? $\endgroup$ – Muphrid Mar 28 '15 at 4:51
  • $\begingroup$ Yes @Muphrid. For $A$ and $B$ both vector-valued functions. $\endgroup$ – Joey Dumont Mar 28 '15 at 21:28
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    $\begingroup$ I can't get anything nicer than what you have. $\endgroup$ – Dan Robertson May 27 '15 at 13:43
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I am not sure if this helps, but I just found that you can actually use the contracted epsilon identity to get a pretty good vector/dyadic representation of the chain rule. The nice thing about this particular identity (at least to me) is that it does what any good chain rule should do, and applies the operator in question to the argument.

Please allow me to preempt this by saying that I am not that familiar with the conventions of dyadic notation, but I will present what I have figured out in index notation form, so that if anyone wants to go in, and fix my notation, they will know how to. Anyway, here is what I found:

\begin{equation} \nabla \times \left(\mathbf{A} \circ \mathbf{B}\right) = -\left(\nabla_\mathbf{B}\cdot \mathbf{A}\right)\left(\nabla\times \mathbf{B}\right) + \left(\frac{\partial \mathbf{A}}{\partial \mathbf{B}}\right)^T\left(\nabla \times \mathbf{B}\right) + \left(\frac{\partial \mathbf{A}}{\partial \mathbf{B}}\right)^T \!\!\!\begin{array}{c} _\cdot \\ ^\times\end{array}\!\!\!\left(\nabla \mathbf{B}^T\right) \end{equation}

The vertical operator notation means that I am crossing the second index of the left tensor with the 1st index of the right tensor, and then contracting the 1st index of the left tensor with the 2nd index of the right tensor.

Working it out in index notation, we start with \begin{equation} \epsilon_{ijk}\frac{\partial A_j}{\partial B_l}\frac{\partial B_l}{\partial X_k} \end{equation}

What we want is to have the levi-civita symbol apply to B. The first step for this is to "free" the l and k indices. We can do this, using the kronecker delta

\begin{equation} \epsilon_{ijk}\frac{\partial A_j}{\partial B_l}\frac{\partial B_l}{\partial X_k} = \delta_{k_0k_1}\delta_{l_0l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} \end{equation}

Now, we can get to the contracted epsilon identity by adding appropriate cancelling terms to the RHS:

\begin{equation} \left(\delta_{k_0k_1}\delta_{l_0l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} - \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}\right) + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} \end{equation}

We can then use the contracted epsilon identity to rewrite this as

\begin{equation} \epsilon_{rk_0l_0}\epsilon_{rk_1l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} \end{equation}

Now, we are halfway there, for $\epsilon_{rk_1l_1}$ applies to $\frac{\partial B_{l_1}}{\partial X_{k_1}}$

Notice also, that when we juxtapose $\epsilon_{rk_0l_0}$ and $\epsilon_{ijk_0}$ we can apply the contracted epsilon identity once again. Combining these two facts gives us \begin{equation} \left(-\delta_{ri}\delta_{l_0j} + \delta_{rj}\delta_{l_0i}\right)\frac{\partial A_j}{\partial B_{l_0}}\epsilon_{rk_1l_1}\frac{\partial B_{l_1}}{\partial X_{k_1}} + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} \end{equation}

The expression in vector/dyadic notation above follows directly from this.

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If you really want it to have $\nabla\times$ and $\nabla_\mathbf{B}$ on it you can rewrite it as: $$ \left(\nabla\times\mathbf{A}(\mathbf{B})\right)_{i}=\epsilon_{ijk}\left(\frac{\partial A_{k}}{\partial B_{n}}\right)\frac{\partial B_{n}}{\partial x_{j}}=\frac{\partial}{\partial x_{j}}\left(\epsilon_{ijk}\left(\frac{\partial A_{k}}{\partial B_{n}}\right)B_{n}\right)-\epsilon_{ijk}\left(\frac{\partial^{2}A_{k}}{\partial x_{j}\partial B_{n}}\right)B_{n} $$ $$ =\epsilon_{ijk}\frac{\partial}{\partial x_{j}}\left(\left(\frac{\partial A_{k}}{\partial B_{n}}\right)B_{n}\right)-\epsilon_{ijk}\left(\frac{\partial}{\partial x_{j}}\left(\frac{\partial A_{k}}{\partial B_{n}}\right)\right)B_{n}=\nabla\times\left(\left(\nabla_{\mathbf{B}}\mathbf{A}\right)\mathbf{B}\right)-\left(\nabla\times\left(\nabla_{\mathbf{B}}\mathbf{A}\right)\right)\mathbf{B} $$ where I've used the "curl" of a matrix, $\nabla\times\left(\nabla_{\mathbf{B}}\mathbf{A}\right)$ (defined as a matrix whose columns are the curl of the columns of the original matrix).

By the way, where I put $\left(\nabla_{\mathbf{B}}\mathbf{A}\right)\mathbf{B}$ I mean the matrix product of the matrix $\nabla_{\mathbf{B}}\mathbf{A}$ ($\mathbf{A}$ changes between rows and $\frac{\partial}{\partial x_j}$ between columns) with the column vector $\mathbf{B}$.

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