The Vector Laplacian: Understanding the Third Term

Suppose ##A = A_i\mathbf{\hat{e}}_i## and ##B = B_i\mathbf{\hat{e}}_i## are vectors in ##\mathbb{R}^3##. Then

\begin{align}
\Delta\left(A\times B\right)
&= \epsilon_{ijk}\Delta\left(A_jB_k\right)\mathbf{\hat{e}}_i \\
&= \epsilon_{ijk}\left[A_j\Delta B_k + 2\partial_mA_j\partial_mB_k + B_k\Delta A_j\right]\mathbf{\hat{e}}_i \\
&= A\times\left(\Delta B\right) + 2\epsilon_{ijk}\partial_mA_j\partial_mB_k\mathbf{\hat{e}}_i + \left(\Delta A\right)\times B
\end{align}

I can't identify the term ##2\epsilon_{ijk}\partial_mA_j\partial_mB_k\mathbf{\hat{e}}_i## on the last line with anything more familiar in terms of standard vector calculus operations. This feels somewhat odd to me, as the additional two terms can be written neatly in terms of "standard" vector calculus operators (namely the cross product and the laplacian) but the third term can't. I hope someone can help me shed light on this matter.

Given [itex](\nabla \mathbf{A})_{ij} = \partial_iA_j[/itex] I think you could write [tex]
2\operatorname{Tr}((\nabla A) \times (\nabla B))[/tex] although this does leave room for confusion concerning which axes the operatiors are being carried out over, and this may be one of those occasions where the suffix notation is much clearer.

Hi thanks for your respons. Isn't the cross-product of two matrices an ill defined notion? At least, I won't call it a "standard" opperation.

Given tensors [itex]A[/itex] and [itex]B[/itex] of ranks [itex]n[/itex] and [itex]m[/itex] respectively, I think it makes sense to define the rank [itex]n+m-1[/itex] tensor [tex]
(A \times B)_{j_1 \cdots j_{n-1} k_1 \cdots k_{m-1}i} = \epsilon_{ij_nk_m} A_{j_1 \dots j_n}B_{k_1 \dots k_m}.[/tex] But this may be another reason why vector notation does not work here.

William Crawford said:

TL;DR Summary: I'm trying to work out different product rules for the Laplacian and I've gotten stuck on the Laplacian of a cross product.

Suppose ##A = A_i\mathbf{\hat{e}}_i## and ##B = B_i\mathbf{\hat{e}}_i## are vectors in ##\mathbb{R}^3##. Then
\begin{align}
\Delta\left(A\times B\right)
&= ...
\end{align}##

Hi. Please humour me if I'm missing the point...

I thought the Laplacian operates only on scalar functions. Since ##A \times B## is a vector (not a scalar function) is ##\Delta (A \times B)## a legitimate expression?

Steve4Physics said:

I thought the Laplacian operates only on scalar functions. Since ##A \times B## is a vector (not a scalar function) is ##\Delta (A \times B)## a legitimate expression?

The Laplacian, or vector Laplacian to be precise, is naturally defined for vector functions as well. In the framework of vector calculus this is often done as
$$ \Delta\mathbf{v} = \nabla(\nabla\cdot\mathbf{v}) - \nabla\times(\nabla\times\mathbf{v}), $$
which has the added benefit of being valid for curvilinear coordinates (i.e. coordinate independant). In Cartesian coordinates the vector Laplacian takes the simple and obvious form
$$ \Delta\mathbf{v} = (\nabla\cdot\nabla)\mathbf{v} = \partial_j\partial_jv_i. $$
In general, the Laplacian can be defined for any differential ##p##-form, but this is a little beside the topic discussed here (please see Hodge Laplacian for more details).

William Crawford said:

The Laplacian, or vector Laplacian to be precise, is naturally defined for vector functions as well. In the framework of vector calculus this is often done as
$$ \Delta\mathbf{v} = \nabla(\nabla\cdot\mathbf{v}) - \nabla\times(\nabla\times\mathbf{v}), $$
which has the added benefit of being valid for curvilinear coordinates (i.e. coordinate independant). In Cartesian coordinates the vector Laplacian takes the simple and obvious form
$$ \Delta\mathbf{v} = (\nabla\cdot\nabla)\mathbf{v} = \partial_j\partial_jv_i. $$
In general, the Laplacian can be defined for any differential ##p##-form, but this is a little beside the topic discussed here (please see Hodge Laplacian for more details).

Many thanks for the clarification. I'd only enccountered the 'basic' Laplacian.