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nabla acting on products

gradient of vector, divergence of dyad product, curl of dyad product
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Mathematics Subject Classification

26B12 no label found26B10 no label found


I have also seen the divergence of a dyad product expanded as
$\nabla\cdot(\vec{u}\,\vec{v}) =
(\nabla\vec{u})\vec{v} + (\nabla\!\cdot\!\vec{v})\vec{u}$
where $(\nabla\vec{u})\vec{v}$ is treated as a tensor acting on a vector. This formula I understand and can prove in an arbitrary coordinate frame.

I cannot seem to work through the divergence of a dyad product given above
$\nabla\cdot(\vec{u}\,\vec{v}) =
Specifically, the $\vec{u}\cdot\nabla\vec{v}$ term gives me trouble. I've read that $(\vec{u}\cdot\nabla)\vec{v}$ is equivalent to

Do you have a pointer to either a generalized definition of $\vec{u}\cdot\nabla$ without requiring Cartesian coordinates or a pointer to how I can correctly interpret
the dot product in $\vec{u}\cdot(\nabla\vec{v})$?

Thank you for your time,

Dear Rhys,
I have taken the formula for the divergence of dyad product from a book of Kalle V\"ais\"al\"a, but unfortunately cannot now prove it. Nevertheless, I believe that the formula is right (K. V. doesn't make mistakes!). Are you sure of your version? I suspect a bit the fact that it implies
\nabla\cdot(\vec{u}\vec{v}) = \nabla\cdot(\vec{v}\vec{u}).

How exactly are you interpreting the
(\nabla\vec{u})v term in my version? I think the confusion may be that your gradient-of-a-vector and my gradient-of-a-vector may be the transpose of each other.

Expanded out, I am interpreting
(\nabla\vec{u})v + (\nabla\cdot\vec{v})u
(\vec{u},k \otimes g^k) v + (\vec{v},k \cdot g^k) u
where ',k' denotes partial differentiation and g is an arbitrary, potentially curvilinear, basis.

- Rhys

Hi Rhys,
Gibbs gave clear rules about how to deal with vector differential operators and dyads (dyadic product)as well. Since you know do it in Cartesian coordinates, let's do it in generalized coordinates.
\vec{g}^r, \vec{g}_s, are contravariant and covariant generalized base vectors, respectively (\vec{g}^r\cdot\vec{g}_s = \delta^r_{s}=\delta^{rs}=\delta_{rs}, the Kronecker delta, invariant at all space vector).
(I choose vector's contravariant components, for example).
Dyads have nine components; it's a nonian maintaining the order of base vectors. Let's take generalized coordinates {\theta^r}; the upper index is an irrelevant convention.
By definition,
where the partial derivatives acts on *any* function following the operator \nabla. Accordingly,
\vec{g}^r\cdot[(\partial{u^s}/\partial\theta^r)\vec{g}_s + u^s(\partial\vec{g}_s/\partial\theta^r)](v^tvec{g}_t) +
\vec{g}^r\cdot(u^s\vec{g}_s)[(\partial{v^t}/\partial\theta^r)\vec{g}_t} + v^t(\partial\vec{g}_t/\partial\theta^r)]=
[(\partial{u^s}/\partial\theta^r)(\vec{g}^r\cdot\vec{g}_s) + u^s\Gamma^q_{sr}(\vec{g}^r\cdot\vec{g}_q)](\vec{v}) +
(\vect{u})[(\partial{v^t}/\partial\theta^r)(\vec{g}^r\cdot\vec{g}_t) + v^t\Gamma^q_{tr}(\vec{g}^r\cdot\vec{g}_q)]=
(\partial{u^r}/\partial\theta^r + u^s\Gamma^r_{sr})\vec{v} + \vec{u}(\partial{v^r}/\partial\theta^r + v^s\Gamma^r_{sr}),
where \Gamma^k_{ij} = g^{kp}\Gamma_{ijp} are the Christoffel symbols,
\Gamma_{ijp} = (\partial^2{x^m}/\partial{\theta^i}\partial{\theta^j})(\partial{x^m}/\partial{\theta^p}),
and the metric tensor
g^{kp} = (\partial{x^m}/\partial{\theta^k}(\partial{x^m}/\partial{\theta^p})
(the {x^i} are Cartesians).
Or if you prefer on term of covariant derivatives
\nabla\cdot(\vec{u}\vec{v}) = u^r|_r\vec{v} + \vec{u}v^r|_r,
where, for instance,
v^r|_r = \partial{v^r}/\partial\theta^r + v^s\Gamma^r_{sr}.
Hope this helps.

..."or a pointer to how I can correctly interpret
the dot product in $\vec{u}\cdot(\nabla\vec{v})$?"

\vec{u}\cdot(\nabla\vec{v} = (u^s\vec{g}_s)\cdot[\vec{g}^r\partial\vec{v}/\partial\theta^r =
[u^s(\vec{g}_s\cdot\vec{g}^r)](\partial\vec{v}/\partial\theta^r) =
u^r(v^s|_r)\vec{g}_s = u^r(v_s|_r)\vec{g}^s (a vector!),
v^s|_r = \partial{v^s}/partial\theta^r + \Gamma^s_{tr}\v^t,
v_s|_r = \partial{v_s}/partial\theta^r + \Gamma^t_{sr}\v_t
the covariant derivatives.

Seems like the root of the trouble was a difference in my interpretation of grad-- seems I'm working with transposes of your definitions.

Thank you,

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