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So, I want to expand this term but facing difficulty with the vector and tensor product. The term is :

$\textbf{p}$ is a vector; and $\mathbb{K}$ is a tensor

$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})]$

There is divergence of all terms and when expanding this i am stuck with how to deal with tensor and vector product.

Thank you for helping me out.

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  • $\begingroup$ Einstein notation generally makes this kind of thing easier: $\partial_i(p_i(K_{jk}\partial_kp_j)) = (K_{jk}\partial_kp_j)\partial_ip_i + p_i\partial_i(K_{jk}\partial_kp_j) = \cdots$ $\endgroup$
    – user856
    Apr 12, 2018 at 10:45
  • $\begingroup$ Thanks a lot for the help. $\endgroup$ Apr 12, 2018 at 10:58
  • $\begingroup$ The first term, on the RHS i.e. $\partial_{i}p_{i}$ in your expression is a divergence but shouldn't it be a gradient of p? $\endgroup$ Apr 12, 2018 at 11:10
  • $\begingroup$ I don't know, maybe I misinterpreted your expression. But you should work it out yourself with the correct semantics. $\endgroup$
    – user856
    Apr 12, 2018 at 11:20
  • $\begingroup$ That's the issue i am facing I get end result something like tensor+vector which is not correct. For example, in your expression from we will get (tensor.tensor).tensor + vector div(tensor.tensor)=== which would give end terms as tensor+vector $\endgroup$ Apr 12, 2018 at 12:37

1 Answer 1

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If

  1. when you write $$\nabla\textbf{p}$$ you mean the 2-tensor that results from taking the "gradient" (covariant derivative) of $\mathbb{p}$ (which is, actually kind of weird),

  2. when you write $$\mathbb{K} \cdot \nabla\textbf{p}$$ you mean the scalar that results from taking the "interior product" of (contracting) $\mathbb{K}$ with $\nabla \mathbf{p}$,

  3. when you write $$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})]$$ you mean the scalar that results from taking the divergence of $\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})$ (which is a vector, since it is the product of a vector with a scalar)

then you can use the identity $$\nabla\cdot(f\mathbf{v}) = \nabla f\cdot\mathbf{v} + f(\nabla\cdot\mathbf{v})$$ to get $$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})] = (\nabla\cdot\mathbf{p})(\mathbb{K}\cdot\nabla\textbf{p}) + \mathbf{p} \cdot \nabla(\mathbb{K}\cdot\nabla\textbf{p})$$

If want to expand it further I would recommend you to rewrite your expression in abstract index notation or something similar

$$\begin{align} \nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})] &= \nabla_a(p^a{K^b}_c\nabla_bp^c)\\ &= \nabla_a p^a{K^b}_c\nabla_bp^c + p^{a}\nabla_a({K^b}_c\nabla_bp^c)\\ &=\nabla_a p^a{K^b}_c\nabla_bp^c + p^{a}[\nabla_a{K^b}_c\nabla_bp^c + {K^b}_c\nabla_a\nabla_bp^c] \end{align}$$

but that doesn't give us anything that can be written in vector calculus notation. (Well, even that $\nabla\textbf{p}$ is not common).

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  • $\begingroup$ Thanks, I also got the same result as you mentioned. $\endgroup$ May 11, 2018 at 9:43

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