How Do You Simplify ν∇⋅(ρν)?

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SUMMARY

The simplification of the expression ν∇⋅(ρν) involves applying the product rule for vector calculus. Specifically, the expression can be simplified using the identity: $$\nabla \cdot (\rho \vec{\nu}) = \rho \nabla \cdot \vec{\nu} + \vec{\nu} \cdot \nabla \rho$$. This results in a scalar that is then multiplied by the vector ν. The discussion emphasizes the importance of understanding vector calculus operations, particularly in the context of tensors.

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echandler
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Hey so probably a really simple question, but I'm stumped. How do you simplify:

ν∇⋅(ρν), where
ν is a vector
∇ is the "del operator"
⋅ indicates a dot product
ρ is a constant.

I want to say to do the dyadic product of v and ∇, but then you would get (v_x)*(d/dx) + ... which would be meaningless, so I'm thinking you do the dot product first, but I can't find an order of operations for algebra involving tensors of any order (except zero order of course).

Thanks in advance.
 
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It's kind of like a product rule for vector calculus:
$$\nabla \cdot (\rho \vec{\nu}) = \rho \nabla \cdot \vec{\nu} + \vec{\nu} \cdot \nabla \rho$$
gives you a scalar. ##\vec{\nu} \nabla \cdot (\rho \vec{\nu}) ## just multiplies the vector ##\vec{\nu}## by that scalar.
 

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