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Product of a scalar and a vector

  • Thread starter croggy
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  • #1
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Homework Statement


I am trying to prove that:

[tex] \nabla \cdot (\psi\mathbf{A}) = \mathbf{A} \cdot\nabla\psi + \psi\nabla \cdot \mathbf{A} [/tex]

Where nabla is a scalar function and A is a vector field

The Attempt at a Solution



I tried expanding both the LHS and RHS, but I think I am getting confused with the order of operations or something.
 

Answers and Replies

  • #2
cristo
Staff Emeritus
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Have you tried putting the components in [itex]\vec{\nabla}=(\partial_x, \partial_y, \partial_z), \vec{A}=(A_x,A_y,A_z)[/itex] and expanded? Try this and show your work.

Alternatively, if you have learnt index notation, the proof is pretty simple.
 
  • #3
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Oh well, this is one way to learn tex.

[tex] \nabla = {\partial \over \partial x}\mathbf{\hat{i}} + {\partial \over \partial y}\mathbf{\hat{j}} + {\partial \over \partial z}\mathbf{\hat{k}}[/tex]
[tex] A = A_1\mathbf{\hat{i}} + A_2\mathbf{\hat{j}} + A_3\mathbf{\hat{k}}[/tex]

[tex]LHS = \nabla \cdot (\psi\mathbf{A}) = \nabla \cdot (\psi A_1\mathbf{\hat{i}}+\psi A_2\mathbf{\hat{j}}+\psi A_3\mathbf{\hat{k}}) = ({\partial \psi A_1 \over \partial x} + {\partial \psi A_2 \over \partial y} + {\partial \psi A_3 \over \partial z} )[/tex]

[tex]RHS = \mathbf{A} \cdot\nabla\psi + \psi\nabla \cdot \mathbf{A}
= A \cdot ({\partial \psi \over \partial x}\mathbf{\hat{i}} + {\partial \psi \over \partial y}\mathbf{\hat{j}} + {\partial \psi \over \partial z}\mathbf{\hat{k}}) + \psi({\partial A_1 \over \partial x} + {\partial A_2 \over \partial y} + {\partial A_3 \over \partial z})[/tex]

[tex]RHS = ({\partial \psi \over \partial x}A_1 + {\partial \psi \over \partial y}A_2 + {\partial \psi \over \partial z}A_3) + ({\partial A_1 \over \partial x}\psi + {\partial A_2 \over \partial y}\psi + {\partial A_3 \over \partial z}\psi)[/tex]

Where the LHS and RHS are remind me of the product rule so:

[tex]RHS = ({\partial \psi A_1 \over \partial x} + {\partial \psi A_2 \over \partial y} + {\partial \psi A_3 \over \partial z} )[/tex](By product rule)

Is that right though? Is the product rule the same as for proper derivatives? WP and wolfram don't seem to give any mention of the product rule for Partial Derivates.
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
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Is that right though? Is the product rule the same as for proper derivatives? WP and wolfram don't seem to give any mention of the product rule for Partial Derivates.
Yes that's fine. The product rule applies to partials just like ordinary derivatives. For example,

[tex]{\partial \over \partial x}(\psi A_1)=A_1{\partial \psi \over \partial x}+\psi{\partial A_1 \over \partial x}[/tex]
 
  • #5
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Would I be right in assuming that the chain rule and quotient rule also still apply?
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
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The quotient rule is just a special case of the product rule, so yes. The chain rule can be a little tricky (it sometimes takes great care to figure out whether a function has an implicit dependence or an explicit dependence on a given variable), but yes it still applies. (a Google search of "chain rule for partial derivatives will probably turn up some useful resources for you)
 

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