# Product of a scalar and a vector

1. May 13, 2009

### croggy

1. The problem statement, all variables and given/known data
I am trying to prove that:

$$\nabla \cdot (\psi\mathbf{A}) = \mathbf{A} \cdot\nabla\psi + \psi\nabla \cdot \mathbf{A}$$

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

3. 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.

2. May 13, 2009

### cristo

Staff Emeritus
Have you tried putting the components in $\vec{\nabla}=(\partial_x, \partial_y, \partial_z), \vec{A}=(A_x,A_y,A_z)$ and expanded? Try this and show your work.

Alternatively, if you have learnt index notation, the proof is pretty simple.

3. May 13, 2009

### croggy

Oh well, this is one way to learn tex.

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

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

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

$$RHS = ({\partial \psi \over \partial x}A_1 + {\partial \psi \over \partial y}A_2 + {\partial \psi \over \partial z}A_3) + ({\partial A_1 \over \partial x}\psi + {\partial A_2 \over \partial y}\psi + {\partial A_3 \over \partial z}\psi)$$

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

$$RHS = ({\partial \psi A_1 \over \partial x} + {\partial \psi A_2 \over \partial y} + {\partial \psi A_3 \over \partial z} )$$(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. May 13, 2009

### gabbagabbahey

Yes that's fine. The product rule applies to partials just like ordinary derivatives. For example,

$${\partial \over \partial x}(\psi A_1)=A_1{\partial \psi \over \partial x}+\psi{\partial A_1 \over \partial x}$$

5. May 13, 2009

### croggy

Would I be right in assuming that the chain rule and quotient rule also still apply?

6. May 13, 2009

### gabbagabbahey

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)