Applying product rule 5 to curl of B field derivation in Griffiths

[Sparky_]

Homework Statement

This question is regarding clarifying some reading in Griffith’s Electrodynamics, page 224.
“deriving the curl of B”
In particular it’s less on electrodynamics and more on some vectors or vector calculus.
The book states: we must check that the second term integrates to zero:

The second term that is referred to is:
$$–(J *\nabla)\frac{\hat{r}}{r^3}$$
• = “dot product”
The x-component is $$–(J *\nabla')\frac{x-x’}{r^3} = \nabla' * [\frac{x-x’}{r^3} J]- (\frac{x-x’}{r^3})(\nabla'*J)$$
“using product rule 5”
Product rule 5 states: $$\nabla*(fA) = f(\nabla*A)+ A*(\nabla f)$$
F is a scalar, A is a vector”

Homework Equations

The Attempt at a Solution

I am having some difficulty matching up the terms and applying “product rule 5” in this case”

If I let the $$f(\nabla*A)$$ term = $$(J*\nabla')(\frac{x-x’}{r^3})$$

And $$\nabla*(f*A)$$ = $$\nabla'*(\frac{x-x'}{r^3} J)$$

The last term does not match.

Meaning, I have $$A*(\nabla f)$$ from the product rule.

The vector A "dot" a divergence.My remaining term does not have a divergence.

How does this product rule fit?

Thanks
-Sparky_

 

[TSny]

You want to use the identity: $\vec{\nabla} \cdot(f \vec{A}) = f(\vec{\nabla} \cdot\vec{A}) + \vec{A} \cdot (\vec{\nabla} f) = f(\vec{\nabla} \cdot\vec{A}) +( \vec{A} \cdot \vec{\nabla}) f$ where I have rewritten the last term in a slightly different form.

If I let the $$f(\nabla*A)$$ term = $$(J*\nabla')(\frac{x-x’}{r^3})$$

Note that on the left you have a divergence of a vector: $f(\vec{\nabla} \cdot \vec{A})$ whereas on the right you do not have a divergence. You'll need to reconsider which term in the product rule should be identified with $$(\vec{J} \cdot \vec{\nabla '})(\frac{x-x’}{r^3})$$