- #1
ozone
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∇Δ
Show that [itex]\nabla \cdot A = 0 [/itex]
Where A is formally defined as
[itex]A(r) = \frac{\mu }{4\pi }\int \frac{J(r')\text{ }}{r} \, dv' [/itex]
I understand that we can distribute ∇ into the integral, and from there we can do a little bit of algebra to get the terms inside the integrand of the form.
[itex]\nabla \cdot \left(\frac{J'}{r}\right) + J'\cdot \nabla \left(\frac{1}{r}\right)[/itex]
Now what happens next is where I get lost. I don't fully understand why we can throw away the first term since ∇ isn't dependent upon J'. I haven't seen anything like this done before and it seems a little fishy. Next griffiths uses some identity to change
[itex] J'\cdot \nabla \left(\frac{1}{r}\right) = -J'\cdot \nabla '\left(\frac{1}{r}\right) [/itex]
Then he repeats this whole process again, and throws away the J term. Finally he finishes with stokes theorem to relate the area integral to a line integrand.
I guess I could just use a little bit of commentary on this proof as it doesn't make much sense to me yet.
Homework Statement
Show that [itex]\nabla \cdot A = 0 [/itex]
Where A is formally defined as
[itex]A(r) = \frac{\mu }{4\pi }\int \frac{J(r')\text{ }}{r} \, dv' [/itex]
I understand that we can distribute ∇ into the integral, and from there we can do a little bit of algebra to get the terms inside the integrand of the form.
[itex]\nabla \cdot \left(\frac{J'}{r}\right) + J'\cdot \nabla \left(\frac{1}{r}\right)[/itex]
Now what happens next is where I get lost. I don't fully understand why we can throw away the first term since ∇ isn't dependent upon J'. I haven't seen anything like this done before and it seems a little fishy. Next griffiths uses some identity to change
[itex] J'\cdot \nabla \left(\frac{1}{r}\right) = -J'\cdot \nabla '\left(\frac{1}{r}\right) [/itex]
Then he repeats this whole process again, and throws away the J term. Finally he finishes with stokes theorem to relate the area integral to a line integrand.
I guess I could just use a little bit of commentary on this proof as it doesn't make much sense to me yet.