Multipole expansion of Vector Potential (A)

  • #1

Homework Statement

So my teacher, as we made the multipole expansion of Vector Potential (\vec A) decided to proof that the monopole term is zero doing something like this:

∫∇'⋅ (J.r'i)dV' = ∮r'iJ ndS' = 0
The first integral, "opening" the nabla: J⋅(∇r'i) + r'i(∇⋅J) this must be equals 0

J = current density vector
n = unit vector

Homework Equations

I didn't get why the last thing (the nabla opened thing) was zero...
He couldn't explain quite well and got confused, then the next day he added one condition to that be zero:
[*] J is defined only in a certain region of space and that's why it's zero.

I didn't get at all...

The Attempt at a Solution

Expanding using Legendre's polinomes the first term comes to: A_0 =mu0 I / {4pir} ∫dl = 0 and I got why, it's because the closed contour.

ps.: Sorry for my bad English :(
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  • #2
Sorry, the latex didn't work. Gonna edit this in one minute.

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