Physics: The vector potential "A"

[SabyE]

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

∫∇'⋅ (J.r'_i) dV' = ∮ r'_i J 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

My question:
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...

 

*Sorry for my bad english

[studiot]

Hello, Saby and welcome.

 

The first term in your expansion is

 

[math]\frac{{{\mu _0}I}}{{4\pi r}}\oint\limits_{} {dl} [/math]

 

Remember what dl is.

 

It is the physical spatial displacement vector. (not the electric displacment vector D)

When taken round a complete loop it must end up where it started.

That is the total displacement must be zero.

 

so

 

[math]\oint\limits_{} {dl} = 0[/math]

 

I didn't get why the last thing (the nabla opened thing) was zero...

 

If you want to display math here go to this (free) site and type it in.

It will generate either Latex or a gif for you to paste in.

Edit forgot the links, sorry.

 

http://www.sciweavers.org/free-online-latex-equation-editor

 

or

 

https://www.codecogs.com/latex/eqneditor.php

Edited May 3, 2016 by studiot