I What is the meaning of Ai Aii Bi Bii in Vector potential?

1. Nov 22, 2016

garylau

Sorry
may i ask a question here~
i don't understand how did GRIFFITHS prove the statement from(C) to (B)
What is his logic in this case?

and

What is Ai Aii Bi Bii in the integral?

thank you

Attached Files:

• 15078666_1009225375866479_7829671217989776891_n.jpg
File size:
45.6 KB
Views:
46
2. Nov 22, 2016

phyzguy

The notations I and II are intended to indicate two different paths from a to b, as in the attached drawing. So the difference between the integral along path I and the integral along path II is equal to the integral along path I plus the integral along path II in the other direction, which is the integral around the whole loop, which is zero.

Attached Files:

• Paths.png
File size:
7.9 KB
Views:
47
3. Nov 22, 2016

Staff: Mentor

The I and II indicate that the line integrals are being taken along two different paths between a and b.

[Edit: like phyzguy just said a moment before me]

4. Nov 22, 2016

Griffths is taking two different paths (I) and (II) in a path integral from point "a" to point "b". He first does a loop (going from "a" to "b" by path 1 and then going from "b" to "a" by path 2) by putting a "-" sign on it. (Notice the position of the endpoints in the integral.) ...edit .. I see 2 very equivalent responses came in right before me.

5. Nov 22, 2016

garylau

it
But it sound unreasonable he gives the proof like this

in fact the equation on the left side are equal to the right side
(1)+(11)(A->B) =(1)-(11)(B->A) ???

What is Griffith trying to doing in the equation

6. Nov 22, 2016

(I)A=>B-(II)A=>B=(I)A=>B+(II)B=>A=0 so that (I)A=>B=(II)A=>B is what he is proving.(Hopefully you can distinguish between the actual equal signs and the equal signs that I used as part of an arrow from A to B.)

7. Nov 22, 2016

garylau

Oh i See thank you

8. Nov 22, 2016

garylau

There is one more question i want to ask

Why the surface integral(II) is inward in this case?

because there are only two possibility for a closed surface integral?

but i think there can are many different paths which correspond to the "outward" closed surface integral are equal to 0

So the proof is not general?

Attached Files:

• 15078666_1009225375866479_7829671217989776891_n.jpg
File size:
25.6 KB
Views:
35
9. Nov 23, 2016