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

[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

 

[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.

 

[Nugatory]

What is Ai Aii Bi Bii in the integral?

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]

 

[Charles Link]

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.

 

[garylau]

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.

it

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.

But it sound unreasonable he gives the proof like thisin 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

 

[Charles Link]

it

But it sound unreasonable he gives the proof like thisin 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

(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.)

 

[garylau]

(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.)

Oh i See thank you

 

[garylau]

(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.)

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?

 

[Charles Link]

The proof is quite general. The integrals are line integrals, not surface integrals.