1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 22, 2016 #1
    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?


    What is Ai Aii Bi Bii in the integral?

    thank you

    Attached Files:

  2. jcsd
  3. Nov 22, 2016 #2


    User Avatar
    Science Advisor

    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:

  4. Nov 22, 2016 #3


    User Avatar

    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]
  5. Nov 22, 2016 #4

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    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.
  6. Nov 22, 2016 #5
    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
  7. Nov 22, 2016 #6

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    (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.)
  8. Nov 22, 2016 #7
    Oh i See thank you
  9. Nov 22, 2016 #8
    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:

  10. Nov 23, 2016 #9

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    The proof is quite general. The integrals are line integrals, not surface integrals.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted