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

  1. Nov 22, 2016 #1
    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
     

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  2. jcsd
  3. Nov 22, 2016 #2

    phyzguy

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    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.
     

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  4. Nov 22, 2016 #3

    Nugatory

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    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

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    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
    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
     
  7. Nov 22, 2016 #6

    Charles Link

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    (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?
     

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  10. Nov 23, 2016 #9

    Charles Link

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