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

Click For Summary

Discussion Overview

The discussion revolves around the notation and logic used in Griffiths' treatment of vector potentials, specifically the meaning of Ai, Aii, Bi, and Bii in the context of line integrals along different paths. Participants are exploring the implications of these notations and the reasoning behind Griffiths' proofs, including questions about the generality of the proof and the nature of the integrals involved.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants seek clarification on how Griffiths proves the transition from statement (C) to (B) and the logic behind it.
  • It is noted that the notations I and II represent two different paths from point a to point b, and the difference between the integrals along these paths relates to the integral around the whole loop being zero.
  • One participant expresses skepticism about the reasonableness of Griffiths' proof, questioning the equality of the left and right sides of the equation presented.
  • Another participant attempts to clarify the proof by breaking down the equations involving the paths and their implications.
  • A question is raised regarding the nature of the surface integral and its orientation, with a participant suggesting that there could be multiple paths leading to an outward closed surface integral that equals zero.
  • In response, one participant asserts that the proof is general and emphasizes that the integrals discussed are line integrals, not surface integrals.

Areas of Agreement / Disagreement

Participants express differing views on the clarity and generality of Griffiths' proof. While some seek to understand and clarify the reasoning, others question its validity and applicability. No consensus is reached on the interpretation of the proof or the nature of the integrals involved.

Contextual Notes

There are unresolved questions regarding the assumptions made about the paths and the nature of the integrals, particularly in relation to surface integrals versus line integrals. The discussion reflects a range of interpretations and uncertainties regarding Griffiths' approach.

garylau
Messages
70
Reaction score
3
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
 

Attachments

  • 15078666_1009225375866479_7829671217989776891_n.jpg
    15078666_1009225375866479_7829671217989776891_n.jpg
    41.1 KB · Views: 564
Physics news on Phys.org
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.
 

Attachments

  • Paths.png
    Paths.png
    3.8 KB · Views: 555
  • Like
Likes   Reactions: garylau
garylau said:
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]
 
  • Like
Likes   Reactions: garylau
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.
 
  • Like
Likes   Reactions: garylau
phyzguy said:
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
phyzguy said:
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
 
garylau said:
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.)
 
  • Like
Likes   Reactions: garylau
Charles Link said:
(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
 
  • Like
Likes   Reactions: Charles Link
Charles Link said:
(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?
 

Attachments

  • 15078666_1009225375866479_7829671217989776891_n.jpg
    15078666_1009225375866479_7829671217989776891_n.jpg
    22.6 KB · Views: 496
The proof is quite general. The integrals are line integrals, not surface integrals.
 

Similar threads

Largest Class Action Lawsuit could upend the nascent AI Industry
  • · Replies 6 ·
Replies
6
Views
1K
Undergrad Principle of Physics: Derivation of -dU/dx=F
  • · Replies 24 ·
Replies
24
Views
3K
Undergrad System, potential energy, and nonconservative forces
  • · Replies 6 ·
Replies
6
Views
2K
Current and Future AI Threats to Humanity and the Human Response
  • · Replies 10 ·
Replies
10
Views
5K
Graduate What Are the Differences Between These Vector Potential Formulas?
  • · Replies 8 ·
Replies
8
Views
2K
Periodic motion -- Potential as a function of a non-linear Force(x)
  • · Replies 2 ·
Replies
2
Views
1K
Graduate Confused about multipole expansion of vector potential
  • · Replies 4 ·
Replies
4
Views
1K
Graduate Vector potential components in small loop antenna
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad How Does the Position of a Free Vector Couple Affect Its Behavior?
  • · Replies 10 ·
Replies
10
Views
3K
Deriving electric and vector potential
  • · Replies 2 ·
Replies
2
Views
1K