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Path independence

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi,I am trying to understand the proof attached.
    My problem is shown by a red arrow.Can someone explain those 2 steps?
    And please answer it as simply as you can...Since I haven't done multivariable calculus..It is a physics course.Thanks

    2. Relevant equations


    3. The attempt at a solution
    Let say the integral of A(r') 。d(r') is L(r),so if you take the definite integral it becomes L(r+dr) - L(r),but in the proof they said dr is infinitesimal so dr = 0,so L(r+dr) - L(r) = 0 =/= A(r) 。d(r) ? path.png
     
    Last edited: Mar 24, 2016
  2. jcsd
  3. Mar 24, 2016 #2

    DrClaude

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    If the integral was path dependent, then you couldn't combine the two integrals together because you could not guarantee that the integral from r to r + dr would pass through r0.

    An infinitesimal dr does not mean dr = 0. It means that it is extremely small, the smallest you can get without being 0. Hence, you can consider A(r') to be constant over the interval r to r+dr and take it out of the integral.
     
  4. Mar 24, 2016 #3
    Sorry, why if A(r') is constant then it becomes A(r) ?
     
  5. Mar 24, 2016 #4

    SteamKing

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    If you have a HW question with ∫ or ∂ in it, it's a safe bet it belongs in the Calculus HW forum. :wink:
     
  6. Mar 24, 2016 #5

    DrClaude

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    I mean that it is constant in the interval r to r+dr, so A(r) = A(r+dr) = A(r+dr/2), etc. By convenience, you take the value to be the one at r. A is still a function of r.
     
  7. Mar 24, 2016 #6

    andrewkirk

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    Hi Davon. The notation in the second last formula with two ##i##s is 'Einstein notation', which is used in differential geometry. Also, the expression ##d\phi(\underline{r})## has a rigorous meaning in differential geometry but in many other uses is either a piece of notation that has no stand-alone meaning, or else just a hand-wave. This leads me to guess that the text might be written in a differential geometry context.
    Is that correct? The way of interpreting the proof is different in diff geom from what would be needed in other contexts.
     
  8. Mar 24, 2016 #7
    Hi Andrew,it was actually part of the notes of my physics course,and I am still in my 2nd year of a physics degree so I guess it is irrelevant to the more advanced maths.My lecturer introduced Einstein notation for proving various vector identities(scalar product,vector product,curl,div,etc...)
     
  9. Mar 24, 2016 #8

    andrewkirk

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    Here's a way you can make the proof rigorous without any high-faluting maths:

    Replace ##d\phi(\underline{r})## by ##\frac{d\phi(\underline{r}+h\underline{e}_i)}{dh}## where ##\underline{e}_i## is the ##i##th basis vector of ##\mathbb{R}^3## (I'm assuming here that the context is 3D Euclidean space).

    Then on the right-hand side, divide by ##h##, replace ##d\underline{r}## by ##h\underline{e}_i##, and enclose both sides of the equation within ##\lim_{h\to 0}\Bigg(...\Bigg)##. Note that the LHS is, by definition, the ##i##th component of ##\nabla(\phi(\underline{r}))##.

    Following steps similar to what's written above, but with these modifications, you should be able to deduce

    $$\lim_{h\to 0}\left(\frac{d\phi(\underline{r}+h\underline{e}_i)}{dh}\right)=\underline{A}(\underline{r})\cdot\underline{e}_i$$

    See how you go.
     
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