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Another proof in electrostatics

  1. Jan 23, 2005 #1

    quasar987

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    In proving [itex]\nabla \times \vec{E} = \vec{0}[/itex] for electrostatic fields, Griffiths switches directly from the equation

    [tex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a} =0[/tex]

    To the conclusion [itex]\nabla \times \vec{E} = \vec{0}[/itex]. As for Gauss's theorem, I am wondering if there is a more precise mathematical proof of this. More precisely, the integral [itex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a}[/itex] can be 0 for three reasons

    1) The vector [itex]\nabla \times \vec{E}[/itex] is perpendicular to [itex]d\vec{a}[/itex] everywhere.

    2) The integral, as a sum, is worth 0 (i.e. some parts are positive, some negative, some nul such that the total is 0.)

    3) [itex]\nabla \times \vec{E} = \vec{0}[/itex]

    Is there a way to exclude the two first possibilities without referring to arguments such as "but this is evidently impossible for an electric field", but only by treating [itex]\vec{E}[/itex] as just another vector field? Many times I tought I had found the answer but later realised, I had not afterall.

    My best attemps, I believe, gets rid of 1) by setting the surface integral as a sphere and supposing it is 0 because the field is everywhere perpendicular to it's surface. Evidently this is not possible, but I can't prove it. (Don't know enough vector field calculus to know where to start) Can this be done?
     
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  3. Jan 23, 2005 #2

    Galileo

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    If I recall correctly, Griffiths concluded [itex]\nabla \vec E \times =0[/itex], because no assumption was made on a particular surface.
    Since the integral is zero for an arbitrary surface, the curl of E must be zero.
    Try convincing yourself of that.
     
  4. Jan 23, 2005 #3

    dextercioby

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    I'm sure u both know that
    [tex] \int\int_{S} \nabla\times \vec{E}\cdot (\vec{n} dS) =0 [/tex]

    results from applying Stokes theorem to a law which asserts that the circulation of the vector field E along any closed loop (closed electric circuit,par éxample) is zero.U know electric field comes from electric force which is conservative (the work does not depend on the path & for a closed path,the work is zero...).

    Daniel.
     
  5. Jan 23, 2005 #4

    dextercioby

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    Yo,guys,this is Stokes theorem:closed loop+open surface bordered by the closed loop...

    :bugeye:

    Daniel.

    P.S.Nothing to do with Gauss whatsoever.
     
  6. Jan 23, 2005 #5

    Galileo

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    Yes, but the book tries to validate the use of a potential function by showing the curl of E is zero.
     
  7. Jan 23, 2005 #6

    dextercioby

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    Of course
    [tex] \vec{F}_{el}=q\vec{E} [/tex] (1)

    And then defining
    [tex] W_{1\rightarrow 2}=:\int_{1}^{2} q\vec{E}\cdot \vec{dl} [/tex] (2)

    This work is path independent...Therefore,\vec{E} is an exact differential form whose circulation along any closed path is zero.Applying Stokes theorem,the curl is zero,therefore u can assume that \vec{E} comes from a potential field.Alternatively:
    [tex] \vec{E} exact----------------->\exists \phi: \vec{E}=-\nabla\phi [/tex]

    Daniel.
     
  8. Jan 23, 2005 #7
    Hey Guys can u tell me the online notes on this topic which explain these well...it wii be appreciated.
     
  9. Jan 23, 2005 #8

    quasar987

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    In 3-D, I can't because I can't visualize well enough. In 2-D, I have concluded that it's not true.

    For exemple, take a field that is tangeant to a circle. Then the integral is 0 for the first reason for a circle (any circle)... and it seems to me it is 0 for any other geometrical figure for the second reason!


    Daniel,

    Basically, you're reminding me that the line integral is path-independant iff [itex]\nabla \times \vec{E} = \vec{0}[/itex], but what I'm asking for is a proof of this statement, since, as I have mentionned, it appears that there are two other plausible reasons we must exclude before this conclusion can be attained.
     
  10. Jan 23, 2005 #9

    dextercioby

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    This is done in any rigurous course of Analysis in the chapter of line integrals and 1-forms...I can't come up with a proof,but i'm sure you can find one in a solid book on the theory of integration...

    Daniel.
     
  11. Jan 23, 2005 #10
    quasar,

    You correctly stated the 3 possible ways that the integral of a vector (in this case, the vector is curlE) over a surface can vanish. Here's why your first two possibilities don't work.

    1) The vector would have to be perpendicular to da everywhere on EVERY possible surface. If you have a surface that's perpendicular everywhere, just bend a little piece of it. Now it's not perpendicular everywhere!

    2) The argument here is similar. If you have a surface where everything adds up to zero, just stretch out a little piece that picks up a positive or negative contribution. Now the total isn't zero anymore!

    So, the only way to ensure that the integral of a vector over ANY surface is zero, is to make the vector itself 0 everywhere.

    I may be missing something (certainly a possibility!) but I don't think so.
     
  12. Jan 23, 2005 #11

    krab

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    I agree with jdavel: the important point, not mentioned by quasar, is that S is an Arbitrary surface.
     
  13. Jan 23, 2005 #12
    ah..... PROVE again? i don't like this word... nothing in physics can be prove without making assumtion (axiom or whatever you wanna call it)
    try use this definition
    [tex](\nabla \times \vec{E})\cdot \hat{n} = \lim_{\Delta S \rightarrow 0} \frac{ \int_{\Delta S} \vec{E} \cdot d \vec{l}}{\Delta S} [/tex] n is the normal vector of the surface delta S

    griffith doesn't give this definition for curl E, but I think this definition is the best... you can really visualize what curl E does...

    really don't wanna argue with you guys which physics law is more fundamental or is one physics law possible be proved by other physics law... however, at least, I hope you guys agree with me... the conservation law is more fundamental than [itex]\nabla \times \vec{E} = \vec{0}[/itex], so what I was doing is proving [itex]\nabla \times \vec{E} = \vec{0}[/itex] from conservation of energy and definition in Mathematics...

    DON"T TELL ME GRIFFITH SAID CONSERVATION OF ENERGY IN ELECTROSTATIC IS A RESULT OF [itex]\nabla \times \vec{E} = \vec{0}[/itex], I DON"T CARE WHAT GRIFFITH SAID AND I DON"T WANNA KNOW WHAT GRIFFITH SAID.... I DON"T KNOW WHY GRIFFITH DO NOT USE THE MOST DIRECT WAY TO DO THIS "PROOF"....
     
    Last edited: Jan 23, 2005
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