Another proof in electrostatics

[quasar987]

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

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

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

1) The vector \nabla \times \vec{E} is perpendicular to d\vec{a} 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) \nabla \times \vec{E} = \vec{0}

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 \vec{E} as just another vector field? Many times I tought I had found the answer but later realized, 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?

 

[Galileo]

If I recall correctly, Griffiths concluded \nabla \vec E \times =0, 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.

 

[dextercioby]

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

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.

 

[dextercioby]

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



Daniel.

P.S.Nothing to do with Gauss whatsoever.

 

[Galileo]

Yes, but the book tries to validate the use of a potential function by showing the curl of E is zero.

 

[dextercioby]

Of course
\vec{F}_{el}=q\vec{E} (1)

And then defining
W_{1\rightarrow 2}=:\int_{1}^{2} q\vec{E}\cdot \vec{dl} (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:
\vec{E} exact----------------->\exists \phi: \vec{E}=-\nabla\phi

Daniel.

 

[heman]

Hey Guys can u tell me the online notes on this topic which explain these well...it wii be appreciated.

 

[quasar987]

If I recall correctly, Griffiths concluded \nabla \vec E \times =0, 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.

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-independent iff \nabla \times \vec{E} = \vec{0}, 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.

 

[dextercioby]

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.

 

[jdavel]

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.

 

[krab]

I agree with jdavel: the important point, not mentioned by quasar, is that S is an Arbitrary surface.

 

[vincentchan]

ah... PROVE again? i don't like this word... nothing in physics can be prove without making assumtion (axiom or whatever you want to call it)
try use this definition
(\nabla \times \vec{E})\cdot \hat{n} = \lim_{\Delta S \rightarrow 0} \frac{ \int_{\Delta S} \vec{E} \cdot d \vec{l}}{\Delta S} 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 want to 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 \nabla \times \vec{E} = \vec{0}, so what I was doing is proving \nabla \times \vec{E} = \vec{0} from conservation of energy and definition in Mathematics...

DON"T TELL ME GRIFFITH SAID CONSERVATION OF ENERGY IN ELECTROSTATIC IS A RESULT OF \nabla \times \vec{E} = \vec{0}, 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"...