Another proof in electrostatics

In summary, Daniel Griffiths concludes that the curl of an electric field, which is defined as the vector sum of all its component vectors perpendicular to the field, must be zero, because there are three ways in which the integral of the vector over a surface can vanish.
  • #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 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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
Yes, but the book tries to validate the use of a potential function by showing the curl of E is zero.
 
  • #6
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.
 
  • #7
Hey Guys can u tell me the online notes on this topic which explain these well...it wii be appreciated.
 
  • #8
Galileo said:
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.

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.
 
  • #9
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.
 
  • #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.
 
  • #11
I agree with jdavel: the important point, not mentioned by quasar, is that S is an Arbitrary surface.
 
  • #12
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
[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 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 [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"...
 
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1. What is electrostatics?

Electrostatics is the study of electric charges at rest, or in other words, the study of stationary electric fields. It involves the behavior and effects of electric charges, as well as the interactions between them.

2. What is another proof in electrostatics?

Another proof in electrostatics is a demonstration or experiment that provides evidence for a particular concept or theory within the field of electrostatics. It is a way to verify the validity of an idea or to further support existing knowledge.

3. How is electrostatics important in our daily lives?

Electrostatics plays a crucial role in many aspects of our daily lives. For example, it helps to explain the operation of electronic devices such as computers and cell phones, and it is also essential in the production of electricity and the functioning of electric motors. Moreover, electrostatics is the basis for various technologies, including air purifiers, photocopiers, and inkjet printers.

4. What are some applications of electrostatics?

Electrostatics has many practical applications in various fields. Some of these include electrostatic painting, electrostatic dust removal, electrostatic precipitation in air pollution control, and electrostatic separation in mineral processing. Electrostatic forces are also utilized in medical technologies like electrocardiography and electrosurgery.

5. How is electrostatics related to other branches of physics?

Electrostatics is closely related to other branches of physics, such as electromagnetism, quantum mechanics, and thermodynamics. It is also connected to other areas of science, including chemistry and biology. Many concepts in electrostatics, such as electric potential and capacitance, have analogs in other fields, illustrating the interconnectedness of different areas of science.

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