Proof of Gauss's Law: Connecting Flux and Closed Surfaces

[dextercioby]

And, as others have said, Gauss' Law can be derived from this principle and Coulomb's Law.

That's just wrong,Krab and i hope u know it.I'm surprised that u can make such an erroneous remark.
I quote from my course of CED:
QUOTE:The fundamental problem of electrostatics in vacuum consists in determining the electrostatic field intensity,if the distribution of electric charge (given by $\rho(\vec{r})$) is completely known.For an electrostatic field,the problem is completely solvable,using the Maxwell equations adapted for this case./QUOTE

This is the course i dream of teaching over the years...

Do you see any other way of solving the problem?

Daniel.

 

[Haelfix]

Dexter, I don't agree. There are many ways to axiomatize a system in physics. The entire discipline is circular in nature, all that matters is that things are experimentally verified.

As a theorist I would tend to agree with you, its easier and more transparent to get Gauss's law the way you describe. I can think of generalizations to relativity as well, and even further to generalized fibre bundles --> K theory, and even to category theory.

But let's talk about the sdm. You know for instance the entire theory can be derived by simply postulating Yang Mills gauge invariance, lorentz invariance and the cluster decomposition principle. Good that's how we learned things.

Less well known is that there are many other ways to derive field theory. The Wightman axioms is another admittedly similar but different way to get field theory.

Another way to get the SDM is by partial wave unitarity bounds and minimality. I can axiomatize my system that way, without ever talking about Gauge invariance, the latter is then a theorem.

 

[Doc Al]

Electrostatics requires only two pieces of information: Coulomb's law and the Principle of Superposition. The whole rest of electrostatics (including Gauss's Law) is just a mathematical elaboration of these two facts. That is what we were taught, and that is what is says in Griffiths.

And this is perfectly true. It turns out that Gauss's Law is more fundamental than Coulomb's law, since it applies to moving charges whereas Coulomb's law does not (except for low speeds). (But that comes later.)

 

[dextercioby]

I agree.I never said that my system of axiomatization (namey postulating Maxwell eq.) was unique,not even at classical level.
I hope u know that i was trying to fight the missconception that Maxwell equations can be derived from "elementary" level...I was referring to Gauss law,but the discussion can be exapanded to include the other three...I believe that these equations can be deduced from other principles,which would use more advanced mathematical concepts and not viceversa...

Daniel.

P.S.Yes,the theory is circular.But my logics tells me that u can't go both ways on this circle...

 

[Galileo]

Starting an EM course with Maxwell's equations is seems very unsound from a pedagogical point of view.

Anyway, I want to give quasar another proof of Gauss' law. All you need is the divergence theorem.

I actually had the whole thing typed out yesterday and just when I wanted to press the 'submit' button, my computer screen went blue (windows, aaargh).
Now I`m bummed so here's the idea of the proof in short:

(I): Show that $\nabla \cdot \vec F =0$ if F has the form: $\hat F=K\hat r/r^2$ ($r\neq 0$).
(II)The divergence theorem holds for simple solid regions, but can be extended to hold for regions that are finite unions of simple solid regions. Any volume region with a 'hole' in it can be viewed as a union of two simple solid regions. (For example: A solid sphere with a cavity in it, the outer surface is the surface of the sphere and the inner one is the surace of the cavity.)
(III) Show that the flux through such a surface is zero (if the inner surface surrounds the origin). This means the flux through both surfaces is equal. Since the surfaces were arbitrary, it means the flux through any connected closed surface that surrounds the origin has the same value. It thus suffices to calculate the flux through a sphere.

 

[dextercioby]

Starting an EM course with Maxwell's equations is seems very unsound from a pedagogical point of view.

Depends on what u mean by "EM course"...If it's the introductory course,i agree...If it's the RELATIVISTIC ELECTRODYNAMICS,i couldn't possibly disagree more...

Daniel.

 

[quasar987]

Ingenious.

(I always make a backup of my massive posts before hitting the dreaded 'Preview Post' and 'Submit Reply' buttons)

 

[heman]

My book starts with the differential form first and then applies the divergence theorm. If you understand how to get to the differential form then I think that's a better way to go.

Hey Fulham what is the name of urs book...

 

[heman]

starting from his equation of the field (which is a direct consequence of Coulom's law)

$$E(r)=\frac{1}{4\pi\epsilon_0} \iiint \frac{r-r'}{|r-r'|^3}\rho(r') d^3r'$$

Let's write is vectorially...

$$\vec{E}=\frac{1}{4\pi\epsilon_0} \iiint \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|^2}(\vec{r}-\vec{r'}) dV'$$

Now, it is done by taking the divergence of this expression. It turns out that

$$\nabla \cdot \left(\frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^2} \right) = 4\pi \delta^3(\vec{r}-\vec{r'})$$

where delta^3 is the 3-D dirac delta function. Therefor

$$\nabla \cdot \vec{E}=\frac{1}{4\pi\epsilon_0} \iiint 4\pi \delta^3(\vec{r}-\vec{r'}) \rho(\vec{r'}) dV' = \frac{\rho(\vec{r'})}{\epsilon_0}$$

The last equality is from the property of the delta function.

Hey quasar you did it well but i am not able to grasp the last two concepts which u applied of Dirac delta function..pls help me out..

 

[vincentchan]

use the difinition
$$\nabla \cdot \ver{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec{A}}{\Delta V}$$
to prove the second last line

the last line is just substitude E by the first line, move the nabla inside the integral and use the second last line...

if you have griffith book, you can find the proof inside... it is not short, but straight forward...

 

[dextercioby]

The last expression (the ratio between the "rho" and the "epsilon0") is wrong..."Rho" should have depended upon $$\vec{r}$$,not $$\vec{r'}$$ .

Daniel.

 

[heman]

use the difinition
$$\nabla \cdot \ver{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec{A}}{\Delta V}$$
to prove the second last line

the last line is just substitude E by the first line, move the nabla inside the integral and use the second last line...

if you have griffith book, you can find the proof inside... it is not short, but straight forward...

Thx Vincent,,i will check Griffiths to make it go inside my head.

 

[quasar987]

The last expression (the ratio between the "rho" and the "epsilon0") is wrong..."Rho" should have depended upon $$\vec{r}$$,not $$\vec{r'}$$ .

Daniel.

Yep, I realized that yesterday while trying to fall asleep. I had to open the lights and note it down .

 

[Kelvin]

I am a year 1 student. My EM teacher told me that Gauss' law and columb's law are almost equivalent, but since gauss' law is always true, it is regarded as more fundamental.

I wonder, if columb's law didn't exist, could gauss derive gauss' law? My teacher derived gauss' law by using columb's law.

I feel strange that columb's law is wrong when the speed is comparable to c, but why gauss' law which is derived from columb's law is valid? I know things like lorentz contraction but logically, if premise is wrong, how can a conclusion drawn can be true?

 

[MiGUi]

I think that using the electromagnetic tensor is a more beautiful way to proof the 4 maxwell equations... That is, varying the action.

Nevertheless.. the theorem of Gauss is the math's theorem of divergence, applied to electromagnetism.

$$\int\int\int_V \vec{\nabla} \cdot \vec{P} dV= \int \int_S \vec{P} \cdot \hat n \, dS$$

 

[dextercioby]

I wonder, if columb's law didn't exist, could gauss derive gauss' law?

We don't know that.If Einstein had not existed,would we have gotten today the theory of relativity...??

My teacher derived gauss' law by using columb's law.

For ELECTROSTATIC FIELDS he can do that...Though i'd do it the other way around...

I feel strange that columb's law is wrong when the speed is comparable to c,

Because DYNAMICAL FIELDS (i.e.varying in space and time) have nothing to do with Coulomb's law...

but why gauss' law which is derived from columb's law is valid?

It is,BECAUSE GAUSS'S LAW IS MUCH MORE GENERAL...

I know things like lorentz contraction but logically, if premise is wrong, how can a conclusion drawn can be true?

And what would be the premise...??

Daniel.

 

[Kelvin]

I mean: we start with columb's law and derive gauss' law. if columb's law is wrong in some cases, then anything derived from it should be wrong in those cases. but the fact is gauss' law is always true but columb's law is not. that's strange

 

[dextercioby]

I mean: we start with columb's law and derive gauss' law. if columb's law is wrong in some cases, then anything derived from it should be wrong in those cases. but the fact is gauss' law is always true but columb's law is not. that's strange

I see what you mean...Well,each of them has a range,a domain of applicability.The one of Gauss's law is simply LARGER,as it includes the domain of Coulomb's law...Nobody ever said that Coulomb's law would be wrong...It is VERY CORRECT,but only in it's domain of applicability,which is STATIC FIELDS.In this case,electrostatic...

It's as simple as that...Maybe for me...

Daniel.

 

[Doc Al]

I mean: we start with columb's law and derive gauss' law. if columb's law is wrong in some cases, then anything derived from it should be wrong in those cases. but the fact is gauss' law is always true but columb's law is not. that's strange

If the only justification for Gauss's law were Coulomb's law, then there would be a gap in explaining how it could apply where Coulomb's law does not. But that's not true.

Also, beware the fallacy of "denying the antecedent":

if p, then q.
not-p;
therefore not-q.​

 

[Haelfix]

Incidentally Gauss's law is not always true, nor is the rest of Maxwell's equations or the Dirac equation. One of the great successes (or downfalls in some peoples minds) of QFT is that it outputs its domain of validity. The theory is simply inconsistent past that point.

For instance, QED and Gravity are clearly wrong as field theories at very high energies, but for different reasons. (Landau pole and renormalization flow respectively)