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

[quasar987]

See the "Gauss's Law Sketch Proof" paragraph .[/URL] Griffiths uses mainly the same argument to prove Gauss's law, but I don't really see the connection between the fact that the flux is Q/epsilon for a sphere and the conclusion that it is worth this quantity too for any closed surface.

Can someone point to a more mathematical proof? Or simply unveil what I'm missing in this argument.

Thanks.

 

[vincentchan]

what kind of prove you want to see? The surface integral of E field = q/epsilon? if yes, why don't you try the simpliest case... a point charge at the center and integrate the sphere with radii R surround it...

 

[Doc Al]

Can someone point to a more mathematical proof? Or simply unveil what I'm missing in this argument.

You might find this helpful: http://www.rpi.edu/dept/phys/Dept2/phys2/activities/gauss/proof.html
It doesn't give a rigorous proof, but it does expand the argument a bit more than the site you gave. The key is that the flux depends on the solid angle, not on the radius (see how the r^2 terms cancel).

 

[FulhamFan3]

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.

 

[vincentchan]

differential form?
you mean divE=charge density?
how could you PROVE that w/o guass law? I really want to know
if you cant, how could you PROVE guass law by the differential form?
are you doing a circular logic here?

Edit:
div E= charge density is the direct result of guass law... you can PROVE it by the definition of "div" & gauss' law,

 

[FulhamFan3]

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

where

$$r = x\hat{i} + y\hat{j}+z\hat{k}$$ (region where you want to find the E-field)

$$r' = x'\hat{i} + y'\hat{j}+z'\hat{k}$$ (region of charge)

From that equation you can use vector calculus to convert to the differential form. I don't want to type down the proof because it involves a lot of typing in TEX. That and a lot of the details aren't clear in my book. For now just accept that the math works out and you get

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$

Then take the volume integral of both sides.

$$\iiint (\nabla \cdot E) dV = \frac{1}{\epsilon_0} \iiint \rho(r) dV$$

$$\iiint \rho dV = Q_{enc}$$ (total charge)

$$\iiint (\nabla \cdot E) dV = \iint E \cdot dA$$ (divergence theorm)

which gives you:

$$\iint E \cdot dA = \frac{Q_{enc}}{\epsilon_0}$$

 

[vincentchan]

$$\nabla \cdot E = \frac{\rho(r)}{\epsilon_0}$$
how could you prove the above equation without using gauss law?

 

[quasar987]

That site if nice Al but FulhamFan3's way satisfies me fully.

 

[FulhamFan3]

$$\nabla \cdot E = \frac{\rho(r)}{\epsilon_0}$$
how could you prove the above equation without using gauss law?

How do you prove it using gauss's law?

 

[vincentchan]

hey, quasar, $\nabla \cdot E = \frac{\rho(r)}{\epsilon_0}$ is the direct consequence of gauss law... you can't get the above formulas without assuming guass law is right in the first place, therefore, if you prove the guass law by the above formulas, you are doing a circular logics...

 

[quasar987]

No, I'll show you vincent, if FulhamFan3 doesn't want to.

 

[vincentchan]

$$\nabla \cdot \ver{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec{A}}{\Delta V}$$
the above is the definition of $\nabla \cdot \vec{F}$

replace F by E and the integral by Q/epsilon, see what you get

actually, people called $\nabla \cdot E = \frac{\rho}{\epsilon_0}$ "the differential form of gauss law", "point form of gauss law" etc... it IS gauss law... what you just did is proving gauss law by gauss law

 

[quasar987]

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.

 

[FulhamFan3]

Just because I didn't type in the proof for converting coulomb's law to the gauss differential form doesn't mean it doesn't exist.

As long as quasar understands I'm happy.

 

[FulhamFan3]

quasar at the end that r isn't a vector, it should be a scalar coordinate

 

[vincentchan]

now you need to prove the inverse square law...
actually, my point is... in physics... not everything can be proved... you can assume the inverse square law is true, and prove the gauss law. Or you can do it the other way around... just want to let you know... if you assume gauss law is true in the first place, the world would look much simpler and make sense...
what guass law said is all charge emit electric field... (imagine E field is a line start at the charge and goes to infinite) and you wrap the charge by a closed surface.. no matter how your closed surface looks like... number of lines passes through the surface is the same...
most of the physicsist use gauss law prove the inverse square law, but not the other way around... how come the there is a 1/4pi epsilon constant in the inverse square law but only a 1/epsilon constant in the gauss law...
and one more thing.. I was wrong... you can prove the gauss law by inverse square law.. but you are making yourself in a mess... try to see it the other way... and understand the definition of div and curl...

 

[quasar987]

Well, Griffiths keeps the vector dependancy of rho... I don't see why that's wrong, as long as rho itself is a scalar.

Actually Griffiths writes

$$\frac{\rho(\vec{r})}{\epsilon_0}$$

(the prime poped out), but that ought to be a typo.. :grumpy:

 

[FulhamFan3]

now you need to prove the inverse square law...
actually, my point is... in physics... not everything can be proved... you can assume the inverse square law is true, and prove the gauss law. Or you can do it the other way around... just want to let you know... if you assume gauss law is true in the first place, the world would look much simpler and make sense...

gauss's law came from coulomb's law, not the other way around.

most of the physicsist use gauss law prove the inverse square law, but not the other way around...

who? I know you can find it that way but that's to validate that gauss's law works, not to prove the inverse square relation.

how come the there is a 1/4pi epsilon constant in the inverse square law but only a 1/epsilon constant in the gauss law...

the original formulation of the inverse square law used "k" instead of $\frac{1}{4\pi\epsilon_0}$. Epsilon is used now because it's more fundamental.

 

[FulhamFan3]

Well, Griffiths keeps the vector dependancy of rho... I don't see why that's wrong, as long as rho itself is a scalar.

Actually Griffiths writes

$$\frac{\rho(\vec{r})}{\epsilon_0}$$

(the prime poped out), but that ought to be a typo.. :grumpy:

I just don't see how a vector position in a formula makes the function scalar. But that's just arguing semantics. As long as we know it's the same thing it's cool.

 

[vincentchan]

who?

All...

gauss's law came from coulomb's law, not the other way around.

do you know which 4 formulas (plus F=qE+qvB) summarize the electromagnetics... I really can't find the coulomb law in maxwell equation...

 

[dextercioby]

U cannot prove (obtain) this equation via theorerical methods.Simple as that.It's like proving the Schroedinger's equation,or the Newton's laws...They are axiomatized...


I can disprove any so-called "proof",for logic reasons...

Daniel.

 

[dextercioby]

gauss's law came from coulomb's law, not the other way around.

WHAT?? Historically,yes,you're right,but physics uses LOGICAL arguments to prove its relations...Maxwell's equations are AXIOMATIZED,you cannot prove them.Any other possible result in nonquantum electrodynamics can be proven via Maxwell's equations.
Here History worths s***,it's logics who comes first.Mawell is the one that did it.With him u may want to argue,not with me...He axiomatized Gauss's law...

Daniel.

 

[cepheid]

now you need to prove the inverse square law...
actually, my point is... in physics... not everything can be proved... you can assume the inverse square law is true, and prove the gauss law.

You don't need to "assume" the inverse-square law. It has been determined experimentally (COULOMB's LAW). The fact that the electric field falls off with the square of the distance from the source is expressed within Coulomb's law. 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.

 

[dextercioby]

You don't need to "assume" the inverse-square law. It has been determined experimentally (COULOMB's LAW). The fact that the electric field falls off with the square of the distance from the source is expressed within Coulomb's law. 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.

Then Griffiths is wrong...Electrostatics requires solving the Poisson equation for the potential...Read more into it.You've been obviously missinformed.

Daniel.

 

[vincentchan]

You don't need to "assume" the inverse-square law. It has been determined experimentally (COULOMB's LAW)

are you saying you are able to use experiment proving a theory is correct? show me how? People thought the PROVED the Newtons law is right hundreds years ago, but they were wrong. Experiment can only DISPROVE a theory, not the other way around... Therefore, if you want to prove gauss law from inverse square law, the best you can do is assume inverse square law is correct, since nothing in nature can be proved right. Once the assumtion is made, the rest is definition and mathematics consequence. And NO ONE can disprove mathematics and definition

 

[FulhamFan3]

Then Griffiths is wrong...Electrostatics requires solving the Poisson equation for the potential...Read more into it.You've been obviously missinformed.

Daniel.

Which book have you written? I'll tell my electrodynamics professor right away to change books.

 

[cepheid]

Dexter,

Please don't be so quick to condescend. I'm well aware of:

$$\nabla^{2}V = -\frac{\rho}{\epsilon_0}$$

But how can you refute what I'm saying when Griffiths begins with Coulomb's law and the Principle of Superposition (experimental results that cannot be derived from any more fundamental results), and does just what he claimed could be done? He derives the whole rest of electrostatics from these two properties of electric charge! Maxwell's equations, as they are presented to us at the undergraduate level, are the end result, the culmnation of (seven chapters in this case) worth of material! With them, the theory is complete, as we are told. In other words, electromagnetism is presented as it was discovered by the fantastic experimental work of all those 18th/19th century guys, Benjamin Franklin, Coulomb, Gauss, Ampere, Faraday, and finally those amazing dudes, Lorentz, Maxwell et al. Presented in this manner, Maxwell's equations arise one by one, as Gauss's law, Ampere's law, Faraday'a law, etc. So how can you blame me for perceiving it otherwise? And since you clearly don't believe me, here it is verbatim.

From Griffiths' Introduction to Electrodynamics, 3rd Ed. Page 59.
Coulomb's law and the principle of superposition constitute the physical input for electrostatics -- the rest, except for some special properties of matter, is mathematical elaboration of these fundamental rules.

I understand what you are saying about Maxwell's equations being axiomatic. After all, $\nabla \cdot \mathbf{B} = 0$ expresses an intrinsic property of magnetic fields that we can observe...they are divergenceless, solenoidal. And as you stated, any result in classical electrodynamics stems from Maxwell's Equations. Since you are far more knowlegable than I am, I'll ask: why is the statement (quoted) not an equally valid stance? Is it "wrong"?

I also recognize that classical electrodynamics is "special"...Maxwell's equations exemplify the perfect consistency and the inherent symmetry. Griffiths even goes so far as to say that subsequant theories strive to emulated this degree of completion.

 

[dextercioby]

I've been taught electrodynamics (classical/relativistic) following an approach similar to the one in J.D.Jackson's "Classical Electrodynamics" and the 2 books by Landau & Lifschitz:"Classical Field Theory" and "Electrodynamics of Continuous Media".(Is that the word?? )

I advise u do the same.Electrodynamics is a theoretical physics discipline whose main objective is to describe the classical electromagnetic interaction using relativistic approach.

Daniel.

 

[vincentchan]

Electrostatics requires only two pieces of information: Coulomb's law and the Principle of Superposition.

can't agree with you more...

 

[dextercioby]

Dexter,

Please don't be so quick to condescend. I'm well aware of:

$$\nabla^{2}V = -\frac{\rho}{\epsilon_0}$$

But how can you refute what I'm saying when Griffiths begins with Coulomb's law and the Principle of Superposition (experimental results that cannot be derived from any more fundamental results), and does just what he claimed could be done? He derives the whole rest of electrostatics from these two properties of electric charge!

I'm sorry,but that's not how a theoretical physicist would do and teach others to do as well...He starts with axioms,uses as much mathematical apparatus he can possibly use and derives all physical properties...
What u're telling me is not electrodynamics.It's not theoretical physics...



Daniel.

 

[cepheid]

Why are you so surprised that the most rigorous approach is not taken in undergrad? I think that the way the material is presented is logical, given that the goal is to teach students the fundamentals of electromagnetism. Maybe you are a theoretical physicist.I am not (yet!) I am a student. This I acknowledge.

We were told Jackson is a graduate textbook. Perhaps I will have a look at it someday.

What you said: "Electrodynamics is a theoretical physics discipline whose main objective is to describe the classical electromagnetic interaction using relativistic approach"

Interestingly enough, just today, in my lecture of the second EM course I'm taking (the "sequel"), we were shown how the Lorentz-force law can be derived by using special relativity, measuring (from a "stationary frame" S,) the force on a charge q moving in some arbitary direction at constant velocity v, due to a source charge Q, moving with velocity V wrt S, in a frame S'.

Magnetism then, is not a separate force at all, but a relativistic effect. Neat!

 

[dextercioby]

Why are you so surprised that the most rigorous approach is not taken in undergrad?

Okay,i agree,but what annoys me is the fact that you & others are mislead into believing that the Coulomb law and the principle of superposition (BS,there's no such thing...) are fundamental and can derive the other results,Gauss law incuded...

We were told Jackson is a graduate textbook. Perhaps I will have a look at it someday.

Do that...

Interestingly enough, just today, in my lecture of the second EM course I'm taking (the "sequel"), we were shown how the Lorentz-force law can be derived by using special relativity, measuring (from a "stationary frame" S,) the force on a charge q moving in some arbitary direction at constant velocity v, due to a source charge Q, moving with velocity V wrt S, in a frame S'.

That equation is one of my favorites:
$$m_{0} \ c \ w_{\mu}=q \ F_{\mu\nu} \ u^{\nu}$$

So elegant,without those E and B's...

Daniel.

 

[FulhamFan3]

the principle of superposition (BS,there's no such thing...)

How do you add vectors then?

 

[dextercioby]

Well,'addition of vectors' is one thing,'fancy words' is another...The former is very mathematical,while the latter is just a fancy way to express the same thing...When i said that "principle of superposition does not exist" i meant that vectors addition exists and not the "principle...".

Vector addition is everything...

Daniel.

 

[krab]

The Principle of Superposition not only exists, it is essential and non-obvious. It means that free space is simple. The electric field from two sources is the sum of the fields from the individual sources. Another way to say it is that free space is linear.

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