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Swapnil Aug28-06 05:33 PM

Proof of Gauss's Law
 
I was wondering, how would you prove Gauss's law (either mathematically or intuitively)? I mean, I know that sometimes people take it as the fundamental law (one without proof) but how would you derive such a law from Coulomb's law?

Any help would be extremely appreciated. I have been fussing over this topic for a month now.

leright Aug28-06 05:44 PM

http://www.rpi.edu/dept/phys/Dept2/p...uss/proof.html

Swapnil Aug28-06 07:49 PM

Thank you very much. That was a good website that gave an intuitive feel for the Gauss's law. Is it possible for you (or someone else) to give me a link that explains the mathematical derivation of Gauss's law (possible using Stoke's theorem or something)?

leright Aug28-06 08:15 PM

Quote:

Quote by Swapnil
Thank you very much. That was a good website that gave an intuitive feel for the Gauss's law. Is it possible for you (or someone else) to give me a link that explains the mathematical derivation of Gauss's law (possible using Stoke's theorem or something)?

That IS the mathematical derivation of Gauss's law. Stoke's theorem has nothing to do with gauss's law. However, we can go from the integral form of gauss's law to the differential form of gauss's law using the DIVERGENCE theorem, but the application of the divergence theorem in order to do this is quite trivial.

Stokes theorem is used to transform the integral forms of maxwell's third equation (Faraday's law) and fourth equation (ampere's law) into their differential forms.

Swapnil Aug29-06 08:46 AM

Quote:

Quote by leright
Stokes theorem is used to transform the integral forms of maxwell's third equation (Faraday's law) and fourth equation (ampere's law) into their differential forms.

Oh yeah. Sorry about that. I was thinking Divergence theorem but I wrote down Stokes' theorem.

Quote:

Quote by leright
That IS the mathematical derivation of Gauss's law. Stoke's theorem has nothing to do with gauss's law. However, we can go from the integral form of gauss's law to the differential form of gauss's law using the DIVERGENCE theorem, but the application of the divergence theorem in order to do this is quite trivial.

Wait. Is that what Divergence theorem is used for, just to tranform Gauss' law from integral form to differential form and vice-versa? This whole time I was thinking that Divegence theorem was used to derive Gauss' law from Coulomb's law. :blushing:

So in the end there is no rigorous proof of Gauss' law (the fact that it works for ANY closed surface)?

Meir Achuz Aug29-06 09:00 AM

Quote:

Quote by Swapnil
So in the end there is no rigorous proof of Gauss' law (the fact that it works for ANY closed surface)?

A simple derivation uses the definition of differential solid angle:
d\Omega=r.dA/r^3. Then \int qdA.r/r^3=4\pi q.
For q outside the closed surface, the sign of the solid angle has to be used.

rbj Aug29-06 09:32 PM

Quote:

Quote by Swapnil
Wait. Is that what Divergence theorem is used for, just to tranform Gauss' law from integral form to differential form and vice-versa? This whole time I was thinking that Divegence theorem was used to derive Gauss' law from Coulomb's law.

the divergence

[tex] \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} [/tex]

is saying the same thing as

[tex] \oint_S \mathbf{E} \cdot d\mathbf{A} = \int_V \frac{\rho}{\epsilon_0} dV = \frac{Q}{\epsilon_0} [/tex]

one is the differential form, the other is the integral form.

Quote:

So in the end there is no rigorous proof of Gauss' law (the fact that it works for ANY closed surface)?
there are plenty of decent E&M fields books that prove it rigorously. can you check one out and look it up?

Swapnil Aug30-06 04:30 PM

Quote:

Quote by rbj
There are plenty of decent E&M fields books that prove it rigorously. can you check one out and look it up?

I have read/looked-over a lot of EM books, but none of 'em offer a satisfying proof of Gauss' law. The most common response is in terms of field lines, which I think is just a clever trick to avoid answering the question.

Galileo Aug30-06 04:38 PM

Here's the idea of the proof in short, you can work out the details yourself:

(I): Show that [itex]\nabla \cdot \vec F =0[/itex] if F has the form: [itex]\hat F=K\hat r/r^2[/itex] ([itex]r\neq 0[/itex]).

(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.

quasar987 Aug30-06 04:41 PM

Quote:

Quote by rbj
there are plenty of decent E&M fields books that prove it rigorously. can you check one out and look it up?

Do you call a proof that uses

[tex] \nabla \cdot \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3} = 4\pi \delta^3(\vec{r}-\vec{r}') [/tex]

rigorous? Just curious.

rbj Aug30-06 08:32 PM

Quote:

Quote by Swapnil
I have read/looked-over a lot of EM books, but none of 'em offer a satisfying proof of Gauss' law. The most common response is in terms of field lines, which I think is just a clever trick to avoid answering the question.

Okay, let's begin with the concept of flux and the natural justification of the inverse-square law in our existance in 3 spatial dimensions.

As an illustrative example, let's think about the concept of power and intensity. At the beach, when you feel the sun's radiation on your skin, what you are measuring is how much energy (in a particular band of frequencies) is impinging upon your skin in a particular amount of time. If you're out in the sun longer, more energy falls on your skin (this is the concept of power: how much energy per unit time) and, in the same intensity of sunlight, the bigger you are, the more of that sunlight you will scoop up: more area exposed to the sun, more power that falls on that area. So, whether it is light or sound waves (or something else), intensity is amount of power that falls on a unit area that is positioned perpendicular to the oncoming wave. it's watts/m2. double the area, you get twice as much power. Double the intensity of radiation, you also get twice as much power.

Okay, consider a [itex]P_0[/itex] watt light bulb and assume we're measuring all of the radiation (infared, too) at a distance of [itex]r[/itex] meters. Imagine a sphere of radius of [itex]r[/itex] surrounding and centered at the light bulb. Conservation of energy says that the entire [itex]P_0[/itex] watts is distributed over the entire surface area of that sphere , which is [itex]4 \pi r^2[/itex] and the argument of symmetry says that every one of those m2 gets the same amount of power. And every little snippet of area is directly facing the light bulb and is perpendicular to the wave front. So, at that distance the power, [itex]P_A[/itex], falling on an area of [itex]A[/itex] is

[tex] P_A = \frac{P_0}{4 \pi r^2} A [/tex]

the factor

[tex] I = \frac{P_0}{4 \pi r^2} [/tex]

is the intensity of radiation and is proportional to the power of the point source, as one would expect. Note also, that with a constant source power, the intensity is also proportional to the reciprocal of the square of the distance from the source. A natural inverse-square behavior. Note also that if one were to integrate the intensity (which is [itex] \frac{P_0}{4 \pi r^2} [/itex] over the entire spherical surface area (which is [itex] 4 \pi r^2 [/itex]) i would be adding up all the little snippets of power falling on the little snippets of area and that would add to the total power [itex] P_0 [/itex] emitted by the point source. Doesn't matter how big the sphere is. If the sphere is twice as big, it has 4 times the surface area, but the intensity is 1/4 as large and it comes out to be the same total power. Adding up all of the little snippets of power impinging upon the surface of the sphere equals the power emitted by the source inside and we can thank the Conservation of Energy law for that simple fact.

Now here's a little example:

Someone, somewhere measured the intensity of radiation from the sun (the power falling on a square meter facing the sun out in a vacuum in Earth orbit) to be: 1360.77 W/m2. We also know that we are 149.6 x 109 meters (about 93 million miles) from the sun. From that we can calculate how big of a light bulb that the sun is:

[tex] I = \frac{P_0}{4 \pi r^2} [/tex]

[tex] 1360.77 \ \mbox{W/m}^2 = \frac{P_0}{4 \pi (149.6 \times 10^9 \ \mbox{m})^2} [/tex]

results in:

[tex] P_0 = 1360.77 \ (\mbox{W/m}^2) 4 \pi (149.6 \times 10^9 \ \mbox{m})^2 = 3.827 \times 10^{26} \ \mbox{W} [/tex]


Now this is intensity, not electric field, but both share the same mathematics. Both are proportional to the source agent (power of point source vs. electric charge of point source) and both are a inverse-square relationship with distance. Now, instead of calling this quantity "intensity" which is power per unit area, for electric charges this natural inverse-square quantity is called "flux density" or "electrostatic flux density", and if we were talking about this other inverse-square phenomenon called gravity, it would be "gravitational flux density". These are those field lines that they like to draw and you don't like. But it doesn't matter, if experimental evidence shows that the electrostatic force follows the inverse-square law precisely (and it does, experiment confirms the exponent of [itex]r^{-n}[/itex] to be be [itex]1.99999999 < n < 2.00000001[/itex], then whether or not you like it, that density of lines of field model is valid, even if it comes from our imagination. If the E field is inverse-square, we can imagine a point source radially emitting a bunch of field lines (a.k.a. "total flux"), the quantity of which is proportional to the amount of charge, and the density of those field lines crossing a perpendicular surfaces (a.k.a. "flux density") is proportional to the number emitted and inversely proportional to the square of the distance.

Now we're moving toward proving:

[tex] \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0} [/tex]

but I want you to get used to these first concepts before moving on. If you don't like any of the above, I ain't gonna waste time continuing.

Another installment to follow.

Meir Achuz Aug31-06 04:13 AM

Quote:

Quote by Meir Achuz
A simple derivation uses the definition of differential solid angle:
d\Omega=r.dA/r^3. Then \int qdA.r/r^3=4\pi q.
For q outside the closed surface, the sign of the solid angle has to be used.

Is anything misssing?

Swapnil Sep1-06 11:42 AM

Quote:

Quote by Galileo
Here's the idea of the proof in short, you can work out the details yourself:

(I): Show that [itex]\nabla \cdot \vec F =0[/itex] if F has the form: [itex]\hat F=K\hat r/r^2[/itex] ([itex]r\neq 0[/itex]).

You mean [itex]\nabla \cdot \vec F \neq 0[/itex], right?

quasar987 Sep1-06 11:44 AM

(No, he meant =0)

leright Sep1-06 01:06 PM

Quote:

Quote by Swapnil
You mean [itex]\nabla \cdot \vec F \neq 0[/itex], right?

No, he meant = 0. The only point where there is a non-zero divergence in that particular vector field is where r = 0, but he excluded that point. Everywhere else the divergence is zero.

Swapnil Sep2-06 11:06 PM

I did the calculation for [tex] \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}[/tex] and I get [tex]\frac{1}{x^2+y^2+z^2}[/tex].

Also, why should the divergence be zero, don't we have a radial field?:confused:

Galileo Sep3-06 03:28 AM

Well, that's one of the special properties of inverse square field. It may look like it has a divergence, but it doesn't and it's the underlying reason for Gauss' law.

You could do it in cartesian coordinates:
[tex]\vec F(x,y,z)=K \frac{\hat r}{r^2}=K \frac{\vec r}{r^3}=K \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{3/2}}[/tex]
but that's really a bit of a hassle.
Try it in spherical coordinates, whhich makes it trivial.

leright Sep3-06 09:24 AM

Quote:

Quote by Swapnil
I did the calculation for [tex] \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}[/tex] and I get [tex]\frac{1}{x^2+y^2+z^2}[/tex].


Also, why should the divergence be zero, don't we have a radial field?:confused:

It seems you're having trouble grasping divergence.

The divergence of a point in space is the NET outward flux through an arbitrarily small volume, per unit volume, enclosing this point, where the small volume approaches zero. BY DEFINITION, a radial field HAS NO DIVERGENCE except at r = 0. All flux going into all arbitrarily small volumes must go out of the small volume, causing a net outward flux of zero, or, a divergence of zero at all points. However, if you determine the divergence at the point where there is charge, you are at the point where flux lines EMANATE, so through an arbitrarily small volume enclosing this point there IS a net outward flux. Gauss's law in differential form states that the divergence of the electric flux density is equal to the volume charge density at that point.

We can apply the integral form of Gauss's law and find the net outward flux through a surface enclosing a charge distribution, and then take the limit of the outward flux as the volume approaches zero. However, you might notice that this limit is zero. In fact, this limit is zero for all charge distributions and all fields, which is quite obvious. This is where divergence of a point comes in. It is, as I stated above, the net outward flux of an arbitrarily small volume, PER UNIT VOLUME, enclosing a point in a vector field. IF we find the limit of the integral form of gauss's law to a volume where the size of that volume goes to zero about a point, the result is zero. However, if we find the divergence, we get the limit of the charge per unit volume, at that point, which is a finite number.

The derivation for Gauss's law, in its most simple form, is a calculation of the net outward flux through a closed surface, where that closed surface encloses a point charge, and this calculation is performed using a closed surface with radial symmetry, and the center of this surface is centered at the point charge. Due to symmetry, E is a constant and da is a constant, and both vectors E and da are at an angle of 0 degrees, so the flux simply becomes Q/ep_0. So, we have proved that for a point charge with a spherical closed surface enclosing it, the net outward flux of the E-field is Q/ep_0. We can rewrite this by finding the net outward flux of the D-field for this configuration, and we simply get Q.

The issue, as you stated, is applying this to all charge distributions and all volumes, ans gauss's law says we can do this. Well, I suppose you could just say that no matter what surface you have enclosing the charge, your net ourward flux will be the same, and therefore enclosed charge will be the same. But this is where I bring in intuition....

Gauss's law is just something that is second nature to me I suppose.


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