# Gauss's law in differential forms

1. Jun 24, 2007

### loom91

Hi,

I'm seeing that many authors like Griffiths and Halliday/Resnick (I've not seen Jackson and Landau/Lif****z) are deriving the differential form of Gauss's law from the integral form (which is easily proven) by using the divergence theorem to convert both sides to volume integrals and then claiming that the integrands must be equal as the integrals are equal over all volumes.

But this argument is flawed. I can change the value of any one integrand over any set of measure zero (such as a countable number of planes) without disturbing the equality of the integrals. The integrands will no more be equal, but the integrals will still be equal, over all volumes.

The derivation by directly calculating the divergence from Coulomb's law also seems dubious. It hinges on writing the derivative of a function that is clearly not continuous, let alone smooth, using Dirac functions (which are of course not functions at all).

How can the differential form be derived rigorously from Coulomb's law/integral form?

Molu

2. Jun 24, 2007

### siddharth

What Griffiths and all says is that, since the integrals are equal for any arbitrary closed surface, the integrands have to be equal for every point in space. I don't see how this argument is flawed?

Last edited: Jun 24, 2007
3. Jun 24, 2007

### loom91

Firstly, they are equating volume integrals, not surface ones.

Secondly, I said why the argument was flawed in my previous post. Equality of n-dimensional integrals, even over all possible n-surfaces, does not rule out pointwise inequality over a set of measure zero (n-1-surfaces, for example).

Molu

4. Jun 24, 2007

### siddharth

Yes, and the volume is bounded by a closed surface.

I don't understand what you're trying to say. Maybe someone with more expertise in math can take a look?

5. Jun 25, 2007

### Dick

Two CONTINUOUS functions that are equal almost everywhere are in fact equal everywhere.

6. Jun 25, 2007

### loom91

That seems reasonable. But is the electric field required to be continuous? That doesn't seem to a commonly applied boundary condition. In fact, in classical electrodynamics, electric fields are assumed to be discontinuous at the surface of perfect conductors in electrostatic equilibrium.

Thanks.

Molu

7. Jun 25, 2007

### Dick

A discontinuity in the electric field can only come from a singularity in the charge distribution. In your perfect conductor example there is a Dirac delta function like sheet of surface charge. In reality its not singular, just very concentrated, but it's a useful approximation.

8. Jun 25, 2007

### loom91

But that means the continuity is not a necessary boundary condition. Then it can not be used to validate the derivation.

Molu

9. Jun 25, 2007

### Dick

If you insist on complete rigor, the way to deal with a delta function is to represent it as a limit of continuous functions (representatives of delta functions). So a delta function can be represented arbitrarily well by a continuous function. As I said, a delta function is a nonphysical (but useful) idealization. Indeed when Heaviside invented the operational calculus it was attacked as being nonrigorous. But they can be dealt with rigorously as distributions.

10. Jun 25, 2007

### Hurkyl

Staff Emeritus
You mean, of course, as a limit of the operators those continuous functions represent.

11. Jun 25, 2007

### Dick

Isn't that what I said? Maybe not. Meant to say it though.

Last edited: Jun 25, 2007
12. Jun 28, 2007

### loom91

So, can you show/point out to me a mathematically valid derivation of the differential form of the Gauss' law? While I like physics far more than mathematics, I also like the mathematics in my physics to be real mathematics, instead of the watered-down hand-waving version used by most physicists. Thanks!

Molu

13. Jun 30, 2007

### loom91

Anyone to help?

14. Jun 30, 2007

### Hurkyl

Staff Emeritus
Before one can rigorously prove Gauss's law, one must precisely state Gauss's law. The proof you reject is a perfectly good proof of one statement of (the differential form of) Gauss's law. Explicitly state what you want to see proved, if that was not it.

Last edited: Jun 30, 2007
15. Jun 30, 2007

### Dick

At this point you don't seem to have any problem with continuous charge distributions. To extend to delta function type charge distributions just do what everyone else does and consider them as limits of continuous distributions.

16. Jul 1, 2007

### loom91

I fail to see how it can be
when it depends on mathematically invalid steps such as differentiating functions that are not differentiable and claiming equality of integrals implies equality of integrands. I do not believe any competent mathematician would consider these to be valid steps in a proof without further justification. I'm not rejecting anything, just looking for an explanation that makes sense.

Molu

17. Jul 1, 2007

### Hurkyl

Staff Emeritus
There's a reason I said it's important to explicitly state the statement in which you're interested.

Some discussions only consider smooth vector fields, and the divergence operator you learned in your elementary calculus classes. That the integrands are equal is a easy consequence of continuity.

Other contexts consider more general classes of vector fields, and even generalizations of that notion! They also use generalizations of the divergence operator. Without knowing what type of objects you are using for your fields, and what definition of divergence you are using, it is essentially impossible to provide a proof that would satisfy you!

18. Jul 2, 2007

### Reshma

David Griffiths' book has all the answers that you are looking for. The differential form of Gauss's law can be found using the divergence theorem dealing with volume and surface integrals. Note that we are dealing with a flux of an electric field, and hence we need to consider a surface area vector in a definite direction not just an integral over all possible areas. Just go through this link to clear up your doubts:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

19. Jul 2, 2007

### pardesi

also almost immediatley after the proof using divergence theorem there is a proof by the dirac delta function in griffith i u think this is illogical(though it isn't) u can refer that proof

20. Jul 2, 2007

### loom91

Well, a general electric field is not required to be differentiable, or even continuous. The integral form of Gauss's law does not impose this requirement. But since the divergence operator is by definition a derivative, it seems inadequate to deal with discontinuous fields. Claiming Dirac deltas to be the derivatives of functions that can not be differentiated seems to be a dubious step. What is this generalised divergence operator you speak of? I only know of the definition using space derivatives. Thanks.

Molu