Gauss's law in differential forms

In summary: Anyone to help?In summary, the author rejects the derivation of the differential form of Gauss's law from the integral form by using the divergence theorem, citing that the integrands will no more be equal, but the integrals will still be equal, over all volumes. Molu
  • #1
loom91
404
0
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 anyone 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
 
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  • #2
loom91 said:
But this argument is flawed. I can change the value of anyone 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.

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?
 
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  • #3
siddharth said:
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?

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
loom91 said:
Firstly, they are equating volume integrals, not surface ones.

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

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

I don't understand what you're trying to say. Maybe someone with more expertise in math can take a look?
 
  • #5
Two CONTINUOUS functions that are equal almost everywhere are in fact equal everywhere.
 
  • #6
Dick said:
Two CONTINUOUS functions that are equal almost everywhere are in fact equal everywhere.

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
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
Dick said:
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.

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

Molu
 
  • #9
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
You mean, of course, as a limit of the operators those continuous functions represent.
 
  • #11
Hurkyl said:
You mean, of course, as a limit of the operators those continuous functions represent.

Isn't that what I said? Maybe not. Meant to say it though.
 
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  • #12
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
Anyone to help?
 
  • #14
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.
 
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  • #15
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
Hurkyl said:
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.

I fail to see how it can be
a perfectly good proof
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
loom91 said:
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

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
loom91 said:
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

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
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
Hurkyl said:
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!

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
 

1. What is Gauss's law in differential forms?

Gauss's law in differential forms is a mathematical equation used in electromagnetism to describe the relationship between electric fields and electric charges. It is a differential form of Gauss's law and is often written as ∇ · E = ρ/ε0, where ∇ is the divergence operator, E is the electric field, ρ is the charge density, and ε0 is the permittivity of free space.

2. How is Gauss's law in differential forms different from the integral form?

The integral form of Gauss's law is used to calculate the total electric flux through a closed surface, while the differential form is used to calculate the electric field at a specific point in space. The differential form is more useful for solving problems involving point charges or continuous charge distributions.

3. What is the significance of Gauss's law in differential forms?

Gauss's law in differential forms is a fundamental law in electromagnetism that allows us to understand the relationship between electric fields and electric charges. It is used extensively in physics and engineering, particularly in the analysis and design of electrical circuits and devices.

4. What are the applications of Gauss's law in differential forms?

Gauss's law in differential forms has many practical applications in physics and engineering, such as in the design of capacitors, electric motors, and transformers. It is also used in the study of electromagnetic waves and in the development of technologies like wireless communication and radar.

5. How is Gauss's law in differential forms derived?

Gauss's law in differential forms is derived from the integral form of Gauss's law using the divergence theorem, which states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that field over the enclosed volume. Through this mathematical manipulation, we can arrive at the differential form of Gauss's law, which is more convenient for certain types of calculations and problem-solving.

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