The integral form and the differential form

It is well known that Maxwell's equations of electro-magnetism have an
integral and a differential form, which are usually considered
equivalent. I have realised that this can not be so.

Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite. Therefore the differential equations can only
be an approximation, while the integral forms are exact. This can also
be checked by noting that the integral form can be proven from the
differential (divergence theorem etc.), while going the other way
requires an assumption of continuity.

Matter, like charge, is discrete and not continuous, so the same must
be true of Einstein's equation of general relativity. If it can not be
written in an integral form, it is wrong !! Can it be? (I imagine one
would have to use the flat-spacetime formulation, which would itself
be interesting in suggesting that spacetime really is necessarily
globally flat.)

Andrew Usher

"Andrew Usher" wrote in message
...
It is well known that Maxwell's equations of electro-magnetism have an
integral and a differential form, which are usually considered
equivalent. I have realised that this can not be so.


http://en.wikipedia.org/wiki/Fundame...em_of_calculus
Read, absorb, and inwardly digest.

On Jul 16, 6:18*pm, Andrew Usher wrote:
It is well known that Maxwell's equations of electro-magnetism have an
integral and a differential form, which are usually considered
equivalent. I have realised that this can not be so.

You are wrong. The integral and differential forms contain the exact
same information - there has been a lot written about this, you should
consider doing some light reading.


Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite.

Your entire argument is based on a silly and incorrect claim. This is
pointless.

[snip conclusion based on a false premise]

On Jul 17, 6:02 am, Eric Gisse wrote:

You are wrong. The integral and differential forms contain the exact
same information - there has been a lot written about this, you should
consider doing some light reading.

I am aware of the derivations and they are as I stated. In particular,
getting the differential from the integral form requires assumptions
of
continuity (in the form that 'charge density' and 'current density'
are
integrable functions).

Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite.

Your entire argument is based on a silly and incorrect claim. This is
pointless.

No, this is correct. What is the 'charge density' at a point? Clearly,
0
unless the point coincides with a charged elementary particle, where
it is infinite. It's as a delta function, which is not differentiable.

Andrew Usher

In sci.astro Andrew Usher wrote:
On Jul 17, 6:02 am, Eric Gisse wrote:

You are wrong. The integral and differential forms contain the exact
same information - there has been a lot written about this, you should
consider doing some light reading.

I am aware of the derivations and they are as I stated. In particular,
getting the differential from the integral form requires assumptions
of continuity (in the form that 'charge density' and 'current density'
are integrable functions).

It's enough for them to be distributions (in the mathematical sense
-- e.g., Dirac delta functions or their derivatives).

Charge, after all, is quantised, and the derivatives therefore are
all either zero or infinite.

Classically, these derivatives are well-defined as distributions. See,
for example, chapter 14 of Jackson, _Classical Electrodynamics_.
(The chapter may vary depending on which edition you have; look
for the section "Lienard-Wiechert Potentials and Fields for a Point
Charge.")

[...] What is the 'charge density' at a point? Clearly, 0
unless the point coincides with a charged elementary particle, where
it is infinite. It's as a delta function, which is not differentiable.

A delta function is differentiable, as a distribution. Derivatives of
delta functions are used all the time.

(Of course, in quantum electrodynamics, the differential form of
Maxwell's equations holds in the Heisenberg picture as a set of
operator equations, for which the issues are different.)

Steve Carlip

On Jul 16, 10:18*pm, Andrew Usher wrote:
It is well known that Maxwell's equations of electro-magnetism have an
integral and a differential form, which are usually considered
equivalent. I have realised that this can not be so.

Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite. Therefore the differential equations can only
be an approximation, while the integral forms are exact. This can also
be checked by noting that the integral form can be proven from the
differential (divergence theorem etc.), while going the other way
requires an assumption of continuity.

Matter, like charge, is discrete and not continuous, so the same must
be true of Einstein's equation of general relativity. If it can not be
written in an integral form, it is wrong !! Can it be? (I imagine one
would have to use the flat-spacetime formulation, which would itself
be interesting in suggesting that spacetime really is necessarily
globally flat.)

Andrew Usher

A lot of people have commented that relativity is a classical physics
theory/mathematics and so cannot be completely right as it is not
quantum mechanical. Hence, the search for "unification."

Andrew Usher wrote:

On Jul 17, 6:02 am, Eric Gisse wrote:

You are wrong. The integral and differential forms contain the exact
same information - there has been a lot written about this, you should
consider doing some light reading.

I am aware of the derivations and they are as I stated. In particular,
getting the differential from the integral form requires assumptions
of
continuity (in the form that 'charge density' and 'current density'
are
integrable functions).

Being integrable is not the same as being continuous. Steve Carlip mentioned
the Dirac delta function. Consider other common discontinuous but integrable
functions like the step and sawtooth functions.

[...]

On Jul 16, 10:18*pm, Andrew Usher wrote:
[snip]
Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite. Therefore the differential equations can only
be an approximation, while the integral forms are exact.
[snip]

You need to study up on what "quantized" means. It does
not mean that the things you are considering taking the
derivative of are not continuous. Go back and read your
intro quantum text again. Get a few more. The functions
are quantum wave functions. Figure out what that means.

Then go away and stop being so silly.
Socks

On Jul 16, 10:18*pm, Andrew Usher wrote:

Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite. Therefore the differential equations can only
be an approximation, while the integral forms are exact. This can also
be checked by noting that the integral form can be proven from the
differential (divergence theorem etc.), while going the other way
requires an assumption of continuity.

1. Maxwell's equations are based upon an incompressible fluid model.
Thus, they fail when functions are not continuous and differentiable.
This is easily shown for example in that switching circuits do not
follow Faraday's law of magnetic induction.

2. Both the differential and integral forms fail at quantized (non-
continuous) levels.

3. Any continuous fluid model will require enough discrete elements so
that they approximate a continuous fluid to some degree or it fails to
give even approximate valid answers.

4. Math is not more real than reality.

5. I hope you've noticed that everyone attempting to "answer" your
question is a moron.

Benj wrote:

On Jul 16, 10:18 pm, Andrew Usher wrote:

Charge, after all, is quantised, and the derivatives therefore are all
either zero or infinite. Therefore the differential equations can only
be an approximation, while the integral forms are exact. This can also
be checked by noting that the integral form can be proven from the
differential (divergence theorem etc.), while going the other way
requires an assumption of continuity.

1. Maxwell's equations are based upon an incompressible fluid model.

No, they are not.

Thus, they fail when functions are not continuous and differentiable.

This sentence proves beyond the shadow of a doubt that you have never taken
- much less passed - a course in electrodynamics.

This is easily shown for example in that switching circuits do not
follow Faraday's law of magnetic induction.

Oh, do we have a retired engineer on our hands?


2. Both the differential and integral forms fail at quantized (non-
continuous) levels.

That's because classical E&M is ... get this ... classical. However, the
covariant Maxwell's equations work just fine for quantum field theory.


3. Any continuous fluid model will require enough discrete elements so
that they approximate a continuous fluid to some degree or it fails to
give even approximate valid answers.

Maxwell's equations aren't based on fluids. Moron.


4. Math is not more real than reality.

Since you understand neither, I suppose we should take your word for it.


5. I hope you've noticed that everyone attempting to "answer" your
question is a moron.

Have you ever set foot inside a classroom that taught the subject?

Have you ever read a book that taught the subject at a level commensurate
with your mouth? Griffiths? Jackson? Hmm?