[Physics] Mathematical proof Maxwell's equations are incorrect?

Ilja Schmelzer ilja.schmelzer at gmail.com
Fri Apr 24 14:57:26 CEST 2020


2020-04-24 16:16 GMT+06:30, Arend Lammertink <lamare at gmail.com>:
> On Fri, Apr 24, 2020 at 8:16 AM Ilja Schmelzer <ilja.schmelzer at gmail.com>
> wrote:

> I really don't see how one could possibly argue that a vector function
> 𝐅 exists, such that Maxwell's eqations form it's Laplacian and
> therewith it's second order derivative.

And I see no reason why one should make such an argument.

First of all, the Laplacian of a function is another function,
and not an equation.  But, whatever this means, why would somebody
claim that, say, solutions of the Maxwell equations are second derivatives
of whatever?

> https://en.wikipedia.org/wiki/Lorentz_ether_theory
> "Lorentz tried in 1899 and 1904 to expand his theory to all orders in
> v/c by introducing the Lorentz transformation."

> It's the very application of the Lorentz transformation to physics
> that is the problem.

No, it is not a problem at all.

> And the reason it exists is because of, you
> guessed it: Maxwell, or at the very least a misinterpretation thereof:
>
> http://www.etherphysics.net/CKT4.pdf

How this confused paper made it through peer review is beyond my understanding.

> https://en.wikipedia.org/wiki/Magnetic_vector_potential#Gauge_choices
> "The above definition does not define the magnetic vector potential
> uniquely because, by definition, we can arbitrarily add curl-free
> components to the magnetic potential without changing the observed
> magnetic field. Thus, there is a degree of freedom available when
> choosing A. This condition is known as gauge invariance."
>
> Would it be far fetched to suggest that the very reason the magnetic
> vector potential has not been defined uniquely is because no vector
> function 𝐅 exists, such that Maxwell's equations form it's Laplacian?

First you would have to suggest something which transform a vector function
into some equations (instead of, say, another vector function like the
Laplacian)
and is nonetheless named "Laplacian".

> The gauge freedom within Maxwell is around the definition of the
> vector potential, as quoted above:
> "we can arbitrarily add curl-free components to the magnetic potential
> without changing the observed magnetic field"
>
> Could it be that because of the vector identity ∇⋅𝐁= 0 that "some
> configurations of the field", namely the addition of curl-free
> components to the field, by definition results in the zero vector?

Again, you have to explain what you mean, given that addition is
an operation, but a configuration is not.

> I respectfully disagree that there's no reason to care about the
> origins of our theories and to check their validity.

Fine, check whatever you like to check,
But what follows if it appears that the origins are nonsense?

The theory may be nonetheless meaningful and make correct
predictions.  Kepler believed in astrology, and his origins were
astrological nonsense.  But his laws of motion for the planets were
nonetheless very good approximations.

> Actually, I don't think it's an exxageration to state that this issue
> defines the difference between a theory of everything and 150+ years
> of trying to find additional equations in order to correct the obvious
> violation of the elementary math defined by the Laplace operator.

I see up to now only confusion. Sorry.

Greetings Ilja



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