Hi Arend, I got stuck at the bit where you said,

"Since v, E and B all have a unit of measurement of velocity in meters per second [m/s],"

Any study in this field that I have ever done considers E to be a Force (force per unit charge) and B to be an angular momentum density. If on the other hand, you begin with A being the primary momentum flow, then Faraday's law falls out automatically as is shown in this paper,

Article Faraday's Law of Electromagnetic Induction

Frederick David Tombe This is more about the mathematical problem and the fluid dynamics domain rather than electroynamics.

What I've found is that the Helmholtz theorem is indeed used within the fluid dynamics domain, but not with a separate component E and B, as is being done within the electrodynamics domain.

This is how it is commonly done within the fluid dynamics domain (eq. 17-19), as far as I can tell:

https://pdfs.semanticscholar.org/9344/48b028a3a51a7567c2b441b5ca3e49ebb85c.pdf

So, what I'm proposing is to define the potential fields for the general case by writing out the terms in the LaPlace operator, which is the 3D generalization of the 1D second derivative. This is slightly different from how they commonly do it, because the scalar potential is negated in my proposal compared to the common definition, which does not make much difference. In addition to that, there's the proposal to also use a separate E and a B component for the two halves of the decomposition within fluid dynamics and this way the definitions become not only general, but also the same as in Maxwell's equations for the "static" case as well as my revision thereof.

The problem for both domains that comes forth this way is the same, namely that we run into the problem that the Helmholtz decomposition has a limit of applicability and that even though it states F should exist and should be uniquely defined by the two potential fields, F cannot actually be defined for the limit when h->0, which is very interesting and has consequences.

I've also posted this question here, where I've edited it a bit further and added some things:

https://math.stackexchange.com/questions/3721666/how-to-compute-divcurl-of-volume-velocity-field-with-helmholtz-decomposition-if

Hopefully, some mathematician finds this problem interesting enough to think about, because it has quite a lot of consequences for the actual applicability of the Helmholtz decomposition in the general case as well. Now that we have shown that the Helmholtz decomposition does not actually hold in this case, it is an interesting question for mathematicians to figure out when this is the case and what consequences this has.

Hi Arend, OK, but sticking exclusively to fluid dynamics, you will have to choose a vector which represents the velocity field, and it would seem that v is that vector. So what exactly does your vector field F represent in fluid dynamics?

Hi Frederick,

Yep, it is common practice in fluid dynamics to work with the flow velocity field v. See for instance:

https://archive.org/details/ACourseInFluidMechanicsWithVectorFieldTheory

It is very interesting to note that in this work, the term "vector potential" appears only twice and there's this interesting statement regarding the scalar potential:

"This serious discrepancy between the predictions of potential flow theory and experiment is known as: d'Alembert'sß paradox - potential flow predicts no drag but experiments indicate drag."

What it comes down to is that the current use of a vector potential within fluid dynamics is rather limited:

"Fortunately, there are several broad classes of flows for which the form of the vector potential is known: TWO-D FLOWS"

And this has to do with "stream functions":

https://en.wikipedia.org/wiki/Stream_function

"The stream function is defined for incompressible (divergence-free) flows in two dimensions – as well as in three dimensions with axisymmetry. The flow velocity components can be expressed as the derivatives of the scalar stream function."

As far as I can tell, the current use of a vector potential within fluid dynamics is limited to this simplification, but there is no notion that the 3D vector potential is actually closely related to the scalar potential via the vector LaPlace operator, as shown in the question text above.

Either way, what F represents is actually one of the questions. Because the curl, div and grad operators all have a unit of measurement in per meter [/m], all we know for certain is that F has a unit of measurement in [m3/s] and that it's a vector, so one is tempted to conclude it represents a volumetric flow, similar to the volumetric flow rate but represented by a vector rather than a scalar.

The problem with this is that in reality, no such thing as a point without a volume actually exists, while with infinitesimal calculus that's exactly how a point is described. So, when one considers each "point" in the vector field to have a volume, as you do with discrete mathematics, F can be computed and represents the volume of flow going trough your "point volume" per second. With infinitesimal calculus, the volume of your "point" goes to zero and that is why F cannot be computed. It seems to be undefined, which is very interesting from a mathematical point of view, because the Helmholtz decomposition also says (slightly edited):

https://en.wikipedia.org/wiki/Helmholtz_decomposition#Fields_with_prescribed_divergence_and_curl

"Let A be a solenoidal vector field and Φ a scalar field on R3 which are sufficiently smooth and which vanish faster than 1/r2 at infinity. Then there exists a vector field F such that:

∇⋅F=Φ and ∇×F=A,

and if additionally, the vector field F vanishes as r → ∞, then F is unique."

This is very confusing. I first thought we needed to go to discrete mathematics, and then I though: No, the solution F=0 is unique and that's very nice.

Until I realied that both the divergence as well as the curl of such an F are, well, zero. :(

So, I have some rewriting to do of my paper and this time I'd like to do it correctly.

The idea is that because both the two potential fields as well as the three velocity fields can be evaluated and F seems to be a multiplication of v times some (infinitesimal) surface area dS, it should be possible to compute the curl F and div F from the velocity field v, even though F itself cannot be computed for a point that has no volume.

Conceptually, this would involve re-arranging the integrals involved in the calculation, but I haven't figured out how to do that yet.

But it is interesting to consider the divergence theorem:

https://en.wikipedia.org/wiki/Divergence_theorem

"the divergence theorem states that the surface integral of a vector field over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence over the region inside the surface."

And the Kelvin–Stokes theorem:

https://en.wikipedia.org/wiki/Kelvin–Stokes_theorem

"Given a vector field, the theorem relates the integral of the curl of the vector field over some surface, to the line integral of the vector field around the boundary of the surface."

So, it seems that F is "the volume integral of the divergence" over a certain volume plus "the integral of the curl of the vector field" over the surface area that encloses said volume. However, Kelvin-Stokes theorem does not handle a closed surface, so we'd have to look at a more general theorem:

https://en.wikipedia.org/wiki/Stokes%27_theorem

"The fundamental theorem of calculus states that the integral of a function f over the interval [a, b] can be calculated by finding an antiderivative F of f [...] Stokes' theorem is a vast generalization of this theorem."

Sine the vector LaPlacian is the 3D generatlization of the second derivative, Stokes theorem, it is rather interesting that Stokes theorem essentially describes the principle of integration in 3D:

"Stokes' theorem says that the integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω."

So, this is a generalization of:

"This classical Kelvin–Stokes theorem relates the surface integral of the curl of a vector field F over a surface (that is, the flux of curl F) in Euclidean three-space to the line integral of the vector field over its boundary (also known as the loop integral)."

That seems to lead to the following:

The surface integral of the curl of a vector field F over the boundary surface of some manifold Ω is equal to the volume integral of the derivative of curl F, which would be B.

But because divergence is also a differential form, it seems one can also write:

The surface integral of the divergence of a vector field F over the boundary surface of some manifold Ω is equal to the volume integral of the derivative of div F, which would be E.

And since v = E + B, it seems that with some clever math it is possible to compute curl F as well as div F from v, even though F itself is undefined in infinitesimal notation and can thus not be computed.

Frederick David Tombe Feynman also has some interesting info on fluid flow:

https://www.feynmanlectures.caltech.edu/II_40.html

There's a paragraph on vortex lines:

"By vortex lines we mean field lines that have the direction of Ω and have a density in any region proportional to the magnitude of Ω. From II the divergence of Ω is always zero (remember—Section 3-7—that the divergence of a curl is always zero). So vortex lines are like lines of B—they never start or stop, and will tend to go in closed loops."

https://www.feynmanlectures.caltech.edu/II_41.html

Hi Arend, Yes, but why go into all this detail? If we start on the premise that A is a momentum density field in the aether, then Faraday's Law falls out, simply from differentiation. This short paper here tells you all you need to know about fluid dynamics where it relates to electromagnetism,

Article Faraday's Law of Electromagnetic Induction

Hi Frederick David Tombe ,

Had a whole reply typed, but hit a wrong key and it got lost. This time I used a text editor and cut/paste. :)

Your paper is very interesting and I think it’s important and your contribution helps to lead to the solution to be able to describe Faraday’s law correctly.

However, you make a few mistakes, most notably eq. 5 and 6, which illustrate why details matter.

Let’s first note that I had to redefine the vector potential to:

Hem = ∇×𝐀em

That way, both the E and H fields are in [m/s] and both potentials are in [m^2/s] and thus do not represent a velocity nor momentum.

The new definition for the vector potential involves division by μ, which has a unit of measurement in [kg/m-s] = [C/m] and is one and the same thing as the viscosity of the medium.

So, then the unit of measurement for the old vector potential, defined as:

B = ∇×𝐀

would be:

[kg/m-s] * [m^2/s] = [kg-m/s^2], which would thus represent the time derivative of momentum.

And the unit of measurement for B, defined as

B = μH

would be:

[kg/m-s] * [m/s] = [kg/s^2] = [C/s], which would thus represent a “current”. However, this is not the kind of “current” we normally call a “current”, since that one is (now) defined as:

Jm = e ω = e ∇×Hem.

It is rather interesting to realize that with the new definitions both B = μH and Jm = e ∇×H lead to one and the same unit of measurement in Amperes and that all electrodynamic units of measurement match seemlessly to the fluid dynamics ones, which is accomplished by just two redefinitions:

Jm = e ω = e ∇×Hem

Hem = ∇×𝐀em

From here, let’s now take a look at your equation 5:

E = −∂A/∂t + v×B − ∇(A∙v)

For the old A you’re working with, we have a unit of measurement in [kg-m/s^2], so it’s time derivative is in [kg-m/s], describing an angular momentum. This is not the same as [m/s], so the units of measurement do not match with this equation.

For the new A, we have a unit of measurement in [m^2/s], so it’s time derivative is in [m^2/s^2], which also does not match the unit for E.

In other words: this equation is incorrect.

Further, when you go from eq. 5 to eq. 6, you forget that E is defined as the gradient of the scalar potential Φ, so it’s curl is zero as well. And thus eq. 6 would become:

−∂B/∂t − (v∙∇)B = 0.

Also, v is (now) defined as:

v = E + H,

so it does denote a vector field and since we know E is curl-free, we can also write:

Jm = e ω = e ∇×v

So, it would be very interesting to work out your eq. 2, also in relation to the question that is the subject of this discussion. While we can write:

∇×v = Jm/e,

we don’t have a similar expression for ∇v. However, it’s unit of measurement would be in [/s], thus denoting some kind of frequency, which would not be an angular frequency as denoted by ω (a vector) in the case of the magnetic field, so let’s denote it by 𝜐, which I think is a scalar.

So, what we get when we multiply ∇v with e, elemental charge, we obtain the same unit of measurement as for Jm in [C/s], [A] or [kg/s^2].

Since momentum is in [kg-m/s], it seems that current is momentum times some distance times some time, so it seems that “current” actually is the integration of (angular) momentum over some distance and over some time. And thus, when we take the time derivative of “current” and we apply either the grad or curl operator, we are left with a unit of measurement representing momentum.

And when we divide momentum by μ, denoting viscosity or permeability, we obtain:

[kg-m/s] * [m-s/kg] = [m^2], so the two (new) potential fields represent an integration over time of momentum divided by viscosity or permeability.

Either way, what is very interesting is that it seems to be possible to define another type of “current”, dielectric current, which is complementary to Ampere’s definition like this:

Jd = e 𝜐 = e ∇E

And since v is (now) defined as:

v = E + H,

and since we know B is divergence-free, we can also write:

Jd = e 𝜐 = e ∇v

And thus we would come to the conlusion that the phenemenon we call “current” actually has two components, one of which has thus far been left undefined and has not been recognized at all, and we could write:

J = Jm + Jd = e ω + e 𝜐 = e ∇×v + e ∇v

The problem is, however, that the gradient of a vector yields a Jacobian matrix:

https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

So, we have to take the determinant of this matrix and then we obtain a "current" that's not a vector but a scalar, so the above equations are not entirely correct and Jd should be:

Jd = e 𝜐 = e det(∇v) = e |∇v|

While it may seem very strange to have a "current" that is not a vector and thus has no direction, it does make sense in some way, because Tesla's longitudinal wave has no current and also no "steady state" DC version and can thus only represent vibrations. I believe this definition is a very important step in being able to understand and describe these FTL longitudinal waves, since we now have a way to describe the translational momentum conservation in an analogue way to the rotational momentum conservation that is described by inductance.

So, even though we still haven't cracked the nut, we are making progress, because both the curl as well as the divergence can also be obtained from the Jacobian. See references below:

https://math.stackexchange.com/questions/2173860/n-dimensional-generalization-of-vector-curl-from-elements-of-jacobian

"In 3 dimensions, the problem is easy: the Jacobian is a 3x3 matrix, and the curl can be obtained from its elements:

[jacobian[2,1] - jacobian[1,2], jacobian[0,2] - jacobian[2,0], jacobian[1,0] - jacobian[0,1]]"

http://math.gmu.edu/~rsachs/math213/curlanddivergence.pdf

"Notice that the curl has components equal to the entries in the matrix J−JT , where J is the Jacobian matrix. The diagonal elements of such a matrix are always zero, while the off-diagonals come in pairs (with opposite signs). Thus a 2-D Jacobian ends up with one component for curl, while a 3-D Jacobian has three independent entries."

As can the divergence:

http://www.math.hawaii.edu/~lee/calculus/Curl.pdf

"The divergence of the vector field F, often denoted by ∇•F, is the trace of the Jacobean matrix for F , i. e. the sum of the diagonal elements of J."

Page 27:

"It is easy to see that the sum of a matrix and its transpose will be symmetric and the difference of a matrix and its transpose will be skew-symmetric. Any n × n matrix can be written as the sum of a symmetric matrix and a skew symmetric one, since we have

A= 1/2 (A+Atr)+ 1/2(A−Atr).

Suppose that a particular point the Jacobean matrix for a vector field F is

[...]

Then

∇ × F = (a32(x) − a23(x)) i − (a31(x) − a13(x)) j + (a21(x) − a12(x)) k .

Then what we notice is that ∇ × F = 0 if and only if J is a symmetric matrix.

[...]

It is very tempting to believe that if F is a vector field with Jacobean matrix J , and we decompose J into the sum of a symmetric and a skew-symmetric matrix, then there would exist a corresponding way of writing F as a sum of two fields, one of which is the field of velocity vectors for a rotation and the other is a field with three mutually orthogonal eigenvectors at every point. This is certainly easily done in the case that F is linear, but in general it is not feasible, because it is usually not possible to find a vector field with pre-assigned Jacobean matrix."

Page 28:

"The best one can say then is that in a sufficiently small neighborhood of a point of interest a vector field F can be reasonably well approximated by a linear field, and therefore it will look pretty much as though it is the sum of two fields, one of which has three mutually orthogonal eigenvectors and has zero curl, and the other of which represents the velocity vectors (within the given neighborhood) of a rotation of all of three space around the axis ∇ × F with an angular speed of 1/2 ||∇ × F|| .

It seems to me that this is the best intuitive interpretation that one can give for the concept of curl. However, such a decomposition of even a linear vector field with F(0) = 0 into the sum of a field with zero curl and one which corresponds to a rotation of 3-space is often not at all apparent visually."

The following is also interesting in this regard:

https://math.stackexchange.com/questions/2515744/observation-on-rot-curl-div-and-grad-on-a-vector-field

So, we haven't yet completed the puzzle, but with the above exercise we found a few new pieces to add. :)

It seems that current density J should not be a vector, but a Jacobian matrix, defined as:

J = e ∇v