How to Recognize Split Electric Fields - Comments

[rude man]

Introduction
In a previous Insight, A New Interpretation of Dr. Walter Lewin’s Paradox, I introduced the fact that there are two kinds of E fields.  One (Em) is generated whenever a source of emf is produced.  The other (Es) is the electrostatic field.  The Es field always terminates on free charges; the Em does not.  Es is a conservative field like gravity; Em is not.  $\nabla x \mathbf Em$ is always non-zero somewhere.
My intent is to reveal the existence of the two fields in a number of situations and, more importantly, show that in most cases this approach is the only acceptable one if conflict with established physical laws is to be avoided.
The Start
The total electric field $\mathbf {E = Em + Es}$.
We define an electric field E as the force F on a stationary charge: $\mathbf E = \mathbf F/q$ with q $\rightarrow 0$.  By this defnition,all  Em and Es in all of...

Continue reading...

 

[Charles Link]

Hello @rude man ,
It may be worthwhile to include a "link" to this homework exercise. This one makes for a very good application of the concepts you discuss in your Insights article. It was very good reading. Thanks. :) See:
https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120

 

[rude man]

Hi again @Charles Link ,

yes, that was a long and memorable, and relevant, thread! Went on for over a year ... cnh1995 finally came up with a very elegant solution using emf and voltage in lieu of Em and Es (the former the respective integrals of the latter of course).

That problem is considered of peripheral interest by some as they question the relevance of "voltage" to begin with but I still think it's of high interest.

My new blog tries to convince that there are many situations where you either accept split fields or violate basic physics. We'll see how things turn out.

Thanks for reading and commenting.

 

[Charles Link]

yes, that was a long and memorable, and relevant, thread! Went on for over a year ... cnh1995 finally came up with a very elegant solution using emf and voltage in lieu of Em and Es (the former the respective integrals of the latter of course).

Perhaps we should "link" @cnh1995 to make sure he sees this, along with your Insights article. I think he would find it interesting.

 

[rude man]

Perhaps we should "link" @cnh1995 to make sure he sees this, along with your Insights article. I think he would find it interesting.

Feel free to do that Charles. I think I'll let him decide; I'm sure he gets notifications of all new Insight articles, just as you & I do.

 

[vanhees71]

I strongly disagree with the claim that there are two kinds of electric fields.

There's one and only one electromagnetic field, consisting in any inertial frame of reference (within special relativity, i.e., neglecting gravity for simplicity here) of 3 electric and 3 magnetic components, being the components of the antisymmetric Faraday tensor $F_{\mu \nu}$.

Alternatively you can represent the em. field in the $\mathrm{SO}(3,\mathbb{C})$ representation of the proper orthochronous Lorentz group, via the Riemann-Silberstein vector components $\vec{F}=\vec{E}+\mathrm{i} \vec{B}$.

There's no physical meaning to any split of the electric an magnetic field components.

 

[Charles Link]

It is perhaps the case that all electric fields, in principle, are one and the same, but for practical calculations, I think @rude man 's approach is a good one: it is often necessary to separate the induced electric field $E_m$ in a conductor from the electrostatic field $E_s$ that immediately arises because $E_{total } \approx 0$. It is also of interest that $\nabla \times E_s=0$, which isn't the case for $E_m$. Therefore, in a practical sense, for computational purposes, IMO, the distinction does have some merit. $\\$ In the homework problem mentioned in post 2, such a distinction made for a very straightforward solution to a homework problem that was somewhat difficult.

 

[PhilDSP]

For that matter, we might very well find in the long run that light, charge, electric fields, magnetic fields, mass, even quantum behavior are all manifestations of the same underlying phenomena each having a particular local (or non-local) context or configuration. One major problem with 4 vector representation of fields is that the specific breakdown of electric versus magnetic fields is destroyed or at least distorted. You cannot take a proper Fourier Transform of the fields represented by 4 vectors, can you? It should be realized that Fourier Transforms of fields expressed in time and position (or rather delta position) provide the needed identities on which the wave characteristics of QM are expressed.

Perhaps there is great value in classifying each separate identifiable manifestation of the fields. Thanks for pointing this out in regard to batteries. The greatest change in the Em field for the battery would likely be when it is charging or discharging, wouldn’t it?

 

[rude man]

Hi @PhilDSP,
as you know, chemical batteries tend to maintain their emf not that far below the fully charged level even as they discharge. The main change of course is the buildup of internal resistance which limits th available current. Which is why batteries are best tested under load.

But yes, emf and Em also decrease with use unlessand until the battery is recharged.

In my log I assumed a zero-internal-resistance battery just as illustration of the existence of the two E fields. If you assume zero internal rsistance then the two E fields exactly cancel another, otherwise you wouldn't reach ion equilibrium.

 

[vanhees71]

You can simply say the electric field is 0 for the zero-internal-resistance battery. Where the heck do you need to split the one and only electric field in two artificial pieces?

 

[rude man]

In order to satisfy Kirchhoff there must be an irrotational E field in the battery to give zero circulation of that field, the Es field..

But this would leave a net E field in the battery, violating equilibrium of the + and - ions.

Furthermore, an emf must be generated to account for the current in the first place.

Both these requirements dictate the existence of a second E field in opposition to the irrotational field. It's called the Em field and is the immediate consequence of the generated emf.

Thus, the circulation of the net E field around the circuit is iR, not zero as some would erronerously believe.

Your assumption of only one E field in the battery is incorrect. I have defined the E field as numerically the force on a unit stationary test charge, assuming that charge does not materially affect the pre-existing E field. So there can't be a net E field in the battery or the test charge would migrate to the - terminal, causing an imbalance of charge between the + and - terminals. This in turn implies finite divergence of the net E field which violates Maxwell $\nabla \cdot \mathbf D = \rho = 0$.

cf. th excerpt from the Skilling text I had previously sent, which totally accords with the above.

 

[rude man]

You can simply say the electric field is 0 for the zero-internal-resistance battery. Where the heck do you need to split the one and only electric field in two artificial pieces?

I'm confused by this question.
Is the E field zero or is it finite?

 

[vanhees71]

It's 0, and of course you cannot understand a battery assuming 0 internal resistance. A very comprehensive discussion about how a battery works can be found here:

https://doi.org/10.1119/1.19327

Another nice one is

https://doi.org/10.1119/1.13128

 

[Charles Link]

See post 193 of https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 for the mathematical detail of how the $E_s$ and $E_m$ electric fields apply. IMO, it is worthwhile in some cases to distinguish these two electric field components as @rude man is doing.

 

[rude man]

You can model the battery as a zero-resistance source of emf in series with a finite resistance $r$ representing the internal resistance. Doing so reduces battery $\int \mathbf Es \cdot d\mathbf l$ by $-ir$ if there is current $i$ drawn, while Em is more slowly reduced over time as the chemical process abates..

I am not a battery expert and my blog was not intended to be an expose of battery operation.

As I said in my blog, if you are unhappy with the battery example I gave three more examples of the need for a split-E model. Enjoy!

 

[vanhees71]

Let's take only the one example with the ring and a time-varying magnetic flux. I don't know, where you need this strange split (though it's mathematically always possible). All you need to calculate the current here is Faraday's Law (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
Since there's no free charge there, we have
$$\vec{\nabla} \cdot \vec{E}=0,$$
and due to Faraday's Law thus $\vec{E}$ is a solenoidal field.

Now you integrate over an area with the circuital wire as a boundary (everything assumed to be at rest within the computational inertial reference frame). Using Stokes's Theorem and the (non-relativistic) Ohm's Law, you get
$$\int_{C} \mathrm{d} \vec{x} \cdot \vec{E}=\int_{C} \mathrm{d} \vec{x} \cdot \frac{1}{\sigma} \vec{j}=-\dot{\Phi}_{\vec{B}}.$$
For a very thin wire the loop integral gets
$$\int_{C} \mathrm{d} \vec{x} \cdot \frac{1}{\sigma} \vec{j}=i (\frac{\sigma_1 L_1}{A_1} + \frac{\sigma_2 L_2}{A_2})=i (R_1+R_2)=-\mathrm{d}_t \Phi_{\vec{B}}.$$
Nowhere do you need this very confusing split of the electric field. I don't even understand what $\vec{E}_m$ and $\vec{E}_s$ should be in this case.

 

[rude man]

I stated in my blog that in this case the split-field approach was neither necessary nor expedient as far as computing voltmeter readings are concerned. It was however included for heuristic purposes, to show that Em ad Es fields both exist and that free charge does also. Which may not be of much practical use but we are supposed to understand physics in the most detailed way possible so for that reason I included it in my blog.

As to the fact that you did not fully understand that particular example I can only apologize for my failure to have explicated adequately.

 

[vanhees71]

No, they do not both exist! There's only one electromagnetic field. This split has no physical significance. It can be a calculational tool in the static case (magnetostatics). It's a bit like with the electromagnetic potentials (or rather the four-potentials), which are just a calculational tool to solve Maxwell's equations but they are not physical fields, which is clear already from the fact that they are gauge dependent.

What I didn't understand in the specific example is, what the split fields are let alone what they are good for in this case to begin with.

 

[andrew s 1905]

No, they do not both exist! There's only one electromagnetic field. This split has no physical significance. It can be a calculational tool in the static case (magnetostatics). It's a bit like with the electromagnetic potentials (or rather the four-potentials), which are just a calculational tool to solve Maxwell's equations but they are not physical fields, which is clear already from the fact that they are gauge dependent.

What I didn't understand in the specific example is, what the split fields are let alone what they are good for in this case to begin with.

What about the Aharonov-Bohm effect due to the EM potentials
Regards Andrew

 

[vanhees71]

The Aharonov-Bohm effect is described not by the em. potentials but a integral along a closed curve, and that's equivalent to a surface integral over the magnetic field, i.e., it's gauge invariant and thus makes sense as a physical observable (and indeed the AB effect has been observed).

 

[hutchphd]

No, they do not both exist! There's only one electromagnetic field.

A semantic issue I believe.
That being said I too am mystified as to what @rude man is trying to do. It seems to me to be equivalent to saying "One can always do a Helmholz decomposition of a vector field. Occasionally it is useful for E". Done?

 

[vanhees71]

Sure, one can always do that. It's a mathematical theorem, but it's not always useful, and the split has no physical significance. Only the electromagnetic field, which is the solution of the complete set of Maxwell equations has a physical interpretation. Mathematically it's useful mostly for the static field. For the full dynamical equations you rather need the retarded solutions (retarded potentials or equivalently Jefimenko's equations for the fields).

 

[hutchphd]

Retarded potentials aside (I'm localized "with" Prof Lewin on the lab table now) are we not tacitly doing that split? We casually talk about "the voltage" in Kirchhoff laws but we really mean the curl free part of the E field and then that associated potential. There are many good electrical engineers who don't get it (as Lewin found out!...too bad about his other exploits). And it confused me for while. So recognizing the split may be a semantic tool for conveying the idea.
But is seems like a few lines is sufficient to show it!

 

[Charles Link]

The voltage from a voltmeter across a resistor in the circuit [edit:i.e. large resistor in the voltmeter as part of the circuit] is found by Kirchhoff's laws by computing the current through the resistor and multiplying by the resistance, taking into account any EMF's from changing magnetic fields. The voltmeter reads $V_{voltmeter}=\int E_{total} \, ds$ across the resistor [edit: i.e. voltmeter resistor], where $E_{total}=E_m +E_s$. See https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 , especially post 193. The voltmeter does not read the electrostatic potential $V_{XY}$. $\\$ I think part of @rude man 's motivation for addressing this topic with an Insight's article is that you get a different perspective when you actually work through the details of solving a homework problem like the one in the "link".

 

[hutchphd]

I am not suggesting that the motivation is impure. But I think the method may not be the most compact and simple. There is no doubt that Prof Lewin's demonstration revealed pathology of thought rampant even in Cambridge (Mass). I am just not sure that this "cure" is less objectionable.

 

[vanhees71]

Retarded potentials aside (I'm localized "with" Prof Lewin on the lab table now) are we not tacitly doing that split? We casually talk about "the voltage" in Kirchhoff laws but we really mean the curl free part of the E field and then that associated potential. There are many good electrical engineers who don't get it (as Lewin found out!...too bad about his other exploits). And it confused me for while. So recognizing the split may be a semantic tool for conveying the idea.
But is seems like a few lines is sufficient to show it!

Well, I think this "split" rather adds to the confusion than helping against it. It's a very clear statement: The electric field in the non-static case is not conserved and thus there's no potential nor a "voltage" to describe it:
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},$$
which implies that the measurement of the electromotive force along a closed circuit (which is what a usual voltmeter, at least an oldfashioned galvanometer-type, indeed measures) can depend on the geometry of the loop as soon as time varying magnetic flux is involved, and that's what Lewin demonstrates in this demonstration experiment. It's known since Maxwell et al not only since Lewin's YouTube video, by the way.

In short: Note that what the voltmeter reads is always EMF, along the loop of the circuit it's connected to. If there are even movable parts you need the complete integral law, which reads
$$\mathrm{d}_t \Phi_{A}[\vec{B}] = \mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B} =-\mathcal{E}=-\int_{\partial A} \mathrm{d} \vec{x} \cdot [\vec{E} + \vec{v} \times \vec{B}].$$
Then all apparent paradoxes with Faraday's induction law in the here discussed context resolve themselves. There's no split of any part of the em. field of physical significant but the one and only em. field!

 

[anorlunda]

I'm sure that you guys will disapprove, but circuit analysis (CA) explicitly excludes charges and fields. They are not needed for most circuits. Thinking about them is a needless complication, so most EEs are better off forgetting about them. IMO, that is the origin of the lack of awareness mentioned in this thread. Those who do work with fields, are often similarly unaware of the QED foundations.

An ideal capacitor is defined by its behavior C dV/dt, without reference to charges. Ditto L dI/dt or an ideal transformer n1:n2, are defined by their behavior without reference to magnetic fields. All those behaviors could be mimicked by a software driven active device instead of L and C, leaving the analysis of the circuit intact regardless of the implementation of the devices.

Lewin's demonstrations show that there are cases where ordinary circuit analysis is not adequate. But then he claims ignorance of the assumptions of CA and trashes the whole idea of CA. If everyone had to use Maxwell's equations to design every circuit, technology would not be where we are today.

 

[vanhees71]

Sure, the circuital laws are nevertheless derived from the quasitationary approximations (electric as well as magnetic!) of the Maxwell equations. For this effective "Kirchhoff theory" you the less need any unphysical splits of fields!

 

[hutchphd]

It's known since Maxwell et al not only since Lewin's YouTube video, by the way.

Of course I agree. It is my understanding that when this demonstration was part of his normal lecture for Intro Physics at MIT there were faculty who accused him of trickery! I don't know this from any authoritative source so it may be apocryphal...but if true it is slightly disconcerting.

 

[vanhees71]

Strange, why should this be trickery? It's well-known standard physics.

 

[rude man]

I'm sure that you guys will disapprove, but circuit analysis (CA) explicitly excludes charges and fields. They are not needed for most circuits. Thinking about them is a needless complication, so most EEs are better off forgetting about them. IMO, that is the origin of the lack of awareness mentioned in this thread. Those who do work with fields, are often similarly unaware of the QED foundations.

An ideal capacitor is defined by its behavior C dV/dt, without reference to charges. Ditto L dI/dt or an ideal transformer n1:n2, are defined by their behavior without reference to magnetic fields. All those behaviors could be mimicked by a software driven active device instead of L and C, leaving the analysis of the circuit intact regardless of the implementation of the devices.

Lewin's demonstrations show that there are cases where ordinary circuit analysis is not adequate. But then he claims ignorance of the assumptions of CA and trashes the whole idea of CA. If everyone had to use Maxwell's equations to design every circuit, technology would not be where we are today.

As an EE I couldn't agree with you more: CA per se practically never invokes fields. That's the game I've been playing all my career. And in fact Kirchhoff never mentions fields, to my knowledge. Just potential rises and drops.

But we are talking physics which of course does involve fields - big time. And I consider it more than just curious that if we equate potential with $\int \mathbf E_s \cdot d \mathbf l$ that Kirchhoff's voltage law is valid as a field expression also, contrary to Lewin's rash statement that "Kirchhoff is wrong". But that requires identifying and separating $Es$ from $Em$ which of course is the subject of my blog, whatever its interest or value.

PS I think microwave plumbers and antenna types would probably disagree with you. I've always been grateful that I didn't have to worry about that area after school was out.

 

[vanhees71]

Of course, Kirchhoff is not wrong in the example demonstrated by Lewin. It's not his fault that people try to use his laws in cases where the underlying assumptions don't apply.

I now the attitude of EEs against "field theory" too well. The most hated lectures in the EE curriculum at my university were "field theory" and "stastical signal theory" (though from that field the most profound contribution of EEs to contemporary physics came out, i.e., Shannon's information-theoretical approach to entropy).

I'm not too surprised that EEs consider field theory obscure, if the subject is confused by artificial unphysical splits of the fields in unphysical pieces... SCNR.

 

[PhilDSP]

An interesting alternate example of where the breakdown of Es and Em leads to both scientific and engineering insights is detailed in Introduction to Plasma Physics: With Space, Laboratory and Astrophysical Applications by Gurnett and Bhattacharjee. If I remember correctly it is described starting about page 70.

The authors derive wave equations for particles traveling through a cold plasma. The wave equations are then translated into dispersion equations. One solution to the dispersion equations yields a modulating Em field and the other solution yields a modulating Es field. An exercise at the end of the chapter is to describe what kind of antenna is needed to generate the modulating Es field.

 

[vanhees71]

I guess that's about the wave modes in the plasma? That's of course another case, i.e., it makes physical sense, as any mode expansion does.

 

[jasonRF]

An interesting alternate example of where the breakdown of Es and Em leads to both scientific and engineering insights is detailed in Introduction to Plasma Physics: With Space, Laboratory and Astrophysical Applications by Gurnett and Bhattacharjee. If I remember correctly it is described starting about page 70.

The authors derive wave equations for particles traveling through a cold plasma. The wave equations are then translated into dispersion equations. One solution to the dispersion equations yields a modulating Em field and the other solution yields a modulating Es field. An exercise at the end of the chapter is to describe what kind of antenna is needed to generate the modulating Es field.

That must be for the unmagnetized case, where the decomposition naturally falls out of the eigenvalue problem. For linear waves in a cold magnetized plasma you still get that Es oscillation parallel to the DC magnetic field (of course thermal effects actually matter for that mode), but the other eigenmodes are mixes of both Es and Em parts except in special circumstances. These "mixed" waves are simply called electromagnetic waves; waves for which $\nabla \times \mathbf{E} \approx 0$ is a good approximation are called electrostatic waves.

 

[PhilDSP]

Yes, you're right. It's the unmagnetized case. Early in the book they start with more simple cases and later derive equations for other conditions. They even analyze shock waves with interesting insights later in the book. It's a great textbook!

 

[rude man]

Thanks guys, way over my head but certainly fascinating!

 

[Dale]

This idea of a non conservative $E_m$ is obviously false in the usual case of a static open-circuit battery. There, according to your own analysis $E_m=-E_s$. Since $E_s$ is conservative it can be represented as the gradient of a potential $E_s=-\nabla \phi$. So $E_m=-E_s=-\nabla (-\phi)$. Therefore $E_m$ is also conservative.

 

[rude man]

This idea of a non conservative $E_m$ is obviously false in the usual case of a static open-circuit battery. There, according to your own analysis $E_m=-E_s$. Since $E_s$ is conservative it can be represented as the gradient of a potential $E_s=-\nabla \phi$. So $E_m=-E_s=-\nabla (-\phi)$. Therefore $E_m$ is also conservative.

$\bf E_m$ is only NUMERICALLY EQUAL to $-\bf E_s$. They are not the same field. One ($E_m$) is the non-conservative, emf-generated E field, the other is the irrotational ($E_s$) field. The fields oppose each other inside the battery, yielding zero net $\bf E$ inside the battery. Vectorially, $\bf E = \bf E_m + \bf E_s.$

The circulation of $\bf E$ around a circuit with battery is not zero. It is in fact the emf.

The above applies even to open-circuit batteries with internal resistance r. If current $i$ flows the battery voltage would be reduced by $i$r. (Voltage is the line integral of $E_s$).

Perhaps the attached files will help.

 

[Dale]

Neither of those pages answer my objection. They do not claim, as you do, that $E_m$ is non conservative in general. That is the specific point that I disagree with, and an open circuit battery in equilibrium seems an obvious counter example

The fields oppose each other inside the battery, yielding zero net $\bf E$ inside the battery. Vectorially, $\bf E = \bf E_m + \bf E_s.$

And therefore since $E_s$ is conservative inside the battery so is $E_m$

 

[rude man]

Neither of those pages answer my objection. They do not claim, as you do, that $E_m$ is non conservative in general. That is the specific point that I disagree with, and an open circuit battery in equilibrium seems an obvious counter example

[/QUOTE]
That is exactly what those pages do claim, that Em is non-conservative. And the line integral around the loop is not zero as it would be were the total E field conservative, but in fact = emf:

$\oint \bf E \cdot d\bf l = \oint \bf( E_m + E_s) \cdot d\bf l = \oint \bf E_m \cdot d\bf l = emf$.

 

[Dale]

$E_s \ne 0$ and $E_m=0$ (for an open circuit battery)

 

[rude man]

$E_s \ne 0$ and $E=0$ (for an open circuit battery)

?

 

[robphy]

I just stumbled upon this thread, which had some good back and forth comments.
I just wanted to chime in.

I strongly disagree with the claim that there are two kinds of electric fields.

There's one and only one electromagnetic field, consisting in any inertial frame of reference (within special relativity, i.e., neglecting gravity for simplicity here) of 3 electric and 3 magnetic components, being the components of the antisymmetric Faraday tensor FμνFμνF_{\mu \nu}.

It is perhaps the case that all electric fields, in principle, are one and the same, but for practical calculations, I think @rude man 's approach is a good one: it is often necessary to separate the induced electric field EmEm E_m in a conductor from the electrostatic field EsEsE_s that immediately arises because Etotal≈0Etotal≈0 E_{total } \approx 0 . It is also of interest that ∇×Es=0∇×Es=0 \nabla \times E_s=0 , which isn't the case for EmEm E_m . Therefore, in a practical sense, for computational purposes, IMO, the distinction does have some merit.

One can always do a Helmholz decomposition of a vector field.

I think it's useful to point out the similarities in viewpoints, but at different levels.

Fundamentally, there is the electromagnetic field tensor field F_{ab} in spacetime.
A particular 3-vector component is the electric vector field \vec E, in space according to an observer.
Arguably,
there's of course nothing fundamental going on here...
but may prove to be conceptually or computationally simpler than using the un-decomposed quantity.
Hopefully, we know the split is artificial and done out of convenience and that there is something deeper, more unified.

According to the Helmholtz decomposition, one can express \vec E in terms
of two components: a curl-free part \vec E_s and a divergence-free part \vec E_m.
Note that this is conceptually similar to writing a vector in terms of rectangular components
E_x, E_y, E_z.
There's of course nothing fundamental going on here...
but may prove to be conceptually or computationally simpler than using the un-decomposed quantity.
Indeed, one may write a vector equation for \vec E as three component equations.
Hopefully, we know the split is artificial and done out of convenience and that there is something deeper, more unified.

It might be worth noting that this Helmholtz split is akin to decomposing the net force on an object
into conservative and non-conservative forces.
Again, nothing fundamental going on here...
but may prove to be conceptually or computationally simpler than using the un-decomposed quantity
[for example, not having to explicitly compute the work done around a loop for the conservative-component... or inventing a convenient quantity "potential energy" which might make work-calculations easier for the conservative-components.].
Again, hopefully, we know the split is artificial and done out of convenience and that there is something deeper, more unified.

It might just be the author and/or the target audience
are more comfortable or find it more convenient
by doing the decomposition at a different level than someone else.

My $0.02.

 

[Dale]

?

You have claimed that inside a battery $0=E=E_s+E_m$ which implies $E_m=-E_s$. It is not possible for the field on the left to be non-conservative if the field on the right is conservative.

 

[etotheipi]

I think a lot of the disagreement here might also be summed up by this claim made in the other thread:

All the examples quoted by Griffith are effectively E fields. A "force per unit charge" IS the definition of an E field.

which must be incorrect. In my frame of reference, I certainly wouldn't call $\mathbf{v} \times \mathbf{B}$, or $\frac{GMm}{qr^3}\hat{\mathbf{r}}$, or $\frac{1}{q} \mathbf{f}_s$ an electric field. Instead, an electric field $\mathbf{E}$ is the electric force (only!) per unit charge.

It appears you are generalising the term electric field to mean any force per unit charge, whilst this is surely a very unconventional (and probably incorrect) usage?

But even then, this still wouldn't explain why you think $\mathbf{f}_s$ is not conservative.

 

[rude man]

Over the years I may have confused some people about all this.

$E_m$ and $E_s$ are both electric fields in every sense of the word. A test charge q experiences a force = q$E_m$ or q$E_s$.

The difference is that the SOURCE of each is different.
$E_m$ fields are associated with production of electricity from a different source of energy. For example, the field around a time-varying B field is pure $E_m$.
$\nabla \times \bf E_m \neq 0 = -\partial \bf B/\partial t. \oint E_m \cdot dl \neq 0 = -\partial \phi/\partial t.$.

By contrast, the source of $E_s$ is always free charge. Flux lines of $E_s$ begin and end on charges. $\nabla \times E_s = 0$ etc.

In my latest blog I gave examples where either the distinction is made or laws of physics are violated. See e.g. my Seebeck effect example.

 

[etotheipi]

$E_m$ and $E_s$ are both electric fields in every sense of the word. A test charge q experiences a force = q$E_m$ or q$E_s$.
The difference is that the SOURCE of each is different.
$E_m$ fields are associated with production of electricity from a different source of energy.

I don't know enough to dispute this, however it seems to directly contradict this earlier post of vanhees (I have bolded the particular part!):

No, they do not both exist! There's only one electromagnetic field. This split has no physical significance. It can be a calculational tool in the static case (magnetostatics).

 

[Dale]

@rude man can you respond specifically to my previous post regarding the battery?

 

[rude man]

You have claimed that inside a battery $0=E=E_s+E_m$ which implies $E_m=-E_s$. It is not possible for the field on the left to be non-conservative if the field on the right is conservative.

How on EARTH do you conclude this?