# How to Recognize Split Electric Fields - Comments

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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...

Dale, cnh1995, weirdoguy and 2 others

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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.

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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.

cnh1995
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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
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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.

weirdoguy
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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.

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?

weirdoguy
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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
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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?

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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.

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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?

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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!

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weirdoguy
vanhees71
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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.

Dale and weirdoguy
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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
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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.

Dale, robphy, zoki85 and 1 other person
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
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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
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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?

Dale, robphy and vanhees71
vanhees71
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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
hutchphd
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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!

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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".

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hutchphd
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