How to Recognize Split Electric Fields - Comments

[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?

 

[etotheipi]

How on EARTH do you conclude this?

$\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi$; Hence $\mathbf{E}_m$ is derivable from a scalar potential.

 

[rude man]

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

I did - twice!
Your basic argument that Es and Em fields can't coexist is baseless.

Look at my blog where I analyze a ring of non-uniform resistance surrounding a time-varying B field. One half of the ring has twice the resistance of the other half. This asymmetry produces both Em and Es fields in the coil.

The Em field is defined solely by the B field irrespective of any presence of a coil or anything else. The non-uniform coil produces Es fields in the two halves.. The circulation of the Em field is the emf. The circulation of the Es field is zero. The two Es fields are equal and opposite in the coil.

 

[etotheipi]

The Em field is defined solely by the B field irrespective of any presence of a coil or anything else. The non-uniform coil produces Es fields in the two halves.. The circulation of the Em field is the emf. The circulation of the Es field is zero. The two Es fields are equal and opposite in the coil.

No that is different; there you are just doing $\oint \mathbf{E} \cdot d\mathbf{l} = \oint (\mathbf{E}_s + \mathbf{E}_m) \cdot d\mathbf{l} = \oint \mathbf{E}_m \cdot d\mathbf{l}$. Arbitrary, but still correct, I guess.

But in the case of the battery, you are saying that the conservative, chemical force inside the battery is actually a non-conservative, electric force $\mathbf{E}_m$ - this is incorrect.

 

[rude man]

$\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi$; Hence $\mathbf{E}_m$ is derivable from a scalar potential.

I have said more than once that Es is NUMERICALLY equal to -Em. That doesn't make them both conservative. I have described their different attributes more than adequately by now IMO.

 

[rude man]

But in the case of the battery, you are saying that the conservative, chemical force inside the battery is actually a non-conservative, electric force $\mathbf{E}_m$ - this is incorrect.

Who said the chemical force is conservative? I said just the opposite.

 

[rude man]

$\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi$; Hence $\mathbf{E}_m$ is derivable from a scalar potential.

Bad conclusion!

 

[Dale]

Your basic argument that Es and Em fields can't coexist is baseless.

I have nothing against a Helmholz decomposition of the fields. My argument is that $E_m$ is not non conservative in this specific case, so a generic statement that it is non conservative to be is false. You have not addressed that.

You said that in a battery $E=0$ and $E=E_m+E_s$. You also said that $E_m$ was non-conservative. These are contradictory statements

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

Es is a conservative field like gravity; Em is not.

Source https://www.physicsforums.com/insights/how-to-recognize-split-electric-fields/

 

[rude man]

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

Correct. E = Em + Es in the battery = 0 (equal and opposite fields).
Es points + to - inside and outside the battery.
Em is present inside the battery only and points - to +.
Circulation of Em = emf.
Circulation of Es = 0.
The defense rests.

 

[rude man]

My argument is that $E_m$ is not non conservative in this specific case, so a generic statement that it is non conservative to be is false. You have not addressed that.

So if you say it's false, that makes it false? I demur.

You said that in a battery $E=0$ and $E=E_m+E_s$. You also said that $E_m$ was non-conservative. These are contradictory statements

Why? Are you remembering that Em and Es point in opposite directions in the battery? Em is conservative, Es is not. What's the problem?

 

[rude man]

$\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi$; Hence $\mathbf{E}_m$ is derivable from a scalar potential.

I've answered this, several times.