How to Recognize Split Electric Fields - Comments

[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.

 

[Dale]

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

The problem is that these two statements are contradictory.

 

[rude man]

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?

I am.
$F = q \bf v \times \bf B$ is equivalent to $q \bf E$.
Change you reference to the moving frame when $\bf v = 0$. Then there is no lorentz foirce, just electric. If you have Resnick & Halliday intro physics text they do a nice job explaining this.

Also, relativity has shown that the Lorentz force is really an electric force. Feynman mentions this somewhere in his lecture notes. I don't pretend to understand it but there it is.
.

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

As I said several times, if Em were conservative the E field circulation would be zero, and it's not, it's the emf.

 

[rude man]

The problem is that these two statements are contradictory.

They're not, and please explain why you think they are.

 

[rude man]

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!):

Vanhees and I have totally disagreed on this subject for a long time. As I told others, you pick who's right.

 

[etotheipi]

I am.
$F = q \bf v \times \bf B$ is equivalent to $q \bf E$.
Change you reference to the moving frame when $\bf v = 0$. Then there is no lorentz foirce, just electric. If you have Resnick & Halliday intro physics text they do a nice job explaining this.

This doesn't change the fact that in my frame of reference I can quite clearly distinguish between the electric force and the magnetic force, and the $q\mathbf{E}$ is quite clearly not the same thing as $q\mathbf{v}\times\mathbf{B}$... you can't just call them the same thing!

The changes in the relative components under a boost into a different frame of reference is a different matter entirely.

As I said several times, if Em were conservative the E field circulation would be zero, and it's not, it's the emf.

No, this is again incorrect. The $\mathbf{E}_m$ (I will call it $\mathbf{f}_s$) you have defined is locally conservative between the terminals of the battery. So a line integral with in this region is dependent only on the start and endpoints within this region, this follows precisely from how we could assign $\mathbf{E}_m = \nabla \phi$. If you integrate it around the entire loop, however, you will still get a non-zero result. It is not conservative around the whole region of integration.

But you are claiming that the two components $\mathbf{E}_m$ and $\mathbf{E}_s$ come from a Helmholtz decomposition of $\mathbf{E}$, whilst this cannot be correct considering that they are both conservative inside the battery!

And with this I will now step back.

 

[Dale]

They're not, and please explain why you think they are.

If some vector field $X$ is conservative then so is $-X$

 

[rude man]

If some vector field $X$ is conservative then so is $-X$

If $-X$ is just $X$ with polarity reversed, sure. But not if they have different sources. Do you admit the fundamental difference between Em and Es fields as I've laboriously and frequently explained?

Did you read my blog on the non-uniform ring? On the Seebeck wire? There I show the coexistence of Em and Es fields; in the wire they even have to be equal and opposite.
.

 

[rude man]

This doesn't change the fact that in my frame of reference I can quite clearly distinguish between the electric force and the magnetic force, and the $q\mathbf{E}$ is quite clearly not the same thing as $q\mathbf{v}\times\mathbf{B}$... you can't just call them the same thing!

I just did. So did Feynman. So did the guy sitting on your moving media.

No, this is again incorrect. The $\mathbf{E}_m$ (I will call it $\mathbf{f}_s$) you have defined is locally conservative between the terminals of the battery.

It is not. Never was, never will be. And I certainly didn't say so.

 

[rude man]

This doesn't change the fact that in my frame of reference I can quite clearly distinguish between the electric force and the magnetic force, and the $q\mathbf{E}$ is quite clearly not the same thing as $q\mathbf{v}\times\mathbf{B}$... you can't just call them the same thing!

The changes in the relative components under a boost into a different frame of reference is a different matter entirely.
No, this is again incorrect. The $\mathbf{E}_m$ (I will call it $\mathbf{f}_s$) you have defined is locally conservative between the terminals of the battery. So a line integral with in this region is dependent only on the start and endpoints within this region, this follows precisely from how we could assign $\mathbf{E}_m = \nabla \phi$. If you integrate it around the entire loop, however, you will still get a non-zero result. It is not conservative around the whole region of integration.

But you are claiming that the two components $\mathbf{E}_m$ and $\mathbf{E}_s$ come from a Helmholtz decomposition of $\mathbf{E}$, whilst this cannot be correct considering that they are both conservative inside the battery!

And with this I will now step back.

Adios amigo.

 

[etotheipi]

So did Feynman.

Could you produce a reference for this?

There is an electromagnetic field $(\mathbf{E}, \mathbf{B})$. In an inertial frame of reference, as @vanhees71 said, it has 3 magnetic and 3 electric components, and the Lorentz force is $\mathbf{F} = q\mathbf{E} + q\mathbf{v} \times \mathbf{B}$. In my inertial reference frame of choice, these two terms $q\mathbf{E}$ and $q\mathbf{v} \times \mathbf{B}$ are not the same, since clearly one is an electric force and one is a magnetic force.

In different inertial frames of reference in relative motion, the measured fields are different in both frames. However in each I can quite clearly distinguish between a magnetic field and an electric field in my frame.

It is not. Never was, never will be. And I certainly didn't say so.

I maintain that $\mathbf{f}_s$ is conservative between the terminals, because I can write it as the derivative of a scalar field. Though I have said this about 6 times, so I have the feeling saying it once more won't change your mind.