How to Recognize Split Electric Fields - Comments

[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.

 

[robphy]

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.

We say a vector field $\vec V$ [in ordinary Euclidean space] is conservative
when $\vec V$ satisfies $\vec 0=\vec \nabla \times \vec V$ (everywhere).

Observe that when "$\vec X$ is conservative", then
"$c\vec X$ is conservative", for any constant $c$,
since $\vec \nabla \times (c\vec X) = c\left( \vec \nabla \times \vec X \right) =\vec 0$.
Thus, "$c\vec X$ is conservative".
In particular, "$-\vec X$ is conservative"

To claim that $\vec X$ is conservative but $-\vec X$ is not conservative",
suggests that you are misusing the definition of a vector field being "conservative".

 

[rude man]

Could you produce a reference for this?

I could but it would take time and I don't think it would do any good.

In different inertial frames of reference in relative motion, the two 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.

If qxB were not an electric field then the net E field would not be zero since we must have an Es field to reach charge equilibrium with the qxB force.

I maintain that $\mathbf{f}_s$ is conservative between the terminals, because I can write it as the derivative of a scalar field.

It isn't.
You can't.
(And I think you meant potential, not field.)

 

[etotheipi]

If qxB were not an electric field then the net E field would not be zero since we must have an Es field to reach charge equilibrium with the qxB force.

Huh? I don't understand. $q\mathbf{E}$ is an electric force, $q\mathbf{v}\times \mathbf{B}$ is a magnetic force. They might act on, say, a single particle. In a different inertial frame of reference the comparative sizes of the two forces might be different.

It isn't.
You can't.
(And I think you meant potential, not field.)

It is.
I can.
A potential is a scalar field...

 

[Dale]

If $-X$ is just $X$ with polarity reversed, sure. But not if they have different sources.

This has nothing to do with the sources, this is a mathematical property of vector fields, irrespective of electromagnetism or any specific application of vector fields. If $X$ is a conservative vector field then $-X$ is also conservative. Full stop. No ifs ands or buts. No caveats.

 

[robphy]

If qxB were not an electric field then the net E field would not be zero since we must have an Es field to reach charge equilibrium with the qxB force.

Suppose there are only two forces being applied on an object,
resulting in static equilibrium of that object via Newton's Second Law.
Those two forces balance... their magnitudes are equal and they are oppositely directed.
However, this "equality in magnitude and opposite in direction"
does NOT necessarily mean that the forces are of the same type.

For a book sitting at rest on the table,
the gravitational force (by the Earth on the book) is balanced
by the normal force (by the table on the book).
The gravitational force is proportional to $G$, $M_{earth}$, $m_{book}$, and $1/r^2_{\tiny\mbox{between their centers}}$.
The normal force is equal to whatever it needs to be to satisfy Newton's Second Law.

If we now push down on the book with force $P$ ,
the gravitational force (by the earth) is unchanged...
but the normal force (by the table) increases...
this new normal force is NOT proportional to $m_{book}$... it is certainly not a gravitational force.

In the Lorentz Force on a charged particle,
$\vec F_{\tiny Lorentz} =q\vec E + q\vec v \times \vec B$.
When $\vec F_{\tiny Lorentz}=\vec 0$,
then the electric force $q\vec E$ balances the $q\vec v \times \vec B$.
But this does NOT mean that $q\vec v \times \vec B$ is an electric force.
The electric force is independent of $\vec v$,
but the magnetic force is proportional to $\vec v$ (and perpendicular to it).
So, $q\vec v \times \vec B$ is certainly not an electric force.


[postscript: Newton's Third Law implies third law pairs
($\vec F_{\small \mbox{A on B}} = -\vec F_{\small \mbox{B on A}}$) are of the same type.
But of course, this is not what is involved in a zero Lorentz force.]

 

[rude man]

This has nothing to do with the sources, this is a mathematical property of vector fields, irrespective of electromagnetism or any specific application of vector fields. If $X$ is a conservative vector field then $-X$ is also conservative. Full stop. No ifs ands or buts. No caveats.

I looked up what the Helmholz decomposition theorem is.
All my electric fields are the sum of an irrotational field $(\nabla \times \bf E_s=0)$ and a solenoidal field $(\nabla \cdot \bf E_m=0)$.

Helmholz says they can coexist.
No ifs, ands or buts.

 

[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.

@robphy has written a good post (#44) that I had overlooked since I was so busy with the more frequent posters.

His math knowledge beats mine after 45 years of aerospace electronics. E.g. I never needed tensors though I did run into them as undergraduate.

To quote him, "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".

Sure might!

 

[vanhees71]

I looked up what the Helmholz decomposition theorem is.
All my electric fields are the sum of an irrotational field $(\nabla \times \bf E_s=0)$ and a solenoidal field $(\nabla \cdot \bf E_m=0)$.

Helmholz says they can coexist.
No ifs, ands or buts.

It's a well-known mathematical theorem. However the physical conclusions you seem to draw from it are at best confusing.

 

[Dale]

I looked up what the Helmholz decomposition theorem is.
All my electric fields are the sum of an irrotational field $(\nabla \times \bf E_s=0)$ and a solenoidal field $(\nabla \cdot \bf E_m=0)$.

Helmholz says they can coexist.
No ifs, ands or buts.

Sure. I understand that what you are doing is Helmholtz decomposition. But what you are still not recognizing is that it is incompatible with your claim that $E=E_s+E_m=0$ for an open circuit battery.

For an open circuit battery the solenoidal $E_m= 0$ and the irrotational $E_s\ne 0$ so $E=E_s\ne 0$. Helmholtz decomposition doesn’t work the way you claim it does in a battery

 

[vanhees71]

For an open-circuit battery you have electrostatics and everywhere $\vec{E}$ is thus a conservative vector field because of $\vec{\nabla} \times \vec{E}=0$. To understand batteries, have a look in some physical-chemistry textbook or a thermodynamics textbook like Sommerfeld, vol. V.

 

[rude man]

Sure. I understand that what you are doing is Helmholtz decomposition. But what you are still not recognizing is that it is incompatible with your claim that $E=E_s+E_m=0$ for an open circuit battery.

For an open circuit battery the solenoidal $E_m= 0$ and the irrotational $E_s\ne 0$ so $E=E_s\ne 0$. Helmholtz decomposition doesn’t work the way you claim it does in a battery

What you are still not recognizing is that $E_m \neq 0$. It is in fact the field that accounts for the battery emf.

 

[Dale]

What you are still not recognizing is that $E_m \neq 0$. It is in fact the field that accounts for the battery emf.

Let’s see if we can start with where I believe we agree.

Do you agree that $\nabla \times E_s = 0$ meaning $E_s$ is irrotational? Do you further agree that this implies that $E_s$ is conservative.

Do you agree that $\nabla \cdot E_m =0$ meaning $E_m$ is solenoidal? Do you further agree that this implies that $E_m$ is not conservative?

Do you agree that any arbitrary vector field can be written (Helmholtz decomposition) as $E=E_s+E_m$, the sum of an irrotational and a solenoidal field?

Finally, do you agree that in a open-circuit battery $E_s \ne 0$ at equilibrium?

I believe that we agree on all those points, but would like to confirm before moving to the points where we disagree.

 

[rude man]

Let’s see if we can start with where I believe we agree.

Do you agree that $\nabla \times E_s = 0$ meaning $E_s$ is irrotational? Do you further agree that this implies that $E_s$ is conservative.

Yes.

Do you agree that $\nabla \cdot E_m =0$ meaning $E_m$ is solenoidal?

Yes.

Do you further agree that this implies that $E_m$ is not conservative?

No, $\nabla \cdot \bf E = 0$ does not imply that $\bf E_m$ is non-conservative. That statement is wrong. Did you mean it?

EDIT: all electric fields free of charge have $\nabla \cdot \bf E = 0$. So quoth Maxwell.

On the other hand, yes, I do claim $\bf E_m$ is non-conservative. But not because $\nabla \cdot \bf E_m = 0.$ It's because $emf = \int \bf E_m \cdot d\bf l$ over the length of the battery.

The criterion for 'non-conservative' is $\nabla \times \bf E \neq 0$.

Do you agree that any arbitrary vector field can be written (Helmholtz decomposition) as $E=E_s+E_m$, the sum of an irrotational and a solenoidal field?

Yes, as I just said.

Finally, do you agree that in a open-circuit battery $E_s \ne 0$ at equilibrium?

Yes. The circulation of $\bf E_s$ in that case goes from the - terminal to the + terminal inside the battery, then goes outside the battery from + back to - to complete the loop. The inside & outside Es fields are equal and in the same direction. And the circulation is zero as required of irrotational fields.

I believe that we agree on all those points, but would like to confirm before moving to the points where we disagree.

Well, not quite! Almost!

 

[Dale]

No, $\nabla \cdot \bf E = 0$ does not imply that $\bf E_m$ is non-conservative. That statement is loony. Did you mean it?

On the other hand, yes, I do claim $\bf E_m$ is non-conservative. But not because $\nabla \cdot \bf E_m = 0.$ It's because $emf = \int \bf E_m \cdot d\bf l$ over the length of the battery.

The criterion for 'non-conservative' is $\nabla \times \bf E \neq 0$.

Yes, you are correct (although the "loony" comment was quite unnecessary, please try to keep the tone professional). My statement was wrong. So we are agreed on the points above with your correction of my mistake.

Here is where I think that we disagree. Specifically for the case of an open-circuit battery in equilibrium, if I understand you correctly you claim that $E=0$ and $E_m \ne 0$. I disagree, I claim that $E_m=0$ and $E \ne 0$.

Do you agree that I am correctly stating your position and that you do in fact disagree with my claim here as I have stated it?

EDIT:

EDIT: all electric fields free of charge have $\nabla \cdot \bf E = 0$. So quoth Maxwell.

Note that when you split an E field you are not guaranteed that the splits follow Maxwell's equations. So while it s true that $E$ follows Maxwell's equations it is not necessarily true that either $E_s$ or $E_m$ do. That is something that must be proven rigorously, which you have not done here.

 

[rude man]

i apologize for using the word 'loony'. I should not have. I was temporarily not in control. And I edited it out.

Yes, absolutely that is where we now still disagree.
The Es field is the consequence of the chemical reaction (the Em field) pushing + charge to the + electrode and - charge to the - electrode. Equilibrium is reached when Es = -Em.

If you like you can consider the Em field postulated. I claim it's electric because (1) it accounts for the above charge balance, (2) when line-integrated over the length of the battery it provides the emf, and (3) because Stanford Prof. H.H. Skilling says so. When dealing with giants of physics and engineering like Dr. Skilling I am hesitant to disagree, especially when it makes 100% sense.

I also mentioned in my last blog that not everyone is comfortable with calling the chemical reaction an E field. Personally I don't know diddly-squat about chemical reactions in batteries. I understand there is quite a number of different ones. But macroscopically at least everything is explained by the Em field.

In that blog I provided three more examples of Em fields balancing Es fields. If you don't have much time, at least look perhaps at the Seebeck effect example which is as simple and uncontroversial as it gets. I look forward to discussing that one with you.

 

[Charles Link]

I haven't studied this discussion in detail, but I think I see one area that I may lend some insight: The field $E_m$ is something that is non zero only inside the battery. The field $E_s =-E_m$ for an open circuit battery, but only inside the battery. Outside the battery $E_s$ is clearly different from the $E_m=0$, so that $E_s$ being irrotational does not imply $E_m$ is irrotational, (which it isn't).

 

[Charles Link]

and a follow-on: Upon a little study of the inductor, I don't know that you would want or need to apply Helmholtz's theorem to the $E_{induced}$. It is clearly not irrotational, but I don't know that it necessarily has $\nabla \cdot E_{induced}=0$. The $E_m$ of a battery can be treated mathematically as being similar to the $E_m=E_{induced}$ in an inductor. I do see some merit to @rude man 's approach.

 

[etotheipi]

and a follow-on: Upon a little study of the inductor, I don't know that you would want or need to apply Helmholtz's theorem to the $E_{induced}$. It is clearly not irrotational, but I don't know that it necessarily has $\nabla \cdot E_{induced}=0$. The $E_m$ of a battery can be treated mathematically as being similar to the $E_m$ in an inductor. I do see some merit to @rude man 's approach.

I think surely the easiest way to treat the lumped inductor is to calculate the voltage via $V = -\int_a^b \mathbf{E} \cdot d\mathbf{l}$ along a path outside the component, and $\mathcal{E} = \oint \mathbf{E} \cdot d\mathbf{l}$ around the closed curve through the inductor and around the outside of the component. And since $\mathbf{E} = \mathbf{0}$ inside the conductor, you get out $V = -\mathcal{E}$. It's also important to note that $\nabla \times \mathbf{E} \neq \mathbf{0}$ inside the inductor even though $\mathbf{E} = \mathbf{0}$ inside the wire and this is why you can no longer constrain $\oint_C \mathbf{E} \cdot d\mathbf{l} = 0$.

With that out the way, I'm not entirely sure how one would go about the decomposition you were referring to inside the wire of an inductor; however you decompose $\mathbf{E} = \mathbf{E}_1 + \mathbf{E}_2 = \mathbf{0} \implies \mathbf{E}_1 = -\mathbf{E}_2$ inside the inductor wire; if $\mathbf{E}_1$ is conservative then so is $\mathbf{E}_2$, and vice versa. And also inside the inductor wire you have $\nabla \times \mathbf{E} = -\partial_t \mathbf{B} \neq \mathbf{0}$. So I don't know how this decomposition can make sense. Perhaps someone else can comment?

The battery is different, since there you have two equal and opposite conservative fields $\mathbf{E} = -\frac{1}{q} \mathbf{f}_{chem}$ and no time varying magnetic fields which might cause rotational induced electric fields. But for the inductor, I'm not sure if it makes sense at all to perform any decomposition into two parts.

 

[Charles Link]

@etotheipi For the case of the inductor, we previously treated it in detail in a couple of threads on this forum, and we were very satisfied with the conclusions that included such a decomposition. See https://www.physicsforums.com/threa...asure-a-voltage-across-inductor.880100/page-5 This one really covered it all, including Professor Lewin and the KVL problem. $\\$ Just to summarize for the inductor, inside there is an $E_{induced}$. We have $\int E_{induced} \cdot dl=\mathcal{E}$. In order to get a finite current in the inductor, we must have an $E_s=-E_{induced}$ in the inductor. Because for the whole loop $\oint E_s \cdot dl=0$, outside the inductor we have $\int E_s\cdot dl=\mathcal{E}$ to match the $\int E_s \cdot dl=-\mathcal{E}$ inside the inductor. It really is very straightforward.

 

[rude man]

and a follow-on: Upon a little study of the inductor, I don't know that you would want or need to apply Helmholtz's theorem to the $E_{induced}$. It is clearly not irrotational, but I don't know that it necessarily has $\nabla \cdot E_{induced}=0$. The $E_m$ of a battery can be treated mathematically as being similar to the $E_m=E_{induced}$ in an inductor. I do see some merit to @rude man 's approach.

Hi Charles,
OK except I just finished reminding @Dale that $\nabla \cdot \bf E=0$.always holds for any electric field $\bf E$ unless charge is present at that point, whch of course it isn't. It's one of Maxwell's four equations.