How to Recognize Split Electric Fields - Comments

[Charles Link]

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

Inside the conductor there is plenty of charge.

 

[rude man]

Inside the conductor there is plenty of charge present.

It's not free charge and besides the conductor is neutral in charge. If div E were not zero it would build up somewhere.

 

[Charles Link]

It's not free charge and besides the conductor is neutral in charge. If div E were not zero it would build up somewhere.

Please study my posts 86 and 87 in detail. I don't have a conclusive answer yet, but I think I am headed in the right direction there.

 

[rude man]

On top of that, we are assuming an ideal conductor so E = 0 so div E = 0 inside the conductor.

 

[Charles Link]

On top of that, we are assuming an ideal conductor so E = 0 so div E = 0 inside the conductor.

Right now we are just considering $E_m$. I don't know that we can say $\nabla \cdot E_m =0$, but I don't think we need to have that for your method to work. The $E_m$ starts in the material, and points in one direction, and stops there. It is not a continuous loop like the $B$ field of a magnet or solenoid. $\\$ To add to this, $E_s$ in the inductor is non-zero, and we will find $\nabla \cdot E_s$ is non-zero in places. Since $\nabla \cdot E=\nabla \cdot (E_s+E_m)=0$, clearly $\nabla \cdot E_m$ is non-zero.
It is possible $E_m$ is constant, but it starts and stops, and thereby we clearly don't have $\nabla \cdot E_m=0$.

 

[rude man]

Please study my posts 86 and 87 in detail. I don't have a conclusive answer yet, but I think I am headed in the right direction there.

No problem with #86.
Ditto #87.
In fact, no problem with your conclusions even though I'm dubious about your rationale.

Right now we are just considering $E_m$. I don't know that we can say $\nabla \cdot E_m =0$, but I don't think we need to have that for your method to work.

Why would you think div Em not equal to zero? There are no net charges in the wire and that's where Em exists.
.

 

[Charles Link]

No problem with #86.
Ditto #87.
In fact, no problem with your conclusions even though I'm dubious about your rationale.

Why would you think div Em not equal to zero? There are no net charges in the wire and that's where Em exists.
.

Please see my additions to the last post (95), including a couple later additions. $\\$ To add to this, your method basically is to treat any EMF's with an $E_m$, and to assume there is inherent in the circuit an $E_s$ that behaves very predictably, as summarized at the bottom of post 89. I do think the method has considerable merit.

 

[Dale]

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.

OK, so for convenience I will write down the statements and number them for easy reference:

(98.1) $\nabla \times E_s = 0$
(98.2) $\nabla \cdot E_m =0$
(98.3) $E=E_s+E_m$
(98.4) $E_s \ne 0$

(98.5) $E=0$
(98.6) $E_m \ne 0$

(98.7) $E_m=0$
(98.8) $E \ne 0$

Out of these you do not agree with (98.7) or (98.8) so I will not use them here, instead I will prove by contradiction that (98.5-6) is incompatible with (98.1-4).

Combining (98.3) with (98.5) we get
(98.9) $E_s=-E_m$

By substitution of (98.9) into (98.1) and multiplying by -1 we get
(98.10) $\nabla \times E_m = 0$

Together (98.2) and (98.10) along with the condition that the field from a battery goes to zero at infinity imply
(98.11) $E_m=0$

which contradicts (98.6). QED

 

[rude man]

I think you are basically misintepreting the equation $\bf E_m = - \bf E_s$.

That equation does not mean that Es is just a negative Em. It means that the effect of the E_m field cancels the effect of the E_s field. (The effect is the force on a test charge). They are equal in magnitude and opposite in direction but they have totally different sources and exist independently of one another, on top of one another.

As I said many times before, the Em field is set up by an emf source, the Es field begins and ends on charges. These conditions can coexist independently.

So in the presence of a finite emf and finite charges $\nabla \times \bf Em \neq 0$ while $\nabla \times \bf Es =0.$

EDITED 6/6/2020. The Em field is along and inside the coil wire, not along the principal axis.
Another example: an ideal inductor L (zero resistance) carrying time-varying current i.
There is the emf generated by its dB/dt field yielding a circular Em field inside the coil wire by $\nabla \times \bf E_m = - \partial {\bf B}/{\partial t}$ There is also an Es field generated by charges at both ends of the coil, $\nabla \times \bf E_s = 0$. The Es field exists inside and outside the coil.

The two disparate fields' effects cancel each other inside the coil but outside there is no Em field. The result is $\oint \bf E_m \cdot \bf dl = L di/dt$ while $\oint \bf E_s \cdot \bf dl = 0$.

cf. post 102.

 

[etotheipi]

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.

But my issue is this; you are saying that $\nabla \times \mathbf{E}_{induced} = -\partial_t \mathbf{B}$ is the non-conservative component of the electric field caused by the time varying magnetic field inside the inductor, and you are saying that $\mathbf{E}_s$ is the conservative component produced by static surface charges on the wire. However, if you want $\mathbf{E} = \mathbf{E}_s + \mathbf{E}_{induced} = \mathbf{0}$, then you cannot have one as conservative and the other as non-conservative. This is the problem with the decomposition you are suggesting.

I think you are basically misintepreting the equation $\bf E_m = - \bf E_s$.

That equation does not mean that Es is just a negative Em.

That is exactly what the equation means... surely it is you who is mis-interpreting it? It follows from Newton's second law.

 

[vanhees71]

What you are still not recognizing is that Em≠0Em≠0. It is in fact the field that accounts for the battery emf.

[EDIT: I retyped the LaTeX, which somehow got distorted. I hope it's not a bug in the forums software?]

The battery emf is a thermodynamic quantity the electrochemical potential.

Your idea of this artificial split of the electric field is highly unphysical and misleading.

At best what seems to be something related to this split is to use the Coulomb-gauged electromagnetic potentials, i.e., you use the homogeneous Maxwell equations to introduce the potentials (in Heaviside-Lorentz units)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}$$
The Coulomb gauge-fixing condition is
$$\vec{\nabla} \cdot \vec{E}=-0.$$
Then you have
$$\vec{\nabla} \cdot \vec{E}=-\Delta \Phi=\rho$$
and thus a split of $\vec{E}$ in a solenoidal field $-(1/c) \partial_t \vec{A}$ and a conservative source field $-\vec{\nabla} \Phi$. But as this analysis clearly shows both parts are gauge dependent and thus unobservable mathematical auxilliary fields. You cannot easily interpret these parts physically in any way. Only the one and only electric field components $\vec{E}$ are physical.

In chemical equilibrium (i.e., for the open battery in the static state) you have a double layer of ions around the electrodes within which there is a static electric field, which is static and thus a pure source field with $\vec{\nabla} \times \vec{E}=0$.

For the closed battery you have a current running through the circuit (including the battery). In the stationary state, where the current is constant, again $\vec{\nabla} \times \vec{E}$. Then you get $\mathcal{E}=U-R I=0$, where is the EMF of the battery (in this stationary non-equilibrium state with a current running through).

For a detailed discussion, see a physical-chemistry book like Morimer or Atkins.

 

[etotheipi]

I came across this treatment of the inductor by Shankar, which unlike Feynman's derivation here (which I personally much prefer), does decompose the electric field:

Source: Fundamentals of Physics II, R. Shankar

This seems bizarre to me, and appears to fall victim to the same argument that @Dale has presented earlier in the thread. Shankar has an induced component $\mathbf{E}_F$ which satisfies $\nabla \times \mathbf{E}_F = -\partial_t \mathbf{B} \neq \mathbf{0}$ making it necessarily non-conservative.

He also seems to be suggesting $\mathbf{E}_C = -\mathbf{E}_F$, with $\mathbf{E}_C$ as a conservative component produced by static charges. This is an obvious contradiction.

I am thus inclined to say that this derivation is invalid, however Shankar is quite an eminent Physicist. I wondered whether someone else could comment on this?

 

[vanhees71]

Well, treatments like this made me sick, when I first learned about AC circuits in high school. Of course Feynman Lect. II is much better.

To treat this circuit you do not need any split. Take the coil be ideal, i.e., think its resistance being lumped into the resistor explicitly drawn. Then you only need to integrate Faraday's Law (in SI units) along the circuit to get
$$\mathcal{E}=\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=R I - U=-\mathrm{d}_t \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}=-L \dot{I}.$$
You get the differential equation
$$\dot{I} + \frac{R}{L} I=\frac{1}{L} U.$$
Since $U=\text{const}$
$$I(t)=c \exp\left (-\frac{R}{L} t \right) + \frac{1}{R} U$$
as the general solution. The integration constant can be determined by an initial condition. E.g., closing the switch at $t=0$ means that $I(0)=0$ and thus
$$I(t)=\frac{U}{R} \left [1-\exp\left (-\frac{R}{L} t \right) \right].$$

 

[etotheipi]

Well, treatments like this made me sick, when I first learned about AC circuits in high school. Of course Feynman Lect. II is much better.

I agree; Shankar's textbook has a lot of good parts but this is not one of them. Not only is this derivation unnecessarily convoluted in invoking the split, it's also flawed (i.e. @Dale's counter-proof). I think the issue mathematically is that he is trying to perform a Helmholtz decomposition on the zeroed out field $\mathbf{E} = \mathbf{0}$, whilst this is surely undefined?

Your derivation in post #103 is on the other hand simple, elegant and valid.

 

[Dale]

I think you are basically misintepreting the equation $\bf E_m = - \bf E_s$.

I am not interpreting anything. I am just doing math. This is a purely mathematical exercise. You accept 98.3 and 98.5, so 98.9 follows strictly by math, regardless of how you interpret it. Similarly with the rest. This is a mathematical exercise.

Surely you are not claiming that your theory is exempt from math. According to the math the set of statements you accept are contradictory, regardless of interpretation of any of the statements.

Does math apply to your theory or not?

They are equal in magnitude and opposite in direction but they have totally different sources and exist independently of one another, on top of one another.

I have no objection to this nor any of your other statements about interpretations. Similarly, for a person standing at rest the downward gravitational force and the upward normal force “are equal in magnitude and opposite in direction but they have totally different sources and exist independently of one another”. And they can be mathematically described by vectors with all of the corresponding vector operations

 

[etotheipi]

I have no objection to this nor any of your other statements about interpretations. Similarly, for a person standing at rest the downward gravitational force and the upward normal force “are equal in magnitude and opposite in direction but they have totally different sources and exist independently of one another”. And they can be mathematically described by vectors with all of the corresponding vector operations

This part is important. We are not saying that since $\mathbf{N} + m\mathbf{g} = \mathbf{0}$ for an object at rest on a flat surface implies that $\mathbf{N}$ is also conservative. The key thing in this case is that $\mathbf{E}_s(\mathbf{r}) = -\mathbf{E}_m(\mathbf{r})$ for all $\mathbf{r}$ inside the battery; i.e. the LHS and RHS are equal as functions. That then implies the dual conservativity.

 

[Charles Link]

$\nabla \times E_s=0$ everywhere. That makes $E_s$ conservative. The $E_m=-E_s$ only inside the battery. We can't in general say that $\nabla \times E_m=0$, or that $\oint E_m \cdot dl=0$. The result is that $E_m$ is non-conservative. $\\$ To me it is somewhat a matter of taste whether you like this splitting as an explanation or not. In any case, some of the mathematics that is being presented to criticize it, IMO, is inaccurate.

 

[Dale]

In any case, some of the mathematics that is being presented to criticize it, IMO, is inaccurate.

Please point out specifically which step in my proof is inaccurate.

 

[Charles Link]

Please point out specifically which step in my proof is inaccurate.

The comment was directed to the $E_m$ being labeled as conservative by @etotheipi .
I also disagree with 98.2. If I am interpreting it correctly, we do not have $\nabla \cdot E_m=0$.

 

[Charles Link]

@Dale Please see the addition to the above post, that 98.2 is, IMO, incorrect. See also my posts 86 and 87.

 

[rude man]

Prof. Shankar is right. He does not violate $\oint \bf E =0$ if $\bf E$ is conservative (or zero) as Feynman does in his lecture notes (vol II chapt 22).

Now I have two illustrious physicists in my camp - Profs.Skilling and Shankar. One at first-rate Stanford, one at a top Ivy college. That's 2-0 by my count.

BTW Prof. Shankar has two excellent introductory physics courses on YouTube.

I have no objection to this nor any of your other statements about interpretations. Similarly, for a person standing at rest the downward gravitational force and the upward normal force “are equal in magnitude and opposite in direction but they have totally different sources and exist independently of one another”. And they can be mathematically described by vectors with all of the corresponding vector operations

That is exactly my point. Your two forces can be written N = mg yet they have as you say " totally different sources and exist independently of one another". In this spirit we write $\bf E_s = -\bf E_m$. So where is the problem?

You might also seriously consider post 102.

 

[Dale]

@Dale Please see the addition to the above post, that 98.2 is, IMO, incorrect. See also my posts 86 and 87.

Well, (98.2) didn’t come from me. That was from @rude man. It is clear that he will need to discard at least one of those equations in (98.1-6), but I am not sure which he will choose.

 

[Dale]

In this spirit we write Es=−Em. So where is the problem?

The proof in post 98 demonstrates the problem conclusively. The equations that you propose are in fact inconsistent with each other. They cannot all be true.

Either you must discard one or you must claim that math doesn’t apply to your theory. You appear to be taking the latter approach which I regard as the worst possible approach. To discard math runs contrary to all of physics since Newton. It would be far better, in my opinion, to fix your equations. I honestly don’t see the point of having equations at all if you are going to discard the rules of math.

Again, does math apply to your theory or not?

 

[Charles Link]

Well, (98.2) didn’t come from me. That was from @rude man. It is clear that he will need to discard at least one of those equations in 98.1-6, but I am not sure which he will choose.

Mathematics may not be his strongest suit. I think he may have made a mathematical error or two in the course of the discussion, but that should not negate the merit of this methodology.

 

[Dale]

Mathematics may not be his strongest suit. I think he may have made a mathematical error or two in the course of the discussion, but that should not negate the merit of this methodology.

As far as I can tell his methodology is a Helmholtz decomposition of the E field, although he uses non standard terminology. That is fine and does have some uses. I do not object to it in general. I am specifically objecting to his treatment of a battery. It is mathematically inconsistent.

When someone uses a standard method incorrectly, then a criticism of the result does not imply a criticism of the standard method. Helmholtz decomposition is fine, but does not produce the result he claims for a battery.

 

[etotheipi]

He does not violate $\oint \bf E =0$ if $\bf E$ is conservative (or zero) as Feynman does in his lecture notes (vol II chapt 22).

You are making things up again . The presence of the time varying magnetic field inside the lumped inductor means that you can no longer constrain the line integral to be zero.

That is exactly my point. Your two forces can be written N = mg yet they have as you say " totally different sources and exist independently of one another". In this spirit we write $\bf E_s = -\bf E_m$. So where is the problem?

Do you just ignore my post #106?

 

[Charles Link]

As far as I can tell his methodology is a Helmholtz decomposition of the E field, although he uses non standard terminology. That is fine and does have some uses. I do not object to it in general. I am specifically objecting to his treatment of a battery. It is mathematically inconsistent.

When someone uses a standard method incorrectly, then a criticism of the result does not imply a criticism of the standard method. Helmholtz decomposition is fine, but does not produce the result he claims for a battery.

I question whether the Helmholtz decomposition works for this case. I can't put my finger conclusively on why it seems to go wrong, but all indications are that it doesn't work. Meanwhile, the separation into $E_m$ and $E_s$ is very straightforward, and the $E_s$ behaves in a very predictable manner.

 

[etotheipi]

Might I propose a treatment of the inductor which makes use of a decomposition of the electric field, that is consistent with @Dale's reasoning? There are some important differences to what you are suggesting.

We consider a lumped inductor, along with a closed curve through the conducting wire from one terminal of the inductor to the other, and out through the vacuum to the first terminal. Inside the conducting wire in the inductor, we introduce a decomposition $\mathbf{E} = \mathbf{E}_{c} + \mathbf{E}_{ind} = \mathbf{0}$. In the low frequency limit $\nabla \times \mathbf{E}_{ind} = 0$ implies $\mathbf{E}_{ind}$ is conservative inside the conducting wire, as is $\mathbf{E}_s$. Then of course the EMF $\mathcal{E} = \oint \mathbf{E} \cdot d\mathbf{l} = -\oint \mathbf{E}_{ind} \cdot d\mathbf{l} = -\int_a^b \mathbf{E}_{ind} \cdot d\mathbf{l} = \int_a^b \mathbf{E}_{c} \cdot d\mathbf{l} = -V$

This is to say that everywhere inside the conducting wire as well as outside the element the electric field is conservative. However, the domain is not simply connected (it is a torus), so the closed curve line integral is still non-zero.

Though I must say this is highly unnecessary, and perhaps even still invalid. Much better to just go with Feynman .

 

[Charles Link]

It seems it is impossible to make $E_{induced}$ to be conservative, and there is no reason it should be, because it isn't. The same is the case with the $E_m$ in general, regardless of the source. It normally seems to exist only in the source of interest, and points in one direction, and starts there and stops there. It also doesn't obey $\nabla \cdot E_m =0$. Perhaps $E_m$ doesn't fit the Helmholtz criterion because it may not be a well-behaved function.

 

[Dale]

Meanwhile, the separation into Em and Es is very straightforward, and the Es behaves in a very predictable manner.

Well, it certainly isn’t as straightforward as it seems since the straightforward analysis for a battery leads to an inconsistent set of equations.

In any case, if what he is describing is not a Helmholtz decomposition then I am exceptionally skeptical of its validity. A much more rigorous treatment is required