How to Recognize Split Electric Fields - Comments

[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

 

[etotheipi]

I'm going to go out on a limb here and just say that I don't think it then makes sense to decompose the electric field inside an inductor. You must either get two conservative fields, or two non-conservative fields (so any talk of a Helmholtz decomposition is out of the question). Of these two choices, both can produce a non-zero closed curve line integral due to the fact the domain is not simply connected. However, these don't make much sense in this context.

Feynman's treatment is the only one I've come across that I understand. I am done messing around with these unphysical splits

 

[Charles Link]

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

Even for the inductor, I don't believe it is a Helmholtz decomposition. It seems to be in the category of a mathematical construction, with the $E_s$ well-behaved and following the rules of electrostatics, (e.g. $\oint E_s \cdot dl=0$), while the EMF's and the associated $E_m$'s, where $\mathcal{E}=\int E_m \cdot dl$, are basically a very different animal.

 

[Dale]

Even for the inductor, I don't believe it is a Helmholtz decomposition. It seems to be in the category of a mathematical construction, with the $E_s$ well-behaved and following the rules of electrostatics, while the EMF's and the associated $E_m$'s, where $\mathcal{E}=\int E_m \cdot dl$, are basically a very different animal.

Then I think some mathematical rigor is necessary. E.g. a definitive formula for calculating these split fields in terms of standard EM quantities. Otherwise any discussion requires consulting an oracle to determine the decomposition

If it is not a Helmholtz decomposition then the only other thing I could see is that it is the decomposition $E_s=-\nabla \phi$ and $E_m= -\frac{\partial}{\partial t} A$ where $\phi$ and $A$ are the scalar and vector potentials in the Coulomb gauge. But who knows, it certainly isn’t described here with enough rigor to say.

 

[Charles Link]

Then I think some mathematical rigor is necessary. E.g. a definitive formula for calculating these split fields in terms of standard EM quantities. Otherwise any discussion requires consulting an oracle to determine the decomposition

It seems to be a model that some like and others don't. In the case of an inductor, there is sufficient symmetry that $E_m$ can be computed. For a battery, such symmetry is absent, and it becomes a somewhat abstract description. It doesn't have tremendous mathematical rigor in that sense. You can postulate $\mathcal{E}=\int E_m \cdot dl$ for some $E_m$, but you can't measure the $E_m$ because it is always offset by a $-E_s$. Personally, I see some merit in the description, but others seem to find it to be lacking in fundamental soundness.

 

[hutchphd]

I do see some merit to @rude man 's approach.

What is the merit? Please be specific.
The open circuit electric potential of a battery is a fixed number, and that is well known and is very useful.
To infer, from that single fact, that there must exist an internal electric field Em (m is for magic) is, charitably, an interesting conjecture. It neither simplifies the circuit analysis nor elucidates any actual underlying physics.
Please can we please stop discussing Chimera. I am certainly finished.

 

[rude man]

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.

I am deliriously happy with 98. 1-6.

BTW 98.3 is a vector addition so $\bf E_s + \bf E_m = 0$ in coil (or battery or ...).
$\bf E_m$ points - to + in battery, $\bf E_s$ points + to - in & outside battery.
|$E_s$| = |$E_m$| in battery or coil wire.
No inconsistency, no sale.

 

[Dale]

I am deliriously happy with 98. 1-6.

Then you have discarded mathematics

 

[Charles Link]

@rude man Clearly 98.2 is incorrect. You do not have $\nabla \cdot E_m =0$ for the $E_m$ that your model gives. I think posts 124 and 125 kind of summarize the general consensus of this methodology.

 

[rude man]

@rude man Clearly 98.2 is incorrect. You do not have $\nabla \cdot E_m =0$ for the $E_m$ that your model gives. I think posts 124 and 125 kind of summarize the general consensus of this methodology.

$\nabla \cdot \bf E = 0$ for ANY electric field. Just ask Maxwell.

 

[rude man]

Then you have discarded mathematics

OK.

 

[Dale]

OK.

With that I think we are done here. There is no point in discussing something that is knowingly so diametrically opposed to all of physics since Newton’s day. This is a very disappointing outcome for this thread.