How to Recognize Split Electric Fields - Comments

[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.

 

[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