etotheipi
rude man said: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 said:How on EARTH do you conclude this?
I did - twice!Dale said:@rude man can you respond specifically to my previous post regarding the battery?
rude man said: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.
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.etotheipi said:##\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi##; Hence ##\mathbf{E}_m## is derivable from a scalar potential.
Who said the chemical force is conservative? I said just the opposite.etotheipi said: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.
Bad conclusion!etotheipi said:##\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi##; Hence ##\mathbf{E}_m## is derivable from a scalar potential.
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.rude man said:Your basic argument that Es and Em fields can't coexist is baseless.
rude man said: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/
Correct. E = Em + Es in the battery = 0 (equal and opposite fields).Dale said:##E_s \ne 0## and ##E=0## (for an open circuit battery)
So if you say it's false, that makes it false? I demur.Dale said: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.
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?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
I've answered this, several times.etotheipi said:##\mathbf{E}_m = -\mathbf{E}_s = \nabla \phi##; Hence ##\mathbf{E}_m## is derivable from a scalar potential.
The problem is that these two statements are contradictory.rude man said:Are you remembering that Em and Es point in opposite directions in the battery? Em is conservative, Es is not. What's the problem?
I am.etotheipi said: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?
As I said several times, if Em were conservative the E field circulation would be zero, and it's not, it's the emf.But even then, this still wouldn't explain why you think ##\mathbf{f}_s## is not conservative.
They're not, and please explain why you think they are.Dale said:The problem is that these two statements are contradictory.
Vanhees and I have totally disagreed on this subject for a long time. As I told others, you pick who's right.etotheipi said: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!):
rude man said: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.
rude man said:As I said several times, if Em were conservative the E field circulation would be zero, and it's not, it's the emf.
If some vector field ##X## is conservative then so is ##-X##rude man said:They're not, and please explain why you think they are.
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?Dale said:If some vector field ##X## is conservative then so is ##-X##
I just did. So did Feynman. So did the guy sitting on your moving media.etotheipi said: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!
It is not. Never was, never will be. And I certainly didn't say so.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.
Adios amigo.etotheipi said: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.
rude man said:So did Feynman.
rude man said:It is not. Never was, never will be. And I certainly didn't say so.
rude man said:Dale said: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.
I could but it would take time and I don't think it would do any good.etotheipi said:Could you produce a reference for this?
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.
It isn't.I maintain that ##\mathbf{f}_s## is conservative between the terminals, because I can write it as the derivative of a scalar field.
rude man said: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.
rude man said:It isn't.
You can't.
(And I think you meant potential, not field.)
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.rude man said:If ##-X## is just ##X## with polarity reversed, sure. But not if they have different sources.
rude man said: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.
Dale said: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 has written a good post (#44) that I had overlooked since I was so busy with the more frequent posters.Dale said: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.
It's a well-known mathematical theorem. However the physical conclusions you seem to draw from it are at best confusing.rude man said: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.rude man said: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.
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 said: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
Let’s see if we can start with where I believe we agree.rude man said:What you are still not recognizing is that ##E_m \neq 0 ##. It is in fact the field that accounts for the battery emf.
Yes.Dale said: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?
No, ## \nabla \cdot \bf E = 0 ## does not imply that ##\bf E_m ## is non-conservative. That statement is wrong. Did you mean it?Do you further agree that this implies that ##E_m## is not conservative?
Yes, as I just said.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. 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.Finally, do you agree that in a open-circuit battery ##E_s \ne 0## at equilibrium?
Well, not quite! Almost!I believe that we agree on all those points, but would like to confirm before moving to the points where we disagree.
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.rude man said: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 ##.
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 said:EDIT: all electric fields free of charge have ## \nabla \cdot \bf E = 0##. So quoth Maxwell.
Charles Link said: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.
Hi Charles,Charles Link said: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.
Inside the conductor there is plenty of charge.rude man said: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.
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 said:Inside the conductor there is plenty of charge present.
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 said:It's not free charge and besides the conductor is neutral in charge. If div E were not zero it would build up somewhere.
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.rude man said:On top of that, we are assuming an ideal conductor so E = 0 so div E = 0 inside the conductor.
No problem with #86.Charles Link said: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.
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 said: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.
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.rude man said: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.
.
OK, so for convenience I will write down the statements and number them for easy reference:rude man said: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.
Charles Link said: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 said: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.