This is exactly why the other "less scientific" definition of voltage (The one about how many electrons there are in one spot) is used in circuit analysis and pretty much everywhere else where you need to actually calculate something.
You keep talking of that as if it were a different physical quantity, measured in different units. It's not. The 'scalar potential' that is part of the (V, A) pair is just the path integral of the conservative part of the total electric field.
It's as if you decomposed a mathematical function into its odd and even parts and than claimed that the odd part is a different, 'less mathematic' definition of function.
You can easily see where that decomposition comes from by writing Faraday's law (use dS for the differential element of area to avoid confusion with the vector potential A, and use E_total instead of E to highlight that it is the resulting field, superposition of coulombian and induced fields). Then express B in terms of the vector potential A, turn the surface integral on the right into a path integral along the surface contour using Stokes (or "the rotor's") theorem. Now take the integral on the right to the left, changing sign. Bring everything inside the same path integral. You are left with a field whose circulation along a closed path is always zero.
Hey, that's a conservative field! Well of course, you have stripped the induced - non conservative - part from the total field.
Congratulations, it can be very useful, but it's only half of the story.
I will come back to this post with formulas and drawings when I will be able to scan.
You say that resistor with voltage drop have zero E field inside. That means that integral E.dl over it is zero meaning that it does not have voltage drop on it! It's paradox, don't you see?
Indeed. But I never said that the resistor has zero E-field inside. I said that the circulation of the conservative part of the E field is zero.
The conservative part of the E field is stronger in the copperas well. It has to be in order to cancel the induced part.
QuoteI will try again.
1) Does KVL hold when inside box is DC battery? 2) Does KVL hold when inside box is battery-powered AC generator? 3) Does KVL hold when inside box is piezo-based 1V AC voltage generator? 4) Does KVL hold when inside box is transformer?
1) yes, outside and inside
2) it depends. Does the generator have a time-varying B field inside ? Is so and if the flux is neatly tucked inside the box, then 'extended KVL' (which is Faraday under disguise) will appear to work outside, but won't work inside when you cross the flux-varying region.
3) I am not familiar with piezoelectric generators, but if there is no dphi/dt involved we probably can treat them as batteries.
4) 'new KVL' which is Faraday under disguise will appear to work outside and fail miserably inside if you attempt to cross the flux-varying region.
Did you just say that Dr.Lewin is mistaken? - Because he claims thatnonconservative field is zero inside copper coil (or secondary of transformer).
So, copper part have *only* nonconservative field, resulting voltage (integral E.dl) equals 1V. You agreed because you did not argue when I said that it is possible to measure 1V AC voltage on transformer secondary w/o resistor connected. (It would be dumb to argue anyway).
In our "box + resistor" case resistor is outside magnetic field of the transformer, so Faraday's law cannot do anything about it, so there cannot be nonconservative fields in form of EMF inside it.
According to logic above, sum of fields in copper and resistor, integral E_conservative.dl and integral E_nonconservative.dl is zero. [edit] At given time of observation obviously.
Fact that you even try to answer those questions is hilarious by it's own We agreed that for voltmeter it does not matter what's inside the box, it cannot sort out electrons - they were moved by class-AB amplifier, little monkeys or Mr.Faraday.
Did you just say that Dr.Lewin is mistaken? - Because he claims thatnonconservative field is zero inside copper coil (or secondary of transformer).He must have meant the total field.
Keep in mind that the left side of Faraday's equation, circulation of E.dl refers to the total field.
Of course it might happen that the conservative field be zero as well: if all fields are conservative, the conservative field is all you've got and in that case the conservative field will be zero inside a perfect conductor. It happens for example in electrostatics: point charge generates a conservative E-field, you place a piece of copper nearby, free charges on the copper surface redistribute to create a contribution that will erase the field inside the copper. So sum of conservative E field due to point charge plus conservative field due to free surface charge equals total conservative E field inside the conductor is zero. And I think it will work with batteries and a copper wire.
It's all about context.
Can you show where Lewin said that?
QuoteSo, copper part have *only* nonconservative field, resulting voltage (integral E.dl) equals 1V. You agreed because you did not argue when I said that it is possible to measure 1V AC voltage on transformer secondary w/o resistor connected. (It would be dumb to argue anyway).No I do not agree and I have already explained how that can happen in one of my previous post (many pages back).
QuoteIn our "box + resistor" case resistor is outside magnetic field of the transformer, so Faraday's law cannot do anything about it, so there cannot be nonconservative fields in form of EMF inside it.
I have to stop you there: as long as you stay outside you can pretend that the contribute of -dphi/dt are either 'battery-like' emf of 'resistor-like' voltage drops. But when you get inside the box - and you have to get inside the box to compute the path integral of the total E-field along the circuit's path, you have to surrender this delusion and come to terms with Faraday.
QuoteAccording to logic above, sum of fields in copper and resistor, integral E_conservative.dl and integral E_nonconservative.dl is zero. [edit] At given time of observation obviously.
The above logic is flawed because it does not keep into account the right hand side of Faraday's equations.
From the outside of a toroidal transformer it does not matter how I place my probes
Did you just say that Dr.Lewin is mistaken? - Because he claims thatnonconservative field is zero inside copper coil (or secondary of transformer).He must have meant the total field.
Keep in mind that the left side of Faraday's equation, circulation of E.dl refers to the total field.
--snip--
Can you show where Lewin said that?If you start "Is Kirchhoff's Loop Rule for the Birds?" video at around 15:10, you will hear it.
QuoteQuoteSo, copper part have *only* nonconservative field, resulting voltage (integral E.dl) equals 1V. You agreed because you did not argue when I said that it is possible to measure 1V AC voltage on transformer secondary w/o resistor connected. (It would be dumb to argue anyway).
No I do not agree and I have already explained how that can happen in one of my previous post (many pages back).LOL. From which alternate universe your physics come from? When I measure 1V on the transformer terminals, you claim that it is not 1V actually?
Don't even say "Faraday" in area where there's no magnetic field!
If you start "Is Kirchhoff's Loop Rule for the Birds?" video at around 15:10, you will hear it.
What I hear is "electric field" and I agree: the total, resulting electric field in copper is negligible. Zero in a perfect conductor. As I have always said.
Where do you hear him saying that (in this particular circuit) the conservative electric field in copper is zero?
Other post[/url] of yours is unfortunately off-topic because it is not specified that there is Dr.Lewin's experiment in the box.
Other post[/url] of yours is unfortunately off-topic because it is not specified that there is Dr.Lewin's experiment in the box.
Oh! I didn't notice that the topic had suddenly become off-topic. My bad.
Other pos of yours is unfortunately off-topic because it is not specified that there is Dr.Lewin's experiment in the box. It was defined that box is magnetically shielded and there is no EMF induced in the wires coming out of the box.
To give things some better perspective i have put together a few images that graphically show the fields.
WTF is the keyboard combination that can make you lose the whole post??
This site does not have a draft saving function? I was copying a ***** line and all went poof!
The ****! What the ****** *****!!!
In the case of the field produced by a primary cylindrical coil here is the induced e field magnitude:
here is plotted with direction in a plane perpendicular to the axis of the infinitely long primary coil (here represented by the orange ring):
Seems perfectly defined (of course it is time-varying but we can express exactly how - this is snapshot frozen in time). The voltage, on the other hand...
I'll skip the part regarding that paper you mentioned back when we had this discussion and that kept using AREA in computing the emf for a single line.
We're going 'full field' now, so forget areas - welcome boundary conditions!
You are able to plot the induced field only when you specify the boundary condition set by the coil generating the field.
After all Maxwell's equations in their differential form are... partial differential equations and if you do not specify boundary and initial conditions, how can you choose the solutions that suits your problem among the infinitely many? In their integral form, since the are essentially integral relations between areas and boundaries you have to specify... well, you know what. Maybe there is some equivalence hidden in there?QuoteYou only need a closed loop when you want to use Faradays law to directly calculate the EMF voltage. It doesn't mean that you can't have EMF without a closed loop just because Faradays law uses a loop area, it just means you have to use other laws to calculate your EMF in those cases.
Except the EMF is only half of the story.
When you put the secondary coil in, the charges in the conductors will rearrange to produce the coloumbian field and the resulting field depends on how you close (or do not close) the loop. After all E_total has the same direction of j. You can end up with a total E that is opposed to E_induced (it helps using a finite albeit big value for sigma). I had written something else in the lost post but **** it! Anyway I still need to scan my drawings so...
And so, what is the true voltage?
Because it seems to me that in your coloring analysis you skipped the Mabilde-McDonald voltage right away and went on calling "true voltage" other voltages.
If I understand correctly, what you call "Charge density" is the voltage definable as potential difference that is associated with the coloumbian aka conservative part of the total field. Why is not that the true voltage?
I can't call the charge density approach true voltage because that is not how voltage is formally defined. But this is the voltage that all voltmeters detect and show as a result and it is never undefined so i tend to use this so called "effective voltage" or "conservative voltage" or "columbian voltage" or whatever you want to call it, just because its more useful to work with, while not breaking any of Faradays or Maxwells math.
In fact, when you bring a voltmeter inside the ring, you can measure any voltage you want between -0.1V and +0.9V
Quote from: BerniYeah forum web software has not advanced much at all in the last 20 years.Well, some web software has advanced. Stack Exchange for example, saves the drafts, upload images directly, has a nice TeX editor... Too bad they pissed me off with their censorship attitude. But, never mind.
I am a bit at a loss, here. I thought you were looking for the true one and only voltage, the one that has is so uniquely defined that it can be expressed as a potential difference, independent of path. And yet you say that potential is undefined?
Are you referring only to the possibility to run more than one time around the core (in which case, we could overlook it, limiting ourselves to a full circle at most) or is this a more fundamental lack of uniqueness?
Because the way you used colors makes me believe you are selecting a particular class of paths to represent your color coded voltages, namely paths along the circle. You fix 0 at one node (A, IIRC) and then compute the path integral along the circle from A to another point P on the circle. Do we agree that this does not exclude the possibility that the voltage from A to P depends on the path?
Which one of the many voltages you have shown is the path-independent one?
(My guess is that it has to be the first one, the conservative one you called "charge density". And yet you say
Which seems to me an admission that... the true voltage is path dependent, as Lewin has always said. But let's forget about this for the moment. The part I am more interested in is the followingQuoteI can't call the charge density approach true voltage because that is not how voltage is formally defined. But this is the voltage that all voltmeters detect and show as a result and it is never undefined so i tend to use this so called "effective voltage" or "conservative voltage" or "columbian voltage" or whatever you want to call it, just because its more useful to work with, while not breaking any of Faradays or Maxwells math.
So to be clear, is this the voltage Mabilde is measuring? The one that Kirchhoffian call the 'true and one voltage'? You just decided to call it with another name (not a problem, we just need to be clear) but can you clarify that this is the case and that what you called "Charge Density" in the picture and call now "effective, conservative, coloumbian voltage" is the Mabilde-McDonald voltage?
Before addressing the meaning of said voltage, let me say that there is one problem with you statement above. You wrote that "that is the voltage that all voltmeters detect and show as a result". I am afraid you are mistaken.
To show the Mabilde-McDonald voltage you need a special measurement setup, consisting in a careful choice of your probes' path. The voltage all voltmeters detect is the one with the operational definition of path integral of E. The one that depends on the path of your probes when there is a changing magnetic field. In fact, when you bring a voltmeter inside the ring, you can measure any voltage you want between -0.1V and +0.9V and that is what the voltmeter show. Even outside you have two different values depending on how you place the probes around the core.
So, no. The 'effective, conservative, coloumbian voltage' is not the voltage that all voltmeters detect and show as a result. You did the experiment yourself!
But I think I know where you got that idea.
I will address that in a separate post, along with my answer to this
QuoteAlright then lets hear how you think one should calculate the charge density on the ends of a open wire in a changing magnetic field.
EDIT: fixed sentence I had left without conclusion.
Stack Exchange for example, saves the drafts, upload images directly, has a nice TeX editor... Too bad they pissed me off with their censorship attitude. But, never mind.Yeah true, but the strict moderation is what makes stack exchange so useful when you just want a quick, concise and correct answer to a technical question. The best answers float up to the top and some answers actually have a significant amount of effort behind them.
Well? How do we calculate it then? Or is there no need to calculate it because it's zero perhaps?
Suppose I put my electron in, with particular initial conditions (like velocity in a particular direction starting from A) and that I have recorded its trajectory from A to B. Would you agree in calling the voltage computed along that path the most meaningful voltage for this particular experiment?
You can read +1V and -1V as well - if you short probe leads that go around solenoid. Some academic scientists and their worshippers may say "Look! I discovered that voltage is path-dependent" while actual "discovery" is just electromagnetic induction.
Oh, come on. I thought we were past that.
Voltage in non-conservative E-field IS path-dependent. This is not even up for discussion.
Nah, the problem is twofold: first, science is not and should not be democratic - so the 'most voted answer is the best' is not necessarily true and the mechanism is flawed from the start. While it might work fine for some 90% of the content, when it comes to specialized stuff, it breaks miserably. This flaw is exacerbated by the editing power that one can exert even if he/she has no clue of what is being discussed (meaning the so called 'expert' in one branch might only have an approximate and sometimes erroneous knowledge of other branches). Imagine if you and ogden had the power to close this discussion because it's crystal clear that this is just a probing error (and you have more yellow square than me and MHz). Or to look if from the other side, if Mhz and I had the power to shut up Kirchhoffians by editing and removing their posts (and once they are removed, if the site has enough traffic there won't be enough people to notice or even care to reopen them).NobodyMany readers in this blog would not have discovered the surprising role of surface charge in keeping the current within a conductor (I will get to that when I will have completed my drawings and scans).
I was happy with the way Physics SE is run: there is highly competent people making the selection there. Not so much in EE. Waste my time once, shame on you. Waste my time twice...
Scientific populism will be a problem a few years ahead.
But enough digressing.
QuoteWell? How do we calculate it then? Or is there no need to calculate it because it's zero perhaps?
We use Maxwell's equations. What else?
Zero? With an abrupt discontinuity at the ends and an induced field of known geometry? What makes you think that?
It is zero in a closed isotropic circular conducting torus perfectly aligned with the circular induced field. But even then, my guts say you just have to move it off axis to see charge pile up on the 'lateral' surface (well, is there any other surface on a torus?). And if you place portion of different resistance, you will certainly see charge pile up at the surface of separation.
I've found plenty of literature supporting my point of view. I need a little bit of time to sift through the best works and select a few images. Keep your popcorns in a warm place, I'm almost done with my flu.
edit: nobody was a truly bold statement. In my defense, it's all Jackson's fault. I'll explain later
Nah, the problem is twofold: first, science is not and should not be democratic - so the 'most voted answer is the best' is not necessarily true and the mechanism is flawed from the start. While it might work fine for some 90% of the content, when it comes to specialized stuff, it breaks miserably. This flaw is exacerbated by the editing power that one can exert even if he/she has no clue of what is being discussed (meaning the so called 'expert' in one branch might only have an approximate and sometimes erroneous knowledge of other branches). Imagine if you and ogden had the power to close this discussion because it's crystal clear that this is just a probing error (and you have more yellow square than me and MHz). Or to look if from the other side, if Mhz and I had the power to shut up Kirchhoffians by editing and removing their posts (and once they are removed, if the site has enough traffic there won't be enough people to notice or even care to reopen them).NobodyMany readers in this blog would not have discovered the surprising role of surface charge in keeping the current within a conductor (I will get to that when I will have completed my drawings and scans).
But enough digressing.
Keep your popcorns in a warm place, I'm almost done with my flu.
If probe tips are shorted meaning voltage is 0V but I get 1V indication, I cannot call it meaningful by any stretch of imagination.
If voltage indication changes depending on how you manipulate with probe wires - you cannot trust your measurements not to mention make big scientific thing out of it.
I already said that and can repeat: "voltmeter reading (but not actual voltage in the connection point) depends on the path of test leads, especially if they are placed in the varying magnetic flux."
If probe tips are shorted meaning voltage is 0V but I get 1V indication, I cannot call it meaningful by any stretch of imagination.
Maybe nature wants to tell you something you are ignoring.
Well, Michael Faraday, a scientist, had the same problem that you do way back in 1831.
If Kirchhoff always held, you couldn't have the modern world and we would be able to have this very conversation. Learn to live with that.
LOL. Have to repeat that you are "arguing against the nonexistent strawman who is apparently suggesting that Farady's law is incorrect, and Kirchoffs is always correct", like broken record.
I might have missed a few but these are the ones i remember right now. Any disagreement on these?