Talk:Displacement current

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[edit] Earlier talk

moved to archives:

copied from earlier talk for ref--Light current 01:40, 10 March 2006 (UTC)

[edit] Definitions of displacement current

I see your still at it LC. Why do you still equate displacement current with electric current in the sense of charge movement? Displacement current is not the transport of electric charge, right? Haven't we agreed on this in the past? No one will be able to measure the movement of (real) electric charge between the plates of a vacuum capacitor. But, this has nothing to do with detecting displacement current. As I've said before, displacement current is the change in electric displacement. Before one could measure displacement current directly, one must define and then measure electric displacement directly, right? Besides, displacement current can be transformed out of Maxwell's equations. Alfred Centauri 03:52, 9 March 2006 (UTC)

Nice to see you are still with us Alfred. When you say 'at it' , Im merely responding to a challenge I posted a few months ago. Yes you and I have agreed on: Displacement current is not the transport of electric charge, right? What Im not sure about agreeing with you on is that nothing 'flows' from one plate to the other. Here we have a User who has performed an experiment that indicates the conduction current in the wires equals the so called dispacement current in the capacitor. In that sense, it appears that User:Mak17f believes these currents to be the same thing and satisfying Kirchoffs law. As you correctly told me some time ago, one must be very careful in applying KL in electric field problems. In fact the field solution will always be correct, wherreas KL may lead you up the garden path (as in this case)! However, the expt does not prove that something is flowing between the plates because I believe that my alternative explanation above for what happens is equally vaild. BTW a lively discussion on all aspects of this subject is in progress at Talk:Ivor Catt.--Light current 14:48, 9 March 2006 (UTC)
I feel that I was extremely accommodating as to your experiment. You stated bluntly “The conduction current in the leads (if there is any) is certainly not equal to the displacement current (if there is any).” I showed there WAS conduction current in the leads as I said I would. I illustrated that this current (through calculations using the geometry of my setup) is equal to what we expected the displacement current to be. What is your explanation as to what happened? Or, are we just argueing to argue? Certainly if you have an alternative explanation, please explain it in detail and support it with calculations.Mak17f 15:24, 9 March 2006 (UTC)
You have indeed been extremely accomodating in your willingness to perform the experiment and in completing it so quickly!. I fear however that you may not have full grasped the purpose of the experiment which was to find out if anything physical flows from one plate of the capacitor to the other. I should perhaps have phrased my assertion slightly differently:
The measured value of displacement current (if there is any) is certainly not equal to the conduction current in the leads.
You have:
  • measured the conduction current in the leads (I agreed that you would- aproximately anyway)
  • illustrated that this current (through calculations using the geometry of my setup) is equal to what we expected the displacement current to be.
You have not:
  • proved that anything physical flows from one plate to the other. You have assumed it by false application of KL.
Im not arguing for the sake of it although it may look like it sometimes.
There has been intense discussion on this subject over the past few months. I feel a lot of the misunderstanding has been about the definition of displacement curent. Is is a current (ie something that flows), or is it something that comes out of Maxwells equations?--Light current 15:47, 9 March 2006 (UTC)
This is the reason why I thought it was necessary to define exactly what I was planning to do before I conducted the experiment. Let me ask you a couple questions on your position:
  1. Can capacitors be charged? Hold charge? What is the fastest they can be charged?
  2. Can the charge polarity be reversed? How quickly?
  3. Do you believe that I=dq/dt? Is there a time varying charge magnitude on the plates?
  4. What current was I measuring? Where did the charge come from?Mak17f 16:05, 9 March 2006 (UTC)
Q1. Yes. Yes. Fastest time is equal to the 2 way transmission time of the equivalent transmission line if the gen is matched to the Z0 of the capcitor. Other values of source Z will give a longer charging time.
Q2 Yes. See ans to Q1
Q3
a)That is one of the definitions of current, yes.
b)Depends what you mean here. There is a time varying electric field. Any apparent charge will be due to the em wave. There cannot be any charge carriers rushing around at these speeds!.
Q4 You seemed to be measuring the conduction current in the lead. Whether this is the total current in the circuit is, of course, not clear with your setup as stray capacitace exists all over the place (even across your CT). You dont need real charge to activate your CT. Its the mag field in and around the conductor that does that.
Sorry if these are not the replies you ewere expecting but these are the only logical answers that fit the facts especially as no one has measured any current flowing between plates (AFAIK). It is also a consequence of KL not being generally applicable in field theory!--Light current 16:31, 9 March 2006 (UTC)
I'll adress your points. But above I also asked that you would tell me exactly what current you propose I measured in the experiment. Also, support the claim with calculations using values from the experiment. Mak17f 17:26, 9 March 2006 (UTC)
Your statement is ambiguous. But if your are asking me what current I wanted you to measure, it is the current actually flowing between the plates. You will need a thin sheet of resistive material (like graphite foil or sheet) with a wire fixed on to each side and fed to a diff amp plugin on the scope. My opinion is that you will not be able to measure any pd across this current probe.--Light current 17:36, 9 March 2006 (UTC)

[edit] Alfred Centauri's Comments

I'm certain I'll be sorry for sticking my nose into LCs and Mak17fs conversation here but I just can't resist. LC, when you say "There cannot be any charge carriers rushing around at these speeds!", I say 'so what?'. The electric current density is the product of charge density and velocity. With the appropriate charge density (e.g., the density of mobile electrons in a conductor), the electron drift velocity can (and is) exceedingly small for any reasonable current. So, what is your point here? Alfred Centauri 16:48, 9 March 2006 (UTC)

Its not the amount of current. The problem is the effective frequency at which they can move around the circuit inside the conductors. In fact at a very few MHz the carriers start to move to the conductor surface. At even more MHz, are there any left at all in the conductor or is it all in the em field?

--Light current 17:12, 9 March 2006 (UTC)

First, it's difficult for me to understand what picture it is you have in your mind when you use phrases like 'the frequency at which they can move around the circuit'. For me, such a phrase conjures up a picture of electrons circulating along a closed path where the frequency refers to the number of round trips made per second. For an alternating electron current within a conductor, I picture the electrons executing a sinusoidal motion that is superposed with their normal random motion. In this, case no electron moves around the circuit. Is this your view, also?

Secondly, it is not my understanding that electrons move toward the conductor surface as the frequency of the current increases. Rather, it is my understanding that the deeper inside the conductor the mobile electrons are, the less they participate in the alternating current. The degree to which this is true depends on the frequency of the current. Is this your understanding, also?

Thirdly, what the heck does your last question mean? What is it that isn't left in the conductor - mobile electrons? What is it that is all in the em field?

Lastly, in the end, it is the amount of current that is at issue here. You said "Because charge carriers in the metal travel very slowly and cannot cause actual charge separtaion in your cct in the times involed (~2us).So any apparent charge (if there is any) on the plates must be due to some other phenomenon". Charge separation is current. To separate charge, charge must be moved. For a given amount of charge to be separated (moved) in a given time, there must be a certain amount of current. In the above, you are claiming there is not enough current to separate the charge in the amount of time given for the reason that the charge is moving very slowly in the metal. Did you mean to say something different? Alfred Centauri 20:29, 9 March 2006 (UTC)

1. The picture I have in my mind is that in the the conductors there are free charge carriers. Lets call them electrons. These poor little electrons can only move in the conductor at a few mph. So how the heck can they transmit anything at 2c/3?. I picture the electrons executing a almost no motion due to the applied fiele. THe motion that is of their normal random motion and no individual electron actually moves very far (or at all even) at rf. So we agree.
2 I should have said: the deeper inside the conductor the mobile electrons are, the less they participate in the alternating current and that this may alos be relevant to the argument about the wires not actually conducting anything
3 At even more MHz, there any very few charge carriers capable of rf conduction at all in the conductor due to skin effect. All the energy then has to be transfered by the em field.
4 You cant achieve full separation of charge just by shuffling a few electrons back an forth at the far end of a long wire. It would takes time (like charging a battery) for the slowly moving charges to have net effect of charge separation.
*There is a way of charging a capacitor v quickly-- but youre not going to like the explanation I have for that!
--Light current 23:17, 9 March 2006 (UTC)
The link I included above is a decent explanation of how current flows. Alfred is correct in how he stated it above. Your fourth point is not correct. We have to overcome this hurdle before moving forward. In the mean time, I will update the article with the accepted textbook definitions. I'll provide references, and I hope opposing views will have the same. Mak17f 00:37, 10 March 2006 (UTC)
Well I looked at that link -its crap! If this is your idea of a good reference, Im very disappointed. It is very simplistic and does not answer the real questions. Its for kids. The top voted answer (tapping your neighbors shoulder) is just nonsense. The 'tapping' velocity has to be about the same as the overall velocity- its obvious. Its the same with electrons nudging each other: they cant 'nudge' faster than they can move. This is so obvious that it defies further explanation. The writer seems to be confusing distance with speed!!!! --Light current 01:09, 10 March 2006 (UTC)
http://www.jimloy.com/physics/electric.htm

Thgis link indicates that its the wave that travels at about the speed of light. But the electrons dont. Also where is the wave?--Light current 00:43, 11 March 2006 (UTC)

http://www.discoverycentermuseum.org/experiments/272.htm
http://answers.google.com/answers/threadview?id=345010Mak17f 16:30, 10 March 2006 (UTC)


(Mak17f, I have indented your entry above to help us keep straight who posted what - LC is quite strict about this kind of stuff ;<))

OK, rather than dispute LCs 4th point, I'm will instead describe my 'picture' of how the separation of charge occurs at 2/3 c or whatever. A conductor, regardless of whether it is the wire connected to a capacitor plate or is the plate itself, has a certain negative charge density and positive charge density. For a charge neutral conductor, these densities are the same. We know that the positive charge density (the protons in the nucleus of the elemental atoms) must remain more or less constant. We also know that the majority of the electrons that constitute the negative charge density are tightly bound to atomic nuclei and this contribution to the negative charge density must remain more or less constant. However, there are a significant number of electrons that are not bound to any particular nuclei but are instead bound to the structure of the solid as a whole. The density of these mobile or conduction band electrons can be changed easily and quickly and without much movement of the actual electrons. To help see this, I like to use the following macroscopic toy model.

Consider a three dimensional regular lattice of identical billard balls connected together by identical springs. When all the springs are in their 'rest' position, the system is in its lowest energy state corresponding to a neutral charge density state in a conductor. Now, imagine that a force is applied to one side of this cube formed by the billard ball/spring lattice. This force accelerates the outside billard balls on one side thus compressing the springs connecting the outside billard balls to their immediate neighbors (in the direction of the force) thus imparting a force on the immediate neighbors thus compressing the springs connected to their immediate neighbors and so on. The result of this force is a positive density wave - a change in the density of the billard balls that propagates through the lattice at a much higher velocity than the velocity of the billard balls themselves. In front of the wave, the density is undisturbed. Behind the wavefront, the density has changed to new value. One can also picture a opposite force that would result in a negative density wave that causes the springs to be stretched rather than compressed.

It shouldn't be hard to see that in this picture, the springs are loosely analogous to the Coulomb force between mobile electrons in a conductor and the density of the billard balls represent the electric charge density. The force that disturbs the lattice represents an external electric field while the energy propagating along the springs represents the energy carried by the EM field. Hopefully, this simple mechanical analogy suggests how charge density (and thus total charge) can change much more rapidly than the actual motion of electric charge would suggest. Alfred Centauri 01:45, 10 March 2006 (UTC)

P.S. I wrote this up before I read LCs critique of the link Mak17f provided. I am now prepared to be ridiculed - go for it LC. Alfred Centauri 01:45, 10 March 2006 (UTC)

I wouldnt ridicule you even if I could because people in glass houses shouldnt throw stones:-)--Light current 01:57, 10 March 2006 (UTC)
How can a wave travel faster than the elements that actually cause the propagation?--Light current 02:24, 10 March 2006 (UTC) Silly question!!--Light current 02:57, 11 March 2006 (UTC)

What does 'elements that actually cause the propagation' mean? What are, in your view, the elements that cause the propagation of, say, a pressure wave in air (or water or rock)? What, in your view, are the elements that cause the propagation of a transverse displacement wave on a taunt string? Is it the matter particles that make up the medium or is it an interaction between the particles in the medium that propagate a disturbance (wave)? Alfred Centauri 04:54, 10 March 2006 (UTC)

In your model the elements are the billiard balls. In air we were taught that it was the molecules of the air undergoing a net non random motion (above and beyond their thermal motin etc) the propagates the pressure wave. On stetched string, again it must be movement of each molecule that drags its neighboring molecule along by molecular attraction.
I would say both elements are essential to the propagation: both the ability of the 'elements' to move at the required velocity of overall propagation, and the ability of the interaction forces to transmit the wave at that velocity. In your ex. the balls must move at the propagation velocity albeit for a short time ; the springs must transmit the forces at the required velocity.
Now in the case of electronic charge carriers, we are told that on average they only move at a few mph (I take this as true altho' I have not measured it). Now whilst interaction forces may be able to act much faster that this, from the above supposition it is plain to see that proper charge separation due to movement of charge carriers is only possible in the lowest frequency cases (inc z.f)--Light current 17:16, 10 March 2006 (UTC)

LC, the particle velocity is not the same as the propagation velocity. For example, the air particle peak velocity for a sinusoidal plane sound wave propagating at a speed of c is given by:

u_{peak} = \frac{2 \sqrt{2} \cdot 10^{\frac{SPL}{20}-5}}{\rho_0 c}

where SPL is the sound pressure level and ρc = 1.18kg / m3 is the density of air at standard pressure and temperature.

As an example, the propagation speed of a sound (pressure) wave in the air at sea level is about 345 m/s. What is the peak particle velocity for a 90dB SPL sinusoidal plane wave? If I've done my calculation correctly, the answer is about 0.0022 m/s.

So, I must disagree with your view that the elements (particles) of the medium must move at the required velocity of overall propagation. Alfred Centauri 22:45, 10 March 2006 (UTC)

OK I'll look at this.--Light current 23:29, 10 March 2006 (UTC)
Having looked at my book on sound, I find that even in longitudinal plane sound waves, the particle velocity is equal to the product of the wave velocity and the slope of the displacement curve (related to amplitude and freuency I presume). Therefore, the particles may move faster than, slower than or at the same speed as, the wave velocity. It is interseting to note that the wave equations for sound waves in air are similar to those for em waves travelling down a TL>
Therefore a sound (or other machanically produced) wave can (counter to intuition) propagate faster than the individual particle velocities. My School course stopped short of the wave equations for sound unfortunately. OK I admit was wrong about sound waves (Ouch!).(lack of proper education!)
Are you saying the same thing happens with em waves where the velocities are much higher? If you are, you can quickly get into trouble when the slope of displacement curve approaches unity as the particle velocity then approches c. --Light current 00:32, 11 March 2006 (UTC)

[edit] Shock waves

When the air particle velocity exceeds the wave velocity, a 'shock wave' is formed - see Sonic boom. In a near perfect analogy, electrons can exceed the local speed of light within a medium and an EM (light) 'shock wave' is produced - see Cherenkov radiation. However, as you point out, such a shock wave cannot be produced in the vacuum as that would imply that the electron velocity is equal to or greater than the speed of light in the vacuum. Alfred Centauri 02:09, 11 March 2006 (UTC)

Yes I was aware of the sonic shock wave and have seen the effect of Cerenkov radiation on water cuased by spent nuclear fuel rods (an eerie blue glow as I remember).Now the problem is whether this knowledge has any bearing on the subject at hand. We have em waves. Particles (charge carriers) may travel locally at high speeds but their average velocity is low, I think you will agree. So this does still leave the question of how charge carriers can get around a macoscopic circuit in the time to move any real charge. The answer, of course, is that they can't and any effects seen at the capacitor plates must be due to the em field. If there is a problem with the source frequency, just increase it until you're sure the carriers cant travel the distance in the time allowed.
Let me ask you a couple of questions:
If the generator is switched off at the peak of a cycle at the capcitor, will the capacitor have any separated charge on its plates? If so how did it get there?--Light current 02:47, 11 March 2006 (UTC)

The answer, as far as I'm concerned, is that the mobile charges don't need to move around the circuit for the charge density on the plates of the capacitor to change. Instead, the mobile charges only need to move closer together or farther apart in order to change the charge density. I'm not sure I can put it any simpler than that. This is the point I attempted to make with the billiard ball analogy. So, let me try again.

The spacing of the billard balls can change very quickly even though the velocity of the billard balls themselves is relatively low. Likewise, in an acoustic wave, the density (or pressure) wave propagates at a much faster speed than the actual particle velocity. Likewise, an electron density wave propagates much faster than the drift speed of the electrons. Finally, do recall that the net electric charge on the plate of a capacitor is determined by the mobile electron charge density there. That is, to 'charge' a capacitor, it is not required to pump electrons from one plate to the other - instead, we simply need to increase the mobile electron density on one plate while decreasing the density on the other - we need to 'push' the electrons closer together on the negative plate and 'pull' the electrons farther apart on the positive plate. To do this requires very little actual movement of charge.

If 1 Coulomb of charge has been separated, then 1 extra Coulomb of electrons has been placed on the negative plate and 1 Coulomb of electrons has been removed from the positive plate. Are these the same electrons? As far as I'm concerned, the answer is no. Can I convince you of this? Apparently not.

For now, that is the best I can do to explain my view of this subject. Oh, and as to your final questions: By switching off the generator, I assume you mean the generator is a current source so that 'switching off' means 'open the circuit'. If your generator is a voltage source, you'd need to open the circuit with a switch, right? Anyhow, it is my opinion that opening the circuit would indeed reveal charge separated onto the plates. Hopefully, you'll accept my explanation above as the answer to the question of how. Alfred Centauri 03:41, 11 March 2006 (UTC)

OK Im happy to accept the idea of the electron density wave (even tho' I havent seen one). How is this different from an em wave? (you hear me ask) and can these EDWs travel at close to the speed of light in the conductors? I agree that to temporarily 'charge' a capacitor, it is not required to pump electrons from one plate to the other but only to increase the electron density on one plate whilst decreasing the density on the other. But this can only lead to a temporary separation of charge on the capacitor unless the charge travels completely round the cct back to a source/sink of carriers. If 1 extra Coulomb of electrons is placed on the negative plate and 1 Coulomb of electrons is removed from the positive plate they dont have to be the same electrons as long as they have somewhere permanent to go/come from (like a battery or dc source).?
Opening the switch to turn off the gen. is correct. We all know that charge can be captured (separated) this way as in a simple 1/2W rectifier and capacitor at the end of a TL, but my point is that this so called charge is not dependent on the charge carriers which only shuttle back and forth (esp at high frequencies), but on the em wave. BTW has the plasmon frequency I have heard of anything to do with this problem?--Light current 04:19, 11 March 2006 (UTC)

This term plasmon is new to me but when I began to read the article, it seemed familiar. Like the photon and phonon, the plasmon is a quantized mode of a field - in this case the field quantity of interest is the (mobile) charge density. As to whether the plasma frequency has anything to do with this problem, I would say that the objection you raise about the electrons not moving fast enough (or is it far enough?) to charge the capacitor would be related to this IF the plasma frequency were low enough. However, as the article on plasmons state, the plasma frequency for most metals is in the ultraviolet range which is why we can't see through metals (excepting of course, transparent aluminum). So, it seems unlikely that objections based on plasma frequency will be valid here - unless your generator is one of those new TeraHertz models (p.s. even higher - if my calculation is correct, the plasma frequency for copper is 2.18 GigaGigaHertz - 2.18E18 Hz!)

Regarding the difference between the charge density waves I keep refering to and an EM wave: the answer is that they go hand in hand. When an EM wave is guided by, e.g., a parallel plate TL, the charge density wave is in lock step with the the guided EM wave - the two cannot be separated. In fact, it is the presence of the charge density wave in the conducting plates that acts to bound the EM wave to the interior region between the plates. Further, it is the propagation speed of the charge density wave in the plates that sets the propagation speed of the EM wave between the plates.

Please explain in more detail why you feel that a charge density wave "can only lead to a temporary separation of charge on the capacitor unless the charge travels completely round the cct back to a source/sink of carriers." This makes no sense to me at all. The source of carriers are the plates of the capacitor and the conductors in the circuit. The mobile carriers of charge are all there just waiting to be redistributed. To "permanently" charge a capacitor we need only to 'pump' the required amount of electrons through the generator and then disconnect the generator or the cap. That's it. Unless you plan on pumping every last mobile electron off of the positive plate and the conductor connecting that plate to the generator (that would make an interesting calculation!), you will not need to move an electron all the way from one plate to the other. Alfred Centauri 22:05, 11 March 2006 (UTC)

[edit] New questions

  • Is an electromagnetic wave capable of achieving steady state charge separation at a point remote from its generation.
  • If so, can we say that the em wave 'carries charge' with it?--Light current 03:23, 11 March 2006 (UTC)
I don't even understand what your are asking in your first question. Can you be more specific by providing more details of your setup and your assumptions?
As far as I'm concerned, the answer to your last question is NO regardless of the answer to your first question, whatever that turns out to be. If an EM wave carried electric charge, EM waves would interact with each other via the Coulomb interaction. This does not happen in classical EM theory (however, in QED, there is a second order photon-photon interaction that occurs when two photons each create a pair of virtual charged particles which then interact before the virtual particles annihilate back into photons - I don't think this effect has been experimentally observed yet). Alfred Centauri 22:01, 11 March 2006 (UTC)

I was asking whether an electromagnetic wave when sent either down a TL or thro' space, can achieve real permanent charge separation at a remote capacitor (connected to the TL or as part of a radio Rx for example). But the main question is the second one. If the answer to the first is true, then how is this charge conveyed? (esp thro' vacuum)--Light current 22:07, 11 March 2006 (UTC)

Ah, now I see. I think I've answered your question to MY satisfaction with my recent comments in the 'Shock wave' section. Recall that we can send an EM wave down a dielectric wave guide (optical fiber) but we cannot charge a capacitor with this wave. However, we can send an EM wave down a two-wire wave guide (twin-lead) and charge a capacitor. The reason we can do this with the twin lead is that the EM wave between the conductors in the twin lead is accompanied by a charge density wave in the conductors. This charge density wave in the conductors is the reason the EM wave can be guided by the conductors in the first place. So, we charge the cap at the end with real electric charge, not the EM wave. Alfred Centauri 22:26, 11 March 2006 (UTC)
----
Having consumed a libation or three, I re-read this section and began to think of a scenario in which what you call permanent charge separation might occur through the action of an EM wave from a remote location. I'm thinking of a conducting rod of a length that matches the wavelength of an EM plane wave incident upon it. Assuming the orientation of the rod in space is correct, the EM wave causes a sinusoidal current in the rod at the same frequency as the EM wave. Of course, this current generates a wave that destructively interferes with the incident wave such that some of the energy carried in the wave is absorbed by the rod. Now, what if this rod is cut in the middle at such time that the E field of the incident wave is at a maximum? Without the benefit of a rigorous analysis, I would think that the charge distribution on the rod at the time of the cut would be such that after the cut, one rod will be net positively charged and the other will be net negatively charged. Of course, for this to work the cut would need to be made VERY quickly. Perhaps a laser pulse could do it.
However, even if the above is not pure BS, I don't see how such a result could be construed to imply that EM waves 'carry' electric charge. Alfred Centauri 01:15, 12 March 2006 (UTC)

Your suggested differentiation betweeen dielectric propagation and 'waveguide' (or TL) propagation of 'em' is very interesting. So we cant charge a capacitor by feeding em energy down an optical fibre, you say? But em energy is transferred from one place to another in both cases, is it not? This implies that you think a TL is charged by the CDW and not by the em wave between the conductors. Am I correct in assuming you believe this?--Light current 00:59, 12 March 2006 (UTC)

You would be correct in assuming that I believe that a TL is charged by charge and not by the EM wave alone. Do keep in mind what I said earlier though, the CDW and EM wave in a TL go hand in hand. We can't have one without the other. Also, please see the section I added before your latest comment but which got involved in an edit conflict. Alfred Centauri 01:15, 12 March 2006 (UTC)

Ah ha! So if we cant have one without the other, no em wave guided by, or in, a pair of conductors can go faster than the CDW. But has this very high velocity (2c/3 say) of CDWs been shown to be possible?--Light current 01:26, 12 March 2006 (UTC)

To quote from my post in the 'Shock wave' section: "Further, it is the propagation speed of the charge density wave in the plates that sets the propagation speed of the EM wave between the plates." BTW, when I speak of a charge density wave, I'm not refering to what are called CDWs in the literature. Instead, I'm speaking of a propagating disturbance in the electric charge density in a conductor, \rho_e(\vec{r},t). At what speed would you expect a change in mobile electron density to propagate in a conductor? Alfred Centauri 02:17, 12 March 2006 (UTC)

The speed I would expect a change in mobile electron density to propagate in a conductor would be faster than I first thought but I dont think as fast as the guided em wave can go. But Im sure you'll prove me wrong on that one. If what you say is true, then there is no Catt (Ivor Catt)anomaly and everything is fine! (apart from skin effect that is!)--Light current 02:24, 12 March 2006 (UTC)

[edit] What is electric displacement?

Also Alfred we never did find out what you meant by a 'flow of displacement' (your apparent defn of DC). How do you define displacement? Displacement of what? Are you referring to vacuum polarisation perhaps when you talk of displacement in vacuo?--Light current 15:22, 9 March 2006 (UTC)

It's not actually my definition, right? After all, an air current is a flow of air - a water current is a flow of water - a displacement current is, literally, a flow electric displacement. What is electric displacement?

It seems that Maxwell thought of electric displacement as a physical displacement of electric charge from equilibrium in a medium - a polarization of the medium. The displacement current, in this view, is the time rate of change of the displacement from equilibrium which amounts to an actual movement of real electric charge - an actual electric current. However, unlike free charge that would accelerate under the influence of an electric field, these bound charges can only remain displaced from equilibrium positions as long as there is an external electric field to counteract the internal electric field between the bound charges. To change the displacement from equilibrium requires a change in the external field. Thus, the displacement current is proportional to the rate of change of the external electric field.

Electric displacement and displacement current as defined above cannot exist in the classical vacuum. However, according to the Standard Model, the vacuum is not empty but actually contains a sea of virtual particles including electrically charged particles. Thus, according to this view, the vacuum can indeed be polarized. It would seem reasonable then that vacuum displacement current of this sort could be detected in the same way an electric current is detected. Alas, virtual particles cannot be directly detected (thus the adjective 'virtual'). However, the effects of virtual particles can and have been detected (see, for example, Casimir effect, and Lamb shift). So, perhaps a changing vacuum polarization is in fact a displacement current. Alfred Centauri 16:20, 9 March 2006 (UTC)

Thank you for outlining your position so succinctly. I actually understand what you are saying! So if we can measure vacuum displacement current perhaps by the exptl method I outlined above, we may have discovered or proved something. I dont think vacuum ploarisation has yet been measured/proved to exist -- or has it?--Light current 16:43, 9 March 2006 (UTC)

I suppose the answer to your question depends on what you are willing to accept as 'proof'. In fact, predicted detectable effects due to the existance of virtual particles and the resulting vacuum polarization have been confirmed experimentally (once again, see Casimir effect and Lamb shift for examples). Of course, if you doubt the principles upon which the Standard Model is constructed, you probably will not be convinced by these results. Alfred Centauri 17:01, 9 March 2006 (UTC)

Well as you know Alfred, I always try to have an open mind and am willing to be convinced! But the real test as always is the experiment.--Light current 17:14, 9 March 2006 (UTC)
Unless of course the experiment cannot be performed -then we have a problem!--Light current 03:25, 11 March 2006 (UTC)


[edit] Removed section

I removed the section 'Displacement Current & Magnetism' as it has some significant problems that must be addressed. The most obvious problem is that the the capacitor equation is a scalar equation but Ampere's Law is a vector equation. The step of determining a vector field from the ratio of two scalars is highly suspect except in a one dimensional problem. However, Ampere's Law cannot be applied to a one dimensional problem (try to define 'curl' in 1-D). I might try to fix this but the author should get first crack. Alfred Centauri 19:16, 14 October 2006 (UTC)

'== Displacement Current & Magnetism == Over the years, many scientists and engineers have questioned whether or not displacement current "causes" magnetic fields. The following simple derivation shows how displacement current and charges interact to form magnetic fields.

The capacitance of a parallel plate capacitor with plates of area A and plate separation of d may be expressed as:

C=\varepsilon_{R}\epsilon_{O}\frac{A}{d} (1)

Where εR and ε0 are dielectric and free space permittivity, respectively.

Another basic equation relates the uniform charge, Q stored in a capacitor, to the capacitance C, and the Voltage across the plates, V.


Q = CV (2)

Divide both sides of equation (2) by the distance between the plates, d:


\frac{Q}{d}=C\frac{V}{d} (3)

Since \frac{V}{d}=E, where E is the electric field between the plates, we may rewrite equation (3) as follows:


\frac{Q}{d}=CE (4)

Substitute the value for C from equation 1 into equation (4):


\frac{Q}{d}=\varepsilon_{R}\epsilon_{O}\frac{A}{d}E (5)

Multiply both sides of equation 5 by d, and take the partial derivative of equation 5 with respect to time. The result is shown below:


\frac{\partial Q}{\partial t} = \varepsilon_{R}\epsilon_{O}A \frac{\partial E}{\partial t} (6)

Using the constituitive relation \varepsilon_{R}\epsilon_{O}E=D, equation (6) may be written as:


\frac{\partial Q}{\partial t}=A\frac{\partial D}{\partial t} (7)

Ampere’s law, as adapted by Maxwell to include the effects of Displacement Current is often written as:


\nabla \times H =J +\frac{\partial D}{\partial t} (8)

If we substitute the appropriate values from equation (7), we are able to rewrite equation (8) as:


\nabla \times H=J+A^{-1}\frac{\partial Q}{\partial t} (9)


But A^{-1}\frac{\partial Q}{\partial t} is another way of expressing J. In this case, since the electrical field is Transverse to the plane of the conductor, we define a new term,


J_{T}=A^{-1}\frac{\partial Q}{\partial t} (10)

The term “J” has traditionally been taken to mean “longitudinal conduction current.” To differentiate it from JT, let us rename J as JL standing for longitudinal current flow. With these re-definitions, we can now rewrite equation (9) in a form that is consistent with the spirit of Ampere:


\nabla \times H=J_{L}+J_{T} (11)


Conclusion Equations 8, 9 and 11 may be used interchangeably, but only if one understands that the source of the magnetic fields is the motion of charges, \frac{\partial Q}{\partial t} .

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