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## Ohm's law, why current induces electric field and cause effect direction

 Quote by Fernsanz It has become clear to me that you are unable to understand the essence, neither the statement of the problem. Now I add: this thread is futile except for cabraham contributions. I will make a last attempt though, in case brains wake up. FACTS: A current can be generated without electric field A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field EXPERIMENT:Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor) Measure the voltage V across the resistor (it will happen by fact 2) CLARIFICATIONS:[LIST=1][*]Step 1 of the experiment involves generating a current withouth electric field. It can be photoemission process, SC, or God moving electrons one by one without electric field.

Perfect.

1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current.

2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?).

3. I then measure the field across it. What do I get?

4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up.

5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!

6. I study the process of secondary emission, which involves electrons with energy ranging from 200 eV all the way to hundreds of keV, hitting materials ranging from conductors to insulators. So the mechanism of electron transport once it hits a material is something I don't just read on, but it is something I experiment on! I truly would like you to do that exact experiment that you just described before you make such claims, because some of us actually HAVE done such experiments!

7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field.

8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort.

Zz.

 Quote by ZapperZ Perfect. 1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current. 2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?). 3. I then measure the field across it. What do I get? 4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up. 5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!
Are you the one who decides what is Ohm's law and what isn't? The current in my scenarios is external? What the hell is a current that travels across the resistor? If i have a current in a SC loop and in t=0 I insert a resistor in the loop, what the hell is external? The E-field that will arise across the resistor is not the one Ohm's law? Which part of the Ohm's law says if an electrical field along a conductor is the Ohm's law or not? Will the resistor violate the relation V=IR while the current in the loop semicondcutor-resistor is dacaying? No, it will not violate, so V=IR, so that's Ohm Law.

Do you even know my job? It will be interesting when I inform my mates that when we anlyze the effetc of space radiation in the circuits of the satellite (for example electrons being pumped into satellite ohmic conductors) we are worngly using the Ohm's law V=IR because the electrons come from outside. Ridiculous

You will never admit you are wrong becasue, as has become clear, any E-field that it is generated you will say it is not the Ohm's law. Stick yourself to explain how the hell that "kind of" electrical field is generated; I will decide if it is Ohm's law or not at my own.

 Quote by ZapperZ 7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field. 8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort.
It seems you can't stop typing "Drude model"!!! Read what I said about it in my previous post because you haven't even read it. Drude model has nothing to say here!!

You can not teach what you ignore. Yes please, end you participation here and let others who have answers participate.

 Quote by Fernsanz The question is clear and direct, yet still unanswered (except for an explanation, right or not, provided by cabraham). It would be useless to "answer" with another question like "could you tell me in the Drude model bla bla bla". Is me who is raising a question; is me who expect an answer which starts something like "the electric field is induced because electrons will accumulate bla bla bla" or "the electric field is induced because a shockwave of electric nature will arise bla bla bla" or "the electric field arise cause God comes down to earth bla bla bla". If someone need explanations or comments on Drude model, harmonic oscillator or turbo engines in cars, please open another thread and feel free to talk about it for ages. The Drude model is of no relevance in this discussion because it is a model to explain the dynamics of electrons once an electric field has been established. My question goes previous to that: how was that electrical field established when you connect the ohmic conductor to a current with no electric field? And you keep answering that the Drude model does not allow a current in an ohmic conductor without electric field. That is not the fact under question. The question is, once again and again, how was that electric field created from a situation when a current started travelling along the ohmic resistor with no previous electric field. When you connect the ohmic resistor to the "no-electric-field-current-source" a electric field arise which was not there before (and which is the E-field the Drude model takes as a starting point to explain electrons dynamics, the model does not address how it was generated, where does it come from or nature of the field). And lastly, is there any people on earth who think that V=IR is not valid for the ohmic resistor? Apparently only kmarinas86; for the rest of the world the Ohm's law is always true even in this situation where the feed to the resistor is a current which conveyed not electric field in its origin (but which subsequently generated an electric field across the resistor by a mechanism that is the subject under question)
I emphasize the following:

Quote by kmarinas86
 Quote by Wikipedia When reactive elements such as capacitors, inductors, or transmission lines are involved in a circuit to which AC or time-varying voltage or current is applied, the relationship between voltage and current becomes the solution to a differential equation, so Ohm's law (as defined above) does not directly apply since that form contains only resistances having value R, not complex impedances which may contain capacitance ("C") or inductance ("L"). http://en.wikipedia.org/wiki/Ohm%27s_law
It becomes even harder to define Ohm's law when your waveforms are neither flat nor sinusoidal. I am not an AC dude. AC circuits are not what I care about. I care about circuits that do not have uniform currents. I care about second-order, third-order, and higher-order electromagnetic effects. I care about circuits that have unusually high inductance values for their resistance which are switched at high frequency. If you have a coil that has 10 henries of inductance and 2 ohms of resistance, how can it reach the full value of current if you switch the circuit on and off every other millisecond?

The above image is from http://www.intmath.com/Differential-...L-circuits.php
The problem with the topic of this thread is that because of its constraints you stubbornly do not consider other physically valid scenarios, and thereby you limit yourself to explanations for which there are better alternatives. The sound analogy is the best one you got, and both cabraham and I both mentioned something of that nature much earlier in this thread. The first seven responses of this thread go exactly this:

Quote by fernsanz
Quote by kmarinas86
Quote by fernsanz
Quote by cabraham
Quote by fernsanz
Quote by cabraham
 Quote by Fernsanz Hi, Lately I've been concerned about the Ohm's law $$J=\sigma E$$ and the physical interpretation of this law depending on what is considered the cause and what the effect. More concretely, it is quite natural and intuitive for me the interpretation of this law in one direction: a electric field $$E$$ over a ohmic conductor will cause a current $$J$$. No one would have problems working out that electric force acting upon electrons will cause them to move, i.e. setting up a current. We can certainly say that the electrical field is the cause in this case and the current is the effect, the result. However, in the reverse direction, namely that a current $$J$$ injected somehow in a ohmic conductor will cause a electric field $$E$$, is by far more counterintuitive. Anyhow the current is generated previously by other means (for instance photoelectric or thermoionic effects) and injected into the conductor an electric field will arise. What is the explanation that acounts for that electric field caused just by movement of electrons in a conductor (which is nothing but a cristaline structure "covered" by a cloud of electrons)? I would gladly read your explanations. Thanks.
It is counter-intuitive indeed, but they are mutually inclusive, neither being the cause nor the effect.

I view it in the following way to see it better. By definition a constant voltage source is one that maintains a constant voltage despite varying load resistance. Let's say a current enters a conductor, the charge carriers in conductors are electrons. They have energy, but lose a portion of that energy when they collide with the lattice structure.

When an electron in the conduction band collides with an electron in the valence band, energy is transferred. So a current is a transfer of energy among electrons, where the actual motion/displacement of an electron is small, but the propagation of energy is near light speed.

The energy loss due to lattice collisions constitutes resistive loss. Thus the conduction electrons carry energy, a portion of which they lose due to resistance i.e. Lattice collisions. By definition, the energy lost per unit charge is the voltage drop. Hence current entering a conductor incurs energy loss due to resistance/lattice collisions, resulting in a drop in voltage.

The converse also holds. A conductor with no current is instantly connected across a small voltage. The e field results in charge motion i.e. Current.

Ohm's law is bi-lateral, where j & e are inclusive. Which one comes first is a moot question. You are thinking the right way.

Claude
The energy lost per unit charge is not the voltage drop because that loss is caused by "mechanical" collision among electrons. Reasoning that way one could take an electron, throw it against a wall and conclude that, since the electron has lost all its energy after hitting the wall, there must be an electric field out of the wall and hence a voltage drop.

So the question is still in the air: Why an electrical field appear just because electrons are moving? (don't lose of sight the "electrical" nature of the force im asking about which is the electrical field that appears in ohm's low, not any other mechanical forces)
Mechanical - electrical?! What is the difference? How do you think resistance takes place? If the electron can travel all the way through the conductor w/o colliding w/ the lattice, there is zero resistance, hence zero voltage drop. Superconductors achieve this property.

The "thermal energy" associated w/ an electron is given by 0.5*m*v^2 = k*t, where m is mass, v is velocity, k is boltzmann's constant, & t is temperature (absolute).

If "resistance" is not a "mechanical" entity, then what is it? I'd like to know. Thanks in advance.

Claude
If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an electric field arises as the result of moving charges in that lattice. You have started talking about resistance, which i did not mention. Resistance is what steal energy from electrons. I'm not asking about where the electrons lose their energy, i'm asking about why an electric field arises; an electrical field that would be measurable by measuring the force exherted on a test charge placed inside the conductor even if such test charge is not moving!!!

Please, i'm not expecting simple or naive answer; i expect a physical explanation on why moving charges in a ohmic conductor induce an electric field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electric force $$f=qe$$.
If moving charges collide into a capacitor, it induces an electrical field due to the collapse of the magnetic field. The collision of like-charged particles which stores potential energy in the electric field is what induces the electrical field in a wire. The potential energy results from the increase of the volume-averaged rms value of the electric field. This is because the volume density of the energy in the electric field varies by (1/2)*(electric constant of permittivity)*e^2. The rms value of e increases as like charges increase their proximity to one another. So the volume-averaged value of e^2 is greater in that case.
Thanks thanks thanks!!! That's the type of reasoning i was looking for. I don't see it clear though: If you get electrons closer to each other than they would be in equilibrium, any of those electron will feel a stronger field from its neighbours, but thats true in all direction!! So the sum of stronger fields at any point would cancel out, even though it is true that the density of potential energy will be higher.

Or do you mean that it should be viewed more like a compresion-rarefaction process like a sound wave? That is, the first avalanche of electrons moving fast get really close to the front-end of standing-still electrons; then, because of proximity this standing still electrons will move forward getting closer to the next ones and so on. Is this a correct visualization?

Thanks again
What do you mean by "Thanks again"? Did you think I was cabraham?

 Quote by Fernsanz Here is an example of a suitable and plausible explanation and the type of answer anyone would expect: at the very moment when the current star entering the ohmic conductor, the leading electrons will get closer to the first electrons in the ohmic conductor; thus, this leading group of electrons in the current will slow down by repulsive force and the electrons in the omhic resistor will start moving. So, the electric field could be the repulsive force between electrons. This can be viewed as a sound wave in which the "slowed down" part and the "repulsed" part are called compression and rarefaction respectively. So, the electric field can be viewed as a lot of pulses, a lot of "electric shockwaves" that all together create a macroscopic E-field that one could measure placing a test charge inside the conductor. It would be also the explanation that the Ohm's law is applicable even if the feeding to the resistor is a non-electric-field-current. You see? that would be something addressing the questions in this thread. An explanation of this type is what Cabraham provided and not a disgression with no constructive attitude showing a total lack of understanding.
Both cabraham and I contributed to this explanation, per above. Yet you only give credit to him. You even responded more thankfully to my explanation, but you thought I was someone else. Nevermind about this then. We've got a plausible explanation that doesn't go into many details as to why electrons behave this way. Do you want another explanation or deepen this one? Do you want to define the differential equations that are required to describe the second-order and higher-order effects of the E-field? Do you want to discussion in this thread to advance? Or do you want some kind of "totally exclusive and equally plausible explanation" for this? Seems like work doesn't it?

The 8th post was from ZapperZ. At this point, all hell in this thread broke loose. After a recent set of questions by ZapperZ, this following conservation was spawned:

Quote by kmarinas86
Quote by cabraham
 Quote by "ZapperZ (questions only)" Sorry, but what experiment is this? You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you? I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?
I've answered this question in spades. The question was "can a current give rise to an E field?" The answer is yes, & I gave an example plus an explanation from peer reviewed uni e/m fields texts. However, the current which gives rise to this conductor's local E field is driven by an external energy conversion source with an associated E field. In other words, the E field produced by the current is a different E field from that of the power source producing the current. Of course the E field produced by the current cannot also drive that same current. Nobody ever implied that.
You are right up to this point.

It turns out that zeroth, second, and fourth order derivatives of the E-field lead back to E-fields. Any even-order derivative does. These derivatives will not all point the same way (for the obvious reason that changes of acceleration are not always aligned with velocity), so these even-order effects are not always "parallel" or "anti-parallel" to each other. Guess what B-fields are generated by? B-fields are caused by odd-order effects of E-fields. ...Duhsville!
ZapperZ thinks an applied field is not restricted to an "internal source", yet cabraham thinks otherwise. If an E-field is external, does that meant it is not "applied field"? It seems really arbitrary to assume either way. This thread is littered with disagreements that stem from mere terminological differences. The above is just one example.

Quote by Fernsanz
 Quote by ZapperZ Perfect. 1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current. 2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?). 3. I then measure the field across it. What do I get? 4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up. 5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!
Are you the one who decides what is Ohm's law and what isn't? The current in my scenarios is external? What the hell is a current that travels across the resistor? If i have a current in a SC loop and in t=0 I insert a resistor in the loop, what the hell is external? The E-field that will arise across the resistor is not the one Ohm's law? Which part of the Ohm's law says if an electrical field along a conductor is the Ohm's law or not? Will the resistor violate the relation V=IR while the current in the loop semicondcutor-resistor is dacaying? No, it will not violate, so V=IR, so that's Ohm Law.

Do you even know my job? It will be interesting when I inform my mates that when we anlyze the effetc of space radiation in the circuits of the satellite (for example electrons being pumped into satellite ohmic conductors) we are worngly using the Ohm's law V=IR because the electrons come from outside. Ridiculous

You will never admit you are wrong becasue, as has become clear, any E-field that it is generated you will say it is not the Ohm's law. Stick yourself to explain how the hell that "kind of" electrical field is generated; I will decide if it is Ohm's law or not at my own.

 Quote by ZapperZ 6. I study the process of secondary emission, which involves electrons with energy ranging from 200 eV all the way to hundreds of keV, hitting materials ranging from conductors to insulators. So the mechanism of electron transport once it hits a material is something I don't just read on, but it is something I experiment on! I truly would like you to do that exact experiment that you just described before you make such claims, because some of us actually HAVE done such experiments! 7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field. 8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort. Zz.
It seems you can't stop typing "Drude model"!!! Read what I said about it in my previous post because you haven't even read it. Drude model has nothing to say here!!

You can not teach what you ignore. Yes please, end you participation here and let others who have answers participate.
Let's face it. You are more an applied physicist than a theoretical physicist. Why would someone like you care to put priority on the opinions of those who lean more toward the science of theoretical physics over those who lean more toward the practice of applied physics? You insist in using the conventions of engineers. Physics has more advanced explanations than are typically used in practice. Aren't you asking about something that your years of experience has not told you yet?
 My two-pennyworth. IMO The classical ideas of fields and the movement of charges (not specifically electrons) don't sit well in the study of electrodynamics. I rather like the approach taken by Prof Mead of 'explaining' electricity as a quantum effect from the outset. What's more, he points out (not in so many words) that when we study physical systems, we inevitably begin with a simple situation. Typically we would avoid friction effects (often by setting ourselves up in a separate universe for the experiment!). For that reason, he elects to only consider superconductors. It sounds crazy but it works. Within the first chapter of his book 'collective electrodynamics' he disposes of the E field and the B field as totally unnecessary artificial constructs and just uses the scalar and vector potentials along with simple QM concepts to completely reconstruct the subject. He doesn't destroy Maxwell's work, he simply redirects it along the lines Maxwell would have gone if he'd been aware of facts we now know. I can't recommend it strongly enough. Just to give you the Flavour:- He points out that the voltage across the ends of a loop (scalar potential) and the Vector potential A around the loop together satisfy the de Broglie relationship of frequency to wavenumber. By considering charge as a wave, it's therefore possible to specify the potentials as a four-vector throughout space - they depend only on J (also a four vector with the charge density.) E and B - you don't need - you can easily calculate a value for them at any point if you want - but it turns out that most of the time you don't need to.
 Above there is a waveform of an R-L circuit whose current builds up to V/R over many time constants. May I re-emphasize that Ohm's law is valid in the resistor from time 0 out to steady state. At t=0, Vr = 0, I = 0. When I = 0.1*V/R, Vr = 0.1*V. When I = 0.5*V/R, Vr = 0.5*V. As the current builds up in the resistor the voltage builds up in unison. At every point in time, Vr = I*R. At t=0, there is no current, I=0. The entire voltage, V, is acoss the inductor, while Vr (voltage across resistor) is zero. Then the current builds up. During this time the voltage across the inductor is decreasing & the voltage across the resistor is increasing. Ohm's law is in effect the entire time, transient as well as steady state. Again, the resistor voltage increases simultaneously w/ the current, such that the ratio of Vr/I is always equal to the resistance R. Regarding E fields & currents, the fact that charges drift in the presence of an E field is indisputable. But whenever an E field imparts kinetic energy to a charge, the E field energy is reduced, as conservation of energy is always happening. The E field energy lost is replaced by an energy conversion process, i.e. chemical reaction in battery (redox), mechanical power input to generator, etc. So E fields when exerting force on charges give up energy, & if not replenished cannot sustain a cuuent. Likewise a current injected into a conductor can produce an E field but again, this E field does not impart force to the network current. Neither J nor E can drive the other. The power source gives rise to both. Without energy conversion, neither J nor E will be sustained. When a battery is connected across a lamp, what drives E & J? It is the redox chemical process. If a cap is charged, then disconnected from the source, the cap can be switched across the lamp & it will momentarily light up, then fade. If a current source like the inductor is switched into a lamp, it will glow momentarily, then fade. Just as J produces E, E can also produce J. But to sustain J & E, energy conversion must take place. The solid state physics aspect is intriguing, but sophomore level EE circuits & junior level EE fields is all that is needed. I think some make it way harder than needed. Any comments/feedback welcome. Claude

 Quote by cabraham Above there is a waveform of an R-L circuit whose current builds up to V/R over many time constants. May I re-emphasize that Ohm's law is valid in the resistor from time 0 out to steady state. At t=0, Vr = 0, I = 0. When I = 0.1*V/R, Vr = 0.1*V. When I = 0.5*V/R, Vr = 0.5*V. As the current builds up in the resistor the voltage builds up in unison. At every point in time, Vr = I*R. At t=0, there is no current, I=0. The entire voltage, V, is acoss the inductor, while Vr (voltage across resistor) is zero. Then the current builds up. During this time the voltage across the inductor is decreasing & the voltage across the resistor is increasing. Ohm's law is in effect the entire time, transient as well as steady state.
Okay, now I see that "Vr" is your resistive voltage drop. OK.

I just discovered that you can use subscripts.

VR.

It's a lot better that show it that way.

 Quote by cabraham Again, the resistor voltage increases simultaneously w/ the current, such that the ratio of Vr/I is always equal to the resistance R. Regarding E fields & currents, the fact that charges drift in the presence of an E field is indisputable. But whenever an E field imparts kinetic energy to a charge, the E field energy is reduced, as conservation of energy is always happening. The E field energy lost is replaced by an energy conversion process, i.e. chemical reaction in battery (redox), mechanical power input to generator, etc. So E fields when exerting force on charges give up energy, & if not replenished cannot sustain a cuuent.
I totally agree with this.

 Quote by cabraham Likewise a current injected into a conductor can produce an E field but again, this E field does not impart force to the network current.
This does not sound right at all. Why should current injected into a conductor be any different than a lightning strike electrocution causing a current to be forced through one's body?

 Quote by cabraham Neither J nor E can drive the other. The power source gives rise to both. Without energy conversion, neither J nor E will be sustained.
What if the power source consists of changing J and changing E?

 Quote by cabraham When a battery is connected across a lamp, what drives E & J? It is the redox chemical process. If a cap is charged, then disconnected from the source, the cap can be switched across the lamp & it will momentarily light up, then fade.
This could be due to changing J and changing E at quantum level.

 Quote by cabraham If a current source like the inductor is switched into a lamp, it will glow momentarily, then fade. Just as J produces E, E can also produce J. But to sustain J & E, energy conversion must take place. The solid state physics aspect is intriguing,
Totally agreed.

 Quote by cabraham but sophomore level EE circuits & junior level EE fields is all that is needed. I think some make it way harder than needed. Any comments/feedback welcome. Claude
Needed for what? People have different plans, so maybe what they need depends on what they need to do.
 Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire? My simple mind always believed the electric field was the stimulus and the current the result. I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.

 Quote by Gordianus Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire? My simple mind always believed the electric field was the stimulus and the current the result. I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.
This type of prejudice is quite common. But consider what it takes to generate an E field. Charges must be displaced, i.e. separated. First, the mere separation of charges involves moving them , which constitutes current. Second, work must be done to achieve this. Third any E field which imparts force & energy to a charge carrier must lose the same amopunt of energy it imparted. CEL is immutable (conservation of energy).

An ac generator is a little more involved than a battery, so we'll look at the latter. If a battery powers a circuit consisting of conductors & a lamp, one can say that the current in the conductors & lamp is related to the E field per Ohm, i.e. J = sigma*E. But the current in the battery is oriented in the opposite direction. Let's use positive current convention.

In a passive element, current enters the terminal that is the more positive of the two, & exits the more negative terminal. The charges drift in the presence of E field per F = q*E.

But in the battery, positive current exits the positive terminal & enters at the negative terminal. This completely opposes the notion that "the E field drives the current". In the battery, the current is oriented in a direction against the E field. Electrons are moving from the pos to neg terminals inside the battery. This is the exact opposite of what happens if the E field "drives" the current.

So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going.

Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer. The cause of anything mechanical, electrical, chemical, nuclear, etc. is the transfer of energy. An inductor energized & shorted has energy per 0.5*L*I^2. If the current is diverted from the short to a resistor, this energy is dissipated as heat. First there was J w/o E, then E appeared, then both decayed to zero. The energy associated with each was converted to heat.

Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy.

Nothing else makes any sense at all. Cheers.

Claude

 Quote by cabraham Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy. Nothing else makes any sense at all. Cheers. Claude
So what caused the delivery of the energy.....

Here's a better idea. Our understanding of cause and effect (i.e. as a before and after thing) is an old and outdated understanding. New understanding of cause and effect is that everything results from change. A cause=A change. Everything that is conserved has always existed and will continue to exist in the same quantities. You must explain the transfer of energy as the result of changes without violating conservation. So this change is actually the change of something or more than one something. So the magnitudes of changes are what determine an effect. Cause is therefore more about the rate of change, the change of that rate of change, and so forth. Cause is therefore an ensemble of change, and therefore it cannot ascribed to an unchanging property or object. All that unchanging properties or objects do (assuming they exist) is put permanent limits on the boundaries of change.

 Quote by cabraham Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer.
J/E is equal to conductivity. Since resistivity is equal to inverse conductivity, the verdict is quite clear. A current is produced by shorting an electric field. This creates a low-resistance path through which charges may travel. If E does not exist, then what is J? On the other hand, E may exist without J in a system with zero conductivity, but without J it cannot increase or decrease. The truth is that they all cause and effect each other simultaneously. What do you think causes the pathway to short? Perhaps its the build up of an electric field at a cathode. Perhaps it is infinite regress.

 Quote by kmarinas86 So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going.
So what makes you think that this by itself explains how much amps are going to come out of the battery? What is the with the obsession with a "singular" cause? The philosophy of "singular" cause is utter baloney nonsense!

Everything that science is capable of explaining is "caused". What do you think triggers the chemical redox reaction? Something must change outside the cathode to trigger the current yo.

Mentor
Blog Entries: 28
 Quote by AJ Bentley My two-pennyworth. IMO The classical ideas of fields and the movement of charges (not specifically electrons) don't sit well in the study of electrodynamics. I rather like the approach taken by Prof Mead of 'explaining' electricity as a quantum effect from the outset. What's more, he points out (not in so many words) that when we study physical systems, we inevitably begin with a simple situation. Typically we would avoid friction effects (often by setting ourselves up in a separate universe for the experiment!). For that reason, he elects to only consider superconductors. It sounds crazy but it works. Within the first chapter of his book 'collective electrodynamics' he disposes of the E field and the B field as totally unnecessary artificial constructs and just uses the scalar and vector potentials along with simple QM concepts to completely reconstruct the subject. He doesn't destroy Maxwell's work, he simply redirects it along the lines Maxwell would have gone if he'd been aware of facts we now know. I can't recommend it strongly enough. Just to give you the Flavour:- He points out that the voltage across the ends of a loop (scalar potential) and the Vector potential A around the loop together satisfy the de Broglie relationship of frequency to wavenumber. By considering charge as a wave, it's therefore possible to specify the potentials as a four-vector throughout space - they depend only on J (also a four vector with the charge density.) E and B - you don't need - you can easily calculate a value for them at any point if you want - but it turns out that most of the time you don't need to.
I know of Carver Mead's work quite well, In fact, if you do a search on PF, I've cited his PNAS paper on his Collective Electrodynamics several times. This is because he cited superconductivity as being the clearest manifestation of QM at the macroscopic scale. His reformulation of E&M fields is actually quite interesting and refreshing.

However, his work cannot derive, for example, Ohm's law, nor can he arrive at an explanation for "resistivity" in metals. Maxwell equations can't either. This is because this is not an issue of electromagnetic field, but rather a materials property (it is why such derivations are very seldom done in E&M classes, but rather, in solid state classes). Mead was able to describe, using his picture, various phenomena of superconductivity because the charge carrier responsible for that phenomenon has long-range coherence. The supercurrent does not interact with the microscopic details of the bulk material, what David Pines termed as a state of "quantum protectorate". So in essence, dealing with just the supercurrent made it "easier".

This is not the case with ordinary metals in the normal state. The question is, how are charges transported from one location to another, resulting in what we call a "current". We know that the charge carriers undergo several interactions: (i) electron-electron interactions, (ii) electron-ion (or phonon) interactions (iii) electron-impurities interactions), etc. (refer to, say, Valla et al., PRL 83, 2085(1999)). At room temperature, in a typical metal, the electron-ion/phonon interactions dominates, resulting in the fact that a free electron trying to move, will not make it way past the mean-free path. That's it. Without any external field, any electrons moving through such a material will thermalize. This is the origin of resistivity. One can derive from First Principles the resistivity of various materials using such a model.

This area is such a well-studied subject in condensed matter physics, because transport phenomenon is one of the most important aspects of that field. I could easily point out to the QFT approach to such a phenomenon (to answer the question from another member who wanted to know if such a thing has to be handled quantum mechanically) using what we call as the propagator, and arrive as the single-particle spectral function. Here it is even clearer that, analogous to the random walk problem, each charge carrier WILL experience multiple interactions impeding its motion. It is why this is a many-body physics problem. It is why the "electron" that we detect in a material does not have the same "mass" as the bare material. In heavy fermionic system, the charge carrier can have a mass more than 200 times the bare mass!

In all of these, there is a clear cause-and-effect, especially in how one gets Ohm's Law, how charges move through a material, etc.

Zz.

 Quote by cabraham This type of prejudice is quite common. But consider what it takes to generate an E field. Charges must be displaced, i.e. separated. First, the mere separation of charges involves moving them , which constitutes current. Second, work must be done to achieve this. Third any E field which imparts force & energy to a charge carrier must lose the same amopunt of energy it imparted. CEL is immutable (conservation of energy). An ac generator is a little more involved than a battery, so we'll look at the latter. If a battery powers a circuit consisting of conductors & a lamp, one can say that the current in the conductors & lamp is related to the E field per Ohm, i.e. J = sigma*E. But the current in the battery is oriented in the opposite direction. Let's use positive current convention. In a passive element, current enters the terminal that is the more positive of the two, & exits the more negative terminal. The charges drift in the presence of E field per F = q*E. But in the battery, positive current exits the positive terminal & enters at the negative terminal. This completely opposes the notion that "the E field drives the current". In the battery, the current is oriented in a direction against the E field. Electrons are moving from the pos to neg terminals inside the battery. This is the exact opposite of what happens if the E field "drives" the current. So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going. Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer. The cause of anything mechanical, electrical, chemical, nuclear, etc. is the transfer of energy. An inductor energized & shorted has energy per 0.5*L*I^2. If the current is diverted from the short to a resistor, this energy is dissipated as heat. First there was J w/o E, then E appeared, then both decayed to zero. The energy associated with each was converted to heat. Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy. Nothing else makes any sense at all. Cheers. Claude
A really good post.

But, for the sake of simplicity and to focus on the essence of the problem, let's get rid off bateries and sources. Just imagine a current flowing in a superconductor loop with no resistance. Then, at some time, let's label it $$t=t_0$$, I insert an ohmic resistor in the loop. At that very instant I will see a voltage i.e. an electric field across the resistor. Of course, both the current and the voltage will quickly decay to 0 since there is no source to keep it flowing but that is irrelevant for our question: how did that E-field arise?

 Quote by kmarinas86 So what makes you think that this by itself explains how much amps are going to come out of the battery? What is the with the obsession with a "singular" cause? The philosophy of "singular" cause is utter baloney nonsense! Everything that science is capable of explaining is "caused". What do you think triggers the chemical redox reaction? Something must change outside the cathode to trigger the current yo.
I never believed in a singular cause. I was just emphasizing that the energy has to be conserved, & must originate somewhere, i.e. redox. The OP question was related to J & E, as to one causing the other. I was merely explaining that they are interactive, mutual, & require energy conversion/source to sustain.

As far was what determines how many amps are supplied by the battery, Sir Oliver Heaviside laid that to rest in the 1870's.

In a previous post you stated

"The truth is that they all cause and effect each other simultaneously"

That has been my position since forever. Reread my posts on this forum or any other & that is what I/ve been saying for years. I've never bought into the belief that 1 specific variable is the cause of another under all conditions.

Claude

I attach this fantastic paper I've found with title: "Electric Field and Plasma Flow: What Drives What?"

It is no exactly the same question that the OP one but very very similar. This is exactly the type of argument I was expecting. As clearly stated by the author, definitions of cause and effect, who goes first and who goes after, can be rigurously obtained with no philosophical burden.

I suspect that, as the subject of the paper is magnetohydrodynamics, something similar can be done for our subject, if it has not been done already. In fact his paper contain the words "Ohm's law" and touch the subject very close.

I would love an analogous paper to exist for Ohm's relation.

PS: The paper concludes, among other things, that flow also drives electrical field; hence, the equation can be interpreted symetrically depending on what is considered the stimulus and what the response. This is just the case for Ohm's law also.
Attached Files
 ElectricfieldAndFlow.pdf (130.9 KB, 14 views)

 Quote by Gordianus Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire? My simple mind always believed the electric field was the stimulus and the current the result. I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.
That is because most people take statements as they are given with little critical attitude towards it, if any. If you dig just a for a moment into many things we think we know well you will end up with more questions than certainties. And that, in turn, will lead you to gain deep insight on subjects most people haven't even realized. Ohm's law is an example as it is the question "why water evaporates below 100ºC when everybody knows that below 100ºC it is liquid". Just an example to comment on your appreciated post.

I hope that, in the end of this thread your view of Ohm's law be really deeper and different. Less simple it seems at first glance. I don't think your mind is simple, by the way.

 Quote by Fernsanz That is because most people take statements as they are given with little critical attitude towards it, if any. If you dig just a for a moment into many things we think we know well you will end up with more questions than certainties. And that, in turn, will lead you to gain deep insight on subjects most people haven't even realized. Ohm's law is an example as it is the question "why water evaporates below 100ºC when everybody knows that below 100ºC it is liquid". Just an example to comment on your appreciated post. I hope that, in the end of this thread your view of Ohm's law be really deeper and different. Less simple it seems at first glance. I don't think your mind is simple, by the way.
Word.

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