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

[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.

 

[cabraham]

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

 

[Fernsanz]

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.
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 I am asking about which is the electrical field that appears in Ohm's low, not any other mechanical forces)

 

[cabraham]

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 loss all his 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 I am 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

 

[Fernsanz]

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.

 

[kmarinas86]

If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an ELECTRICAL field arise 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 ELECTRICAL field arise; 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 ELECTRICAL field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electrical 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.

 

[Fernsanz]

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 that's 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

 

[ZapperZ]

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.

Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.

Zz.

 

[Fernsanz]

Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

In an Ohmic conductor the Ohm's law says that J=\sigma E. So, according to this constitutive relation whenever a J exists in the conductor you have a electric field E=\frac{1}{\sigma}J. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction J implies E but we can read it in the direction E implies J? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case

 

[Fernsanz]

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 that's 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?

Interestingly enough I have come across this post from the thread "Why current attain steady state": https://www.physicsforums.com/showpost.php?p=2778731&postcount=5
It is quite clarifying and supports my previous picture of sound wave to explain the propagation of the electric field.

This also explain that electric field induced when the current is the "first cause" is exactly the same as the one obtained from Ohm's law, because in any case the current has to be the dynamic solution of an existing electric field. So now it's clear for me: the Ohm's law is a true constitutive relation which relates two quantities no matter which one is the "first cause"

Thanks AJ Bentley

 

[cabraham]

The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Did this help?

Claude

 

[Fernsanz]

The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.

 

[Fernsanz]

The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.

 

[cabraham]

I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.

I used "holes" in a loose sense. Also, peer-reviewed e/m fields texts insist that the charge distribution inside the conductor is indeed a source of E field. Where did you read otherwise? Of course a single e- radiates an E field in all directions. But a group of e- at one end, & a void of e- at the other end of a conductor has an E field directed from +ve to the -ve end. There is an E field thast extends outward as well, & it does slow down e- "before it".

Why would peer reviewed uni texts say that if it were not so? What are your references? I never expalined a directed current based on charge accumulation. The charge accumulation accounts for the E field inside the good conductor (not SC). Please don't put words in my mouth.

I try to help you by giving free advice which takes years to acquire, & you rudely rebuke me with straw man arguments, correcting me on issues I never raised. Just out of curiosity, how far along did you get regarding education. Are you a practicing scientist/EE? Just curious. You have a lot of sass & moxie with little chops to back it up. The problem with these forums is that everybody thinks they know more than everyone else. You exemplify that attitude.

Claude

 

[ZapperZ]

In an Ohmic conductor the Ohm's law says that J=\sigma E. So, according to this constitutive relation whenever a J exists in the conductor you have a electric field E=\frac{1}{\sigma}J. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction J implies E but we can read it in the direction E implies J? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case

In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.

 

[kmarinas86]

Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

He can't. It cannot be done. Ohm's law considers no such things. The "E" in Ohm's law is not the "E" of the real world. It is fictitious. It does not even necessarily equal the projection of the "E" over a differential length of the conductor. That only works if the electric field inside the wire is uniform and if the electric field outside the wire approaches each wire element at the same angle. Only in very a contrived circumstance with a bizarrely shaped electric field could you meet those conditions in a curved wire element. If the real "E" varies over the length of the conductor, speaking of the "E" in Ohm's law is like talking about the tooth fairy.

Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.

And because an electric field (changing or not) is needed to generate a current, the "E" of Ohm's law is not the electric field.

In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

Some people allege that a changing electric field does not "generate" a magnetic field and that it's just a correlation. Many of those same people allege the opposite to be the case as well.

Ohm's "law" is very quaint. For example, \mathbf{J}=\sigma\mathbf{E} applies only for a uniform field in the direction of the current. What kind of "law" is that?

There is no "right-hand law" in physics. Left-handed materials exist. We have something called "the right-hand rule". Ohm's "law" should be called "Ohm's rule". I did a Google search, and it turns out many concur.

I've just shown you an example where I could have a current without the use of an external electric field to generate that current.

Want to give an example of a situation without an electric field that can generate a current? You could try gravity, but you have to assume it is not fundamentally due to electromagnetism. You could also hook up a battery to a remote control. One could argue that it is external to the conductor, but it is part of the same circuit, so it still works because its field is "internal" to that circuit just as the current it produces will be.

Want another example? Look at the current from a photoemission process. Look ma! No electric field!

The photon has an electric field. It is changing, but so what? It is still an electric field.

I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.

You're right, it's not identical.

 

[jtbell]

\mathbf{J}=\sigma\mathbf{E} applies only for a uniform field in the direction of the current.

Neither the field nor the current density need to be uniform. If the conductor is isotropic (conducts equally well in all directions), then at each point the current density and the electric field have the same direction, but possibly different directions at different points.

If the conductor is anisotropic, the current density and electric field can be in different directions at a given point. In this case you have to use a tensor for the conductivity.

 

[cmos]

In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

I think this is one of the best quotes I have ever seen on this forum. I encourage everybody reading this thread to really let ZapperZ's quote sink in. It is directly relevant to the topic at hand as well as to many of the misconceptions that commonly occur to students (both formal and informal) of physics.

Quite often, equations will be written in an "aesthetic" form. The left hand side (LHS) may sometimes be the cause, other times it may be the effect, yet other times the equation will simply be a mutual relation.

In the present case of Ohm's law,
J=\sigma E,
J is the effect and E is the cause. I suppose this makes perfect sense writing the equation this way because it places the dependent variable on the LHS and the independent variable on the RHS, i.e. it is in the from y(x)=mx.

Not to get off-topic, but I believe a few examples might be helpful to some...

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

The most famous equation in all of physics:
E=mc^2
This is simply, what I called above, a mutual relation. Rest energy is mass, mass is rest energy.

 

[Fernsanz]

I used "holes" in a loose sense. Also, peer-reviewed e/m fields texts insist that the charge distribution inside the conductor is indeed a source of E field. Where did you read otherwise?

Of course that charge distribution is a source of E field, nobody is telling otherwise. But it can not be the source of net current in a preferred direction.

Of course a single e- radiates an E field in all directions. But a group of e- at one end, & a void of e- at the other end of a conductor has an E field directed from +ve to the -ve end. There is an E field thast extends outward as well, & it does slow down e- "before it".

So, are you suggesting that the two ends of the wire act as a capacitor where electrons accumlate at one end and fault at the other end and that this situation is what causes the whole electric field?

Why would peer reviewed uni texts say that if it were not so? What are your references? I never expalined a directed current based on charge accumulation. The charge accumulation accounts for the E field inside the good conductor (not SC). Please don't put words in my mouth.

Most engineering books give too much simplified views of physicis situations, they are not intended to give a deep insight of physics and so there is no reason to think they are totally right when one tries to go beyond "little balls called electrons" and consider "classical electrodynamics". So, raising the authority principle as a valid argument I think is not the type or reasoning a scientist should accept. Anyway, one of my references is Jackson's "Classical Electrodynamics"

It was no my intention to put words on your mouth.

I try to help you by giving free advice which takes years to acquire, & you rudely rebuke me with straw man arguments, correcting me on issues I never raised. Just out of curiosity, how far along did you get regarding education. Are you a practicing scientist/EE? Just curious. You have a lot of sass & moxie with little chops to back it up.

I'm a physicist and telecomunication engineer (something like a mix of Electrical Engineering and Computer Engineering in Spain). So I think I have gone through all those years to acquire the knowledge. Yet I have questions no matter how many years I have studied; don't you?

The problem with these forums is that everybody thinks they know more than everyone else. You exemplify that attitude.

I agree with both quotes. Literally.

 

[Fernsanz]

In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A..

If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, B=\mu_0 H, so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

It seems you don't read my posts. First, it was precisely me who mention the photoemission process as a differente way of obtaining current. But the big big mistake you are making is that none of your examples refer to an OHMIC medium. I stick to, and my question is about, ohmic media. So yes, I want just a single example where you have a current in a ohmic medium without E-field.

I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process.

Once again you haven't read my post where I provided the experiment (so simple and usual) you are demanding. I will repeat it: take a circuit composed of a photodiode (current source which shoot I amps) conected to a resistor (the ohmic conductor). As you yourself have pointed, the generation of the current in the photodiode requires no electric field; however, once that this current cross the resistor you have an electric field (or a voltage V, as you prefer). So, from the current generated with no electric field you end up having a electric field. And I bet you wil apply Ohm's law V=IR to know the voltage across the resistor, right? So this is a clear proof that the Ohm's law is valid no matter which magnitude is the cause.

It is so funny the picture of an electrical engineer analyzing a circuit with a current source and thinking "wait wait, I have to take into account how this current is generated, because if it is being generated in a way that does not require an electric field and then pumped into my circuit then I can not apply Ohm's law V=IR in my circuit because the "first cause", "the driving force", is I and not V. I will have to find out myself the relation between V and I for the resistor in this case... I better change my job", hahaha. So ridiculous that should be enough for you to think twice what you are claiming.

In the present case of Ohm's law,
J=\sigma E,
J is the effect and E is the cause.

Wrong. It can be read the other way around also. Just read the reasoning above

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

Again, no matter which one is the cause I can assure that if a mass m is moving with an acceleration a I can solve for F and the other way around.

 

[cabraham]

I think this is one of the best quotes I have ever seen on this forum. I encourage everybody reading this thread to really let ZapperZ's quote sink in. It is directly relevant to the topic at hand as well as to many of the misconceptions that commonly occur to students (both formal and informal) of physics.

Quite often, equations will be written in an "aesthetic" form. The left hand side (LHS) may sometimes be the cause, other times it may be the effect, yet other times the equation will simply be a mutual relation.

In the present case of Ohm's law,
J=\sigma E,
J is the effect and E is the cause. I suppose this makes perfect sense writing the equation this way because it places the dependent variable on the LHS and the independent variable on the RHS, i.e. it is in the from y(x)=mx.

Not to get off-topic, but I believe a few examples might be helpful to some...

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

The most famous equation in all of physics:
E=mc^2
This is simply, what I called above, a mutual relation. Rest energy is mass, mass is rest energy.

Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Of course, capacitors tend to behave as voltage sources, the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of

F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

Claude

 

[cabraham]

Of course that charge distribution is a source of E field, nobody is telling otherwise. But it can not be the source of net current in a preferred direction.



So, are you suggesting that the two ends of the wire act as a capacitor where electrons accumlate at one end and fault at the other end and that this situation is what causes the whole electric field?



Most engineering books give too much simplified views of physicis situations, they are not intended to give a deep insight of physics and so there is no reason to think they are totally right when one tries to go beyond "little balls called electrons" and consider "classical electrodynamics". So, raising the authority principle as a valid argument I think is not the type or reasoning a scientist should accept. Anyway, one of my references is Jackson's "Classical Electrodynamics"

It was no my intention to put words on your mouth.



I'm a physicist and telecomunication engineer (something like a mix of Electrical Engineering and Computer Engineering in Spain). So I think I have gone through all those years to acquire the knowledge. Yet I have questions no matter how many years I have studied; don't you?




I agree with both quotes. Literally.

The accumulated charge only was brought up to account for the small E field in the conductor. No, this charge accumulation is not "driving" the network current. You asked if current entering a conductor can give rise to an E field, & why so. I answered. The current entering the conductor is part of a larger network driven by an independent power source, i.e. generator, battery, photodiode, etc. The energy conversion in the prime mover produces the current & associated E field.

The charge density in the conductor accounts for the internal E field, but does not motivate the network current.

As far as my peer-reviewed textbooks not being sophisticated enough, what texts do you suggest I read? Even solid state physicists refer to resistance as the result of collisions. This is their position in conductors as well as semiconductors. How do you account for heating loss due to resistance if collisions are not the right answer? Many would love to know.

I realize that undergrad EE/physics texts cannot go as deep into the sub-atomic world as needed, but they do not preach heresy. They gloss over things & give simplified models, but they don't outright propagate untruths. Again, if resistance is not mechanical, owing to collisions, please enlighten us as to what it really is.

You claim that I exemplify the "know it all attitude", but I rely on the work of countless others for my claims. The person with the know-it-all attitude is the one who defies peer-reviewed texts & journals, stating "here's how it really is", w/ no proof to back it up. Just show me references where resistance is not mechanical, heat is not the result of collisions, & I will be grateful, as will many others.

Nothing personal at all. You just don't seem comfortable with my explanation, but I researched this topic carefully to assure that I do not post heresy. Please clarify. Thanks in advance.

Claude

 

[cmos]

If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, B=\mu_0 H, so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

...

Again, no matter which one is the cause I can assure that if a mass m is moving with an acceleration a I can solve for F and the other way around.

Fernsanz, you're both completely missing the point and hitting it right on the spot. Of course, if you can mathematically manipulate any equation to solve for whatever quantity you want, then it must be true. But this doesn't change the physical interpretation of which is cause, which is effect, and which is neither.

 

[cmos]

Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The setup you speak of requires that the inductor is connected to a DC voltage source. The "termination by a perfect short" implies that the inductor is placed across the source. The voltage source drives the current through the inductor.

The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

See above.

It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

How then would you make a constant current source? The only way to do it is to do it the way actual current sources are made, by putting a feedback mechanism into a variable voltage source.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Yes, equations can be inverted to solve for whatever you want. That is what makes mathematics so useful, especially for investigating physical phenomena. But the physical interpretations do no change with how you manipulate an equation.


At the risk of going off topic...

Force & acceleration are mutually dependent.

Yes, since forces cause bodies to accelerate.

In modern physics, F = m*a is discarded in favor of
F = dP/dt.

If by modern you mean since 1687. The quantity m*a is simply a special case of dp/dt; the latter being the most general case. So in general, forces cause bodies to change their momentum (Newton's 1st and 2nd Laws).

Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that.

You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.

Ask any physicist.

I should give you that advice, but it seems you would rather argue with them with a closed-mind and erroneous claims.

 

[kmarinas86]

Neither the field nor the current density need to be uniform. If the conductor is isotropic (conducts equally well in all directions), then at each point the current density and the electric field have the same direction, but possibly different directions at different points.

If the conductor is anisotropic, the current density and electric field can be in different directions at a given point. In this case you have to use a tensor for the conductivity.

Which E are you talking about?

Read what ZapperZ said to Fernsanz:

Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

Zz.

ZapperZ's implied answer is "no". You can read the rest of what said. He was testing "Fernsanz".

So what is the answer to my question?

Signed,
Kmarinas86

 

[kmarinas86]

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that.

You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.

Of FORCE and ACCELERATION:
1. Claude Abraham says they are mutually dependent.
2. Claude Abraham says they are not cause/effect related.
3. Claude Abraham says they cannot exist independently.

IMPLICATIONS:
1. Claude Abraham does not believe acceleration can be observed without force.
2. Claude Abraham does not believe force can be observed without acceleration.
3. Claude Abraham believes that no acceleration means no force.

ONLY ONE PROBLEM FOR CLAUDE ABRAHAM:
1. This only works if you equate "force" as the "vector sum of all forces" on a given object.

SEVERAL PROBLEMS FOR YOU:
1. I quote the following from you: "You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other."
2. "You're logic" is wrong.
3. Claude Abraham said, "Force & acceleration are mutually dependent."
4. Claude Abraham said, "Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other."
5. Claude Abraham did not say, "Force and acceleration are independent of each other."
6. You did not prove that mutual dependence of events implies causation.
7. You did not understand that Claude Abraham used the term "mutual dependence" to make a point about mutual existence (i.e. one does not exist without the other), not causality.
8. You say, "If force and acceleration were independent of each other, then neither could be a cause of the other." You imply that this proves his logic is wrong. However, if (A), then (B), then Claude Abraham says of this quote (A="force and acceleration were independent of each other"=false) (B="neither could be a cause of the other"=null). The word "could" in (B) is a direct reference to "force and acceleration...independent of each other" in (A). Per points (3.) and (5.), Claude Abraham must find your statement irrelevant to his arguments. (B) is (null) rather than false because in order to evaluate it requires knowledge of (A). This says to Claude Abraham: (B) requires (A) to be meaningful. If (A) is false, then (B) can only be known if (A), which is false, is known. That's why his position on (B) is "null". In consideration of all points above as what they really mean, his conclusion does not violate your proposition.
9. You were corrected by 23 year-old business student who does not major in science or liberal arts.

Signed,
Kmarinas86

 

[ZapperZ]

Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Of course, capacitors tend to behave as voltage sources, the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of

F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

Claude

This is a rather strange post because it appears to be self-inconsistent.

First of all, as a condensed matter physicist, I could derive Ohm's Law using statistical mechanics using nothing more than the model for free electron gas. With NO applied E-field, you get no net drift velocity in the electron gas, thus, no net current. It is only UPON the action of such a field do you get, within the Drude model, such a current, resulting in the ever-familiar Ohm's law. Please see Chapter 1 in Ashcroft and Mermin's Solid State text.

Now, all you have to do to falsify this is to show that there is some spontaneous current that does not depend on the external field, and that upon the generation of such a current, the SAME E-field as above is the result. If you are successful, I can use it as our new accelerating mechanism at our particle accelerator.

Till then, all I have is E-field causes the current in Ohm's Law.

Zz.

 

[ZapperZ]

If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, B=\mu_0 H, so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

Again, you are considering only the MATHEMATICAL aspect of the equation. Mathematically, you can rearrange it any way you like. But it doesn't mean that you can switch around the interpretation. The "interpretation" part, which gives it meaning, depends on the PHYSICS.

It seems you don't read my posts. First, it was precisely me who mention the photoemission process as a differente way of obtaining current. But the big big mistake you are making is that none of your examples refer to an OHMIC medium. I stick to, and my question is about, ohmic media. So yes, I want just a single example where you have a current in a ohmic medium without E-field.
Once again you haven't read my post where I provided the experiment (so simple and usual) you are demanding. I will repeat it: take a circuit composed of a photodiode (current source which shoot I amps) conected to a resistor (the ohmic conductor). As you yourself have pointed, the generation of the current in the photodiode requires no electric field; however, once that this current cross the resistor you have an electric field (or a voltage V, as you prefer). So, from the current generated with no electric field you end up having a electric field. And I bet you wil apply Ohm's law V=IR to know the voltage across the resistor, right? So this is a clear proof that the Ohm's law is valid no matter which magnitude is the cause.

It is so funny the picture of an electrical engineer analyzing a circuit with a current source and thinking "wait wait, I have to take into account how this current is generated, because if it is being generated in a way that does not require an electric field and then pumped into my circuit then I can not apply Ohm's law V=IR in my circuit because the "first cause", "the driving force", is I and not V. I will have to find out myself the relation between V and I for the resistor in this case... I better change my job", hahaha. So ridiculous that should be enough for you to think twice what you are claiming.

I believe cmos has responded to your scenario. Furthermore, the current generated in the photodiode DOES require a field. It is why you need a PN junction inside one of those things. And you should look up why one needs such a junction (read about the depletion zone). The phenomena of a photodiode is NOT the same as a typical photoemission process.

The microscopic aspect of Ohm's Law came out of the Drude Model. Could you please start with that and show me where you can get a current that, in turn, causes the identical E-field to be generated? The relevant reference will be any Solid State text, such as Ashcroft and Mermin or Kittel.

Zz.

 

[kmarinas86]

Neither J nor E is the "cause" or "effect". Either one can take place first.

Did the chicken cause an egg and an egg cause a chicken? What about the chicken's diet and the diet of its young? What about their requirements to exist?

The cause of J is the sum and derivatives of all externally-originated fields extending through that point. The cause of E is the sum and derivatives of all externally-originated fields extending through that point.

The cause of an E may be an external E and/or an external B.

The cause of an B may be an external E and/or an external B.

And so forth...

Observers beware,
Causes are everywhere,
But their effects are local,
So put on your bifocals.

An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short.

Inductive voltage drop = Inductance * change of current / change in time

Then the short is opened via a switch, and the inductor de-energizes into the resistor.

Origin of I (after short):
Inductive voltage drop < 0 volts

The current pre-existed the voltage.

Origin of current (before short):
Inductive voltage drop > 0 volts

When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Origin of current (before short):
Inductive voltage drop > 0 volts

Origin of I (after short):
Inductive voltage drop < 0 volts

V = Inductive voltage drop

Of course, capacitors tend to behave as voltage sources,

So does shorting an inductor.

Inductors and capacitors are storage devices. They can dissipate energy in the form of electricity, which means they are capable of producing voltage and current (as well as V/m and A/m fields).

the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either.

Well of course they can.

But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

True that.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

No kidding, isn't that what they teach in high school?

I have not looked yet, but I would not be surprised if anyone here disagrees with this.

Someone might correct you by saying there are more than 3 ways to write Ohm's law *gasp - omgish!*. It utterly pointless to argue with people who respond like that except to embarrass them by replying with common sense.

Next...

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of F = dP/dt.

True enough.

Force & acceleration are not cause/effect related.

If you consider the time delay of propagation of fields at the speed of light and their dissipation with distance, this is not true at all.

Does the Sun's light cause the Earth to be exposed to radiation (relatively non-lethal doses at that)? Of course. Does space weather (not limited to solar weather) transfer force to Earth weather? Of course! Is the reverse true? Many orders of magnitude less!

For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

You mean:

For one to be a cause at the exclusion of the other, it must exist independently of the other.

A force is "applied" at the same point as the acceleration of a thing it causes. Does this mean they are "dependent"?

Observers beware,
Causes are everywhere,
But their effects are local,
So put on your bifocals.

Claude

 

[kmarinas86]

This is a rather strange post because it appears to be self-inconsistent.

First of all, as a condensed matter physicist, I could derive Ohm's Law using statistical mechanics using nothing more than the model for free electron gas. With NO applied E-field, you get no net drift velocity in the electron gas, thus, no net current. It is only UPON the action of such a field do you get, within the Drude model, such a current, resulting in the ever-familiar Ohm's law. Please see Chapter 1 in Ashcroft and Mermin's Solid State text.

Inductors and capacitors are both sources of electrical power.

Disagreement exists when one alleges that inductors are without voltage and capacitors are without current.

Without voltage, what is a V/m field?

Without current, what is an A/m field?

Without both, how do you obtain electrical power?

What does one mean when saying "power without E-field". They could mean discharging an inductor. But an inductor can have a voltage when the current is changing! It is called an inductive voltage drop (or a gain if we are talking about coils in an electrical generator). A charged particle moving inertially dissipates no power. Therefore specific kinetic energy of a particle must change for power to be a relevant factor.

Now, all you have to do to falsify this is to show that there is some spontaneous current that does not depend on the external field, and that upon the generation of such a current, the SAME E-field as above is the result. If you are successful, I can use it as our new accelerating mechanism at our particle accelerator.

Till then, all I have is E-field causes the current in Ohm's Law.

Zz.

What is an external field? It is any field external to what you got. If you are measuring the field at a point in space, external fields are fields which exist outside that point. How many points do you think space has? If you are measuring the field in a conductor, then your "external" field can include the battery connected across that conductor. Your external field can be "internal" in your circuit and "external" to the part of the circuit you are measuring.

 

[Fernsanz]

Furthermore, the current generated in the photodiode DOES require a field. It is why you need a PN junction inside one of those things. And you should look up why one needs such a junction (read about the depletion zone). The phenomena of a photodiode is NOT the same as a typical photoemission process.

You have reached a point where you say absolute no senses: the deplection zone is what causes the E-field that generates the current? Oh my God, that's too much. The electric field in the deplection zone opposses to the current! The deplection zone is only a result of equilibrium between diffussion and conduction, it has nothing to do with a net current. Have you ever seen a diode giving current on its own? lol.

So, according to you there is no way of generating a current without an electric field? It is funny how you are getting inconsistent because some post ago you named photoemission as a method to generate current without E-field. As for photoelectric effect, I would like to see how you try to explain it in terms of electrical field, lol.

Hence, being obvious that we can generate a current without electrical field (for instance by mean of photoemission or photoelectric effect, thermoionic effect, SC, etc, and if you claim otherwise it would be so ridiculous that I would leave the thread), take one of this methods and pump that current into a resistor. Every single people on Earth will use the Ohm's law V=IR to obtain the voltage across that resistor.

The microscopic aspect of Ohm's Law came out of the Drude Model. Could you please start with that and show me where you can get a current that, in turn, causes the identical E-field to be generated? The relevant reference will be any Solid State text, such as Ashcroft and Mermin or Kittel.

Could you please start with telling me if there is any case in which you wouldn't use the Ohm's law V=IR when a current cross a resistor?

Could you please start with telling me if a person analyzing a circuit which has a current source has to ask about the nature of the current in order to know if it he/she can apply the Ohm's law V=IR?

Could you please start with telling me why my solar panels give a voltage across my house lamps?

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always applicable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.

 

[ZapperZ]

You have reached a point where you say absolute no senses: the deplection zone is what causes the E-field that generates the current? Oh my God, that's too much. The electric field in the deplection zone opposses to the current! The deplection zone is only a result of equilibrium between diffussion and conduction, it has nothing to do with a net current. Have you ever seen a diode giving current on its own? lol.

No, I didn't say anything that you said here. I was pointing out that such a device is NOT devoid of any E-field, contrary to what you were claiming! Read what you wrote and read the particular point that I was responding to.

So, according to you there is no way of generating a current without an electric field? It is funny how you are getting inconsistent because some post ago you named photoemission as a method to generate current without E-field. As for photoelectric effect, I would like to see how you try to explain it in terms of electrical field, lol.

Now it is my turn to tell you that you haven't been reading my post.

Note that I emphasized that the E-field that I was referring to is the identical E-field as described within Ohm's Law, NOT any other E-field. I already described many situations where I can have current without any external E-field. But these situations are not the same as the Ohmic E-field that is in question.

Could you please start with telling me if there is any case in which you wouldn't use the Ohm's law V=IR when a current cross a resistor?

Yes, when the resistor becomes a superconductor, or when the resistor is a semiconductor.

Could you please start with telling me if a person analyzing a circuit which has a current source has to ask about the nature of the current in order to know if it he/she can apply the Ohm's law V=IR?

No. And why is this relevant? The nature of a current source doesn't negate the fact that a component must also have an applied potential across its terminals. I work with constant current power supply all the time. There's nothing here to indicate that these things only pump current into a circuit, and the E-field comes later.

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always appliable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.

Ohm's law is valid under the majority of cases. This is because the material that we use for most circuits are metals that obey the Drude model under the majority of the situation that they are used, i.e. the APPROXIMATION that the conduction electrons can be modeled as free electron gas. This explicitly described in Chapter 1 of Ashcroft and Mermin. However, go a bit further and look at Chapter 3, and you'll see a whole chapter titled "Failure of the Free Electron Model"! There are materials and situations in which the Drude Model, i.e. the foundation for Ohm's Law, no longer works! This is not my opinion, it is a FACT!

I still don't understand why we are debating if Ohm's Law works or not. I don't ever recall in my original post of contesting the validity of Ohm's Law. I did contest the validity of your interpretation, i.e. that any kind of interpretation is valid based on mathematics. Since you claim that you can reverse the "order" of the cause-and-effect in this, I asked for you to go back to the http://people.seas.harvard.edu/~jones/es154/lectures/lecture_2/drude_model/drude_model_cc/drude_model_cc.html" , and show where, at the microscopic level, that you can get the electric field (the identical field in Ohm's Law) as a consequence of the drift velocity of the electrons. To me, this is the only way to show a convincing argument of what you are claiming.

Zz.

 

[ZapperZ]

What does one mean when saying "power without E-field". They could mean discharging an inductor. But an inductor can have a voltage when the current is changing! It is called an inductive voltage drop (or a gain if we are talking about coils in an electrical generator). A charged particle moving inertially dissipates no power. Therefore specific kinetic energy of a particle must change for power to be a relevant factor.

That is certainly true. An inductor can have its own potential difference.

What is an external field? It is any field external to what you got. If you are measuring the field at a point in space, external fields are fields which exist outside that point. How many points do you think space has? If you are measuring the field in a conductor, then your "external" field can include the battery connected across that conductor. Your external field can be "internal" in your circuit and "external" to the part of the circuit you are measuring.

I don't know if I mentioned the phrase "external field". Since I was talking within the context of the Drude model, then I certainly was using the E-field within that scenario. It certainly doesn't mean external to the conductor or the circuit.

Zz.

 

[Fernsanz]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage V across the resistor (it will happen by fact 2)

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage satisfies the Ohm's law: V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity or semiconductors". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors? Are you avoiding by all means facing the problem?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.

 

[cabraham]

The setup you speak of requires that the inductor is connected to a DC voltage source. The "termination by a perfect short" implies that the inductor is placed across the source. The voltage source drives the current through the inductor.



See above.



How then would you make a constant current source? The only way to do it is to do it the way actual current sources are made, by putting a feedback mechanism into a variable voltage source.



Yes, equations can be inverted to solve for whatever you want. That is what makes mathematics so useful, especially for investigating physical phenomena. But the physical interpretations do no change with how you manipulate an equation.


At the risk of going off topic...

Yes, since forces cause bodies to accelerate.



If by modern you mean since 1687. The quantity m*a is simply a special case of dp/dt; the latter being the most general case. So in general, forces cause bodies to change their momentum (Newton's 1st and 2nd Laws).



You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.



I should give you that advice, but it seems you would rather argue with them with a closed-mind and erroneous claims.

I don't believe what I'm reading. The inductor terminated in a perfect short has 0 volts across it. Not a voltage source, but a dead short. Please research this. It is elementary. Ask a professor. Inductors terminated in a SC short retain their current indefinitely. When the path is switched so that said inductor is terminated in a resistance, the current at the instant of transition remains as it was, then decays. The initial voltage across said resistor is I*R, then it decays. In other words I determines V.

How would I make a constant current source? The same way the power company makes a constant voltage source, except that I drive the turbine at constant torque. To obtain constant voltage, the power company uses constant speed drive. It's pretty easy. They use CVS because it offers lower losses than CCS. Also, CVS provides constant frequency, useful for synchronous machines.

A photodiode terminated in a short or virtual short (op amp inputs) outputs a constant current source behavior proportional to incident light intensity. This is a good example of a CCS. The feedback in the op amp loop does not control the constant current, the illumination does. The feedback forces the 2 op amp inputs to remain at virtual zero potential difference. The light controls the magnitude of the current.

Also, batteries can be constructed for CCS operation, but CVS works better for common cell materials (carbon-zinc, alkaline, NiMH, Li ion, etc.). But research on nuclear cells indicates that they lend themselves to CCS operation. Interestingly, not only are nucells CCS devices, but AC as well. A nucell outputs a constant ac current.

Seriously, this notion that voltage is the cause of current, force causes acceleration, & that CCS devices do not exist, is just a groundless prejudice. A generator can be operated for CVS or CCS output. Inductors terminated in a superconductor hold constant current indefinitely. If the SC has a resistance across it, & the SC is switched out, the resistor will dissipate the inductor energy. The initial voltage is forced by I*R. I determines V in this case, vice-versa for a cap.

Not too hard.

Claude

 

[cabraham]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage across the resistor (it will happen by fact 2)

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage is V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.

Very good explanation. We agree on these points, but you still don't like my collision explanation of resistance. Any references that you can recommend?

Claude

 

[Fernsanz]

I don't believe what I'm reading. The inductor terminated in a perfect short has 0 volts across it. Not a voltage source, but a dead short. Please research this. It is elementary. Ask a professor. Inductors terminated in a SC short retain their current indefinitely. When the path is switched so that said inductor is terminated in a resistance, the current at the instant of transition remains as it was, then decays. The initial voltage across said resistor is I*R, then it decays. In other words I determines V.

Exactly. In fact this is the reason to use what is called "freewheeling" or "flyback" diodes: they protect against the peak voltage observed at the very moment of switching.

 

[Fernsanz]

Very good explanation. We agree on these points, but you still don't like my collision explanation of resistance. Any references that you can recommend?
Claude

I'm still trying to get that references. I'm in talks with AJ Bentley, which is the one who provide the explanation in term of "E shockwaves" in the thread I referenced. I will let you know when I have a good reference.

What is clear now is that we are the only one who, up to now, are aware of the problem: to explain why a current generates a E-field. So I'm afraid you and me are the only one to work out the problem.

Thank you for explanation (even if I don't like). A pleasure discussing with you.

 

[kmarinas86]

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always appliable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.

Ohm's law is valid under the majority of cases. This is because the material that we use for most circuits are metals that obey the Drude model under the majority of the situation that they are used, i.e. the APPROXIMATION that the conduction electrons can be modeled as free electron gas. This explicitly described in Chapter 1 of Ashcroft and Mermin. However, go a bit further and look at Chapter 3, and you'll see a whole chapter titled "Failure of the Free Electron Model"! There are materials and situations in which the Drude Model, i.e. the foundation for Ohm's Law, no longer works! This is not my opinion, it is a FACT!

I still don't understand why we are debating if Ohm's Law works or not. I don't ever recall in my original post of contesting the validity of Ohm's Law. I did contest the validity of your interpretation, i.e. that any kind of interpretation is valid based on mathematics. Since you claim that you can reverse the "order" of the cause-and-effect in this, I asked for you to go back to the http://people.seas.harvard.edu/~jones/es154/lectures/lecture_2/drude_model/drude_model_cc/drude_model_cc.html" , and show where, at the microscopic level, that you can get the electric field (the identical field in Ohm's Law) as a consequence of the drift velocity of the electrons. To me, this is the only way to show a convincing argument of what you are claiming.

Zz.

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

 

[Fernsanz]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Lol, lol. Please inform of this to universities where nobody teach that the laws to analyzing circuits depend on the values of inductances and resistances, lol.

Also inform to all engineers around the world (including me) who always use Ohm's law. I will inform my mates that we have wrongly used the Ohm's law in the magnetotorquers of the satellite to rate the freewheeling diodes.

Lol, after this claim of yours (strongly supported by a lot of reasoning in your post by the way) it is pointless to keep arguing.

 

[cabraham]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Don't be offended when I say that you have much to learn about circuit theory. There is no shame in not knowing it, the shame is claiming to know more than the experts & not backing it up. I know absolutely nothing about veterinary medicine. I'm not a problem as long as I let the world know that. But if I were to claim to be a vet, & people bring their pets to me in full trust, that is a problem for which I am liable.

Ohm's law is always valid in an R/L network. When the switch is closed, inductor current starts at zero for example, & the source & resistor are in series with L. As the current increases in L & R, the voltage across R, Vr, increases likewise. At every point in time, Vr = R*I. Ohm's law is valid from the start until the steady statevalue of I = V/R. No exceptions have ever been found.

Who says otherwise. Again, when I ask how much formal circuit theory you've had, I'm just asking what your sources are, not trying to be confrontational. What credible reference claims that Ohm's law is valid only some of the time? Please share.

Claude

 

[ZapperZ]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage V across the resistor (it will happen by fact 2)

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage satisfies the Ohm's law: V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity or semiconductors". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors? Are you avoiding by all means facing the problem?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.

So was my assertion that you cannot have such a thing via the Drude model. You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you? So now, if we AGREE on that, and we are sticking by the idealized view of a conductor and resistor per the Drude model that PRODUCES Ohm's law, I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

One could confirm this another using a more generalized approach by applying the Boltzmann transport equation, which contains more than one mechanism for charge transport. Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?

Zz.

 

[cabraham]

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.



So was my assertion that you cannot have such a thing via the Drude model. You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you? So now, if we AGREE on that, and we are sticking by the idealized view of a conductor and resistor per the Drude model that PRODUCES Ohm's law, I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

One could confirm this another using a more generalized approach by applying the Boltzmann transport equation, which contains more than one mechanism for charge transport. Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?

Zz.

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.

Also, photoelectric effect describes currents produced due to photon energy transfer. The current in a photodiode is photon induced, not field induced.

Does this help?

Claude

 

[Fernsanz]

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.

? You are fantastic: instead of answering, you ask (pretending to be surprised) about this generic experiment, and the whole thread deal with this experiment from the very beginning!, lol. By the way, who said that the generation of the current has to take place in the ohmic conductor? Read the experiment again please, is clear enough.

[...]I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

Answered by yourself:

I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

Are you reaching contradictions with yourself?

 

[kmarinas86]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Lol, lol. Please inform of this to universities where nobody teach that the laws to analyzing circuits depend on the values of inductances and resistances, lol.

Also inform to all engineers around the world (including me) who always use Ohm's law. I will inform my mates that we have wrongly used the Ohm's law in the magnetotorquers of the satellite to rate the freewheeling diodes.

Lol, after this claim of yours (strongly supported by a lot of reasoning in your post by the way) it is pointless to keep arguing.

Don't be offended when I say that you have much to learn about circuit theory. There is no shame in not knowing it, the shame is claiming to know more than the experts & not backing it up. I know absolutely nothing about veterinary medicine. I'm not a problem as long as I let the world know that. But if I were to claim to be a vet, & people bring their pets to me in full trust, that is a problem for which I am liable.

Ohm's law is always valid in an R/L network. When the switch is closed, inductor current starts at zero for example, & the source & resistor are in series with L. As the current increases in L & R, the voltage across R, Vr, increases likewise. At every point in time, Vr = R*I. Ohm's law is valid from the start until the steady statevalue of I = V/R. No exceptions have ever been found.

Who says otherwise. Again, when I ask how much formal circuit theory you've had, I'm just asking what your sources are, not trying to be confrontational. What credible reference claims that Ohm's law is valid only some of the time? Please share.

Claude

You both assume that "V" stands for the resistive voltage drop.

If V is the resistive voltage drop, then what letter is used your inductive voltage drop? Do you even care about the inductive voltage drop?

What letter do you use for applied voltage?

What about the capacitive voltage drop?

Haven't you ever heard of the analogy between RLC circuits and mechanical harmonic oscillators?

Voltage is not limited to I*R. I*R is a voltage drop (V_R). It is not voltage related to stored energy. It is voltage related to dissipated energy. Voltage related to stored energy can use symbols (V_C) or (V_L) depending on what is discussed.

There is such a thing as static charge inside a non-shorted capacitor. Where is the current in that? What R in a capacitor has any meaning? A capacitor stores voltage does it not?

Why are other meanings of V ignored in this discussion? It may be fine in the field to assume that V means resistive voltage drop, but this is utterly misleading to outsiders. How can I believe even 50% of what I read here when so much goes around being equivocated?

The topic of the thread is "Ohm's law, why current induces electric field and cause effect direction".

The truth is that current induces a magnetic field. The magnetic field is a first-order effect of the electric fields of the charges which the current is comprised of. If this current induces an electric field, it must be a secondary-order effect of those electric fields, meaning that the magnetic fields they produce must have a changing magnitude. A macroscopically viewable example of combined second-order effects can be easily demonstrated with a coil electromagnet. The inductive voltage drop which exists in the inductor is in fact a second-order effect of those electric fields. That voltage drop behaves like an applied voltage (in this case, an induced electrical field) in the opposite direction, and this creates a delay of the current in reaching its final value determined by V_(source)/R. A charge undergoing acceleration produces such a secondary-order effect. The same is true for deceleration of charge.

The truth is that current inducing electric fields has absolutely nothing to do with a constant DC current, so any analysis in this thread that assumes a constant DC current is utterly beside the point and totally irrelevant. The so-called "constant" AC current may cause an oscillating electric field, but what a surprise that is! NOT. "Why current induces electric field and cause effect direction" is about circuit laws[/url] because to quantify from scratch the electric fields due to changes of currents at a local level requires the use of differential equations.

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-equations/5_RL-circuits.php

This is really basic stuff, and I always try to keep it on the top of my head. Every model I question on these forums I connect with fundamental basics which must go on in any situation, regardless of its magnitude, and even if the "scenario" denies, contradicts, or fails to mention that such effects must exist!

 

[kmarinas86]

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

Equilibrium requires that forces must be in equilibrium. If you attain it exactly, you have no leak. Otherwise, you can have a leak of charge.

You have electrons and things that contain electrons. Without something to contain electrons, there is no equilibrium for them because then the electrons just repel each other away.

any of those electron will feel a stronger field from its neighbours, but that's 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.

Well that is what happens when charge is stored in a bottle yo.

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

This part really does answer the question, "Why do currents induce electric fields and cause/effect their direction?" In fact, one could care less whether you're using an ohmic conductor or not because either way the strength of the electric fields can be produced by the interaction of electric charges. What should worry you more are the differential equations for the electric and magnetic fields.

Talking about the basics is obviously sufficient to show that the question posed does have a valid approach to an answer. In this case, it is far too simple to arrive at this answer. You don't even need decades of experience to do it. You just need knowledge of the laws of physics.

Imposing an "ohmic" view on this scenario "currents causing electric fields" is ridiculous.

 

[ZapperZ]

? You are fantastic: instead of answering, you ask (pretending to be surprised) about this generic experiment, and the whole thread deal with this experiment from the very beginning!, lol. By the way, who said that the generation of the current has to take place in the ohmic conductor? Read the experiment again please, is clear enough.

So now we are going out of Ohmic conductor after you insist that we HAVE to deal with one?

Answered by yourself:


Are you reaching contradictions with yourself?

No, because the examples I gave were NOT in a conductor. There were in vacuum as my initial attempt to show you that one can have current without an applied field (i.e. not via Ohm's law).

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it. I take it that, since you have continued to not address this model, that you either have never heard of it, or not aware of it. Therefore, before you continue to insist that you can simply rearrange Ohm's Law and make a different interpretation of it simply on your ability to do algebraic manipulation, I strongly suggest you look up the link I've given you on the Drude model and the free electron gas.

If you don't wish to do that, then there's nothing I can do to correct what I know to be a mistake in your view. If you are OK with that, then there's nothing I need to care about.

Zz.

 

[kmarinas86]

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!

 

[kmarinas86]

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it.

A coherent summary of a truth.

 

[Born2bwire]

So now we are going out of Ohmic conductor after you insist that we HAVE to deal with one?



No, because the examples I gave were NOT in a conductor. There were in vacuum as my initial attempt to show you that one can have current without an applied field (i.e. not via Ohm's law).

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it. I take it that, since you have continued to not address this model, that you either have never heard of it, or not aware of it. Therefore, before you continue to insist that you can simply rearrange Ohm's Law and make a different interpretation of it simply on your ability to do algebraic manipulation, I strongly suggest you look up the link I've given you on the Drude model and the free electron gas.

If you don't wish to do that, then there's nothing I can do to correct what I know to be a mistake in your view. If you are OK with that, then there's nothing I need to care about.

Zz.

I won't be weighing in on some of the other discussions but in terms of the OP's question, ZapperZ has hit upon it several posts back and this is just another good summarization of the answer.

Physics is not the same as mathematics in that we can not simply rearrange an equation at our leisure and find that all rearrangements, while mathematically viable, are still physically valid. Ohm's Law is a macroscopic description of the movement of carriers through a medium as given by the Drude Model. The idea being that as the carriers move through the material, they collide with the constituent particles of the material (atoms, molecules, lattice). In doing so, they give up some of their kinetic energy in the form of vibrational energy in the material bulk (phonons = heat) and also undergo a change in momentum.

Now we could of course have this happen without any applied electric field. We have an electron gun that shoots a stream of electrons into a bulk. But this is not going to give rise to the same physical phenomenon because the electrons will continually bleed off their energy (thus allowing us to simply make a bulk thick enough such that no appreciable amount of carriers can penetrate through). In addition, the constant changes in directions from the collisions will hamper any real net movement of the carriers. Yes, the charge carriers themselves will give rise to an electric field, but it will not be the same electric field as an applied field because it will obviously be one that is not conducive to the propagation of the charges through the material (by virtue of the fact that the individual fields will try and separate the carriers).

Only if we have an applied field will we see a continuous injection of energy back into the carriers so that they can recover from the energy lost to the collisions and progress through the bulk. In addition, the applied field also reasserts the drift direction which would eventually be lost if we allowed the carriers to continually collide randomly.

So Ohm's Law does not make physical sense without the applied electric field because it is an assumption that is required by the underlying physics that gave rise to the relation. Of course, the above is nothing more than another restatement of ZapperZ's post.