Superconductivity at room temperature??

[W.D.Clinger]

The crux of my question still remains, how do you reconcile the division by zero mathematically?

There is no division by zero when Ohm's law is stated as V = I * R.

If you somehow know that the three quantities V, I, and R are nonzero, then and only then are you allowed to rewrite Ohm's law as I=V/R or R=V/I. In practical situations, where you're trying to calculate I or R, those are often the most useful forms of Ohm's law, so many engineers learn Ohm's law in one of those forms and forget (or never learn) that the most general forms of Ohm's law do not involve any division.

If you take a 12 V battery capable of sustaining a 100Amp current and you connect it to length of zero resistance superconductor.

Then something's likely to go blooey.

The current flowing is 100Amps and your voltmeter across the positive and negative terminals will read zero.

Your ammeter which is in series with the conductor reads 100 amps, assuming the conductor does not evaporate.

The voltage of 12V which is known prior to the test and the resistance is zero by defintion, cannot be used used to calculate the current.

Hence my question.

In the situation you describe, the battery's internal resistance becomes the total resistance of the circuit. Furthermore the voltage measured across the terminals is likely to drop from 12 volts to essentially zero volts, because the 12-volt drop moves inside the battery and would be measured across that internal resistance. (In reality, it's more complicated because batteries aren't really constant-voltage sources.)

ETA: See also ziggurat's answer in post #1078.

 

[Skwinty]

In practical situations, where you're trying to calculate I or R, those are often the most useful forms of Ohm's law, so many engineers learn Ohm's law in one of those forms and forget (or never learn) that the most general forms of Ohm's law do not involve any division.

You are correct here.

When I studied electrical engineering 40 years ago, I just memorised the formulae and never gave much thought to their derivation.

I sucked at math and still do.

 

[ben m]

If you want to take Skwinty's scenario with an "ideal" battery---one with no internal resistance---then you can do so, but realizing how the "100A" limit comes in.

A voltage source with a current limit does not maintain its "rated" voltage at all times --- i.e. it is by definition not a constant-voltage source. It behaves as follows:

1) If the current is below 100A, the voltage is set to 12V and the current is set to whatever the circuit draws at 12V.
2) If the current is 100A, the voltage drops to whatever the circuit requires at 100A and the current is set to 100A.

An "ideal" current-limited battery, connected to a superconductor, will crank its current to 100A, switch from voltage-regulated to current-regulated behavior, and consequently set its voltage to zero.

 

[Olowkow]

A current source outputs a few hundred millivolts, at high current into heavy cables to the superconducting coil. All the voltage is dropped across the cables, and the superconducting current is simply this voltage drop divided by the (non SC) cables and connector resistances, since all series elements see the same current. What is so difficult?

If a "battery" is involved, some active means of sourcing a current must be used. A battery is a voltage source (albeit not "constant").

One confusing thing that has been said many times is "connect a battery to a superconducting junction..." Apparently, some just take for granted that there is some sort of current limiting element, but unless this is stated it can be very misleading and has been a source of great misunderstanding here.

 

[ceptimus]

Superconductors only have zero resistance up to a certain current density - then the superconductivity stops.

So for a given diameter of superconducting wire, you might be able to get, say, 30 amps flowing, but if you try to exceed that then it stops being a superconductor until the current falls back below 30 amps.

Superconducting electromagnets normally have a short section of the winding passing through a heater. The heater is used to warm the superconductor to the point where it becomes a normal conductor with a measurable resistance. A battery can then be connected to the winding - either side of the heated section - to get the current flowing. Once the current is established, the heater is switched off so that the whole winding becomes superconducting. At this point the battery can be disconnected.

The battery voltage can be very low. Even a tiny voltage will cause the current flowing in the superconductor coil to gradually increase - there is no resistance to oppose the voltage, but the inductance of the winding does provide a balancing voltage for the battery while the current, and therefore magnetic field, is increasing.

 

[ben m]

One confusing thing that has been said many times is "connect a battery to a superconducting junction..." Apparently, some just take for granted that there is some sort of current limiting element, but unless this is stated it can be very misleading and has been a source of great misunderstanding here.

There's nothing wrong with this for an SIS junction, which indeed develops an EMF (~millivolts) which does a perfectly good job of regulating current---you can indeed plug an ideal millivolt battery into an SIS junction and it'll find a sensible equilibrium.

 

[ben m]

What confuses me is your claim that the behavior is totally determined by the acceleration and trajectory, and nothing else; but in the very next breath you claim that the behavior is not totally determined by the acceleration and trajectory on their own, since in some cases the electron radiates and in other cases not. Thus you are clearly arguing that even though the acceleration and trajectories are the same, the electron does not always radiate:

Let's repeat one more time. I will be as pedantic as humanly possible, and if it doesn't get through I will give up.

The fields Ei(x,t) and Bi(x,t) due to any point charge i, depend only on the position xi(t), velocity vi(t), and qi of that charge. Sometimes there is a nonzero acceleration ai = dvi/dt, and if so there must be a force responsible. The Ei(x,t) and Bi(x,t) DO NOT depend on the nature of the force responsible; they depend only on xi(t) and vi(t).

Understand? Read carefully. Every word of that is in there for a reason; do not skip any of them, nor assume any additional ones. The reason we make this distinction is that you (erroneously) kept pretending there was some distinction between electric and mechanical forces, as a reason to reject the "spinning ring of charge" calculations.

The radiation pattern of a charge distribution, with N charges in it, depends on the total E and B fields of the entire charge distribution---in free space, the radiation depends only on the form of the fields at asymptotically large distances. It does not, in any case, depend individually on the individual components Ei and Bi due to the individual source charges---only on their sums E and B.

• It is possible to find ensembles of charges, each one accelerating, but for which E = B = 0. Since E=B=0 is a nonradiating field, such ensembles don't radiate and neither do the component charges.
• It is possible to find ensembles of charges, with each one accelerating, but for which E is constant in time and B is constant in time. Since E=const and B=const is a nonradiating field, such ensembles don't radiate.
• It is also possible (indeed easy) to find ensembles of charges for which E and B are time-dependent. Such ensembles usually do radiate, as is well known.

Understand? Read carefully. Every word of that is in there for a reason; do not skip any of them, nor assume any additional ones. The reason we must make this distinction is that you keep erroneously saying that all accelerating charges radiate; any time anyone objected to that, you pretended we were saying that no accelerating charges radiate. Both are wrong.

Again, modulo typos, the above is a boring and essentially trivial statement about utterly standard E&M; it reflects the derivation from Maxwell's Eq. found in (e.g.) Jackson chapter 14 or Griffiths chapter 9. You can derive it from material in Serway.

If you think any of the above is wrong, be prepared for another embarrassment when you finally get around to checking the textbook.

If you think that the above contradicts what we've said before, please illustrate with direct quotes.

 

[Olowkow]

There's nothing wrong with this for an SIS junction, which indeed develops an EMF (~millivolts) which does a perfectly good job of regulating current---you can indeed plug an ideal millivolt battery into an SIS junction and it'll find a sensible equilibrium.

Well this is of course true. Is it really relevant to an honest and accurate explanation of the matter at hand? Or is it just "I'm right", nerney nerney? I'm trying to explain to technicians, not to PhD's. This is like arguing about hooking up a doorbell, and how much force will be applied to the button. How many ideal "millivolt" batteries are there on the market? You folks are just going too far. This is trivial simple stuff, and it is being tortured to the point of madness.

 

[ben m]

Well this is of course true. Is it really relevant to an honest and accurate explanation of the matter at hand? Or is it just "I'm right", nerney nerney? I'm trying to explain to technicians, not to PhD's. This is like arguing about hooking up a doorbell, and how much force will be applied to the button. How many ideal "millivolt" batteries are there on the market? You folks are just going too far. This is trivial simple stuff, and it is being tortured to the point of madness.

And yes, it's all utterly trivial. We dwell on it only because Prins thinks there's a hitch that both (a) proves BCS theory wrong and (b) proves his new theory right. If you can find a way to disabuse him of that, without simplifying things to the bare minimum, be my guest. But be prepared for heated objections to simple statements like "we set the power supply voltage to 1mV".

ETA: I think page 5 of this thread is around where Johan fell out of reach of the normal language of circuit analysis.

 

[Olowkow]

And yes, it's all utterly trivial. We dwell on it only because Prins thinks there's a hitch that both (a) proves BCS theory wrong and (b) proves his new theory right. If you can find a way to disabuse him of that, without simplifying things to the bare minimum, be my guest. But be prepared for heated objections to simple statements like "we set the power supply voltage to 1mV".

ETA: I think page 5 of this thread is around where Johan fell out of reach of the normal language of circuit analysis.

Well, I understand. I have read every post. But why bother dwelling on something that is trivial? Sort of like the cartoon, "someone is wrong on the internet!" Anyone who understands the arguments can decide for himself who is right and who is wrong.

I see the problem now as similar to the rebuttal of my posts. "Here is an exception to what you said, and I wish to imply that you therefore did not know beforehand. So your claim is void."

Dr. Prins is seeing this also, and it is true in many cases, hence his frustration. The arguments must be made purely within physics, and stated clearly, which has been done very often, but they are now becoming annoyingly repetitious. The insults are especially annoying on both sides.

I happen to agree generally with the physicists who are disagreeing with Dr. Prins, but I also see the occasional silliness that arises when they come up with a basically irrelevant exception to a claim which Prof. Prins really apparently had taken into consideration without having explicitly said so. That is pettiness.

I do not yet grasp what Dr. Prins means when he keeps insisting that one can somehow superimpose a voltage from a capacitor or another supply (battery) on an already superconducting segment or junction. I am not a PhD in physics, but this does not make any sense to me.

Nor do I understand Prof. Prins' problem with the acceleration issue, which I will not even attempt to reconstruct here. I do believe that one or more of his antagonists, as well as Johan himself, has misspoken on several occasions on this subject, further complicating the matter. It is now hopelessly confused for me at least.

I really hope this thread can relax the snarkiness a bit, and get into the science of the matter. If someone is wrong, then that is pretty much the end of the matter if he will not yield.

 

[sol invictus]

I happen to agree generally with the physicists who are disagreeing with Dr. Prins, but I also see the occasional silliness that arises when they come up with a basically irrelevant exception to a claim which Prof. Prins really apparently had taken into consideration without having explicitly said so. That is pettiness.

The problem is that Johan's grasp of the basics is so weak and so erratic that it's very difficult to know what he's leaving out because it's obvious, and what he's leaving out because he has something completely wrong.

Nor do I understand Prof. Prins' problem with the acceleration issue, which I will not even attempt to reconstruct here. I do believe that one or more of his antagonists, as well as Johan himself, has misspoken on several occasions on this subject, further complicating the matter. It is now hopelessly confused for me at least.

If you have any questions, just ask and we'll clear it up for you.

I really hope this thread can relax the snarkiness a bit, and get into the science of the matter.

There is no science to the matter, not unless we simply discuss superconductivity. Johan's ideas are utter nonsense - as far as I can tell they're based on a crude and vague mental picture of physics, with no math, no self-consistency, and incorrect predictions for many phenomena far more basic than superconductivity. When you couple that to his arrogance, paranoia, and inability to comprehend what he's told, it makes for an exchange that gets old very quickly.

 

[johanfprins]

comments on post 1055 by Ziggurat

Yes. You claim so every time that you claim polarization energy is stored inside a superconductor when there is no field.

PLEASE do the experiments outlined in post #964 and see for yourself that this is occurring within a superconductor. But of course God Ziggurat does not allow experimental results to interfere with what he wants to believe.

JFP: "I only claim that the field is cancelled at the positions of the charge-carriers"
Zig: If the field isn't canceled throughout the superconductor, then there's still a voltage. But you're claiming that there is no voltage, which means that the field is canceled everywhere.

There cannot be voltage once a current is flowing since the polarized charge-carriers short circuits the superconductor. Whether there is a voltage when no current is flowing, I do not know since I have not been able to do such an experiment.

Plus, of course, it still doesn't rescue you from the fact that even if cancellation happens only at the position of those charge carriers, those charge carriers still must have zero polarization energy, or else the field would not completely cancel at their location.

So you claim that when the force on the charge of the charge-carriers, caused by the applied electric-field, is balanced by a polarization force, the charge-carrier must have zero polarization energy. This is of course nonsense since if the polarization energy of the charge-carrier becomes zero there will not be a polarization force to oppose and cancel the applied electric force at the position of the charge-carrier.

Quote JFP: "Now you want to claim that the electric force at the position of the entity is another spring?'
Indeed: its energy scales as the square of its value (just like a spring's displacement), and its value varies in linear opposition to the value of the polarization (just like a spring compressing another spring). Mathematically, the relationship between electric field and polarization energy is exactly like the relationship between one spring compressing another spring. This isn't exactly hard, how do you keep failing at it?

And you then say after compression the polarization energy becomes zero. So you argue that if one uses a spring to compress another spring, then after reaching equilibrium you will have no energy stored in the spring that you compressed. Amazing logic!

If the total energy doesn't change, why would anything happen?

So when I step over the edge of a precipice I will be able to stay at that height standing in the air, since by falling down the "total energy will not change and why would anything then happen". Thanks I will not be so scared of heights anymore!

Without a change in energy, there is not force acting to make a change.

Are you now still talking about total energy as you have just been talking? Then you are talking BS.

Quote JFP: When stretching out an ideal spring with a force, equilibrium will be reached when the force generated by the elasticity of the spring equals the applied force".
Zig: And when does that happen for two springs in contact? Why, it happens when the total energy of the two springs is minimized! The two conditions are the same thing.

So you believe that when you apply a force to an object, it will only move when the total energy (potential plus kinetic energy) becomes minimized by this movement. Incredible!.

Quote JFP: " The work done by the applied force is then equal to the stored energy in the spring. No minimization of total energy is needed".Zig: You don't need to minimize energy if you can assume that one of your springs is infinitely stiff. That's clearly not relevant here, though.

So you are arguing that an applied electric field will never be able to pull negative and positive charges from each other since it is a "soft spring". Incredible!

Quote JFP: "The localized entities do not have to have zero spring constants for the applied field to be cancelled at their positions".
They do if you want them to be polarized while the field is canceled. No field, no polarization unless the spring constant is zero.

How can a dipole with a zero spring constant balance and cancel the applied field on it? No opposing force is possible with a zero spring constant which can balance the applied electric force. A zero spring constant requires that the force between opposite charges must be zero. God Ziggurat's new Coulomb interaction!?

Quote JFP: Whatever the polarized potential energies of the charge-carriers is on an electron-energy diagram, once a super-current flows, it short-circuits the superconductor and therefore the voltage must become zero".
Zigg: Which means no field. And it also means no polarization, unless the spring constant is zero.

You are thus arguing that once a current is flowing an externally applied field along the direction of current flow will not be able to polarize the charge-carriers. Your argument is contrary to experiment: Do the experiment outlined in post #964, and convince yourself. Personally I believe experimental results and not the logic of God Ziggurat.

In no case can you get a nonzero polarization energy when the voltage is zero.

But according to experiment you do; and therefore you are wrong. Not that any experimental result will ever swing your viewpoint: You are GOD and just know better.
__________________

 

[johanfprins]

Comments on post#1063 by Ziggurat

Originally Posted by johanfprins "So maybe it is not even possible: Maybe just as impossible as to expect from you to give a definition of what it actually is".
I just GAVE a definition. It's quite simple: superconductivity is zero resistance with a Meissner effect. Perfect conductivity is zero resistance without a Meissner effect.

Thus a wild dog is a dog which lives in nature and hunts. A house dog is a wild dog that does not live in nature and does not hunt. This does not tell me anything about what a dog actually is.

You are the only person who has a problem with this definition, because you choke on the concept of zero resistance for some strange reason, even though there is no reason to object to it.

Yes I do "choke" when a scientist does not define what he is talking about: You use "zero resistance" to define superconduction and ideal conduction without first defining what zero resistance is.

To try and bring some reality to this endless discussion, I am sure that you must all accept that when it comes to conduction and electrical resistance Drude's equation must be a starting point (The London brothers did so and their work is still highly appraised); i.e. that for charge carriers with mass m moving with a speed v at a time t, one can write that:

m*(d/dt)v+(m/(tau))*v= q*E.............(1)

Where (tau) is the relaxation time. Indeed London's so-called first equation is obtained by assuming that in a superconductor the second resistive-friction term can be set to zero (i.e. (tau)=infinity). But it is just dangerous to set terms in a differential equation equal to zero before having solved such an equation; and only then determining what happens when (tau) approaches infinity. Assuming that the speed is v=0 before the electric field is switched on, it is easy to derive that at time t the value of v must be:

v=((q*E*(tau))/m)*(1-exp(-t/(tau))....... (2)

In terms of the current density J for a charge-carrier density N, this equation leads to as:

J=((E*(tau)*N*q^2)/m)*(1-exp(-t/(tau)) (3)

Thus when t=0 J=0; and after equilibrium is reached for t>>>(tau) the equilibrium current is given by:

J_eq=E*((N*(tau)*q^2)/m) (4)

Which is Ohm's law with a conductivity:

(sigma)=((N*(tau)*q^2)/m) (5)

So the London brothers, and all superconduction experts since then, claim that superconduction with zero resistivity occurs when (tau) becomes very large; Which seems correct. But Eq. 3, will not settle down to reach an equilibrium value for the current. It is exactly for this reason why zero resistivity cannot be defined by Ohm's law. For large (tau) Eq. 3 cannot become Eq. 5: it becomes the following:

J=((N*E*q^2)/m)t (6)

The current will just keep on increasing indefinitely with time unless you add an external resistor to the superconductor. But the fact remains that without external resistors the current will not settle down to be a current with a constant drift speed : Therefore one cannot define a resistivity for the material by using Eq. 5 (i.e. by using Ohm's equation). And when using an external resistor, there is no physics reason why the electric-field within the conductor must become cancelled. Just as in any vacuum diode the charge-carriers will still accelerate all the way from the injection contact to the end-contact: And this demands the presence of an electric-field or else there will be no current.

Which is exactly what happens when charges are free to move about without resistance. They move to cancel the applied field. There is no mystery to this.

When the charges transport a current it is impossible for them to cancel the applied electric-field. As I have pointed out time and again, these charges then move in an attempt to cancel the applied electric-field but can never achieve such a cancellation since they cannot achieve a static redistribution of charges to generate a large enough polarization field to oppose the applied electric field..

It's exactly the same thing that happens to an ordinary conductor in an electric field, except it can happen in the presence of a current as well.

Not if the charge-carriers are the same charges that has to redistribute to cancel the applied electric field..

Well, yes. The charges also have to be free to move. And the charges here are NOT free to move except within the gap. There is a barrier to entering the gap (the work function) which prevents them from flowing fast enough to cancel the field.

The source of electrons at the cathode are present above the work function of the cathode and you can increase this source by heating the cathode to higher temperature. So there is no barrier involved to the number of charge-carriers.

Furthermore, you are arguing that the "Cooper pairs" have more freedom to move inside a superconductor than the electrons within a vacuum gap. Why would this be the case? I thought that Cooper pairs have even less freedom to move individually since they supposedly move "coherently".

 

[Dilb]

Thus when t=0 J=0; and after equilibrium is reached for t>>>(tau) the equilibrium current is given by:

J_eq=E*((N*(tau)*q^2)/m) (4)

Which is Ohm's law with a resistivity:

(rho)=((N*(tau)*q^2)/m) (5)

Yes, that well known Ohm's law, current equals resistance times voltage (I = V*R). Are you really making this mistake?

 

[W.D.Clinger]

J=((E*(tau)*N*q^2)/m)*(1-exp(-t/(tau)) (3)

...snip...

OK one can argue that I am being devious since when allowing (tau) to become large in Eq. 3, this equation will not settle down to reach an equilibrium value for the current. It is for exactly for this reason why zero resistivity cannot be defined by Ohm's law. For large (tau) Eq. 3 it becomes the following:

J=((N*E*q^2)/m)t (6)
The current will just keep on increasing indefinitely with time unless you add an external resistor to the superconductor. But the fact remains that without external resistors the current will not settle down so that one cannot define a resistivity for the material.

No. There's a mistake in the highlighted equation (6).

From (3):

[latex]
\begin{align*}
\hbox{{\bf J}} &= \frac{\hbox{{\bf E}} \tau N q^2}{m}\left( 1 - e^{-t / \tau} \right) \\
&= \frac{N q^2 \tau}{m} \hbox{{\bf E}} \left( 1 - e^{-t / \tau} \right) \\
&< \frac{N q^2 \tau}{m} \hbox{{\bf E}}
\end{align*}
[/latex]

So J does not increase indefinitely after all. As t becomes large, J converges to that last value.

At equilibrium, therefore, J equals that last value. Oddly enough, that's the second equation shown in the current Wikipedia article on the Drude model. Go figure.

 

[johanfprins]

Comments on post #1064 by Ben m

Have you ever heard of F=ma? And maybe F=qE? This covers all cases. Pick any electric field direction you want. Calculate F. That tells you the 2nd time derivative of the position vector. thereafter you look at the velocity, etc., and figure out for yourself what is orthogonal to what. (ETA: F, a, and E are vectors, but the latex tag was borked.)

Thanks! The way you argued I was wondering whether you know this. This is why I asked.

I think (but this is a question for an expert) that electron falls into a potential well when it forms a pair at the injection contact; this potential energy gets converted into kinetic energy.

Usually when an electron changes its energy by "falling" into a potential well the energy is radiated away.

This kinetic energy carries the electron through the superconductor at high speed.

Why will the two electrons then stay bonded as a Cooper pair?

At the extraction contact, the kinetic energy is just enough to get out of the pairing-based potential well again.

Really!? According to the accepted explanation of tunneling between a superconductor and a normal conductor it is argued that the Cooper pair breaks up into two normal charge-carriers when the electric-field in the junction accelerates the Cooper pair so that this increase in kinetic energy cancels the binding energy. Why is this needed in the junction if the Cooper pair already has a high enough kinetic energy within the superconductor before entering the tunneling junction?

Because Faraday's Law (invoked during the inductive part of the transient) does give you an electric field in a superconductor, and that electric field can accelerate charge carriers.

And after the transient is gone? What then accelerates the charges being injected?

If you want to know how the charge carriers get accelerated, you have to pay attention to the phenomena available to accelerate them.

Which phenomena? I have been asking for weeks on this thread for you to explain "these phenomena" to me; but you have not. An induced electric field during the initial transient (just after switching on the current) cannot be responsible for acceleration once an equilibrium current has been established.

 

[johanfprins]

Comment on post #1095 by W. D. Clinger

No. There's a mistake in the highlighted equation (6).

From (3):

[latex]
\begin{align*}
\hbox{{\bf J}} &= \frac{\hbox{{\bf E}} \tau N q^2}{m}\left( 1 - e^{-t / \tau} \right) \\
&= \frac{N q^2 \tau}{m} \hbox{{\bf E}} \left( 1 - e^{-t / \tau} \right) \\
&< \frac{N q^2 \tau}{m} \hbox{{\bf E}}
\end{align*}
[/latex]

So J does not increase indefinitely after all. As t becomes large, J converges to that last value.

You are wrong; To find the limit one must notice that the exponential stays near to unity for (tau) approaching infinity. Thus to obtain the limit you must expand the exponential function into its series expression: And then take the limit. This is a routine procedure in calculus: Do you not know this?

I did make a more serious mistake since I was a rush. I, however, corrected this mistake. The argument is still correct that Ohm's law cannot define zero resistance since the current cannot settle down for Ohm's law to become valid when (tau) goes to infinity.

 

[johanfprins]

Comments on post 1065 by Ben m

So, Johan, you've been arguing for 30 pages about a circuit whose IV curve you didn't know. I based my counterarguments on the correct IV curve and you basically called me an idiot nonstop.

What happened? Were you pretending to know the IV curve? Had you seen an IV curve once long ago and misremembered it? Were you assuming your theory was right, assuming BCS was wrong, and guessing what the IV curve must have been to jibe with those assumptions?

No I just made a mistake about Fig. 44-16. For an SIS to be superconducting while a voltage appears across the junction, this voltage appears for an "external reason". You can prove it to yourself by doing the experiments in post #964.

All the DC IV curves I have ever seen for a Josephson junction only show a super-current through the junction when V_J=0. You state that there will be an AC current as soon as a voltage appears over the junction, no matter how. Can you refer me to an IV diagram that shows the latter current for V_J being not zero anymore? All the IV characteristics I have ever seen show that the super-current immediately falls to zero as soon as a voltage appears across the junction.

And in fact most books I have read implies that you must place an "external voltage" over the junction while the SIS can still super-conduct in order to get radiation being emitted. So I am still of the opinion that one will deduce from EMF-balance that the voltage V_J across the junction implies a voltage 2V_J seen by the power supply. I am fairly sure that experimental measurements will vindicate my deduction that this must be so. Of course, anybody can make a mistake; even I can

 

[johanfprins]

Comment on post #1066 by sol invictus

Neither Zig nor ben nor I have said the part I italicized anywhere in this thread. In fact you've been told the contrary over and over and over and over and over and over. You think otherwise only because you are incapable of comprehending what you read. If you disagree, why not find a quote, Johan? It should be easy.

Well what do you then mean by saying that the behavior is only determined by the acceleration and not anything else? It implies that when the acceleration is the same as in a case where the charge-carriers radiate, then in the latter case the charge-carriers must also radiate. Does it not?

You've only now checked the basic I-V curve? The curve that defines the DC and AC Josephson effects? One of the most basic pieces of evidence for the theory of superconductivity? You didn't know what it was?

I wrongly assumed that Fig. 44-16, shows the AC effect on its own; since I have not seen a single IV curve that did show me a super-current for a voltage larger than zero; except Fig. 44-16. But in retrospect Fig. 44-16, does show the ac effect with an applied external EM radiation. It might thus still be relevant to my argument, unless you can refer me to an IV curve that shows a current higher than I_max after a voltage appears over the junction. The evidence I have is that the current will drop.

You're a complete fraud, Johan. Go play with your grandkids - but please, don't try to teach them any physics.

People on this thread have complained about insults being thrown at each other. I will thus refrain from responding to you in kind.

 

[W.D.Clinger]

No. There's a mistake in the highlighted equation (6).

From (3):

[latex]
\begin{align*}
\hbox{{\bf J}} &= \frac{\hbox{{\bf E}} \tau N q^2}{m}\left( 1 - e^{-t / \tau} \right) \\
&= \frac{N q^2 \tau}{m} \hbox{{\bf E}} \left( 1 - e^{-t / \tau} \right) \\
&< \frac{N q^2 \tau}{m} \hbox{{\bf E}}
\end{align*}
[/latex]

So J does not increase indefinitely after all. As t becomes large, J converges to that last value.

You are wrong; To find the limit one must notice that the exponential stays near to unity for (tau) approaching infinity. Thus to obtain the limit you must expand the exponential function into its series expression: And then take the limit. This is a routine procedure in calculus: Do you not know this?

I know the difference between taking the limit as t increases without bound and taking the limit as τ increases without bound. I stated the correct limit for t increasing without bound.

You seem to think you were taking the limit as τ increases without bound, but that limit does not exist: J goes to infinity as τ does. (ETA: The highlighted statement is incorrect; see correction in post #1106.) I guess you were trying to use that to argue that Ohm's law does not hold, but you got the resistance wrong, as dilb pointed out and as you admitted here:

I did make a more serious mistake since I was a rush. I, however, corrected this mistake. The argument is still correct that Ohm's law cannot define zero resistance since the current cannot settle down for Ohm's law to become valid when (tau) goes to infinity.

You corrected your mistake by changing the word "resistivity" to "conductivity".

Ohm's law says V=I*R at equilibrium. If R is zero and I finite, then V must be zero. If V were positive and R were zero, then I would be infinite, which is not physically possible.

(You could have said that without using Drude's model, taking limits, or talking about an irrelevant series expansion, but you like to make everything as complicated as possible. Some of us prefer to make everything as simple as possible, but no simpler.)

The conclusion I draw from Ohm's law is that there is no equilibrium for which V is positive and R zero. The conclusion you draw is that Ohm's law does not hold when R is zero. I think you're drawing the wrong conclusion.