Superconductivity at room temperature??

[johanfprins]

Again. Gauss' law always applies, Johan. Always. It's a law of physics.

I have never said anything else.

Edited by LashL: 

Removed personalization

If you deny that, which you do, you deny the validity of Gauss' law, and hence Maxwell's equations.

I have NEVER denied the validity of Gauss' law as long is it is NOT applied by nincompoops!

Edited by LashL: 

Removed personalization

 

[sol invictus]

Again. Gauss' law always applies, Johan. Always. It's a law of physics.

I have never said anything else.

Really?

Gauss' law does not apply in this case, as you have applied it, since there is no charge accumulation at the interfaces

Gauss' law is always true

But not at the surface of two metals with different resistances through which a current I flows.

'Nuff said.

 

[johanfprins]

'Nuff said.

Note how dishonest you are:

Originally Posted by johanfprins
Gauss' law does not apply in this case, as you have applied it, since there is no charge accumulation at the interfaces

As usual you lie by omission. That is the worst kind of lie ever: Lawyers do it and therefore everybody worldwide cheered whan T Rex devoured the lawyer in Jurrasic Park. I wish I could send you a T Rex.

 

[sol invictus]

Note how dishonest you are:

Originally Posted by johanfprins
Gauss' law does not apply in this case, as you have applied it, since there is no charge accumulation at the interfaces

As usual you lie by omission. That is the worst kind of lie ever: Lawyers do it and therefore everybody worldwide cheered whan T Rex devoured the lawyer in Jurrasic Park. I wish I could send you a T Rex.

No, Johan. From just the previous page:

Gauss' law is always true

But not at the surface of two metals with different resistances through which a current I flows.

And in any case, "the way you have applied it" was simply to apply it to a particular case - the interface of two metals with different resistivities and a current. You assert unequivocally that Gauss' law does not apply to that case. Thereby, you deny the laws of physics, and you are wrong.

 

[ben m]

You are posting senseless gobbledegook. In both cases you can calculate a divergence: And you have claimed that a divergence is ONLY possible when charges accumulate. Now you are backpedalling like hell!

Johan, you not only don't know how to calculate the divergence, you don't know what it is.

You can always calculate what the divergence is. Sometimes you find out that it is zero. The divergence can only be nonzero where there is a net charge density. The divergence, which remains well-defined and calculable, is equal to zero when the charge density is zero.

I (like everyone else) sometimes describe div.F = 0 using words like "there is no divergence". I did not mean "divergence equations don't apply" or "there is no such thing as a divergence."

It's like solving a force law problem and saying "there is no force"---do you interpret this as nonsense? "You claimed to use force laws, now you say there is no force!"

Johan, I repeat: is there a real live human who knows calculus that you can get to explain this to you?

 

[ben m]

It is quite clear what I meant when considered in context: i.e. that the change in electric-field is not caused by charge-accumulation:

Yep. In the Fe-Cu-Fe case, the change in electric vector field (considering both strength and the direction) computes as div.E != 0, i.e. that the divergence is not zero. According to Gauss's Law, this happens only when rho != 0, i.e. when there is a nonzero charge accumulation.

Are you still saying that div.E = 0 in this case? If so, you're just making a vector calculus mistake (and not for the first time). Are you saying that div.E doesn't say anything about charge accumulation? Then you're saying Gauss's Law is wrong (and not for the first time.)

 

[Ziggurat]

We are back to the same dishonest tricks you played at the beginning of this thread. It is quite clear what I meant when considered in context: i.e. that the change in electric-field is not caused by charge-accumulation

Evidently it's caused by magic.

What exactly do you think creates electric fields?

And if you use Gauss' law to derive that this is the case, you are using Gauss' law in an invalid manner.

There are applications of Gauss's law which may not be useful, or may not tell you what you need to know, but the law itself is always valid, so the only way to use Gauss's law in an invalid manner is to do the math wrong.

Which you did. Because you couldn't calculate a divergence correctly.

Unfortunately I was born with an affliction: My long term memory is severely impaired.

 

[sol invictus]

. It is quite clear what I meant when considered in context: i.e. that the change in electric-field is not caused by charge-accumulation: And if you use Gauss' law to derive that this is the case, you are using Gauss' law in an invalid manner.

I think you really, honestly, truly just don't get it.

If the electric field has a non-zero divergence somewhere, there is always a charge accumulation there. Always. No exceptions, ever. Why? Because that's what Gauss' law says.

If the electric field has zero divergence somewhere, there is no charge accumulation there. Ever. No exceptions.

And yes, you have repeatedly denied this, and I have repeatedly quoted you doing so.

 

[Myriad]

I tried asking this before, but I was ignored. I still think it might be a good way to establish some basics, so I'll try again.

Johan, you're familiar with what Faraday cages do, and how they work, correct?

So my question is, if you run a small DC current through a Faraday cage, does it stop functioning as a Faraday cage? (Let's say, for example, that the Faraday cage is cubic, and ungrounded, and the current is applied from one corner of the cube to the extreme opposite corner.)

If your claim that no local accumulation of charge can occur where a current is flowing is correct, then a Faraday cage, which normally cancels out externally applied electric fields by the local relocation and accumulation of charge within the conductive cage to offset the external field, should stop working, and no longer shield the inside from external electric fields, when the DC current through the cage is turned on. Do you predict that's what would happen, in an actual test?

Respectfully,
Myriad

 

[johanfprins]

I tried asking this before, but I was ignored. I still think it might be a good way to establish some basics, so I'll try again.

Johan, you're familiar with what Faraday cages do, and how they work, correct?

I hope so: As an aside I hope that I understand a Farady cage better than Aharanov and Bohm did.

So my question is, if you run a small DC current through a Faraday cage, does it stop functioning as a Faraday cage? (Let's say, for example, that the Faraday cage is cubic, and ungrounded, and the current is applied from one corner of the cube to the extreme opposite corner.)

Here I need more information from you. To run a current from one corner to the diagonal opposite corner, there must be a potential difference between the corners. The idea of a Farday cage is that the potential energy is everywhere the same on the Faraday cage. So before I try and answer the rest of your question, can you be more specific on how you intend to generate such a current. Are you suggesting that one should generate electrons in a vacuum at one corner and then accelerate them to the opposite corner? If this is so, how do you propose that it should be done?

 

[Myriad]

I hope so: As an aside I hope that I understand a Farady cage better than Aharanov and Bohm did.

Here I need more information from you. To run a current from one corner to the diagonal opposite corner, there must be a potential difference between the corners. The idea of a Farday cage is that the potential energy is everywhere the same on the Faraday cage. So before I try and answer the rest of your question, can you be more specific on how you intend to generate such a current. Are you suggesting that one should generate electrons in a vacuum at one corner and then accelerate them to the opposite corner? If this is so, how do you propose that it should be done?

Well, if I actually did the test I was planning to use a regulated current source, let's say at 100 mA. But while we're still in the thought experiment domain, let's say a battery and a resistor.

And rather than go to all the trouble of making a hard vacuum around my test bench, and have to go digging out my space suit and all, I was planning to use some wires to connect the current source (regulated supply or battery) to the Faraday cage. (I studied electronics in Horowitz' course, and if I recall correctly he was a great advocate of using elongated pieces of conductive material for moving electric currents from place to place.)

Respectfully,
Myriad

 

[johanfprins]

Well, if I actually did the test I was planning to use a regulated current source, let's say at 100 mA. But while we're still in the thought experiment domain, let's say a battery and a resistor.

And rather than go to all the trouble of making a hard vacuum around my test bench, and have to go digging out my space suit and all, I was planning to use some wires to connect the current source (regulated supply or battery) to the Faraday cage. (I studied electronics in Horowitz' course, and if I recall correctly he was a great advocate of using elongated pieces of conductive material for moving electric currents from place to place.)

I still do not quite understand: Is the source and resistor outside the Faraday cage and are you then connecting the source to a corner of the Faraday cage? Is there then a connection at the other corner of the Faraday cage so that the current is flowing within the Farady cage itself from one corner to the other? Not directly from corner to corner through the "interior" of the Faraday cage?

 

[Myriad]

I still do not quite understand: Is the source and resistor outside the Faraday cage and are you then connecting the source to a corner of the Faraday cage? Is there then a connection at the other corner of the Faraday cage so that the current is flowing within the Farady cage itself from one corner to the other? Not directly from corner to corner through the "interior" of the Faraday cage?

Yes.

Sorry, I thought that was clear. The DC current is flowing through the metal of the Faraday cage itself, from an external power supply, not through the interior. (I'm not sure why "interior" is in scare-quotes in your question, since the definition of the interior of a Faraday cage should be clear-cut. If it helps clarify, this cage can be made of solid metal sheets instead of mesh. Assume for now the interior is filled with air at standard temperature and pressure; later we will also place electric field measuring apparatus inside the cage if needed).

Respectfully,
Myriad

 

[LashL]

We have already had to edit and/or split multiple posts in this thread earlier today and have now had to split several more to Abandon All Hope again. Please, folks, cut out the personalizations, name-calling, and bickering and stick to the topic - which is not other members of the forum. Thank you in advance for your anticipated cooperation.

Replying to this modbox in thread will be off topic  Posted By: LashL

 

[Reality Check]

I have claimed that there cannot be any charge accumulation at the interfaces while an ohmic current is flowing and that it is then wrong to invoke Gauss' law (as if a free electron current is flowing through vacuum) to explain the change in electric-field when passing through the boundary.

You would be still be wrong

• Gauss's law does not really explain any 'change in electric-field'.
  It relates a distribution of electric charge to a spacially varying electric field. I would say that the electric field is explained by the charge distribution.
• Gauss's law does not just apply when 'a free electron current is flowing through vacuum'.
• Gauss's law can be invoked no matter what current is flowing through what.
  Saying that there is a situation where charges exist and Gauss' law cannot be evoked is the same as saying that Gauss' law is invalid because Gauss's law can be evoked whenever charges exist.
  Gauss's law can even be evoked when charges do not exist but that is rather trivial .

The simple science that everyone is trying to teach you is that Gauss's law can always be evoked.

 

[johanfprins]

You would be still be wrong

• Gauss's law does not really explain any 'change in electric-field'.
  It relates a distribution of electric charge to a spacially varying electric field. I would say that the electric field is explained by the charge distribution.
• Gauss's law does not just apply when 'a free electron current is flowing through vacuum'.
• Gauss's law can be invoked no matter what current is flowing through what.
  Saying that there is a situation where charges exist and Gauss' law cannot be evoked is the same as saying that Gauss' law is invalid because Gauss's law can be evoked whenever charges exist.
  Gauss's law can even be evoked when charges do not exist but that is rather trivial .

The simple science that everyone is trying to teach you is that Gauss's law can always be evoked.

Please stop posting the obvious. You are not amusing at all.

 

[johanfprins]

Yes.

Sorry, I thought that was clear. The DC current is flowing through the metal of the Faraday cage itself, from an external power supply, not through the interior. (I'm not sure why "interior" is in scare-quotes in your question, since the definition of the interior of a Faraday cage should be clear-cut. If it helps clarify, this cage can be made of solid metal sheets instead of mesh. Assume for now the interior is filled with air at standard temperature and pressure; later we will also place electric field measuring apparatus inside the cage if needed).

Respectfully,
Myriad

OK. Now I understand your question. Usually, as I understand it, a Faraday cage must be everywhere at the same potential in order to screen an outside electric-field. How a current from one corner to the other will affect the screening is not clear to me. I expect that there will not be any accumulation of charge anywhere owing to a DC-current flowing through the cage itself, even if the cage consists of mesh. But if you add an external electric field while the current is flowing I do not know. I suspect that the geometry of the cage and the direction of the external field should play a role. It will be an interesting experiment though! If you do it let me know the result.

In this regard, it might be in some respects a similar situation than for a MOS transistor; where the current flows from the source to the drain without accumulation: But when switching on the gate, you can pinch off the current. I hope this is of help?

I did not intend "scare" quotation marks. Obviously the interior of a Faraday cage is the holy part of the cage.

 

[ben m]

OK. Now I understand your question. Usually, as I understand it, a Faraday cage must be everywhere at the same potential in order to screen an outside electric-field. How a current from one corner to the other will affect the screening is not clear to me.

Just as an exercise, to demonstrate that you're serious: please for a moment adopt the mainstream assumptions, like what your E&M textbooks tell you:

a) Gauss's Law is true everywhere, without exceptions
b) Maxwell's Equations are linear and obey the superposition principle

and re-analyze the situation under those assumptions. Can you do that? It's pretty easy.

 

[johanfprins]

Just as an exercise, to demonstrate that you're serious: please for a moment adopt the mainstream assumptions, like what your E&M textbooks tell you:

a) Gauss's Law is true everywhere, without exceptions

I have NEVER denied this: But it must be correctly applied. What you call charge-accumulations at interfaces might be dynamic polarisation right through the conductor. I do not want to go further into this, because we will then again spend endless posts insulting each other. What is more important is to return to modelling superconduction. Questions like: What are the charge-carriers and how do they gain kinetic energy to convey a current without dissipating this energy; are more important.Iit is also important to admit that within a conductor these issues are largely determined by quantum mechanics. We have not yet grabbed this nettle completely when it comes to conducting materials (in fact any materials: Wannier functions and the Hubble model are far from being REALLY satisfactory): Not even in the case of the most perfect metals like gold and copper. Although there are very good clues in the work of Mott and Anderson which I am trying to follow up when I do not spend my time trying to defend my integrity on this thread.

b) Maxwell's Equations are linear and obey the superposition principle

I have never denied that Maxwell's wave equations obey the superposition principle.

and re-analyze the situation under those assumptions. Can you do that? It's pretty easy.

I have, and am still doing so, but it is not as "pretty easy" as you are thinking. I am still of the opinion that most of your analyses on conducting materials are wrong since you think that it is similar to free electrons moving through a vacuum. I can, however, promise you that if I find that you are correct, I will acknowledge it, no matter how many more insults you, Sol and Ziggurat will then throw at me. I want to understand the REAL physics involved; and physics is HARD WORK! You need opposition! But you, Sol and Ziggurat are sometimes going overboard. Nonetheless, I admit that I do owe you thanks since you have raised arguments which are very challenging indeed. It is good physics to argue these differences. But as a Solid State Physicist who have spend many hours thinking about conduction in materials (and as my citation index proves have made very good contributions in this field), I do not agree with the way in which you have invoked Gauss' theorem within a conductor. It is a bit more complicated as you want to believe. If it is not, I will be the first to concede and lick my wounds.

I need some rest: Im am only one person, and you are at least three on this thread who are in opposition.

 

[ben m]

Can I suggest a simple setup with the same physics as the Faraday cage, but for which the analysis is simpler?

            _/\/\/\__       
 Vext--||--|   R     |--||-- ground          
           |         | 
           |---||----|
              "cage"

We have a capacitor-gap which I call the "cage"; a conducting path around it has a small resistance "R" (like the walls of a Faraday cage). We may confirm that this is indeed a normal Faraday cage. You can put whatever electric field you want to outside of the cage (Vext), but since such fields are only capacitively coupled to the isolated RC loop, the field across the "cage" is zero no matter what Vext is. That's the point of a Faraday cage.

You can confirm that the "conducting path around the cage" is necessary: if you remove the R from the circuit, breaking that path, you're left with

 Vext--||--|         |--||-- ground          
           |         | 
           |---||----|
              "cage"

where the center capacitor is no longer a Faraday cage, and which DOES feel the effects of Vext. So we confirm that the Faraday cage works only when its walls are conductive.

Now we'll add a battery Vbatt, connected to the cage, such that current flows through the walls. This is Myriad's current-carrying Faraday cage, in circuit form:

           ___(+|-)__
           |  Vbatt  |
           |         |
           |_/\/\/\__|      
 Vext--||--|   R     |--||-- ground          
           |         | 
           |---||----|
              "cage"

Let's note the details: if (Vbatt=0) it's a normal old Faraday cage. If Vbatt is nonzero, it looks like current flows through the "wall" R but also creates a field inside the "cage". The question is: NOW, what is the effect of Vext? Does it change the field in the central capacitor? ("because the current is flowing, the Faraday cage stops screening external fields")? Does Vext have the effect of turning off the loop current ("it's like a field effect transistor, the current cannot go past the stationary charges that screen Vext")?

Analyze away. This is obviously simpler than any sort of 3D field cage you might be considering, and I hope you agree that the electromagnetic principles are the same.