Superconductivity at room temperature??

[Ziggurat]

I did not see an acceptable derivation at all anywhere!

You don't find them acceptable because they don't give the answer you want, not because you can find any errors in them. Meanwhile, your own derivations, which are riddled with errors, you found acceptable because they DID give the answer you wanted, even though it wasn't even the same answer each time.

 

[Myriad]

The incontrovertible physics fact reamains that charge cannot accumulate enough anywhere within a conductor to cancel an appliecd electric-field while a current is flowing through any conductor which has "free" charge-carriesr that can be accelerated.

So, if you put a small DC current through a Faraday cage (say, from one corner to the diagonally opposite corner of a cubic metal enclosure, at one amp), it would no longer act as a Faraday cage?

Respectfully,
Myriad

 

[ben m]

No you are using "physics" which nobody has ever heard of

Look back at the thread, Johan. I use Gauss's Law (everywhere), Ohm's Law (in the iron), Newton's Laws of Motion (for charges that can't scatter, i.e. in the superconductor) and that's it.

I remind you that the posited solution obviously obeys Gauss's Laws and all of the boundary conditions---you've made no objection to that. (ETA: although you tried to.) It's obviously an equilibrium solution---you don't even seem to care. Your objections seem to be hoping to find "Newton's Laws will prevent you from arriving at this solution" but you certainly haven't shown that.

(Even your weird calculation, had it been correct, wouldn't have shown that. If you have a calculation that says "under force X, current diverges" then that would be what happens. Current divergence would cause a time-dependent charge density, which would modify the fields, and you'd iterate your calculation.

You don't say "ahh! I found a calculation that says the current is not constant in space, therefore I reject the premises of the calculation" so the setup is irrelevant.

 

[johanfprins]

Post by reality check 1907

Now is is ben m, Sol, Ziggurat and me who are pointing out the obvious and physically possible

• A continuous function has a gradient.

• I agree, but (rho) is NOT actually a continuous function since the charge-carriers scatter; and therefore (rho) can only be approximated as being a continuous function when the distances that matter during differentiation are FAR, FAR larger than the scattering length.
  

  Thus a continuous resistivity-function even right up to the interface has a gradient.

  It is NOT continuous right up to the interface since the assumption of continuity fails long before the distance to the interface is equal to the scattering length s from the interface.
  

  A step function (through the interface) has a gradient, i.e. the Dirac delta function.

  Not in this case!
  

  The situation does not change if there is a weld between the interfaces. It is even more physically possible because there is no Dirac delta function.

  Then calculate for me the change in (rho) in terms of the scattering length within the depth of the weld. You are being ridiculous.

This is basic mathematics that is taught in undergraduate courses: You can find the gradient of both a continuous function and a step function.

No! Basic mathematics is based on the assumption that (rho) is really continuous with position. The fact is, however, that the (rho) function within a conductor is NOT really continuous for scattering distances larger than zero. And can thus never be really continuous since scattering distances are never zero.

It seems your professors in New Zeeland never taught you when a function is REALLY continuous and under which conditions it can be approximated as being continuous even if on the micro-scale it is not continuous. And I thought that our academic institutions in South Africa are bad!

 

[ben m]

Y'know, I was about to start typing up a detailed analysis of the stability of the standard solution. But I'm not going to waste my time.

 

[Ziggurat]

Look back at the thread, Johan. I use Gauss's Law (everywhere), Ohm's Law (in the iron), Newton's Laws of Motion (for charges that can't scatter, i.e. in the superconductor) and that's it.

Indeed. One can treat the interface region as a black box, never calculate what the resistivity or its gradient is within this region, and still arrive at the incontrovertible conclusion that johan is wrong, based purely on what happens outside this region.

 

[sol invictus]

I did not see an acceptable derivation at all anywhere!

There isn't one acceptable to you in this thread, that much is clear. I suspect no such derivation exists at all. No matter how explicit the math, how detailed the explanations, and how perfectly consistent with the known laws of physics and the experimental data it is, you will not believe it if it goes against your faith.

There is no point in discussing physics with you, Johan, because you are not a physicist, or even a scientist. You have the opposite mindset - you will not accept any evidence contrary to your viewpoint. You are embittered and angry to the point that you cannot think rationally at all.

I think you once were an intelligent person, so it's actually rather sad. As I have said before, I urge you to go and play with your grandchildren, and leave science to people still capable of rational thought. Your one-man crusade is unhealthy for you, deeply misguided, and most likely painful and embarrassing to your family. Let it go.

 

[Reality Check]

I agree, but (rho) is NOT actually a continuous function since the charge-carriers scatter; and therefore (rho) can only be approximated as being a continuous function when the distances that matter during differentiation are FAR, FAR larger than the scattering length.
...snipped gibbersh and insults...

There is no such thing as 'distances that matter during differentiation'. Differentiation is a mathematical operation that uses infinitesimals.

The examples that have been given have specific functions for rho. They can all be differentiated.

Give me your function for rho (including the step function for abutting interfaces or another function for a weld).

The simple physics is that rho is continuous withing the materials and so can be differentiated. At the interface between the materials

• If they abut then rho is a step function and can be differentiated.
• If they are welded then rho is a continuously changing function and can be differentiated.

The rest of your post is gibberish because what you are stating is that rho can never be differentiated because of some waffle about scattering distances. You are basically saying that Gauss's law can never be applied to any real situation (maybe even Maxwell's equations).

 

[ben m]

Johan, let me help you at least clarify your line of argument. The (mainstream) claim is that this configuration can exist for an iron/SC/iron device with a constant current flowing through it.

(iron, field E) ( - charge) (SC, field 0) (+ charge) (iron, field E)

My claim is that this has the following properties:

• It obeys Maxwell's Equations, if the charge sheets have area density E/epsilon0. The charge sheets are responsible for changing the field from E to zero.
  
• It is a stable configuration. No force wants to either stop or accelerate the moving charge carriers from "cruising" through the zero-field superconductor, and no force wants to displace the charge sheets.
  
• This configuration is actually arrived at when turning on the E field for the first time. No matter what the details are, any nonzero E field generates a larger current in the SC than in the iron. Therefore there's a current divergence (div.J) at each boundary, and therefore there's dQ/dt with the correct sign to generate the E-field-cancelling charge excess. This remains true until the E field in the superconductor is zero.

Can you clarify: what point of this are you disagreeing with? Do you think that charge sheets of E/epsilon0 do not cancel the E field? Do you think that there's a force on the charge sheets, or the charge carriers in the SC, while it's at E=0? Or do you think that this configuration would work if you could get there, but that you never actually get there?

 

[johanfprins]

I did not run away; but unfortunately picked up a very bad bout of flu. Thus, as the Terminator said: "I'll be back!'. I believe that the correct analysis of Ben's IRON-COPPER-IRON sandwich will prove where his, Sol's and Ziggurat’s misconceptions are when it comes to electric-currents and Ohm's law. The latter formula is of course NOT A LAW but a purely an empirical relationship which is only valid, especially in its intrinsic format, under very special conditions. So see you all later after I feel better.

 

[Ziggurat]

The latter formula is of course NOT A LAW but a purely an empirical relationship which is only valid, especially in its intrinsic format, under very special conditions.

At yet, you appealed to Ohm's law as absolute plenty of times in the past. It's funny how physics becomes either immutable or an invalid approximation, depending on your needs of the moment.

 

[ben m]

Before Johan launches in to his sure-to-be-interesting objections to Ohm's Law, let me remind him that the iron-SC-iron sandwich has the electric fields I claim it does (nonzero-zero-nonzero), while a DC current is flowing, as a matter of experimental fact. I do not need Ohm's Law to tell me what fields to plug into Gauss's Law. The fields I plug into Gauss's Law are the ones that are experimentally known to be there.

I would also point out that Ohm's Law agrees with the experimentally-known fields. (E = J rho, so rho=0 implies E=0.) If Ohm's Law doesn't apply because the microphysics is wrong, I guess we got lucky because all of the field analysis is correct anyway.

I would also point out that Newton's Laws agrees with the experimentally-known fields. (absent scattering, F = qE = ma = m/q dI/dt, so dI/dt = 0 implies E=0) If Newton's Law doesn't apply because the microphysics is wrong, I guess we got lucky because all of the field analysis is correct anyway.

 

[sol invictus]

I did not run away; but unfortunately picked up a very bad bout of flu. Thus, as the Terminator said: "I'll be back!'. I believe that the correct analysis of Ben's IRON-COPPER-IRON sandwich will prove where his, Sol's and Ziggurat’s misconceptions are when it comes to electric-currents and Ohm's law. The latter formula is of course NOT A LAW but a purely an empirical relationship which is only valid, especially in its intrinsic format, under very special conditions. So see you all later after I feel better.

Ohm's law is only valid under special conditions? What an amazing coincidence that the last century of development of electronics wasn't hampered in the slightest by using wrong laws of physics!

I guess it's just pure dumb luck that this computer I'm using functions at all. After all, it contains lots of junctions between materials of different conductivities, and the engineers that design it assumed Ohm's laws hold. I guess we just keep winning the lottery over and over.

How many laws of physics are you willing to abandon to cling desperately to your belief that you're right and we're wrong, Johan? You already abandoned conservation of energy, Maxwell's equations, calculus, and now Ohm's laws (how many did I miss?).

 

[Reality Check]

How many laws of physics are you willing to abandon to cling desperately to your belief that you're right and we're wrong, Johan? You already abandoned conservation of energy, Maxwell's equations, calculus, and now Ohm's laws (how many did I miss?).

The laws of quantum mechanics

• The Dirac equation which he has asserted is not relativistically invariant and does not describe an electron correctly.
• He also has asserted that the solution to Schroedinger's equation for a single particle is not a solution for a single particle!
  As I recall, he thinks that the wave equation is an actual physical wave and so a particle is an actual wave packet.

 

[Ziggurat]

(how many did I miss?).

Conservation of momentum and basically all of standard quantum mechanics. And probably more.

 

[johanfprins]

SUMMARISING THE ARGUMENTS BASED ON BEN’S IRON-COPPER-IRON

To make sure that I understand the arguments, which have been raised on this thread in the case of the iron-copper-iron device, I am going to summarise them as I “understand” them from your “impeccable” posts.

I have been consistently accused on this thread as “having had it” and that I must accept that “you (meaning I) will never recover” from my Alzheimer’s fate. Since this seems to be the overwhelming opinion on this thread I must obviously consider such a majority-opinion seriously. Thus, I am now asking you (the people on this thread who have made these accusations) to help me to correct my “deteriorating understanding” of physics: Please read carefully what I, in “my mentally retarded state”, concluded from your arguments and amend my exposition where you do not agree with it. Your help will hopefully stop me from becoming a total nincompoop; like the majority of modern physicists are at present: At least all of those who are in control of modern physics. If you can de-ossify my mind, it must surely be a service to physics. So please indulge an old man in his dotage.

We have a sandwich consisting of iron-copper-iron. There are two boundaries A (when moving from the left-hand side out of the iron into the copper) and B (when moving from the right-hand side out of the copper into the other iron element).

In terms of the internal electric field one has within the iron elements that the electric field is E_IR and within the copper that the electric-field is E_CU, so that:

E_IR >E_CU----------------------(1)

Since there is an electron-current with density J_e flowing from left to right, the electric fields within the materials are pointing from right to left: Thus, the electric-field E_IR leaving the iron at boundary B is reduced to E_CU within the copper. You claimed that this is a result of Gauss’ law which thus mandates that there must be a sheet of negative charge at the boundary B in order for this reduction to occur. When the electric-field within the copper E_CU leaves the copper, at boundary A, it is increased to become E_IR on the other side within the iron. You claimed that in this case Gauss’ law also applies, and now mandates that there must be a sheet of positive charge at boundary A. Is this correct?

Furthermore, within the two materials one can write in terms of the intrinsic format of Ohm’s law that:

For copper (rho)_CU*J_e=-E_CU-----(2)
And
For iron (rho)_IR*J_e=-E_IR--------(3)

You claimed that the “resistivity-field” (rho) is differentiable right up to, and through the boundaries A and B; so that one can calculate a gradient (d/dx) of the resistivity (rho) when passing through such a boundary; even if this calculation requires one to invoke Dirac’s delta-function. Thus, you claim that:

[(d/dx)(rho)*J_e=-(d/dx)E------------------------------------------(4)

Is this correct?

And since (d/dx)E must be caused by a sheet of charge at such an interface, this charge-accumulation is actually caused by the gradient of the resistivity (rho): i.e. if this gradient in resistivity were not there, there would be no charge-accumulation at all; and thus no change in the electric field across the interfaces A and B. When replacing the copper by iron, there will be no gradient in the resistivity at the interfaces A and B, and therefore no accumulation of charge at these interfaces; and thus no change in the electric-field. There will thus only be a change in the electric-field with position if one can “tune down” the resistivity of the central element to be below the resistivity of iron: The increase in sheet-charge, causing the concomitant change in the electric-field, must thus increase proportionally to the reduction in resistivity of the central element below the resistivity of the iron. Is this correct?

Do you all agree with this summary? Or am I too senile to understand your “spot-on” logic? If I am too retarded, then please amend my exposition in this posting so that we can reach an agreement on what you are actually, really arguing.

 

[ben m]

We have a sandwich consisting of iron-copper-iron. There are two boundaries A (when moving from the left-hand side out of the iron into the copper) and B (when moving from the right-hand side out of the copper into the other iron element).

In terms of the internal electric field one has within the iron elements that the electric field is E_IR and within the copper that the electric-field is E_CU, so that:

E_IR >E_CU----------------------(1)

Good so far. Don't know what you mean by "elements that".

Since there is an electron-current with density J_e flowing from left to right, the electric fields within the materials are pointing from right to left: Thus, the electric-field E_IR leaving the iron at boundary B is reduced to E_CU within the copper.

In the past I've been talking about the electric field going from left to right, which means the current goes from left to right, which means the electrons are moving right to left. When I say "current" I always mean the ordinary electromagnetic current, the one that appears in Maxwell's Equations. Since electrons are negatively charged, the direction of electron motion is the opposite of the direction of the current.

This is the mirror image of the situation you describe above. Either one works, I'll try to keep them clear.

You claimed that this is a result of Gauss’ law which thus mandates that there must be a sheet of negative charge at the boundary B in order for this reduction to occur. When the electric-field within the copper E_CU leaves the copper, at boundary A, it is increased to become E_IR on the other side within the iron. You claimed that in this case Gauss’ law also applies, and now mandates that there must be a sheet of positive charge at boundary A. Is this correct?

Given your electric field definition (E points leftwards) this is correct.

Furthermore, within the two materials one can write in terms of the intrinsic format of Ohm’s law that:

For copper (rho)_CU*J_e=-E_CU-----(2)
And
For iron (rho)_IR*J_e=-E_IR--------(3)

The solution is the same whether or not you assume this to be true.

You claimed that the “resistivity-field” (rho) is differentiable right up to, and through the boundaries A and B; so that one can calculate a gradient (d/dx) of the resistivity (rho) when passing through such a boundary; even if this calculation requires one to invoke Dirac’s delta-function. Thus, you claim that:

[(d/dx)(rho)*J_e=-(d/dx)E------------------------------------------(4)

Is this correct?

Not quite.

a) Can you draw a Gaussian surface with one face to the left of A, and the other face to the right of A? Can you put both of these surfaces somewhere where you know the fields? (I hope so, because you just told me that the field in the iron is E_IR and the field in the copper is E_CU, so you claim to know the fields somewhere.) Stop right there. Draw your Gaussian surface such that its faces are both in known fields. Now you're ready to evaluate the integral form of Gauss's Law. The functional form of E right at the interface never comes into it.

b) IF the E field in fact drops as a perfect, mathematical step function, THEN indeed the charge density has to be a perfect delta-function sheet. Notice the words "if" and "then". If you think the delta-function-shaped sheet is unphysical, then stop supposing that the E field drops as a perfect step function. If the E field drops over any finite distance, then the charge volume density is finite-but-nonzero over that finite distance.

c) If there were a delta function, and you were too lazy to zoom in and see the rapidly-changing-but-not-actually singular function close to the interface: it doesn't make a difference to Gauss's Law. The only thing you will ever do with that charge-density is to integrate it over a volume, and that integral is just as sensible (and finite) over a delta function as it is over a smooth function.

d) I don't care whether the E field comes from Ohm's Law or not. Gauss's Law works on E fields and charges, not currents and resistivities. If it's not possible for the resistivity to be a step function, such that the E field is not a step function, fine. Not one iota of the analysis requires a step function.

And since (d/dx)E must be caused by a sheet of charge at such an interface, this charge-accumulation is actually caused by the gradient of the resistivity (rho): i.e. if this gradient in resistivity were not there, there would be no charge-accumulation at all; and thus no change in the electric field across the interfaces A and B.

Fine. (Again, with the caveat that any source of an E-field-difference---whether perfectly-Ohms-Law-like or otherwise---has the same effect.)

When replacing the copper by iron, there will be no gradient in the resistivity at the interfaces A and B, and therefore no accumulation of charge at these interfaces; and thus no change in the electric-field.

True.

There will thus only be a change in the electric-field with position if one can “tune down” the resistivity of the central element to be below the resistivity of iron: The increase in sheet-charge, causing the concomitant change in the electric-field, must thus increase proportionally to the reduction in resistivity of the central element below the resistivity of the iron. Is this correct?

If you mean that the charge areal density Q/A = epsilon0 * (E1-E2) = epsilon0 * J *(rho1 - rho2), then yes. The first equality is just Gauss's Law, the second equality is Ohm's Law. If Ohm's Law doesn't apply for some reason, Gauss's Law still does.

 

[johanfprins]

Good so far. Don't know what you mean by "elements that".

I simply stated that within the two iron elements you have that the electric-field is E_IR and within the copper it is E_CU; and at, or within the surface this field must change. The rest of your arguments do not seem to contradict what I have stated: Namely that you are claiming that the accumulation of charge at the interfaces is caused by the gradient of the resistivity; which, in turn. causes the electric-field to change from E_IR within the iron to _CU within the copper at boundary B and that E_CU changes into E_IR when crossing boundary A. So state it again, do you agree with this?

In the past I've been talking about the electric field going from left to right, which means the current goes from left to right, which means the electrons are moving right to left.

This did confuse me since we are talking about negative charges being injected into the copper which then, according to you, coast from the injection contact to the target contact. Can we thus keep the left-hand contact, between the iron and copper (interface A), as the injection contact?

a) Can you draw a Gaussian surface with one face to the left of A, and the other face to the right of A? Can you put both of these surfaces somewhere where you know the fields? (I hope so, because you just told me that the field in the iron is E_IR and the field in the copper is E_CU, so you claim to know the fields somewhere.) Stop right there. Draw your Gaussian surface such that its faces are both in known fields. Now you're ready to evaluate the integral form of Gauss's Law. The functional form of E right at the interface never comes into it.

I have not contradicted this from the time you finally stated it cogently: A rare experience when trying to follow your arguments.

b) IF the E field in fact drops as a perfect, mathematical step function, THEN indeed the charge density has to be a perfect delta-function sheet. Notice the words "if" and "then". If you think the delta-function-shaped sheet is unphysical, then stop supposing that the E field drops as a perfect step function. If the E field drops over any finite distance, then the charge volume density is finite-but-nonzero over that finite distance.

I have not suggested anything else, did I?

c) If there were a delta function, and you were too lazy to zoom in and see the rapidly-changing-but-not-actually singular function close to the interface: it doesn't make a difference to Gauss's Law. The only thing you will ever do with that charge-density is to integrate it over a volume, and that integral is just as sensible (and finite) over a delta function as it is over a smooth function.

I thought that this is what you tried to say in your clumsy manner all along.

d) I don't care whether the E field comes from Ohm's Law or not. Gauss's Law works on E fields and charges, not currents and resistivities.

But your BIG argument is that you must have a gradient in the resistivity for such charges to accumulate at the interfaces. And that the larger the gradient, the larger the accumulation of charge will be.

If it's not possible for the resistivity to be a step function, such that the E field is not a step function, fine. Not one iota of the analysis requires a step function.

OK so the charge need not be a singular sheet: Correct?

If you mean that the charge areal density Q/A = epsilon0 * (E1-E2) = epsilon0 * J *(rho1 - rho2), then yes. The first equality is just Gauss's Law, the second equality is Ohm's Law. If Ohm's Law doesn't apply for some reason, Gauss's Law still does.

Well, I thought you used this argument to explain how it is possible to have enough charge-accumulation so that the electric-field is cancelled and THEREFORE Ohm's law for "zero-resistivity" applies. Are you now backtracking on this explanation?

 

[ben m]

I simply stated that within the two iron elements you have that the electric-field is E_IR and within the copper it is E_CU; and at, or within the surface this field must change. The rest of your arguments do not seem to contradict what I have stated: Namely that you are claiming that the accumulation of charge at the interfaces is caused by the gradient of the resistivity; which, in turn. causes the electric-field to change from E_IR within the iron to _CU within the copper at boundary B and that E_CU changes into E_IR when crossing boundary A. So state it again, do you agree with this?

Fine. (Just don't get hung up on the Ohm's Law detail when we get back to superconductors for which Ohm's Law is of limited use.)

This did confuse me since we are talking about negative charges being injected into the copper which then, according to you, coast from the injection contact to the target contact. Can we thus keep the left-hand contact, between the iron and copper (interface A), as the injection contact?

Fine. Henceforth the left contact is the "electron injection" contact, and the field (and current) are pointing right to left.

But your BIG argument is that you must have a gradient in the resistivity for such charges to accumulate at the interfaces. And that the larger the gradient, the larger the accumulation of charge will be.

Careful. In this 1-D setup, the gradient of resistivity (times the current density) is the volumetric charge density. Coulombs per cubic meter in SI units. I'd reserve terms like a "large accumulation of charge" for the area charge density, Coulombs per square meter, which is the integral of the volumetric density in the direction normal to the boundary.

Gauss's Law, for any Gaussian surface enclosing the entire boundary, cares about the area charge density, not the volumetric charge density.

Therefore Gauss's Law cares about the integral of the gradient of the resistivity. The integral of the gradient is just the total change in resistivity.

So: the larger the gradient at any point, the larger the volumetric charge density at that point. The larger the net change in resistivity, the larger the area charge density (i.e the "accumulation of charge") in the boundary region.

Well, I thought you used this argument to explain how it is possible to have enough charge-accumulation so that the electric-field is cancelled and THEREFORE Ohm's law for "zero-resistivity" applies. Are you now backtracking on this explanation?

I'm not backtracking at all. I'm trying to be excruciatingly careful with definitions so that you don't have any confusing points to latch onto.

The point on which I am being pedantic is this: please do not argue using statements like "you said Ohm's Law is generally valid in superconductors! Drude and London say you are a moron!". Really, don't do it. I am making a Gauss's Law argument involving electric fields. Ohm's Law is a useful way of setting up that argument. In the iron-copper-iron situation,, in equilibrium, Ohm's Law has valid microphysics and gives the right fields. In the iron-SC-iron situation, in equilibrium, Ohm's Law gives the right fields whether or not it is valid microphysics.

You are likely to find this confusing, which is why I'd much rather talk about the fields.

 

[johanfprins]

Fine. (Just don't get hung up on the Ohm's Law detail when we get back to superconductors for which Ohm's Law is of limited use.)

Thanks for admitting this!

I am making a Gauss's Law argument involving electric fields. Ohm's Law is a useful way of setting up that argument. In the iron-copper-iron situation,, in equilibrium, Ohm's Law has valid microphysics and gives the right fields. In the iron-SC-iron situation, in equilibrium, Ohm's Law gives the right fields whether or not it is valid microphysics.

This is where we have a difference in opinion since there is never an electric-field within a superconductor while a current is flowing through it: And this must also be the case when a current is increasing through the superconductor.

You are likely to find this confusing, which is why I'd much rather talk about the fields.

Good let’s do exactly that: Let us at first ignore the iron-SC-iron case and see whether we can get some agreement on the iron-copper-iron case. I decided to proceed with thought experiments in future. This post is thus the first thought experiment. Other supporting ones will be posted as the discussion progresses.

Let us now return to the iron-copper-iron sandwich: Since the copper has a resistivity (rho)_CU it also has a resistance R_CU, which for a length L and cross-sectional area A₀ follows as:

R_CU=(L/A₀)*(rho)_CU

We are now going to replace the copper element, which has a resistivity (rho)_CU, with an iron element which has a resistivity (rho)_IR>(rho)_CU. BUT we modify the iron element to have a total resistance R_IR which is exactly equal to the resistance of the copper element R_CU that it replaces.

One way to do this would be to simply increase the cross-sectional area A_IR of the iron to be larger than the cross-sectional area A₀: Or to decrease its length. It is, however, more informative to keep the length L the same and to increase the cross-sectional area. If one does the latter uniformly, the interfaces at positions A and B between the external iron-wings and the new central iron element do not match; as was the case before replacing the copper-element. Although this can be handled mathematically, it complicates the discussion: We will here rectify this mismatch by shaping the new central iron element so that the interfaces do match at positions A and B. There are many ways in which this can be done.

Consider generating a bulge in the new iron-element: Assuming that for the iron-copper-iron sandwich one had a circular cross-sectional area A₀ with a radius R₀, then the new, bulging central iron-element can be chosen to have a radius R(x) where x measures the distance along the sandwich. If one chooses the origin of the x-axis at the centre of the bulging iron element, one must have at the interface position A, where x=-(L/2), that the radius must be R(-L/2)=R₀, and at the interface position B, where x=(L/2), that the radius must also be R(L/2)=R₀. At x=0, the bulge is chosen to reach a maximum so that R(0)=R_MAX. How the bulge is shaped can be freely chosen, as long as one ends up with R_IR-BULGE=R_CU. For example, one can choose a parabolic function, which we will do here, so that:

R(x)=R_MAX-(ALHA)*x²Where
(ALPHA)=(2/L)²*(R_MAX-R₀)

Thus, within the new bulged iron-element the cross-sectional area is given by:

A(x)=(PI)*R².

When a current I flows, the electron current-density J_e(x) at position x within the bulged iron-element is given by:

J_e(x)=I/A(x)

From the intrinsic format of Ohm’s law one must thus have for the electric-field E(x) at position x that:

E(x)=-(rho)_IR*J(x)

When now following the electric-field from the right-hand contact, it is E_IR until it reaches the interface B, whence it starts to “diverge” after it enters the central iron element. This “divergence” steadily decreases to become zero at x=0, whence it increases again (converges) to again become E_IR at the interface A. According to your reasoning invoking Gauss’ law, this mandates that there is an excess of negative charge for x>0 while there is an excess of positive charge for x<0. So, are you arguing that the bulged element is a colossal dipole that forms around the position x=0?

I, however, doubt it very much that such a distribution of charge can form. It is unlikely that free electrons will stack themselves as sheets when x>0 while being absent so that positively-charged sheets will be stacked when x<0: Especially while they are conveying a current I through the whole bulged element. What is the real reason why the electric-field is “diverging” and “converging” within the bulged iron element? This happens since charge cannot accumulate when a current I is flowing: The reason why the electric-field has to change with position x is thus exactly the opposite to the one you are claiming. Since the total current I must be the same at every position x, the current density J_e must change with position: And since the density of charge-carriers is the same everywhere, the current density can only change by a change in the drift-speed: And since the drift speed is determined by the applied electric-field within the material the electric-field E must decrease when the drift speed must decrease, and it must increase when the drift speed must increase to keep I the same at all positions x.

Let us now calculate the “divergence” (d/dx)E(x), from the last equation above. It is given by:

(d/dx)E(x)=-(rho)_IR*(d/dx)J_e

Note that this “divergence” is not caused by any gradient in the resistivity; since the resistivity is now everywhere the same. Thus, one can generate a “divergence” of E without having a gradient in the resistivity. One only needs a change in overall geometry. Your argument that one must have a gradient in resistivity is thus wrong! And, in fact, as I have pointed out time and again above, the resistivity function (rho) is an “unphysical” creation to obtain and intrinsic expression for Ohm’s law, but it is not a real continuous function which can be differentiated with position under all conditions: especially not near a boundary.

This mistake of assuming a really continuous resistivity (or conductivity) has also been incorporated into Maxwell’s equations and this has led to the misconception that a current carried by “electrons’ through a conductor is the same as a current of equal intensity carried by free electrons through a vacuum under the action of an electric-field. As I have stated time and again on this thread, these currents are not the same.

In fact, the “divergence” or “convergence” of an electric-field within a conductor cannot be compared with the divergence or convergence of an electric-field in vacuum.

There must thus be another reason why the electric field changes when moving from the iron into the copper. Since at the iron-copper interface, the sudden decrease in resistance will cause an increase in the total current I if the electric-field does no change, this increase can only be prevented by a decrease in the electric-field E. This has nothing to do with the accumulation of any charge at the boundary. It has to do with the cooperative flow of wave-packets (pseudo-electrons) through a metal-conductor. The decrease in electric-field within the copper is caused by the very fact that when a current I flows through a circuit, consisting of different conductors, this current must have the same value I at every point x along the circuit: Therefore, just as much charge per unit time must always appear on the other side of a boundary-volume than is simultaneously flowing into this boundary-volume. To repeat: Your so-called “divergence” in E within a conductor is caused by a cooperate effect which is, in turn, mandated by the requirement that the current I must remain the same along the circuit. It has nothing-nothing to do with an accumulation of charge anywhere: Gauss’ law does not play any role in this case.