Superconductivity at room temperature??

[ben m]

Yes, but you must apply Gauss’ law correctly, and then you will find that at the iron-copper interfaces there CANNOT be any accumulation of the charge carriers at all.

You're wrong. I applied it correctly and got the correct answer, which is that there is a charge accumulation.

You seemed to have decided, long before you ever thought about this iron-copper issue, that there would be no charge. Now you find that plain-vanilla Gauss's Law doesn't confirm your expectation, so you're trying to make it complicated.

Now let us enclose this interface within a Gaussian volume and measure the amount of charge that flows per unit time from the copper into this volume and the amount of charge per unit time that flows out of this volume into the iron. Guess what? The amounts are the same. Thus according to Gauss’ law, THERE CAN BE NO ACCUMULATION OF CHARGE-CARRIERS WITHIN THIS VOLUME.

Too bad this wasn't Gauss's Law. The divergence theorem for charge, which we've stated many times, is div.J = drho/dt. When the current in and the current out are the same, the enclosed charge is constant in time.

(ETA: for example, draw a loop bisecting any capacitor in a DC circuit. Draw a surface around a a Van de Graaff that's leaking corona current. Etc. It's utterly trivial to get div.J = 0 and rho > 0.)

Gauss's Law is div.E = rho/epsilon0.

So why does the electric-field change across the interface: Exactly for the same reason why the pressure changes when water flows from a larger diameter pipe into a smaller diameter pipe:

Nope, we've been through that too. Flow from a large pipe to a small pipe is a divergence = 0 flow, as expected.

When you apply Gauss’ theorem correctly you will find that this drop in electric field-strength has NOTHING to do with the accumulation of charge-carriers at the interface.

Clarify: are you saying div.E = 0, or are you saying div.E doesn't matter?

 

[johanfprins]

I'm not clear on how your argument would deal with the simpler case of a parallel-plate capacitor in an RC circuit. If I draw a surface enclosing only one of the plates, what happens?

We are not talking about a displacement current between two capacitor plates: We are talking about a DC current flowing through an interface between two different metals.

Depending on my attitude towards displacement currents, either I'm forced to conclude that something other than the flow of of physical charge can balance things out or that not every volume needs to be balanced at all times. Whatever the interpretation, charge accumulates on the plate.

In this case yes, since there are not actual free charge-carriers flowing through the space between the two capacitor plates. Only the two poles of a battery can act like two oppositely-charged capacitor plates through which a current can actually be injected through one pole into the space between the two poles and then ejected at the other pole.

When you have conductors with different properties in series between two battery-poles, the free charge-carriers within these conductors cannot accumulate at any surface through which they flow between the two battery-poles: This is so since the total current I reamains the same along the whole circuit between the battery-poles. This means that at any interface between the two battery-poles through which the current I flows, just as many charge-carriers must be entering such an interface as are leaving this interface on the other side. Thus constructing a Gaussian surface around any such interface demands that charge-carriers cannot accumulate within such a Gaussian volume.

The change in electric-field when moving through an interface between two different metals can thus not be caused by an accumulation of charge-carriers at the interface between the two conductors. Thus, the change in electric field strength from one conductor to the other while a current is flowing, is not caused by Gauss' law for electric-field lines.

 

[sol invictus]

The fact is that if the inflow of any substance per unit time into a Gaussian volume is equal the outflow of the same substance per unit time out of the Gaussian volume, then Gauss' law mandates that there can be no accumulation of that substance within the Gaussian volume: Whether it is incompressible water or charge or "what have you".

You have confused electric field lines with electric current. It is not Gauss' law that guarantees what you say above, it is conservation of charge (or water, or whatever substance).

Gauss' law is something else - it relates the charge to the electric field, not the charge to the current. It says nothing about the current.

 

[johanfprins]

You're wrong. I applied it correctly and got the correct answer, which is that there is a charge accumulation.

You did not apply it correctly since, when applied correctly, you will find that just as much charge flows into the Gaussian volume as is flowing out of the Gaussian volume. This means that according to Gauss’ law there can be no accumulation of free charge-carriers at the interface.

You seemed to have decided, long before you ever thought about this iron-copper issue, that there would be no charge.

I did not “decide” it since it must be so from Gauss’ law itself that when the same amount of charge is flowing into a volume per unit time as is flowing out of a volume per unit time, there can be no charge accumulation within the volume.

Now you find that plain-vanilla Gauss's Law doesn't confirm your expectation,

The plain-vanilla Gauss’ law is exactly confirming what I said and what everybody knows who understands real physics.

so you're trying to make it complicated.

What is so complicated to understand? It is plain logic that when the inflow and outflow into a Gaussian volume match, there can be no accumulation of what is flowing througg the Gaussian volume.

Too bad this wasn't Gauss's Law. The divergence theorem for charge, which we've stated many times, is div.J = drho/dt. When the current in and the current out are the same, the enclosed charge is constant in time.

Correct! This is exactly what you have when a current is flowing around a circuit. The charge stays the same within every volume. What is so strange about this?

Gauss's Law is div.E = rho/epsilon0.

For electric field lines, yes: But Gauss’ law is not just valid for electric field lines.

Nope, we've been through that too. Flow from a large pipe to a small pipe is a divergence = 0 flow, as expected.

A non-zero divergence without accumulation of the material that flows into and out off a Gaussian volume, or without a sink or source within the Gaussian volume through which the material is flowing? This is not possible.

Clarify: are you saying div.E = 0, or are you saying div.E doesn't matter?

What I am saying is that the non-zero change in the electric field across such an interface is not caused by the accumulation of charge at the interface. Thus, claiming that it is caused by a “divergence” which is, in turn, caused by charge-accumulation within the Gaussian volume, as one will have for an electric field in vacuum entering and leaving a Gaussian volume with a stationary charge distribution within it, is wrong. Gauss' own law, applied to charge-flow, does not allow the required accumulation of extra charge at the interface.

 

[ben m]

Johan, let me give you some resources to understand your mistake.

There are two laws of physics we're talking about.

The conservation of charge ( http://en.wikipedia.org/wiki/Continuity_equation ) tells us a true fact about current or charge entering and leaving a volume. Gauss's Law ( http://en.wikipedia.org/wiki/Gauss's_Law ) tells us a true fact about electric field flux entering and leaving a volume.

Both are laws of physics; both are true all the time. They're separate, one does not override the other or replace it or cancel it.

 

[johanfprins]

Gauss' law is something else - it relates the charge to the electric field, not the charge to the current. It says nothing about the current.

Oh but it most surely does: If the charge density within the volume does not change with time, Guass' law for the current states that (div)J=0. When you have (div) of anything you have Gauss' law for that thing.

 

[johanfprins]

Johan, let me give you some resources to understand your mistake.

There are two laws of physics we're talking about.

The conservation of charge ( http://en.wikipedia.org/wiki/Continuity_equation ) tells us a true fact about current or charge entering and leaving a volume. Gauss's Law ( http://en.wikipedia.org/wiki/Gauss's_Law ) tells us a true fact about electric field flux entering and leaving a volume.

Both are laws of physics; both are true all the time. They're separate, one does not override the other or replace it or cancel it.

Yes they are both laws of physics: But if extra charge cannot accumulate within your Gaussian volume across the interface while a current is flowing, then the change in electric field across the interface while a current I is flowing cannot possibly be caused by a non-zero divergence as mandated by Gauss' law. The reason for this change has another explanation.

 

[ben m]

Oh but it most surely does: If the charge density within the volume does not change with time, Guass' law for the current states that (div)J=0. When you have (div) of anything you have Gauss' law for that thing.

I will only use the words "Gauss's Law" to refer to the specific one in Maxwell's Equations.

There are other laws that also use the Gauss-Ostrogradsky Divergence Theorem. One of them is the charge continuity equation.

 

[sol invictus]

Oh but it most surely does: If the charge density within the volume does not change with time, Guass' law for the current states that (div)J=0. When you have (div) of anything you have Gauss' law for that thing.

Nonsense. Find one reference anywhere that refers to the continuity equation as "Gauss' law".

The continuity equation says

div.J=-d(rho)/dt

while Gauss' law says

div.E=rho.

 

[johanfprins]

I will only use the words "Gauss's Law" to refer to the specific one in Maxwell's Equations.

There are other laws that also use the Gauss-Ostrogradsky Divergence Theorem. One of them is the charge continuity equation.

I am happy with that. It makes more sense than Sol's typical claptrap: i.e.

Nonsense. Find one reference anywhere that refers to the continuity equation as "Gauss' law".

The continuity equation says

div.J=-d(rho)/dt

while Gauss' law says

div.E=rho.

I am just repeating that no matter what you want to call the law, say Gauss-O, it mandates that there can be no accumulation of free charge carriers at the interface of two metal conductors in series. Thus the change in electric-field cannot be ascribed to Gauss' law for electric field-lines. It is more akin, but not totally so, to the drop in pressure when water flows from a larger pipe into a smaller pipe, causing a drop in pressure: It relates to energy conservation.

 

[sol invictus]

I am just repeating

Indeed, you are simply repeating the same false and unsupported assertion over and over again.

 

[ben m]

Johan, I think you're also getting (as you've done many times in this thread) the transient case and the DC case confused. (ETA: and I think you're getting them confused on purpose, because you need to find something to get confused about in order to avoid admitting you were wrong.)

How did charge get onto this interface? When you turned the E field on, there was (briefly) a time when J_FE < J_CU. This is div.J != 0, so during this time drho/dt !=0, and thus charge accumulates. (ETA2: During this time div.E = rho is still true. There are no exceptions, remember.)

By the time we get to the equilibrium, the charge is already there. In equilibrium, we find that rho != 0 and div.E != 0 (Gauss's Law) and div.J = 0 and drho/dt = 0 (charge continuity).

the change in electric field across the interface while a current I is flowing cannot possibly be caused by a non-zero divergence as mandated by Gauss' law. The reason for this change has another explanation.

Sorry, Johan, we calculated the divergence of E. It was nonzero. There are no "other ways to do it".

You might as well be saying, "I know that f(x) changes, from f(0) = 10 to f(1) = 11, but this must not be due to a derivative." Or "I know that there is area between f(x) and the x-axis, but this must not be due to the integral." It's utter lunacy to think that there is a loophole here.

 

[ben m]

But if extra charge cannot accumulate within your Gaussian volume across the interface while a current is flowing, then the change in electric field across the interface while a current I is flowing cannot possibly be caused by a non-zero divergence as mandated by Gauss' law. The reason for this change has another explanation.

I'm still just floored by this.

You're sitting there looking at a trivially simple vector field, and you're looking at 10 reposts and restatements of correct calculations of the divergence, and ... well, I have a hard time even describing what you're doing.

You're flatly denying that the calculation means anything.

You're accepting all of the premises and denying that they lead to the result.

In fact, I believe this is my first sighting in the wild of Achilles-Tortoise logic

"Well, now, let's take a little bit of the argument in that First Proposition -- just two steps, and the conclusion drawn from them. Kindly enter them in your notebook. And in order to refer to them conveniently, let's call them A, B, and Z: --
(A) Things that are equal to the same are equal to each other.
(B) The two sides of this Triangle are things that are equal to the same.
(Z) The two sides of this Triangle are equal to each other.
Readers of Euclid will grant, I suppose, that Z follows logically from A and B, so that any one who accepts A and B as true, must accept Z as true?"
"Undoubtedly! The youngest child in a High School -- as soon as High Schools are invented, which will not be till some two thousand years later -- will grant that."
"And if some reader had not yet accepted A and B as true, he might still accept the sequence as a valid one, I suppose?"
"No doubt such a reader might exist. He might say 'I accept as true the Hypothetical Proposition that, if A and B be true, Z must be true; but, I don't accept A and B as true.' Such a reader would do wisely in abandoning Euclid, and taking to football."
"And might there not also he some reader who would say 'I accept A and B as true, but I don't accept the Hypothetical '?"
"Certainly there might. He, also, had better take to football."

Johan accepts that div.E = rho is true, and that (surface integral of E flux) = (volume integral of div E) is true, and that a surface integral of E was nonzero, but doesn't accept that these three add up to rho being nonzero.

Football, Johan?

 

[johanfprins]

Johan, I think you're also getting (as you've done many times in this thread) the transient case and the DC case confused. (ETA: and I think you're getting them confused on purpose, because you need to find something to get confused about in order to avoid admitting you were wrong.)

No it is you who have been hiding behind the transient field and non-transient field all along. When one switches on the electric-field it spreads with the speed of light through the conductor: This is so fast that for conductors, of normal lengths, the charge-carriers can be assumed to start moving simultaneously. Thus, even if the current is still transient, it is transient simultaneously along all positions along the conductor. Charge cannot accumulate anywhere: The charge-carriers all have the same charge and, even if some might try to accumulate, they cannot since they will be pushed along by the charge-carriers behind them; even during the time that the current has not yet reached an equilibrium value.

The most important fact is that through any interface normal to current flow, even within a single material, just as much current is flowing into the interface as out of the interface: The free charge-carriers can never on average accumulate to be at a higher density than they were before the electric-field was switched on; not even during the transient stage.Thus, along the whole circuit, no matter out of how many different resistive elements it consists and no matter what the cross-sectional areas of these elements are, charge cannot accumulate anywhere ever.

How did charge get onto this interface? When you turned the E field on, there was (briefly) a time when J_FE < J_CU. This is div.J != 0, so during this time drho/dt !=0, and thus charge accumulates. (ETA2: During this time div.E = rho is still true. There are no exceptions, remember.)

Unadulterated BS.

By the time we get to the equilibrium, the charge is already there.

The charge that is within any volume is the same right from the start: It is the same as it was even before the current starts to flow. It remains the same whether a transient current or an equilibrium current is flowing. It does not accumulate at any position to be more than it was before a current started to flow.

In equilibrium, we find that rho != 0 and div.E != 0

No we find that E changes across the interface between two different conductors even though there is no excess charge-accumulation at the interface.

Sorry, Johan, we calculated the divergence of E. It was nonzero. There are no "other ways to do it".

Are you sure that you calculated the "divergence", or just the change in E across the interface? For example, when you have the same material which suddenly changes its cross section to a much smaller value, the current density changes across the interface, but this change is NOT a divergence of the current since you still have the same total current flowing into and out of the interface-surface.

You might as well be saying, "I know that f(x) changes, from f(0) = 10 to f(1) = 11, but this must not be due to a derivative." Or "I know that there is area between f(x) and the x-axis, but this must not be due to the integral." It's utter lunacy to think that there is a loophole here.

We are talking about "divergence" and Gauss' law, not about normal derivatives and integrals. It seems that you do not understand what "divergence" really means! Why does this not surprise me?

Bye for now

 

[ben m]

Are you sure that you calculated the "divergence", or just the change in E across the interface?

Yes. I calculated the divergence div.(E), taking into account the magnitude and the direction of E correctly, and it was nonzero.

For example, when you have the same material which suddenly changes its cross section to a much smaller value, the current density changes across the interface, but this change is NOT a divergence of the current since you still have the same total current flowing into and out of the interface-surface.

Yes. In this case I also calculated the divergence of E, taking into account the magnitude and the direction, and the divergence was zero. Again, this is absurdly simple, standard vector calculus.

You've already seen the explicit equations---for both cases, in fact---and you couldn't find an error in them. (ETA: http://www.internationalskeptics.com/forums/showpost.php?p=7528633&postcount=1991 and http://www.internationalskeptics.com/forums/showpost.php?p=7528937&postcount=1992 ) Try again.

 

[sol invictus]

Are you sure that you calculated the "divergence", or just the change in E across the interface? For example, when you have the same material which suddenly changes its cross section to a much smaller value, the current density changes across the interface, but this change is NOT a divergence of the current since you still have the same total current flowing into and out of the interface-surface.

Yes, Johan - that was explained to you a few posts back, after you illustrated that you didn't understand it.

Suppose we have an electric field E in the x-direction, with zero y and z components. The x-component of E depends on x as in the following diagram:

   E = E0    E=E0/2       E=E0
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ---->

In symbols, E_x=E₀ for x<0, E_x=E₀/2 for 0<x<L, and E_x=E₀ for x>L, and E_y=E_z=0 for all x, y, and z.

Math question: what is div.E?

Are you capable of answering that, Johan?

 

[dasmiller]

deleted

 

[johanfprins]

Yes, Johan - that was explained to you a few posts back, after you illustrated that you didn't understand it.

Suppose we have an electric field E in the x-direction, with zero y and z components. The x-component of E depends on x as in the following diagram:

   E = E0    E=E0/2       E=E0
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ----> 
 ---> --->     -->    ----->   ---->

In symbols, E_x=E₀ for x<0, E_x=E₀/2 for 0<x<L, and E_x=E₀ for x>L, and E_y=E_z=0 for all x, y, and z.

Math question: what is div.E?

Are you capable of answering that, Johan?

How do you expect me to answer anything if you do not tell me why E_x = E₀/2 between x=0 and x=L. You see why I abhor your posts.

Ben's last post has some flesh, but it still does not prove that free charge-carriers accumulate at the interface A of the iron-copper-iron sandwich; which is what you all have claimed all along: The latter is not possible when a current is flowing through the interface, whether the current is transient or not! The reason for the divergence of the electric-field must be something else than the accumulation of free charge-carriers at interface A. That is what we have to solve in this discussion.

 

[ben m]

How do you expect me to answer anything if you do not tell me why E_x = E₀/2 between x=0 and x=L. You see why I abhor your posts.

Why do you need to know that? Divergence is just a certain sum of derivatives of vector components.

If I asked you for the derivative df/dx of f(x) = x^2, do you have to ask "why" f changes value? No, df/dx = 2x no matter whether f(x) represents a spring energy, a trajectory, a hill, a lens, etc.

The reason for the divergence of the electric-field must be something else than the accumulation of free charge-carriers at interface A.

ETA: in spite of your insistence to the contrary, this IS SAYING THAT GAUSS'S LAW IS WRONG. Clear as a bell. Gauss's Law says that rho = div.E, you are saying that "something else than rho" = div.E, i.e you are saying rho != div.E, i.e. you are saying Gauss's Law is wrong.

 

[dasmiller]

When one switches on the electric-field it spreads with the speed of light through the conductor: This is so fast that for conductors, of normal lengths, the charge-carriers can be assumed to start moving simultaneously. Thus, even if the current is still transient, it is transient simultaneously along all positions along the conductor. Charge cannot accumulate anywhere: The charge-carriers all have the same charge and, even if some might try to accumulate, they cannot since they will be pushed along by the charge-carriers behind them; even during the time that the current has not yet reached an equilibrium value.

The most important fact is that through any interface normal to current flow, even within a single material, just as much current is flowing into the interface as out of the interface: The free charge-carriers can never on average accumulate to be at a higher density than they were before the electric-field was switched on; not even during the transient stage.Thus, along the whole circuit, no matter out of how many different resistive elements it consists and no matter what the cross-sectional areas of these elements are, charge cannot accumulate anywhere ever.

So, charge cannot accumulate anywhere ever if a current is flowing. And if current is not flowing, then charge carriers are (by definition) not moving and therefore not accumulating anywhere. Does it follow that the charge of a piece of metal can never change?