Superconductivity at room temperature??

[sol invictus]

I have not denied the validity of Gauss' law when there is an accumulation of charge. I just claim that it does not apply when there is not an accumulation of charge.

Gauss' law is a law of physics, Johan. It is always true. If you deny that it applies to any situation whatsoever, then you deny Gauss' law.

There really isn't any point in continuing a discussion about physics with someone that doesn't accept physics. And of course in your case, you only don't accept things when they contradict your beliefs, which is a particularly unscientific (and, frankly, stupid) approach.

 

[Ziggurat]

I have not denied the validity of Gauss' law when there is an accumulation of charge. I just claim that it does not apply when there is not an accumulation of charge.

It always applies, johan. Always. Even when there is no charge anywhere, Gauss's law remains valid. That's what makes it a law.

Once again, you are denying the fundamentals of physics, without even understanding that you're doing it.

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[ben m]

I have not denied the validity of Gauss' law when there is an accumulation of charge. I just claim that it does not apply when there is not an accumulation of charge.

Why not narrow it further, Johan? "Gauss's Law applies when it is not disproving my theory. It does not apply otherwise." That takes care of edge cases, like my office. (My office is chock-full of "not an accumulation of charge", and Gauss's Law works just fine, thanks.)

 

[johanfprins]

It always applies, johan. Always. Even when there is no charge anywhere, Gauss's law remains valid. That's what makes it a law.

Then why did Ben claim that it does not apply within my bulging iron element? If it always applies when the electric-field changes with position, then surely it must also be the case within my bulging iron element. It is really you guys who are claiming that you have the inalienable right to decide when Gauss' law applies and when not.

Now answer this question, when the electric-field within a conductor changes with position, for example changes in strength and direction, can it only happen when Gauss' law applies? YES or NO?

Once again, you are denying the fundamentals of physics, without even understanding that you're doing it.

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It is YOU who are doing this since you cannot even understand what an ohmic circuit is ...

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[D'rok]

Then why did Ben claim that it does not apply within my bulging iron element?

WTF?

 

[ben m]

Since the change of electric-field across the interface of your copper-iron thought experiment is not caused by a "divergence" which is caused by accumulated charge. In fact the change in electric-field is specifically caused by the fact that there cannot be an accumulation of charge.

Good heavens, I thought we had gotten you to concede this one. Is it really as simple as that? You simply don't know what divergence means. When someone says "this field has a divergence and this one does not", you can't tell the difference.

Do you have access to a human being who knows vector calculus, Johan? Go ask that person to explain it to you.

 

[ben m]

Then why did Ben claim that it does not apply within my bulging iron element? If it always applies when the electric-field changes with position, then surely it must also be the case within my bulging iron element. It is really you guys who are claiming that you have the inalienable right to decide when Gauss' law applies and when not.

Johan: the divergence of a vector field is a specific quantity

div F = dFx/dx + dFy/dy + dFz/dz.

It is not "any field that changes strength".

For a clean, neutral example: Look at this field:

F = 1/r^2 in the outward, radial direction. (Look familiar? It's the field we expect from a point charge.)

That field changes strength everywhere---it's strong for small r and weak for large r---but has zero divergence except at r=0. Gauss's Law therefore says that the charge density is nonzero at r=0, and zero everywhere else.

In your iron-wire-with-a-bulge example, the field varies both in magnitude and in direction. Both the magnitude changes and the direction changes go into that divergence calculation.

In my iron-abutting-copper example, the field varies in strength but is in a constant direction. The magnitude change goes into the divergence calculation and the direction drops out.

They're different fields, Johan. They have different shapes and different vector-calculus properties, including different divergences, like I said.

What the heck did you think I was doing? I pulled the word "divergence" out of a hat, then made up values for it just to shoot down your theory?

 

[Ziggurat]

Then why did Ben claim that it does not apply within my bulging iron element?

If he had, he would have been wrong. But he never made any such claim. In contrast, you have repeatedly made the claim that it doesn't always apply, you've been wrong every single time you make the claim, and we have repeatedly pointed to quotes where you explicitly make such a claim.

If it always applies when the electric-field changes with position, then surely it must also be the case within my bulging iron element.

It does apply. You just haven't done the calculations correctly. As Ben pointed out, the divergence of a field with x, y and z components cannot be calculated by looking at (d/dx)E_x. This is basic vector calculus, and you failed.

Now answer this question, when the electric-field within a conductor changes with position, for example changes in strength and direction, can it only happen when Gauss' law applies? YES or NO?

This is a nonsensical question. Gauss's law applies always, period.

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[D'rok]

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This thread is starting to make me sad. In the last page or so, Johan seems to have...

1) denied that Gauss' law applies universally;
2) denied doing so; and
3) claimed that Ben did so

...in that order and all in the space of a few posts.

Yikes.

 

[Reality Check]

I never denied this: All I am saying is that it is not responsible for the change of the electric-field across that iron-copper boundary since there is not an accumulation of charge at the boundary while a current is flowing through the two ohmic elements.

Well Duhhh !
Gauss's law does not cause the distribution of charge. No one thinks that.

It is a law about any distribution of charge caused by anything.
Your continuous mention of ohmic elements and accumulation of charge does not maiter.
Gauss's law still applies (you should know this but just in case):

Gauss's law may be expressed in its integral form:

where the left-hand side of the equation is a surface integral denoting the electric flux through a closed surface S, and the right-hand side of the equation is the total charge enclosed by S divided by the electric constant.
Gauss's law also has a differential form:

where ∇ · E is the divergence of the electric field, and ρ is the charge density.

See: just charge and current density and electric fields. No mention of the nature of the material (no 'ohmic element').

Thus you need to either

• Find an error in ben m's use of Gauss's law or
• Accept that his use is correct.

And no ranting about 'accumulation of charge acros the boundary', etc. You should know that Gauss's law still applies. Otherwise you are denying that Gauss's law applies, i.e. that Maxwell's equations are wrong.

 

[ben m]

If Johan wants a math, we're talking about the following two fields. I represent strong fields as "=" and weaker fields as "-". I type in "X" to show the Gaussian surface.

    XXXXXXXX
    X      X
====X===---X----
====X===---X----
====X===---X----
====X===---X----
====X===---X----
    X      X
    XXXXXXXX
[QUOTE]

We've done this a bunch of times.  The electric flux integrated over the left-hand surface is -E[SUB]Fe[\SUB] x A.  The flux integrated over the right-hand surface is E[SUB]Cu[/SUB] x A.   The flux integral over the entire closed surface is A*(E_Cu - E_Fe), which is not zero.  Via the divergence theorem, the volume integral of the divergence is therefore not zero.

[CODE]
    XXXXXXXXXXXXXXX 
    X        /----X--
    X       /     X
====X===---' /----X--
====X====---'     X
====X=====--------X--
====X====---.     X
====X===---. \----X--
    X       \     X
    X        \----X--
    XXXXXXXXXXXXXXX

The electric flux integrated over the left-hand surface is -E0[\SUB] x A_L. The flux integrated over the right-hand surface is E_R x A_R. You yourself specified that the "weakening" of E is inversely proportional to the area: E_R = E₀ x A_L/A_R. Therefore the right-hand surface's flux is AR * ER = AR*E0*AL/AR, which is E0*AL. The flux integral over the whole surface is now -E0*A0 + E0*A0 = 0. Via the divergence theorem, the volume integral of the divergence is therefore zero. If you were tempted to keep the Gaussian surface entirely inside the metal, that's fine too:

             /-------
    XXXXXXXXXXXXXXX      
====X===---' /----X--
====X====---'     X
====X=====--------X--
====X====---.     X
====X===---. \----X--
    XXXXXXXXXXXXXXX
             \-------

Gauss's Law is valid for any surface whatsoever, this one is just more complicated to analyze. Note that there is now flux (in, negative, strong field) through the left surface, and flux (out, positive, weak field) out through the right surface---this perhaps tempts one to say the divergence is nonzero. But there's also flux out through the top and bottom surfaces---this counts towards the integral as well. How do I know that the divergence actually adds up to zero? Because you specified that E = rho*J. Div.(E) = rho x Div.(J) + J x grad(rho) where rho is resistivity You also specified grad(rho) = 0---it's all iron---so Div.(E) = rho x Div.(J). You also specified Div.(J) = 0, because Div.(J) = d(charge density)/dt, which is also zero---getting this to zero is exactly why you had to make J inversely proportional to A. Therefore Div.(E) = 0.

 

[ben m]

Sorry, messed up the font on the 1st diagram.

    XXXXXXXX
    X      X
====X===---X----
====X===---X----
====X===---X----
====X===---X----
====X===---X----
    X      X
    XXXXXXXX

 

[Vorpal]

Johan: the divergence of a vector field is a specific quantity
div F = dFx/dx + dFy/dy + dFz/dz.
It is not "any field that changes strength".

So if (d/dy)F_y =0 and (d/dz)F_z is zero, while (d/dx)F_x is non-zero, then F_x is not changing in strength along the x-direction? Brilliant Ben: You have done it again: Failed elementary calculus.

You can have the field strength be varying over space and yet have div F = 0. A very simple example is F = xx̂ - yŷ. That was ben's point, with the more physically relevant example of F = (1/r²)r̂, which satisfies this property everywhere except at the origin.

 

[johanfprins]

You can have the field strength be varying over space and yet have div F = 0. A very simple example is F = xx̂ - yŷ. That was ben's point, with the more physically relevant example of F = (1/r²)r̂, which satisfies this property everywhere except at the origin.

Thanks for explaining Ben’s incomprehensible motives. This is of course exactly correct that if there is no charge density at a point you still can have (div)F=0. IN A VACUUM!!!!!. BUT what Ben Sol and Ziggurat cannot understand is that within a conductor one is NOT working in a vacuum. My first year students know this: Why can these three not know it?

The fact that one can use Drude's equation to model the current as if there are N "free" electrons within the metal is misleading, since there are NOT N free electrons within a conductor. This is easily demonstrated by measuring the heat capacity of a metal. The number of "free" particles that can gain heat energy are n<<<<<N. When a current flows, there are, however, suddenly N wave packets (pseudo-particles) to carry the charge. THESE WAVE PACKETS ARE NOT FREE ELECTRONS IN A VACUUM The motion along the length of the metal along the direction of the current is by cooperative motion of the wave packets PAST THE STATIONARY POSITIVE CHARGES. It is a quantum-mechanical effect. Only a total fool will argue that it is the same as free electrons moving through a vacuum.

The fact that the current is caused by cooperative motion of wave packets PAST POSITIVE CHARGES allows the effective electric-field to “diverge” and change direction without requiring any charge-accumulation. As I have just now posted above, when you have a 90 degrees bend in the copper, the electric field will bend with the copper. If you calculate (div)E you will find that it is not zero around the bend EVEN THOUGH THERE IS NO ACCUMULATION OF CHARGE. Similarly there cannot be any accumulation of charge at the interfaces even though you have a change in field strength from E_IR to E_CU. It is easy to prove that if there could be charge accumulation at interfaces Ohm’s law will no longer be valid. It is, however a waste of time to try and explain this to Ben, Sol or Ziggurat: They just do not know their physics and definitely not any Sold State Physics.

Let me conclude with the following thought experiment: Assume that the lengths of the iron elements and the copper are the same: i.e. equal to L, and a current I is flowing through the sandwich. The voltages will then be V_IR across the iron elements and will be V_CU < V_IR across the copper: This is caused according to Ben, Sol and Ziggurat by the accumulation of charge at the boundaries which is mandated by the change in resistivity when going from iron to copper. The electric field within the iron element is thus E_IR=V_IR/L and within the copper E_CU=V_CU/L. Now replace the copper with another identical iron element of length L. For the same current, the voltages are now the same across all three iron elements. According to Ben, Sol and Ziggurat this is so since the resistivities are now the same, and therefore there is now no charge accumulation at the boundaries. Now replace the iron again with copper which has a smaller diameter so that R_CU is the same as R_IR and send the same current through. The voltage across the copper is now the same as when the original copper was replaced by the iron element: Since there was no charge-accumulation for the iron element, this must also mean that in this case there is no charge-accumulation at the iron-copper interfaces even though the resistivity changes across these boundaries are still the same. Please people!! Stop believing charlatans.

 

[Ziggurat]

Thanks for explaining Ben’s incomprehensible motives. This is of course exactly correct that if there is no charge density at a point you still can have (div)F=0. IN A VACUUM!!!!!. BUT what Ben Sol and Ziggurat cannot understand is that within a conductor one is NOT working in a vacuum. My first year students know this: Why can these three not know it?

Vacuum, solid, liquid, gas, silly putty, it doesn't matter. Gauss's law remains true in all cases. You claimed it didn't, then claimed you never said it even though the quotes are there for everyone to see.

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Let me conclude with the following thought experiment

Because your other ones have worked out so well for you...

I don't think I'll bother this time.

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[johanfprins]

Vacuum, solid, liquid, gas, silly putty, it doesn't matter. Gauss's law remains true in all cases. You claimed it didn't,

You are blatantly lying again. 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. Correct, Gauss' law does not apply in this case, as you have applied it, since there is no charge accumulation at the interfaces

then claimed you never said it even though the quotes are there for everyone to see.

I did not say anywhere that Gauss' law is wrong: Only that in the hands of incompetents it can be wrongly applied: As you, Ben and Sol have demonstrated so aptly on this thread.

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Because your other ones have worked out so well for you...

I don't think I'll bother this time.

Of course not since even you, with your minute knowledge of physics, must be starting to realise that you are wrong.

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[Ziggurat]

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[sol invictus]

Gauss' law does not apply in this case

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

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

 

[ben m]

Correct, Gauss' law does not apply in this case, as you have applied it, since there is no charge accumulation at the interfaces I did not say anywhere that Gauss' law is wrong:

That's not how physics works. If one of the quantities relevant to the equation is not there, you don't just skip it: you set that term to zero.

Gauss's Law says div.E = rho/epsilon0. This is true everywhere. In regions of no charge excess, it's simply that rho=0, and therefore div.E=0.

How do you handle Newton's Laws, Johan? "F=ma. But here is an experiment with no forces! Therefore Newton's Law does not apply!" No, it's just that if F=0 then a=0---that's how you apply it.

So we have two problems: First, you don't know what divergence means, nor how to calculate it, nor how to recognize a correct calculation. Stop guessing at the answers---go talk to a human being and ask them to explain it to you. Second, you don't seem to care what the divergence is, because if it turns out to be nonzero you are prepared to reject Gauss's Law. I can't help you with that, it's just crazy.

 

[johanfprins]

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