Superconductivity at room temperature??

[johanfprins]

Now I can again get back to Ben's little games based on being deliberately leading the discussion away from the main topic.

In case it helps, which I doubt, let me challenge you with another thought experiment. Consider a gold block and a sodium block; neutral and NOT in contact.

What has this to do with a metal to n-type semiconductor contact?

It is clear that their Fermi levels are different---if I want to take an electron out of the gold into the vacuum, I have to expend 5.1 eV; if I lower this electron into the sodium, I only gain 2.3 eV.

How do you lower this electron into the sodium? Nowhere have I claimed that I first extract an electron from one material into the vacuum and then "lower" this electron into another material.

According to your mental picture, this difference must represent an electric field.

When you make contact between the two materials there MUST be a field owing to the momentary offset between the Fermi-levels. This has NOTHING to do with extracting an electron into the vacuum from one material and then "lowering" it into another material.

Where is the field? In the vacuum between the blocks?

Why will there be a vacuum between the two materials when they are making contact?

No it isn't, the field here is zero. (It's a vacuum, I can use Gauss's Law.) At the surface of the metal? No it isn't; if I drop a proton or a positron onto either block, it doesn't feel the Fermi level difference at all.

You are obviously again being deluded, or delibertaely trying to confuse the discussion with irrelevant nonsensical examples. I have so far not talked about a vacuum between two materials, but only about the momentary electric field across the interface between them just after having made contact.

 

[johanfprins]

What's <p²> in the wavefunction I gave, Johan? And I don't want an answer like "It MUST be zero because [fill in nonsense here]", I want a calculation.

Good luck, you'll need it.

I went back and looked at the wavefunction as you posted it. I do not see that the expression <p²> forms part of it as you are claiming? Where does it form part of the expression; and why do YOU claim that it forms part of the expression? Neither does <p>.

 

[johanfprins]

And... you walked right into ben's trap. If it's got kinetic energy when inside the cavity, and you just remove a wall (an action which DOES NOT provide the photon with more energy), then why does it suddenly now have kinetic energy?

To have kinetic energy a light wave must move relative to an inertial refrence frame. When it is within the cavity, it forms a standing wave because the boundaries of the cavity do not allow the light wave to leave the cavity. Thus the wave is not moving relative to the inertial reference frame of the cavity. Thus it now has no kinetic energy within this reference frame at all. Its energy is all stationary energy, and therefore you can weigh standing light waves within a cavity.

When you remove the one wall, the light waves are not contained to be stationary within the cavity anymore, and they now move out into space with a speed c relative to the inertial reference frame of the cavity: In fact, they now move with speed c relative to all inertial reference frames. Therefore the conservation of energy mandates that the total stationary energy of a standing light wave within the cavity, must now become kinetic energy when this light wave is being emitted from the cavity.

You have violated conservation of energy, johan.

Where have I done this? The stationary energy becomes kinetic energy without violating energy conservation.

 

[johanfprins]

Go on, try it. Take the ground-state wave function of the hydrogen atom, and actually calculate the kinetic energy and the potential energy. Tell us what you get. This is standard introductory textbook quantum mechanics, it's not really hard.

I have done that and find that the total energy of the ground-state electron of a hydrogen atom is 13.7 eV below its rest mass energy. So how in HELL can it have kinetic energy?

 

[johanfprins]

Even if you believe this, you ought to see that it conflicts with your statements about contact potentials. Your false understanding of standing waves disagrees with your false understanding of Fermi levels.

Where does contact potentials play a role when it comes to standing light waves? Are you really all there?

How is an electron-in-sodium different than a photon-in-a-cavity? You say they're both standing waves. But if I chop one of the boundary conditions off of the cavity, you recognize that the photon gets out.

A light wave gets out. It need not be photon it can be single laser beam.

If I chop one of the boundary conditions off the sodium, by bringing in the gold, you somehow think that the electron can't get out without a "force" to accelerate it out. Why the difference, Johan?

The difference is simple: A light wave is NOT CHARGED, while an electon wave is charged. This makes quite a difference you know. Or do you think that an uncharged wave should react with matter in the same manner as a charged wave?

 

[johanfprins]

What energy? There's no potential within the cavity. If your energy isn't potential energy, then it's kinetic energy. It sure as hell isn't rest mass energy, because photons have no rest mass.

Photons, modelled as "separate free particles", obviously has no rest mass: At last you got something right: Congratulations. But they are not photon-particles flitting around within a cavity. The light forms standing waves exactly as Planck has modelled cavity radiation correctly. A standing wave has NO kinetic energy, so that its energy must be STATIONARY ENERGY. This is obviously the same as mass-energy. Thus, when you stop a light wave in its tracks, its kinetic energy morphs into mass energy within the inertial reference frame within which it is now stationary.

Everytime when an electron orbital wave "absorbs" a photon wave, the latter waves stops in its tracks and entangles with the electron wave. Since the electron wave now has an increase in its mass-energy, it has to morph instantaneously into a higher energy wave-shape. If you are not to clever, you will call the latter morphing a "quantum jump by an electron-particle".

You can even do this on a macro scale: See experiments by Lene Hau.

 

[johanfprins]

Johan, let me ask you for the 12th time: do you know an actual, in-person scientist of any sort?

Many and I am myself one with a very successful career: See my CV on my website www.cathodixx.com. I would dearly want to compare my standing in physics to yours but you are obviously not a physicist with any credentials at all.

You have nothing to lose. If you're right, and if Zig and I are morons, surely your random acquaintance will back you up---maybe they'll suggest a devastating proof that will put us in our place.

This is impossible. You three are so dogmatic that it will be just as difficult as to convince a "flat earther" that the world is a sphere .

If Zig and I are right,

You are wrong.

you'll learn something and stop being humiliated.

All the dogmatic arguments you are raising I already know and some of them I have, to my great shame, taught to many students. Thus you cannot teach me any physics since you are not yet where I am. But you have been a great help to sharpen my arguments and to point out some mistakes in them. To date the latter mistakes do not exonerate you, Ziggurat and Sol from being severely deluded. You do know quite a bit of physics, but you obviously do not understand what you know.

 

[marplots]

Quick question.
What is "STATIONARY ENERGY"?

I am unfamiliar with the term.
(But I'm also not a physicist.)

 

[Ziggurat]

I went back and looked at the wavefunction as you posted it. I do not see that the expression <p²> forms part of it as you are claiming? Where does it form part of the expression; and why do YOU claim that it forms part of the expression? Neither does <p>.

Really, johan? Really? Of COURSE <p> and <p²> aren't part of the wave function. They're expectation values that you calculate from the wave function. Do you know how to do that calculation? You show no signs of having any idea. Yet this is really, really, really basic quantum mechanics. As in, first chapter of an intro textbook simple. I can do the calculation for you, and even show my work, if you admit you don't know how to do it.

I have done that and find that the total energy of the ground-state electron of a hydrogen atom is 13.7 eV below its rest mass energy. So how in HELL can it have kinetic energy?

I already told you. The kinetic energy is +13.7 eV, the potential energy is -26.4 eV, and so the sum is -13.7 eV. And I seriously doubt you actually did the calculation, I'm quite confident you merely looked up the result. But go on, prove me wrong: calculate the kinetic energy, and calculate the potential energy, and show your work. Or admit you can't, and I'll do the work. Or if you think that's unfair, I'll even do one half first (your choice, kinetic or potential) if you promise to then do the other half.

 

[Ziggurat]

Quick question.
What is "STATIONARY ENERGY"?

I am unfamiliar with the term.
(But I'm also not a physicist.)

It's johan's own little invention. It has no standard meaning in physics.

 

[sol invictus]

I went back and looked at the wavefunction as you posted it. I do not see that the expression <p²> forms part of it as you are claiming? Where does it form part of the expression; and why do YOU claim that it forms part of the expression? Neither does <p>.

You just failed quantum mechanics 101.

<p²> is the expectation value of the momentum squared. In slightly less compact notation, it's [latex]$\langle \psi | p^2 | \psi \rangle$[/latex], where the wavefunction I posted is [latex]$\psi(x) = \langle x | \psi \rangle$[/latex]. Computing <p²> for such a trivial wavefunction is one of the simplest and most basic tasks in quantum mechanics. The result, if you can learn how to get it, will correct one of the fundamental errors you are making.

So, what is <p²> for the wavefunction I posted, Johan?

 

[ben m]

Johan, never mind the metal-metal contact. The energetics and quantum mechanics of such a contact are the same as those of a Schottky diode, missing only the bandgap. This is apparently too complex for you to understand. However, it pushed you into revealing that you don't actually understand Fermi levels at all.

Johan, never mind the light-wave example. There is a close analogy---extremely close, in fact---between the quantum mechanics, boundary conditions, wave equations, and energetics of "light escapes from cavity when end is cut off" and of "valence electron escapes from sodium when another metal is put in contact". This analogy is apparently too hard for you to understand. However, it pushed you into revealing that you don't understand kinetic energy.

Johan, never mind the atom. The atom is a nice unambiguous quantum system in which kinetic, potential, and mass-energy issues are easy to calculate explicitly, and which I wanted to use by analogy to show the distinction between <p> and <p^2> for electrons in metals. Never mind, this is too complex for you to understand. However, it has pushed you into revealing that you don't understand Freshman quantum mechanics.

 

[Ziggurat]

However, it has pushed you into revealing that you don't understand Freshman quantum mechanics.

That was revealed long ago, and probably more than once. johan has previously demonstrated complete ignorance of expectation values and how to calculate them. But this is yet another glaring example, as well as proof that he hasn't learned anything from those previous failures. johan would clearly fail pretty much any physics class, were he to take one now. He will claim it's because everyone else is wrong, but the thing is, whether or not we're wrong, we're all consistent. Which means that we're all working from self-consistent theories, so anyone with a clue would be able to figure out how to get the same answer from those theories that we're all getting.

Einstein, for example, knew perfectly well how to work with Galilean relativity. In fact, facility with Galilean relativity, wrong though it was, was pretty much necessary in order to figure out the right way to do things. Einstein couldn't have developed special relativity if he wasn't able to do Galilean relativity. johan's inability to figure out how to get the same answers that every other physicist is getting with standard theories proves his own incompetence, regardless of whether or not those standard theories are correct. And his unwillingness to even try, while at the same time appealing to those standard theories (though in a self-contradictory manner), demonstrates his deep dishonesty. He is, in short, an incompetent liar. And that judgment stands regardless of whether or not the rest of us are correct about the physics.

 

[johanfprins]

You just failed quantum mechanics 101.

<p²> is the expectation value of the momentum squared. In slightly less compact notation, it's [latex]$\langle \psi | p^2 | \psi \rangle$[/latex], where the wavefunction I posted is [latex]$\psi(x) = \langle x | \psi \rangle$[/latex]. Computing <p²> for such a trivial wavefunction is one of the simplest and most basic tasks in quantum mechanics. The result, if you can learn how to get it, will correct one of the fundamental errors you are making.

So, what is <p²> for the wavefunction I posted, Johan?

My but you are slow to catch on: A stationary wave HAS no momentum and no kinetic energy. Of course I know the simple calculation you have just done since I have taught it for many years (over 40 years) to my students. It is based on a fallacy since it has NOTHING TO DO WITH MOMENTUM. When you take the square root you obtain (hbar) times the standard deviation of the harmonic wave in k-space (reciprocal space). ALL harmonic waves have this spread, and this spread has NOTHING to do with MOMENTUM OF THE WAVE OR ANYTHING ELSE. It HAS NOTHING TO DO WITH MOMENTUM SINCE A STANDING WAVE DOES NOT HAVE A DE BROGLIE WAVELENGTH. A standing wave's momentum is zero so that its de Broglie wavelength must be INFINITE if it has one.

So I ask again what has this calculation, which even you can do (congratulations: I would never have guessed that you are capable of doing it) of a fictituous momentum-squared parameter to do with the wavefunction you have quoted: Note that for your wave function the argument of the sine function is k_n* x. To multiply this argument with (hbar)/(hbar) so that the argument of the sine function becomes [p_n*x]/(hbar) is contrived nonsense; since the relationship p=(hbar)*k IS ONLY VALID WITHIN AN INERTIAL REFERENCE FRAME RELATIVE TO WHICH AN ELECTRON IS MOVING WITH MOMENTUM p. A stationary electron wave CANNOT model a moving electron having a momentum p. Ehrenfest's theorem will then be wrong.

Even if you want to believe the paranormal Voodoo of wave-particle duality, it still means that when the particle is moving the wave must move along. And since the wave is NOT moving relative to the inertial reference frame in which it is stationary, this MUST mean that the particle MUST also be stationary. So were does your kinetic energy come from? More Voodoo?

 

[johanfprins]

Let us return to my statements

As I expected, we have gone off on a tangent. Since I know and have taught the Vodoo, paranormal metaphysical Copenhagen interpretation of quantum mechanics, and since my proof that I discovered superconduction at room temperature can be done without having to resort to a realistic interpretaion of quantum mechanics, let me try and frame my statements so that they are not heresies against your superstitious, dogmatic beliefs.

So Since you have agreed with my statement 1, I am posting it again:

Statement 1: When creating an interface between an n-type semiconductor and a metal, where the latter has a Fermi level at a lower energy than the n-type semiconductor, and when neither of the two materials are charged up before generating the interface, then, after creating the interface, a dipole layer forms across the interface, since negative charge-carriers move across the interface from the n-type semiconductor into the metal, thus leaving a positively charged depletion-layer behind (in contact with the interface), and forming a negatively charged layer (in contact with the interface) by accumulation within the metal. Yes or No?

Now let me try and add a statement 2, which will not have you burning me at the stake:

Statement 2: If the density of donors within the n-type semiconductor is sufficiently high an equilibrium dipole will manifest when the Fermi-level at the metal-dipole interface has the same energy as the Fermi-level at the semiconductor-dipole interface. Yes or no?

 

[sol invictus]

My but you are slow to catch on: A stationary wave HAS no momentum and no kinetic energy. Of course I know the simple calculation you have just done since I have taught it for many years (over 40 years) to my students. It is based on a fallacy since it has NOTHING TO DO WITH MOMENTUM. When you take the square root you obtain (hbar) times the standard deviation of the harmonic wave in k-space (reciprocal space). ALL harmonic waves have this spread, and this spread has NOTHING to do with MOMENTUM OF THE WAVE OR ANYTHING ELSE. It HAS NOTHING TO DO WITH MOMENTUM SINCE A STANDING WAVE DOES NOT HAVE A DE BROGLIE WAVELENGTH. A standing wave's momentum is zero so that its de Broglie wavelength must be INFINITE if it has one.

I asked you to do a calculation:

What's <p²> in the wavefunction I gave, Johan? And I don't want an answer like "It MUST be zero because [fill in nonsense here]", I want a calculation.

Good luck, you'll need it.

Instead of getting a calculation, I got "It MUST be zero because [fill in nonsense here]".

Can you do the calculation, Johan?

 

[johanfprins]

I missed this question by Marplots

Quick question.
What is "STATIONARY ENERGY"?

I am unfamiliar with the term.
(But I'm also not a physicist.)

Kinetic energy is energy that moves relative to an inertial reference frame and therefore it is not stationary energy. Rest-mass is energy which is not moving; so it is stationary energy. Light has no rest mass energy since it is moving with the speed c relative to all inertial reference frames.

When light forms a standing wave within a cavity, it is not moving with a speed c relative to the cavity (which defines an inertial reference frame). If it were to be outside the cavity, this same light will have only kinetic energy, since the light-wave is then transporting energy with speed c relative to the inertial reference frame of the cavity. When this light is, however, trapped as a standing wave within the cavity, its energy cannot move with speed c relative to the inertial reference frame of the cavity. The light now has no kinetic energy relative to this inertial reference frame. So the light energy is now stationary energy; and it does then actually manifest as mass-energy. The latter can be proved by weighing a box that forms a black-body cavity while heating the box.

If the light within the box consisted of photons flitting around with speed c relative to the box, there would not be this mass increase; since such "particles" only has kinetic energy and no stationary (mass) energy. They cannot form an equilibrium centre-of-mass as in the case for gass molecules within a cavity.

 

[johanfprins]

I asked you to do a calculation:



Instead of getting a calculation, I got "It MUST be zero because [fill in nonsense here]".

Can you do the calculation, Johan?

Of course I can by cheating; since I have to assume that I can multiply k with (hbar) to get a "de Broglie momentum" which you can then Voodoo-interpret as defining a kinetic energy. The fact is that the wave-function you have posted ONLY depends on k and, as is the case for all harmonic waves, a single value of k is ONLY proportional to the momentum of the wave when the wave is a coherent wave that actually moves through space. Your wave IS NOT MOVING THROUGH SPACE and therefore to assume that k is proportional to momentum is Voodoo-physics!

The de Broglie wavelength is only valid when an electron-entity actually moves relative to an inertial reference frame with momentum p. And as I have stated, when this happens, the electron-wave must move along; even when you believe in the metaphysics of wave-particle duality. When the electron wave is a stationary wave within a metal, this wave cannot move relative to the metal's reference frame: So it cannot have kinetic energy. Just as in ths case of a standing light wave, its total energy within the metal is mass-energy.

Now it is quite clear that you Ben and Ziggurat will never be able to think outside the dogmatic interpretation of modern physics that you want to believe in. I have thus asked that we return to my proof that I have discovered superconduction at room temperature: After all, this is what this thread is supposed to be about: I will try to make my statements to not be contrary to your holy dogma which I know by heart, but which I now also know is paranormal metaphysics.

 

[sol invictus]

Of course I can by cheating

There's no need to "cheat": just do the calculation.

 

[johanfprins]

There's no need to "cheat": just do the calculation.

Why should I? If YOU do not know how to do it why do you not ask an undergraduate to do it for you? It is nonsensical to calculate a function defined by p for a standing wave since such a wave does NOT have any momentum and thus also not any kinetic energy; or any relationship involving momentum.

If you asked me to calculate <k²> you would have asked me to calculate real physics, since all harmonic waves have k-vectors even when they are not coherent waves. Only when the wave is a running coherent wave are p and k directly related. Your wave-function is not a coherent running wave at all; so why are you asking me to calculate metaphysics on the Jeff Randi website which specialises in exposing paranormal claims?