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Superconductivity at room temperature??

Where is Farsight's logic flawed? You always post bombastic statements like this without any proof whatsoever. Consider the following: A metal with a workfunction W. Thus the energy of an electron at the Fermi-level is W below the vacuum level; and since the energy of a stationary electron at the vacuum level is equal to me0*c2, where me0 is its rest mass energy,

Nope. The equation e=sqrt(p^2 + m^2) is a statement of kinematics. It is a description of how the electron's energy depends on its motion. It does NOT include other forms of energy, like electrostatic energy, or gravitational potential, or spring constants, or what-have-you. You are incorrect (or, rather, Farisght is incorrect) to try to squeeze non-kinematic energies into the kinematic term.

Never mind, though, Johan! All of the experiments that disagree with you are fraudulent, right?
 
no, Johan, the two experiments are using the SAME prediction, they both agree with each other and with Einstein's original paper. They both agree with the two postulates you are quoting.
I am not going to argue with you, since you are setting me up against Einstein, and then making it a cradinal sin to disagree with Einstein on certain points. Einstein was not God: He did make mistakes which he himself admitted. So let me try again:

Postulate 1 of the Special Theory of Relativity according to Ray D'Inverno in his text book entitled "Introducing Einstein's Realtivity":

All inertial observers are equivalent.

Do you agree or do you NOT agree?
 
Nope. The equation e=sqrt(p^2 + m^2) is a statement of kinematics. It is a description of how the electron's energy depends on its motion
Correct, but only when the electron is not bound within an atom or a material. It is also valid when the free electron is not moving and then it gives for a SOLITARY electron within its own inertial reference frame that e=m . This is the energy that a free electron has at the vacuum level. Thus, equation cannot give the energy when the electron has an energy that is less than its free energy: i.e. when e<m.
It does NOT include other forms of energy, like electrostatic energy, or gravitational potential, or spring constants, or what-have-you.
Exactly! But when you include these energies and calculate the energy of the electron to be e=m/ where m/ is smaller than m, then according to this equation the electron cannot have kinetic energy.
You are incorrect (or, rather, Farisght is incorrect) to try to squeeze non-kinematic energies into the kinematic term.
No Farsight is not doing this: This is what you people are doing: You are claiming that for an electron energy e less than m, the kinematic term p2 is still valid. We are saying that when the energy of an electron e is less than m, there can be no kinetic energy whatsoever: Just as Einstein has proved. The equation e=sqrt(p2+m2) does then not apply.
Never mind, though, Johan! All of the experiments that disagree with you are fraudulent, right?
You are again projecting yourself, Sol and Ziggurat. It is really you who ignore experimental results that we all agree are correct. The fact remains that for a hydrogen atom with rest mass MH, formed from a proton with rest mass mP and an electron with rest mass me one has that:

(mP+me)*c2-MH*c2=h(nu)=13.6 eV.

Do you deny that this equation is correct? Do you deny that this equation clearly proves that the rest masses of the proton and the electron must have decreased during bonding? So e=sqrt(p2+m2) cannot apply anymore since it would require that p2 must be negative.

Now of course you want to wriggle out of this dilemma by postulating that e consists of both potential energy as well as kinetic energy, and that for this reason it only seems that the rest mass is decreasing. Why do you claim kinetic energy and potential energy? Because the Hamilton operator that is used can be written in opertor format as Hop=-pop2/2m +Vop(r)=Top+Vop.

When you solve the Schroedinger equation you obtain a spectrum of possible eigen-energies of which the lowest for a hydrogen atom is the s-state, es. Next you claim that this energy must consist of the sum of kinetic energy eT and potential energy eV. But this is quantum mechanically impossible, since neither the operator for kinetic energy Top nor the operator for the potential energy Vop commute with Hop. This means that these two energies cannot manifest simultaneously with the total energy es. Thus there are no kinetic and potential components for es.

To wriggle yourselves out of this dilemma you tried to bully me to calculate <Top> and <Vop> in order to argue that since these expectation values are not zero, there must be kinetic energy and potential energy. But this is absurd since these average values can only be obtained when one measures the kinetic energy of an ensemble of s-states and with each measurement you detroy the s-state and generate a new quantum state that is an eigen-state of the kinetic energy. This is further proof that the energy of the s-state does not have a kinetic energy component. The same for the potential energy.

I can generate at randon many operators which do not commute with Hop and calculate their expectation values. Say I call an operator a cat operator CATop, which does not commute with Hop and find that <CATop> is not zero: Does this mean that the s-state of the hydrogen atom has a cat within it? Stop being absurd!
 
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I've gone back a few pages and cleaned out (or edited) breaches of rule 0, rule 11, rule 12; the posts that have been moved can be found here: http://www.internationalskeptics.com/forums/showthread.php?t=226944

If you have any questions as to why your post was moved, you are welcome to PM me, post a question in FM: Questions, or file an Appeal.

In the mean time...please stay on topic and stop the bickering; especially johanfprins and D'rok.
Replying to this modbox in thread will be off topic  Posted By: Locknar
 
What? A common synonym for rest mass is invariant mass. Look it up. It doesn't change under the Lorentz transformation. (Another quantity---formerly called "relativistic mass" but now more commonly called "energy"---does change under a Lorentz transform, and I hope that's what you're thinking of.)
No, that's not what I'm thinking of. Invariant mass varies. Surely you know this? Rest mass is a measure of a system's energy content when you're motionless with respect to it. A cannonball in free space has a greater rest mass than the same cannonball on the surface of the earth. That's because it has gravitational potential energy equivalent to the kinetic energy of a cannonball falling at 11 km/s. I'm not kidding about this. Think it through and do some research, see for example http://en.wikipedia.org/wiki/Mass_in_general_relativity and http://www.ncbi.nlm.nih.gov/pmc/articles/PMC336961/

In electromagnetism, the binding energy is energy missing from the total electromagnetic field configuration, not from either of the particles. Work out (i.e., do a volume integral of E^2) the field energy of the 1/r^2 field of an isolated proton; work out the 1/r^2 field of an isolated electron. (The renormalization cutoff is irrelevant here.) Then work out the field energy of a classical proton+electron dipole separated by d. There's a small region with increased E^2, but a very large region with decreased E^2.
I disike this approach because it suggests that the particle is something separate from its field. We made the electron in pair production, from an electromagnetic wave, an electromagnetic field variation. Now we've got a standing wave, a standing electromagnetic field. Call it something else if you like, but remember the wave nature of matter, we don't have some point-particle which is distinct from its electromagnetic field.
But that said, yes, a hydrogen atom doesn't attract a positron like an electron would, the energy distribution in the space around the locus is different, the emitted photon is proof positive that it's different.

The difference in total field energy between the isolated+isolated and the "dipole of size d" is precisely U = q1 q2/(4 pi e0 d), the potential energy gained by bringing the two particles together. It's true that some of the mass of the isolated electron (or proton) is due to the energy in this field, so it's indeed possible to ascribe the binding energy to "mass"---but if you were to treat it that way, there'd be equal amounts of mass/self-energy missing from each particle.
It isn't equal amounts because the electron plays the part of the cannonball while the electron plays the part of the Earth. The electron accelerates towards the proton nearly two-thousand times as much as the proton accelerates towards the electron. Discount the latter and it's the electron potential energy that's converted into kinetic energy then radiated away. Setting the dipole aside too, don't think that all the electromagnetic field energy has simply vanished just because there's no net electromagnetic force on a positron.

The gravitational case is much subtler---that field integral difference, if you treat it like an attractive EM field, has the wrong sign---and I wouldn't be on very firm ground discussing it.
Do check it out though, it is important.
 
What quantity do you call "rest mass"?
The amount of energy in a system that is at rest with respect to me. Divided by the speed of light squared of course.

If you've reserved "rest mass" as a term for some other quantity, what do you call the absolute, Lorentz-invariant, gauge-invariant quantity that is predicted by theory and confirmed by experimental observation, that everybody else calls "rest mass"?
I call it rest mass, but don't think that it's absolutely invariant. The mass of the cannonball in free space is greater than its mass on the surface of the Earth. It really is. You gave it kinetic energy when you fired it upwards at 11km/s, and this was converted into gravitational potential energy. In the cannonball. Conservation of energy applies.
 
I call it rest mass, but don't think that it's absolutely invariant. The mass of the cannonball in free space is greater than its mass on the surface of the Earth. It really is. You gave it kinetic energy when you fired it upwards at 11km/s, and this was converted into gravitational potential energy. In the cannonball. Conservation of energy applies.

As I said:

As far as I can tell, for you "mass" is synonymous with "energy".

Regarding general relativity, this is supposed to be a thread about Johan's claims for room-temperature superconductivity. His misunderstandings of physics are relevant to that, but yours aren't.
 
Yes, it really is irrelevant. First off, we can do ALL of this non-relativistically, because NOTHING here is moving fast. So why drag relativity into this when you really don`t need to?
To explain it to you. And Johan was talking about a light wave in box, which rather suggests that EVERYTHING here is moving fast.

Second, and more importantly, this tells us absolutely nothing about the split between potential energy and kinetic energy.
I've already said that all energy is in the end dynamical. You know why a cannonball potential energy is converted into kinetic energy? Because the coordinate speed of light varies in a non-inertial reference frame. Kinetic energy at the subatomic level is transformed into macroscopic kinetic energy. The cannonball hits the ground, that kinetic energy is radiated away, and the mass of the cannonball is now less than it was.

You`re arguing against something nobody claimed.
I said It really isn't more energy Zig, it's less energy. That radiated photon proves it. It's that simple. Sol was banging on about the electron having positive kinetic energy despite and ignoring the mass defect.

How would you know? All you have referred to is total energy, and we agree on that. Your only claim to zero kinetic energy amounts to hand waving. But I can do better than that. I can calculate the kinetic energy. And I get a positive value. I can use the same techniques to calculate the total energy, and guess what: I get the right answer. Can you do the same? Can you do a calculation to show what the kinetic energy is? Do you even know how to do that in quantum mechanics? Because if all you`ve got is what you`ve offered so far, well, that`s not enough.
I can't calculate it offhand. I could look it up and give you a calculation, sol will tell you that I'm no dummy, but I won't because it still wouldn't be enough. No calculation can get it across to you that kinetic inside the system is exhibited as the mass of the system, and the electron is such a system. You'll continue to think of the system itself moving and exhibiting kinetic energy whilst retaining an invariant mass.

No you haven`t.
Yes you have. This is the whole point Johan was making. The photon in the box doesn't have any mass. It has kinetic energy instead. The box doesn't have kinetic energy, it has mass instead.

Why not? Because you don`t like calling it kinetic energy?
You can't have your cake and eat it because it's double counting.

That`s the way kinetic energy is defined in quantum mechanics. And it`s defined that way for a damned good reason. You can call it something else if you like, but the truth is that you don`t actually have a good reason to call it anything else. All you have is some weird hand-waving attempt to wrap up non-relativistic quantum mechanics in some superficial semblance of relativity.
E=mc² is not some superficial relevance of relativity, Zig. Go and look at Einstein's 1905 E=mc² paper Does the inertia of a body depend upon its energy content which refers to a radiating body losing mass. The body loses kinetic energy. It loses mass. It doesn't gain kinetic energy and keep the same mass.

Farsight said:
It really isn't, Ziggurat. The electron in the hydrogen atom is the equivalent of rapidly jiggling a box containing a light wave.
No, actually, it isn`t equivalent to that at all, for a whole host of reasons. For example, electrons are spin-1/2 particles, while photons are spin-1 particles. You can`t construct a spin-1/2 particle out of a spin-1 particle.
You construct two spin ½ particles out of a spin-1 particle in pair production. You split the incident photon over a nucleon to get an electron and a positron.

There`s also the little problem of charge, namely that electrons HAVE charge and photons don`t.
It isn't a problem. It's just chiral frame dragging, the electron is a bit like a roton in a BEC, and conservation of charge applies in pair production.

This is a really bad analogy meant to hand-wave away your confusion about the complexity of pair creation. But it doesn`t actually work.
It ought to be good enough to get you interested. The electron isn't the same as a photon, but you can split a photon into two other photons with twice the wavelength. And you can annihilate the electron and the positron, which you made in pair production, to yield two photons. So think again about a photon in a box. Don't dismiss it along with 4π/nc^1½ and the depiction below just becuse it doesn't tie in with what you've been taught. Think of the proton as something much smaller, right in the in the middle. Remember there's no surfaces, so add more contours to fatten the torus to a near-spherical symmetry. Enjoy the pleasure of finding things out.

toroid2.jpg
 
Do you all agree or do you NOT agree?

The definition of "equivalent" is a very specific one; I would agree with the statement "all inertial observers are equivalent" from someone who intended to use the appropriate "equivalence". What are you going to do with it, Johan?

Your turn: Do you agree, or not, that the Lorentz transformations are consistent with the two principles of relativity?
 
I've already said that all energy is in the end dynamical.

Irrelevant word salad. Kinetic energy and potential energy have standard definitions. But even if you choose not to use the standard definitions, the fact remains that electrons, even bound electrons in stationary states, have energy which is not affected by electric potential. That is what johan denies (because it demolishes his misconception of the significance of the Fermi energy), and that is the point you keep simply avoiding. Goodness knows why you're intent on avoiding this point. Perhaps you think you can somehow rescue johan from being completely wrong, but you really can't.

Yes you have. This is the whole point Johan was making. The photon in the box doesn't have any mass. It has kinetic energy instead. The box doesn't have kinetic energy, it has mass instead.

Um... now you have directly contradicted johan. He has directly stated that a photon in a resonating cavity does NOT have kinetic energy.

You really don't get what's under debate here.

E=mc² is not some superficial relevance of relativity, Zig. Go and look at Einstein's 1905 E=mc² paper Does the inertia of a body depend upon its energy content which refers to a radiating body losing mass. The body loses kinetic energy. It loses mass. It doesn't gain kinetic energy and keep the same mass.

This is completely irrelevant to EVERYTHING I have been discussing with johan. You really just don't get it.
 
The definition of "equivalent" is a very specific one; I would agree with the statement "all inertial observers are equivalent" from someone who intended to use the appropriate "equivalence". What are you going to do with it, Johan?
OK let us become more specific: As far as I could ascertain in text books the consistent picture is that equivalent means that when two competent (not an observer who will squash an anode into the surface of a diamond), equivalent observers do the same experiment without making any mistakes, they must get the same result: Do you agree with this?

Your turn: Do you agree, or not, that the Lorentz transformations are consistent with the two principles of relativity?
In terms of my present knowledge I agree absolutely!
 
Um... now you have directly contradicted johan. He has directly stated that a photon in a resonating cavity does NOT have kinetic energy.
It has been very rare that Ziggurat posted anything on this thread that is relevant and correct. This is, however, one of those rare surprising occasions. He is correct: I claim that when there is a light wave within a totally enclosed resonant cavity, it must be a standing light wave that, like any and all standing waves, does not have any kinetic energy. Thus the light energy is now completely stationary energy: And stationary energy is rest mass energy.

Similarly, when an incoming light wave entangles with a standing electron wave within a metal, its kinetic energy, before entanglement, becomes mass energy which adds to the rest mass energy that the electron-wave has had before entanglement. If this increase of mass energy causes the electron wave to have an increased mass energy which is lower than the rest mass energy of a free electron in space, the electron wave remains within the metal: When, however, the entangled mass energy of the electron wave is larger than the rest mass energy of a free electron, the increase in mass ONLY NOW has a kinetic energy component and therefore the electron is ejected from the metal: This is the correct model for the photo-electric effect. There are no "particles" involved.

Farsight, don't allow these bullies to chase you from this thread. As I have already posted, your posts are like breaths of fresh air. We do not have to agree on every single issue to be appreciative of each other's viewpoints.
 
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The definition of "equivalent" is a very specific one; I would agree with the statement "all inertial observers are equivalent" from someone who intended to use the appropriate "equivalence". What are you going to do with it, Johan?
OK let us become more specific: As far as I could ascertain in text books the consistent picture is that equivalent means that when two competent (not an observer who will squash an anode into the surface of a diamond), equivalent observers do the same experiment without making any mistakes, they must get the same result: Do you agree with this?
Probably not, unless you're using some strange definition of "do the same experiment" and/or "get the same result".

Two inertial observers may disagree about whether some particular cannonball is moving.

Two inertial observers may disagree about whether two lightning strikes were simultaneous.

Two inertial observers may disagree about whether a magnetic field is present.

Et cetera.

Your turn: Do you agree, or not, that the Lorentz transformations are consistent with the two principles of relativity?
In terms of my present knowledge I agree absolutely!
:thanks

ETA: I think we could agree that, in a situation for which special relativity is appropriate (which means, among other things, that gravity is negligible), the results seen by two inertial observers will be equivalent after an appropriate Lorentz or Poincaré transformation has been applied to transform those results into a common inertial frame.
 
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Probably not, unless you're using some strange definition of "do the same experiment" and/or "get the same result".

Two inertial observers may disagree about whether some particular cannonball is moving.
As usual you are displaying your enormous ignorance when it comes to physics. If both observers do the SAME experiment within their respective inertial reference frames they must get the same result. This is what Einstein's first postulate means.

THIS MEANS THAT IF ONE FIRES A CANNON BALL WITHIN HIS/HER REFERENCE FRAME AND THE OTHER FIRES A CANNON BALL IN AN IDENTICAL MANNER WITHIN HIS/HER REFERENCE FRAME, BOTH WILL MEASURE THE SAME SPEED AND PATH FOR THEIR RESPECTIVE CANNON BALLS.

What you are talking about is the case where only one of the observers fire a cannon ball within his/her reference frame and the other observer is an onlooker looking in on the experiment while moving relative to the experiment. This is not the same as when both observers do exactly the same experiment within their own respective, inertial reference frames.

Are you able to understand the difference? Yes or No?

I do not know why I waste my time to even respond to your irrelevant comments!
 
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What you are talking about is the case where only one of the observers fire a cannon ball within his/her reference frame and the other observer is an onlooker looking in on the experiment while moving relative to the experiment. This is not the same as when both observers do exactly the same experiment within their own respective, inertial reference frames.

Are you able to understand the difference? Yes or No?
Yes.

I do not know why I waste my time to even respond to your irrelevant comments!
You have the right to remain silent.
 
Originally Posted by johanfprins
What you are talking about is the case where only one of the observers fire a cannon ball within his/her reference frame and the other observer is an onlooker looking in on the experiment while moving relative to the experiment. This is not the same as when both observers do exactly the same experiment within their own respective, inertial reference frames.

Are you able to understand the difference? Yes or No?
Yes.

Good: Thus, when two inertial observers have the same radio-active material with them and they BOTH measure the decay rate of this material on the samples they have with them, will they get the same answer?
 
Ziggurat said:
Um... now you have directly contradicted johan. He has directly stated that a photon in a resonating cavity does NOT have kinetic energy.
Sheesh Ziggurat, a massless photon doesn't have mass, it has kinetic energy. But confine it in a box and that system now has mass rather than kinetic energy. It's so simple and so rooted in the relativity of a radiating body loses mass that I cannot believe you don't understand this. And I'm frankly appalled that nobody has conceded that Sol's moving box is something different to a light wave in a box, which is what Johan referred to. The standing wave doesn't have kinetic energy because it's standing. It exhibits inertia instead of momentum. It isn't moving, but it still resists your attempts to change its state of motion, just as it did when it was moving through space at c.

Johan: I'm now seeing dismissal of patent evidence here along with words like "irrelevant" and "word-salad" plus denial of the bleedin' obvous along with evasion and dishonesty. It's exactly like discussions I've had with young-earth creationists. And we aren't progressing at all on the superconduction theme. Since I'm pushed for time, let's take it offline. I was wrong about these guys. Sorry. The thing that persuaded me was the lack of a response to The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves.
 
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Johan: I'm now seeing dismissal of patent evidence here along with words like "irrelevant" and "word-salad" plus denial of the bleedin' obvous along with evasion and dishonesty. It's exactly like discussions I've had with young-earth creationists. And we aren't progressing at all on the superconduction theme. Since I'm pushed for time, let's take it offline. I was wrong about these guys. Sorry. The thing that persuaded me was the lack of a response to The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves.
I am glad to see we are in total agreement about standing light waves when light is confined. You cannot imagine what I went through on this thread for more than a year.

Maybe you have missed it but I did post all the reasons why my experiment proved superconduction at room temperature in post 2647. I think the logic is overwhelmingly clear. So far nobody attacked it on this thread. If you have anymore questions contact me by e-mail.

Thanks, and best wishes for the New Year!
 
I'll take a look Johan, and email you. Happy New Year.

I can imagine what you've been through. I understand that not invented here attitude and the science advances one death at a time resentment that goes with it. Nick Kim, an environmental chemist in New Zealand, didn't do his "peer review" cartoon for nothing!

http://www.lab-initio.com/screen_res/nz060.jpg

Edited by LashL: 
Please see Rule 5 regarding hotlinking.
 
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