Superconductivity at room temperature??

[johanfprins]

No, actually, you cannot get that number by calculating the classical potential between an electron and a proton.

I have NEVER said that you can get it by just using the classical potential on its own. This is the type of mistake that only you, Sol and Ben have made consistently on this thread. You are the ones who claim that you can apply classical electrodynamics to a state that has been obtained by solving the Schroedinger equation.

What I say is that you use the classical potential between the electron and the proton as input to the Schroedinger equation. After the solution it has no direct meaning anymore; as you Sol and Ben want to argue.

If you try, you will not get -13.6 eV, you will get -27.2 eV. I've done it.

Of course you have done it since you are believing in doing Voodoo calculations.

You clearly haven't, and don't actually know how to despite claiming you can.

Exactly! I am not so
stupid to calculate irrelevant physics: I leave that to you Sol and Ben.

Sure, you can do that, EXACTLY like you can do that in Newtonian mechanics, because both Newtonian mechanics and the Schrodinger equation are explicitly non-relativistic.

The Lorentz transformation at low speeds applies. So how can they be "non-relativistic"?

I don't agree. The Schrodinger equation is explicitly non-relativistic. And mathematically, it's provable that the choice of reference potentials makes ZERO difference to any expectation value calculation.

You have calculated the expectation value of the "potential energy" <V(x)> and you claim that you will get the same expectation value using ANY reference potential? I can assure you that <V(x)+K> will not be the same as <v(x)>. But maybe it is just that I cannot do the calculation. I will thus appreciate it if you can prove to me that <V(x)> is equal to <V(x)+K>.

Not that it matters, since this also means we're free to pick the reference potential equal to the vacuum level.

Different reference potentials will give different values for the "expectation value" of the potential; and this will give you different solutions to the total energy of the electron-wave. But maybe it is just I who cannot do the calculation. Can you prove to us that for different reference potentials you will get the same expectation value for the potential energy? I suppose i am not as good as you when it comes to doing this mathematics.

No, johan, this is most definitely NOT a possibility. It's a mathematical impossibility. The only possible k-space distribution which will give <k²>=0 is a delta function at zero, but that also has <k>=0.

Really? Please calculate k using a deltafunction (delta)(k-kp) around kp. Do you get zero or do you get kp? And then calculate <(k-kp)²> and show that it is larger than kp.

But as you maintain: Maybe I am just too stupid to do the calculation correctly. So please correct me.

In fact, it's an absolute MATHEMATICAL requirement that <k²> >= <k>². That's not even peculiar to k space, that's true of any distribution. This is a really basic and obvious mathematical fact, and you screwed it up completely.

Not so for a travelling, coherent wave which has only ONE unique value for k. Its represenation in k-space is the delta function and in this case <(k-kp)²> is according to my mathematics zero. But of course I am dumb and stupid when it comes to these calculations: Please help me from my stupidity and post the correct calculation here so that I can learn at the feet of the master.

I don't care if you don't want to call it kinetic energy. <p²>/2m is still part of the total energy of the wave function. Call it whatever you like, it's still a necessary component.

It still cannot be kinetic energy since kinetic energy, if it exists must be given by <p>²/2m. But for a stationary wave <p>=0, thus there cannot be any kinetic energy. Or is it again that I am just stupid?

Furthermore, your logic is quite obviously absurd. For instance, if a free particle wave function is a superposition of a left-moving wave packet and a right-moving wave packet, it's not a free particle at all. It's quite clearly moving. It has kinetic energy, doesn't it?

It is not a superposition of two weave packets. Within its own inertial refrence frame it is a localised stationary wave, or else it will violate Galileo's inertia. When you are within another reference frame, this localised stationary wave (within its own reference frame) will move at a speed v past you. If you now apply the Lorentz transformation you will find that this localised wave is now longer and that along its length the "time changes": i.e. the phase angle of the wave is now different with position along the wave. For this reason the wave is now a running wave with a de Broglie wave length. ONLY NOW there is a momentum and a de Broglie wave length: And only now does the wave have kinetic energy. But when the wave is not moving relative to you, it does not have a de Broglie wavelength since it does not have a momentum at all: And therefore there is also no component of its energy that is kinetic energy.

And yet, <p> = 0.

Correct: No momentum - no kinetic energy. It has been like this since the time Newton postulated his laws and no physics which came afterwards (including quantum mechanics) has, and I believe can, change this fact.

And once again, you prove you cannot do math.

Well then why do YOU not do the correct derivations and post them here.

If <F>=(d/dt)<p>, then <F> ds = (d/dt)<p>*ds. And that is NOT the same thing as (ds/dt)*d<p>.

There is a very simple book called Teach Yourself Calculus. Try an read it and you will find that one of the fundamental laws of calculus is that you can write that for two functions y and x who both depend on t that (d/dx)y= (d/dt)y*(d/dx)t.

Furthermore, it doesn't even make sense in either form, since expectation values are not functions of position.

Only when the wave is stationary and the classical potential does not change with time. When the wave moves, the expectation values do change with time, just as Ehrenfest has proved.

There's also the additional irony that you have previously fought tooth and nail against using Ehrenfest's theorem in this very thread.

I did not fight tooth and nail against using Ehrenfest's theorem. I do fight tooth and nail when you Sol and Ben use correct theorems to argue Voodoo physics: As is your custom.

And yet, you just said in this very post that you could get the energy from the classical potential.

What potential do you insert into the Schroedinger equation before solving it, if not the classical potential? Plesae solve the hydrogen atom for me without using e²/r.

Oh, but then, that's me expecting consistency from you, and that's wrong of me.

You are the one who is inconsistent and who does not even know that (d/dt)y*ds can be written as (d/dt)s*dy. And you want to tell me that I cannot handle calculus?

Apparently everything to do with the Schrodinger equation is nonsense.

Where have I stated this? What I keep on stating is that the Schroedinger equation is incorrectly interpreted in terms of paranormal concepts like "wave-particle duality" and "complementarity" which lead to the oxymoron conclusion that a standing harmonic wave can have kinetic energy. No stationary harmonic wave can ever have kinetic energy: Only a harmonic wave that moves relative to an inertial reference frame has kinetic energy within the reference frame relative to which it is moving.

And yet, you demanded we use it.

Why not? It gives results: But what I really demand is that you should interpret these results correctly so that you do not make the paranormal conclusion that a stationary entity has kinetic energy.

I know you've thrown the very idea of consistency to the wind,

YOU claim that a stationary entity has kinetic energy and then has the arrogance to accuse ME of inconsistency?

but there are limits,

Yes I agree: You, Sol and Ben have long ago crossed over these limits into paranormal never-never land.

now said both terms in <H> = <p²>/2m + <V> have no meaning.

Can you give me an experiment where either one of these terms has been measured separately from the other?

I have given you the meaning. So here I will give it again: For a time independent potential well measured from the vacuum level, the Schroedinger equation gives fairly accurate solutions of the energy-intervals of standing waves as measured from the vacuum level. These energy-intervals are reductions in the mass-energy of the waves, so that their total energies (which are purely mass energies) are all below the rest mass energy.

It only seems as if one can model the reduction of mass by using parameters which remind one of potential and kinetic energy. This is an artifact because the Schroedinger equation uses rest mass as an input, and therefore it seems to be an eigenvalue equation of the Hamiltonian of a "particle" with mass m. It is this mass as input that is misleading since a differential wave equation that models matter should not have mass as input: Its solutions must be mass energy m, which will have a kinetic energy component ONLY when m>m_e; BUT NOT when m<m_e as you claim that one can have have for bound electrons around a nucleus.

 

[Cuddles]

Farsight, perhaps you should go back to one of your old threads about disagreeing with relativity rather than derailing this one. This thread is supposed to be about Johan's beliefs. Just because it tangentially touches on various subjects does not mean this thread is open for everyone to jump in with their own pet theories.

 

[johanfprins]

Again, another irony: you were the one who violated energy conservation with your theory (remember the Josephson junction mess?), finally conceding that you were wrong after pages and pages of denial.

We can all make mistakes. At least I am able to admit when I made one and then to correct it. I note that you, Sol and Ben never attacked me after I have posted the correct explanation of Josephson-"tunnelling" on my website which proves that Josephson's model also, and still, violates energy connservation.

You, Sol and Ben will never admit that you are wrong when consistently claiming that a stationary object has kinetic energy. The latter claim is the grossest violation of the conservation of energy ever! A stationary object (matter and/or wave) does not move and therefore it does not have kinetic energy.

My position, right or wrong, doesn't violate energy conservation. My position is that the electron has 13.6 eV (that's the right number, not 13.7) less energy when bound to a hydrogen atom than when free. Yes, my position is that it has kinetic energy, but the potential energy is LESS than -13.6 eV, so the total energy I calculate matches observation. Your criticism is thus nonsense.

Have you ever measured this "potential" energy and "kinetic energy" separately? Where and when? And if you have not, how do you know that they really exist as separate entities?

But you can't calculate the energy.

Neither can you. I have pointed out time and again that we all agreed that when an electron with mass m_e bonds with a proton with mass m_P, one obtains a hydrogen atom with mass M_H which is less thatn the sum of the separate masses of the electron and proton: i.e. the mass energy decreased by (delta)m_eP*c²=(m_P+m_e)*c²-M_H*c²: And have asked you time and again to calculate (delta)m_eP. You claim all along that you can calculate it, while I cannot, but your calculation is not forthcoming. It seems to me that you are lying again. If not, then post you calculation.

You can't do it using standard quantum mechanics, because you don't believe in it and don't have the competency to pull it off.

Again a serious distortion of the truth. I do not believe in the Copenhagen interpretation: This does not mean that I do not believe that quantum mechanics have been very successful: But I also note that this has been in those fields where the Copenhagen interpretation can mostly be ignored.

And you can't do it with any alternative method, since all you have as an alternative is hand-waving which cannot be turned into quantitative predictions.

OK, I will bow to your greater wisdom and magnificence and sit at your feet while you calculate (delta)m_eP*c². But then GodDammit do it and prove that you can do it. Stop your childish whining!

 

[sol invictus]

OK, I will bow to your greater wisdom nad magnificence and sit at your feet while you calculate (delta)m_eP*c². But then GodDammit do it and prove that you can do it. Stop your childish whining!

That is done in the first semester of every QM course, Johan. You claimed to have taught QM - and yet you don't know how to do that calculation? Are you admitting you lied? Or is it just that your memory is so bad that you cannot remember?

Part of that calculation - which again correctly reproduces the results of an incredibly broad range of experiments, and is rightly considered the most successful and powerful theory in the history of science - is to compute <p²> for the electron. If that were zero, the calculation would be wrong - it would not agree with experiment. End of story.

 

[johanfprins]

I can think of one person who prefers to ignore/dismiss/hide/fudge experimental results, because he's not able to admit that the results prove that his theory is wrong. That person is you, Johan.

Let me try and explain again: Assume that you have a moving clock which moves relative to an inertial reference frame with a speed v₁ as well as relative to a second inertial reference frame with a speed v₂: In other words the two reference frames relative to which the clock is moving also move relative to one another.

A person within reference frame 1 will note a time dilation of (gamma1)*(delta)t(1) and the person within reference frame 2 will measure another time dilation (gamma2)*(delta)t(2). These time dilations must be different. Now what you people claim is that time dilation IS ACTUALLY occurring on the moving clock within its own inertial reference frame. This will require the clock to simultaneously run at TWO different time-dilated rates. This is impossiible unless you believe in the paranormal like you, Sol and Ziggurat do. If you do not believe in the paranormal, it can only mean that the time dilations observed within reference frames 1 and 2, do not actually occur on the moving clock itself

This in turn must mean that when a clock is flown around the earth and the change in clock rate caused by gravity is subtracted after the clock has been brought back, then there cannot be any dilation of time on the clock itself which has been caused by Special Relativity. Hafele and Keating concluded that there is such a time dilation; even though Einstein himself stated that the dilation is only happening within the reference frame relative to which the clock is moving: It can thus clearly not really happen on the clock itself within its own reference frame within which it is stationary.

Thus if Hafele and Keating did not fudge their results and did get such time dilations, it really means that Einstein's Special Theory of Relativity must be wrong. I do not believe that this theory is wrong, and therefore I proposed an alternative experiment which will prove that this theory is not wrong, but will give the result as demanded by Einstein's theory: i.e. that the clock itself will not keep slower time within its own inertial reference frame within which it is stationary while moving relative to another inertial reference frame in which time-dilation is experienced (stationary means that it does not have any kinetic energy; even though you Sol and Ziggurat would want to claim that the clock can be stationary and still have kinetic energy)

 

[johanfprins]

No, you won't. It takes approximately two seconds to apply the Lorentz transformation and see that it predicts length contraction.

Then do it: I will even help you. At time t^/ within reference frame K^/, we will choose the rear end of the rod to be at x^/=0 and the front end of the rod to be ar x^/=L^/ at the same instant in time. Thus the coordinates of the rear end are (x^/=0, t^/=t^/) and the front end (x^/=L, t^/=t^/).

Now transform these coordinates into K by using the Lorentz transformation and then derive the "contracted" lenght L. After all it will only take you two seconds will it not?

 

[excaza]

You do realize it will you less time to actually answer the question than to write your foot stamping 8000 paragraph response every time you're asked to do something? For someone who claims to be vastly more enlightened than everyone else, you sure are stingy with your lessons.

 

[johanfprins]

No, Johan is wrong. His nonsense-theory to predict "no age difference" in the muon-storage-ring experiments---flat wrong---and no non-gravitational effect in the Hafele-Keating experiment---also flat wrong. Protip: Someone who's "not wrong" doesn't have to accuse entire fields of experimenters of deliberate fraud.

I did not accuse them of deliberate fraud. It is just human to find what you set out to find even when it is not really what you set out to find. It is not that Hafele and Keating committed fraud. But as I explained above, if their results are really correct (which I doubt), they did not prove Special Relativity to be correct, but just the opposite: They proved it to be wrong.

A clock travelling with the muons and brought back to the laboratoty will not show a time-dilation relative to a clock that did not travel with the muons.

 

[johanfprins]

You do realize it will you less time to actually answer the question than to write your foot stamping 8000 paragraph response every time you're asked to do something? For someone who claims to be vastly more enlightened than everyone else, you sure are stingy with your lessons.

Well, if you follow the thread you will notice that I admitted that I am "incompetent" and asked the master to do the calculation and to post it. What more do you want me to do? BTW, I like yout avatar: It is very appropriate you know!

 

[Small Bones]

I did not accuse them of deliberate fraud.

Dictionary reference:

fraud (frɔːd) [Click for IPA pronunciation guide]

—n
1. deliberate deception, trickery, or cheating intended to gain an advantage
2. an act or instance of such deception
3. something false or spurious: his explanation was a fraud
4. informal a person who acts in a false or deceitful way

[C14: from Old French fraude, from Latin fraus deception]

Collins English Dictionary - Complete & Unabridged 10th Edition

Fraud is a deliberate act therefore when you used the word fraud you implied a deliberate act. That's the dictionary definition.

 

[johanfprins]

Not being an expert, I need some help understanding the notation.

To wit, what is the difference in meaning between "p" and "<p>", or "V" and "<V>"? I don't understand the use of <angle brackets> here.

Let us rather call the so-called momentum-operator p_op: It supposedly means that when you apply it to the wave-function (psi) it is supposedly renders one of the possible momentums that can be measured say: p_n: i.e. p_op(psi)=p_n(psi). The statistical average of these possibilities, as calculated using the wave-function as a "probability distribution", supposedly gives us <p>. This is interpreted as follows: If you have a collection of identical wavefunctions (psi) and measure the momentum on each one of them, this is the value that you will measure most of the time. Obviously, such waves cannot be stationary waves but must be travelling waves. When applying this concept to a standing wave shoukld only measure one value namely zero unless you live in a paranormal univers in which momentum has no meaning.

The operator which supposedly "measures" the different values of the momentum is interpreted to be the following calculus operator: -i*(del) where i is the imaginary number square root of minus unity and del is the gradient detivative. When you apply it to the ground-state wave function of the s-electron you get an imaginary number. This is not surprising since this wave is stationary: NOT MOVING. It cannot have a momentum.

It is, however, now incorrectly assumed and vigorously defended by Sol, Ziggurat and Ben, that the product of these imaginary numbers actually does give the stationary wave kinetic energy, even though the wave is not moving at all: If this is not Voodoo physics then we are living in a crazy world. I do not believe that qunatum mechanics is actually "bizarre", as have been claimed for nearly 100 yeras by now; but that it only seems bizarre since it it is incorrectly interpreted.

 

[W.D.Clinger]

Farsight, perhaps you should go back to one of your old threads about disagreeing with relativity rather than derailing this one. This thread is supposed to be about Johan's beliefs. Just because it tangentially touches on various subjects does not mean this thread is open for everyone to jump in with their own pet theories.

Farsight thinks he is supporting Johan's beliefs. Johan needs all the help he can get, and "ton of bricks" Farsight is offering the kind of help Johan deserves.

For example...

"If you apply the Lorentz transformation correctly for a passing rod, you will find time dilation and length-stretching".

No, you won't. It takes approximately two seconds to apply the Lorentz transformation and see that it predicts length contraction.

Then do it: I will even help you. At time t^/ within reference frame K^/, we will choose the rear end of the rod to be at x^/=0 and the front end of the rod to be ar x^/=L^/ at the same instant in time. Thus the coordinates of the rear end are (x^/=0, t^/=t^/) and the front end (x^/=L, t^/=t^/).

johanfprins omitted a prime on the highlighted L. Unless he was trying to sneak something by us, he meant to write (x^/=L^/, t^/=t^/).

Now transform these coordinates into K by using the Lorentz transformation and then derive the "contracted" lenght L. After all it will only take you two seconds will it not?

Within reference frame K^/, the proper length of the rod is L^/-0=L^/.

Within any inertial reference frame K that's moving along the x axis of K^/, the observed length of the rod is its proper length (in K^/) divided by gamma, where gamma is greater than 1. (The exact value of gamma depends on the relative velocity of the two inertial frames, which johanfprins did not specify. That actually simplifies the problem, because all we need to know is that the relative velocity is positive, hence gamma is greater than 1.) Hence the observed length of the rod is less than its proper length, so observers who are moving with respect to the rod do indeed observe the contraction Farsight denied and johanfprins appears to be denying.

sol invictus and most of the other active participants here are very familiar with relativistic length contraction, so that reasoning probably took us less than two seconds.

It took longer to read johanfprins's formulation of his question and to spot his omission of the prime than it took to derive the length contraction Farsight had denied.

 

[sol invictus]

Let us rather call the so-called momentum-operator p_op: It supposedly means that when you apply it to the wave-function (psi) it is supposedly renders one of the possible momentums that can be measured say: p_n: i.e. p_op(psi)=p_n(psi). The statistical average of these possibilities, as calculated using the wave-function as a "probability distribution", supposedly gives us <p>. This is interpreted as follows: If you have a collection of identical wavefunctions (psi) and measure the momentum on each one of them, this is the value that you will measure most of the time.

No!

Dear god, Johan, how thick can you be? How many times have we explained to you that the average is NOT the same as values that you measure most of the time? The average of the rolls of a 6-sided die is (1+2+3+4+5+6)/6=3.5. How often do you roll 3.5 on a 6-sided die, Johan?

I give up. Goodbye.

 

[ben m]

Where have I stated this? What I keep on stating is that the Schroedinger equation is incorrectly interpreted in terms of paranormal concepts like "wave-particle duality" and "complementarity" which lead to the oxymoron conclusion that a standing harmonic wave can have kinetic energy.

Johan, you don't have a coherent position. Whenever you're asked to use the Schrodinger equation, you invent all sorts of reasons not to. Then you come back to Schrodinger and claim to be "defending" the true version against "misinterpretation". Then you go back to refusing to calculate anything, in any interpretation at all. Are you having fun?

Who cares? Young (of double-slit experiment fame) might have falsified results, right?

 

[johanfprins]

Dictionary reference:


Fraud is a deliberate act therefore when you used the word fraud you implied a deliberate act. That's the dictionary definition.

I did not use the charge fraud but wrote that they probably "fudged" the result. This happens more regularly in physics to obtain the result that the physicist wants, than we want to know of: For example, the epicycles in Ptolemy's model, Einstein's Cosmological Constant, Renormalisation,ertc. etc. etc. It does not imply that the fudgers are fraudulant.

Since, according to your quote, fraud imples "deliberate" I should probably not have used the word "deliberately" with "fraud": However, I can point out cases where a person was charged with fraud while the person had been unaware that what he/she were doing is fraudulent in the eyes of the law. So I think you are off base. Try something else!

 

[excaza]

What more do you want me to do?

Answer the question and stop acting like a child. One would think you could have inferred that from the post you quoted.

 

[johanfprins]

johanfprins omitted a prime on the highlighted L. Unless he was trying to sneak something by us, he meant to write (x^/=L^/, t^/=t^/).

For the time being I will lower myself and answer you since for the first time on this thread you posted something that is not nonsense: Yes I made a typo: It should have been x^/=L^/. Well done! Although I think that my goldfish could also have picked this up.

Within reference frame K^/, the proper length of the rod is L^/-0=L^/.

Within any inertial reference frame K that's moving along the x axis of K^/, the observed length of the rod is its proper length (in K^/) divided by gamma, where gamma is greater than 1.

This is the regular BS that you keep on posting to foul up this thread. I gave you the coordinates, and if you are capable of doing so (which you have demonstrated time and again you are not) you can look up the Lorentz transformation, and if you are capable of doing so (which I sincerely doubt) you can apply the Lorentz transformation to these coordinates: And if you do this ( which I doubt that you are capable of doing) you will find that you, as usual, posted nonsense. Must I send my goldfish to help you? sol invictus and most of the other active participants here are very familiar with relativistic length contraction, so that reasoning probably took us less than two seconds.[/quote] Do the Lorentz transformation; or get a high school student to do it for you! Please stop being under the impression that you can think and reason. With all due respect, take it from me, you do not have the intellectual capacity to do so.

 

[Small Bones]

I did not use the charge fraud but wrote that they probably "fudged" the result. This happens more regularly in physics to obtain the result that the physicist wants, than we want to know of: For example, the epicycles in Ptolemy's model, Einstein's Cosmological Constant, Renormalisation,ertc. etc. etc. It does not imply that the fudgers are fraudulant.

This is what you wrote:

Thus, Hafele and Keating, and those who followed in their footsteps must have fudged their results to obtain what they wanted to obtain. There are many dishonest scientists who are willing to do this when the stakes are high.

You didn't say they probably "fudged" you wrote they must have fudged
Fraud is "deliberate deception, trickery, or cheating intended to gain an advantage" and you wrote "must have fudged their results to obtain what they wanted to obtain". Are you really denying that you accused them of fraud?

If you refrained from resorting to hyperbole you wouldn't be caught out making unsubstantiated accusations.

 

[johanfprins]

Johan, you don't have a coherent position. Whenever you're asked to use the Schrodinger equation, you invent all sorts of reasons not to.

What you want me to do is derive parameters which you interpret incorrectly; and which do not have the physics relevance you assign to them. Why shoudl I play along to derive parameter which you assin paranormal; significance to? I have now "conceded" that I am according to you "too stupid" to do this, and requested you all, as I have requested you all along, to derive (delta)m_Pe for me. You all claim that you are experts that can do this easily. So why do you not derive what I ask YOU to derive, but attack me if I do not want to derive what you want me to derive.

Then you come back to Schrodinger and claim to be "defending" the true version against "misinterpretation". Then you go back to refusing to calculate anything, in any interpretation at all. Are you having fun?

Are you having fun? You never derive what I ask YOU to derive, so why must I derive what you want ME to derive. So if you are so extremely capable as you claim behind anonymity,

1. derive for me from the Schroedinger equation what (delta)m_Pe is. After all we all must agree that it is a relevant physics parameter.
And
2. use the coordinates of a stationary rod at a time t^/ to derive the "length contraction" of the rod within a passing inertial reference frame. It has been all along you, Sol annd Ziggurat who have been playing games: Not I.

Who cares? Young (of double-slit experiment fame) might have falsified results, right?

He did not have to fudge his results and he did not claim paranormal nonsense such as stating that a stationary entity has kinetic energy.

 

[ben m]

Thus if Hafele and Keating did not fudge their results and did get such time dilations, it really means that Einstein's Special Theory of Relativity must be wrong.

That's right, Johan, you are the only person on Earth capable of doing these calculations correctly. If you say that "the Hafele and Keating observation disagrees with SR", well, how could you ever do a physics calculation wrong? An uninterrupted conspiracy of identical 'mistakes' stretching back to 1905, sound much more likely than Johan F. Prins getting confused.