Superconductivity at room temperature??

[Gods Big Toe]

My problem is with the so-called twin paradox. Although each twin is under the impression that the other twin\'s clock is going slower and has to take it into account when, for example, communicating with the other twin, the clocks with the two twins are actually keeping exactly the same time within their own reference frames respectively. Thus, one twin cannot actually age slower than the other twin. My conclusion is that when they meet up one day they will find that they still have the same age. I might be wrong, and therefore I have proposed a simple experiment:

My proposed experiment, which you claim is the same as the muon experiment, is not the same as the muon experiment since I am using two identical shuttles of radio-active material to act as the two twin-clocks: Then send the one radio-active shuttle on multiple journeys to and fro, and then in the end compare the two shuttles within the reference frame of the shuttle that stayed at home. I predict that they will still have the same age: i.e. that the shuttle that was sent on the journey will not have more undecayed radio-active material than the one that stayed behing: They still have exactly the same amount of non-decayed material within them.

It is a simple experiment that can be done by using a weaving loom.

Is the Hafele-Keating experiment not good enough for you?
http://en.wikipedia.org/wiki/Hafele–Keating_experiment

 

[johanfprins]

Calculate the energy of hydrogen.

WE all know that the ground state energy is E₀=-13.7 eV as measured from the vacuum level and that at the vacuum level the stationary energies of the proton and the electron are m_p*c² and m_e*c² while the stationary energy of the hydrogen atom is M_H*c². Thus the energy of the hydrogen atom is clearly the latter which can be obtained from the stationary energies of the proton and the elerctron as:

M_H*c²=(m_P*c²+m_e*c²)-13.7.

What is your problem?

 

[Ziggurat]

Then do it.

That's not the deal, johan.

I'll do half the calculation, IF you promise to do the other half. You even get to pick which half, <V> or <p²>/2m.

Or admit that you can't, and I'll do the whole thing.

Well we know that we need a photon of approximately 13.7 eV to break up a hydrogen atom.

But you can't predict this number, can you? You can't calculate this energy. You can't tell me why it's 13.7, not 13.8, or 26.4, or any number, can you?

Now go ahead and calculate the masses (delta)m_p and (delta)m_e as you claim that you can do.

Not until you agree to my terms above. They're quite fair. I have to do the work first, but I won't bother unless you're going to do your part too.

The solution derived from Schroedinger's equation for Sol's famous standing waves in a box

This is the hydrogen atom we're talking about. The solution isn't the wave Sol gave for the particle in a box.

gives the energies E_n=(hbar*k_n)²/m_e)-V, which must be interpreted as an approximation for (delta)m_P+(delta)m_e: So where is your kinetic energy?

It's +V, not -V (though V may be negative). And there's no k anywhere in the hydrogen atom solution.

What calculation do you want me to do?

Either <p²>/2m or <V> for the ground state of hydrogen, take your pick. I'll even do the other one first, if you agree to do one of them.

Or you could do the whole thing, if you have some alternative method for calculating the ground state energy of the hydrogen atom. Or you could admit that you can't do the calculation, and I'll do the whole thing for you.

 

[johanfprins]

Is the Hafele-Keating experiment not good enough for you?
http://en.wikipedia.org/wiki/Hafele–Keating_experiment

No it is not: I have discussed this in detail on Polywell and am not willing to again go through all the arguments here. In the Hafele-Keating experiment, and other flying clock experiments since then, the time difference measured was not a dilation but always a time expansion owing to the fact that the clocks flew at heights were they must tick faster owing to lower gravity. Time dilation as predicted by the Special Theory of Relativity has not been directly measured in these experiments.

And in fact, time dilation does not actually occur on the moving clock itself, but is only observed when compared to another clock which is not moving along. This is a relativistic-effect, just like a parabolic path can be a relativistic effect of a linear path within a passing inertial refrence frame. Although we see a paraboloc path, the path is NOT parabolic within its own inertial reference frame.

Similarly, although according to us a clock moving past us is keeping slower time, this not actually happening on the moving clock itself. The Hafele-Keating experiment only measured a time expansio which I believe is caused solely by the lower gravity that the clock experiences while it is up in the air.

It is for this reason that I have proposed the loom experiment since the moving shuttle and the remaining shuttle experience the same gravity field all the time. There is no extra interpretation required, The result will either show a time dilation on the moving shuttle or show no time-dilation at all. I am 100% sure that it will show no time dilation. It is an easy and cheap experiment to do and the result should settle this issue for all time. So let's do it!

 

[Ziggurat]

WE all know that the ground state energy is E₀=-13.7 eV

Yes, but quantum mechanics can actually calculate this value independently of the measurement. That's the whole point of a theory: getting something which can make quantitative predictions that are accurate. Standard quantum mechanics can do that. Standard quantum mechanics can produce this number from the theory. But you don't believe in standard quantum mechanics.

Can you do a calculation to derive this number? Because you certainly haven't.

 

[ben m]

It is for this reason that I have proposed the loom experiment since the moving shuttle and the remaining shuttle experience the same gravity field all the time. There is no extra interpretation required, The result will either show a time dilation on the moving shuttle or show no time-dilation at all. I am 100% sure that it will show no time dilation. It is an easy and cheap experiment to do and the result should settle this issue for all time. So let's do it!

Ha ha ha! Cheap and easy? Johan, the fastest solid objects ever accelerated by mankind---the Pioneer probes---moved at 10^4 m/s, or 0.00003c. At this speed, the gamma factor differs from 1 by only 5x10^-10. The fastest controlled land speeds are of order 10^3 m/s (ultracentrifuge rotors, for example) putting gamma at 1 + 1e-12.

Meanwhile, the most precisely-known radioactive decay rates (the muon, in fact) have an experimental uncertainty of 2x10^-5. Decay rate precision for radioactive nuclei are 10^-3 at best. It would be utterly impossible to detect a one-part-in-10^-10 change in the muon decay rate, so the experiment you propose is utterly impossible. (Think for a moment about the statistical error on any decay-rate measurement and you'll see why these numbers are what they are.)

Meanwhile, yes, the muon storage ring does exactly this, except with a fast bunch of ions rather than a slow solid object. A bunch of muons is sent out-and-back at high speed in the lab frame. During the trip, a comparable bunch of muons in the lab experiences 10 half-lives, as is obvious from their decreased numbers. When the traveling muons get back, 65% of them are still alive, indicating that the bunch aged by less than one half-life during the trip. The standard prediction was right and your prediction is wrong.

The only difference between this and your proposal is that we don't *actually* set up the comparable-bunch-of-muons-in-the-lab. But we've observed the decay rate of such muons a zillion times, and "2.2us passed on the lab clock" means exactly the same thing as "a bunch of muons at rest in the lab went through one half-life".

 

[sol invictus]

WE all know that the ground state energy is E₀=-13.7 eV as measured from the vacuum level and that at the vacuum level the stationary energies of the proton and the electron are m_p*c² and m_e*c² while the stationary energy of the hydrogen atom is M_H*c². Thus the energy of the hydrogen atom is clearly the latter which can be obtained from the stationary energies of the proton and the elerctron as:

M_H*c²=(m_P*c²+m_e*c²)-13.7.

What is your problem?

Calculate that -13.7eV, Johan.

You can't, because you're incompetent. If you could, you'd learn that the electron has non-zero, positive kinetic energy when bound to a proton. Go ahead, calculate it and prove me wrong.

And by the way, your nonsense "logic" applies to planets orbiting the sun just as well as it applies to hydrogen. The mass of the earth-sun system is less than the mass of the earth plus the mass of the sun. Does the earth on its orbit have negative kinetic energy, Johan? Is it impossible to even define its kinetic energy? Of course not, to both.

 

[sol invictus]

It is for this reason that I have proposed the loom experiment since the moving shuttle and the remaining shuttle experience the same gravity field all the time. There is no extra interpretation required, The result will either show a time dilation on the moving shuttle or show no time-dilation at all. I am 100% sure that it will show no time dilation. It is an easy and cheap experiment to do and the result should settle this issue for all time. So let's do it!

You're 100% wrong about this, just as you are about every other aspect of physics. Variations on this experiment have been done hundreds of times, and every single one came back consistent with relativistic time dilation, both gravitational and velocity-induced. The GPS system relies on both corrections, if either wasn't there, it wouldn't work.

 

[Gods Big Toe]

No it is not: I have discussed this in detail on Polywell and am not willing to again go through all the arguments here. In the Hafele-Keating experiment, and other flying clock experiments since then, the time difference measured was not a dilation but always a time expansion owing to the fact that the clocks flew at heights were they must tick faster owing to lower gravity. Time dilation as predicted by the Special Theory of Relativity has not been directly measured in these experiments.

Since the equations predicted both types of time dilation and the results were virtually what was predicted, you must be saying that both are wrong. It then must be an incredible coincidence that the the equation for gravitational time dilation was wrong by virtually the amount calculated for velocity induced time dilation.

Can you show how you derived this number?

 

[ben m]

WE all know that the ground state energy is E₀=-13.7 eV

Every physics undergrad in the world can derive that number, Johan. Start with the electron mass, the proton mass, and electron charge. Apply Coulomb's Law and the Schrodinger equation.

Let me point out two aspects of your apparent lack of quantum skills: (a) not knowing how to carry out a basic quantum calculation, and (b) not recognizing that this is calculable. If I were giving an exam to an undergrad class, and asked "Calculate the binding energy of ground-state hydrogen", a student who answered like you ("you measure the mass difference, everyone knows this is -13.7") would get a worse grade than someone who said, say, "well, I started an eigen-energy calculation but I got lost in the algebra and ran out of time."

 

[Reality Check]

The simple fact is that for objects, like a cannon ball, its rest mass energy is the same at every position in space and also within any gravitational field. Therefore, one can neglect the rest mass energy as a baseline energy.

The simple fact (as this "yelping doggie" Reality Check) knows is that by definition the rest mass energy is called the rest mass energy because the body is at rest!
The energy of an electron is its rest mass emergy + any relativistic mass - any potential enerygy.
Thus an electron within the lowest energy state of a hydrogen atom has LESS energy than its rest mass energy because it is in a potential well.

The rest of the post is a display of ignorance.
A quantum particle in a potential well has energy. It is usual to call this kinetic energy to match the same usage for a classical particle in a potential well.

 

[Reality Check]

....
You then find from the Lorentz transformation that the rod does not change in length at all.

You have obviously made a fatal error in your calculations since it is an undergraduate exercise to show that the length of a moving rod does change in SR.

 

[Reality Check]

...
Within its own inertial reference frame it keeps the same time as any other clock keeps within its own inertial reference frame.
...
I then derive that owing to the validity of the relativity of simultaneity, one cannot see the front and rear of a passing rod at the same isnstant in time and therefore one cannot see the length of a passing rod at all
...

And this seems to be your fatal, undergarduate error.
You have done your calculations the frame of a rod where there is no time dilation or length contraction or relativity of simultaneity !
The fact is that in the undergraduate exercise of calculating the length of a passing rod, one can see the length of a passing rod every time.

 

[Reality Check]

Having made those concessions, however, I'm going to borrow Farsight's "ton of bricks" to throw at johanfprins the next time Dr Prins suggests someone is "a total moron", "an utter fool", "oaf", "idiotic", "criminal", "fundamentalist zealot", "yelping doggie", "cowering anonymous troll", "crackpot", "insane", "totally incapable and incompetent", "DECEITFUL", "BLATANTLY lying", "STUPID", "a shamaan", "Voodoo", "paranormal", "flat earthers", "absurd", or "deviously dioshonest" [sic] when their only offense was to point out that some aspect of Prins's argument is inconsistent with accepted physics.

You missed out my new favourite insult from johanfprins: apparently I am a "yelping doggie" for pointing out undergraduate physics (the derivation of length contraction in SR) to him!

 

[Reality Check]

My problem is with the so-called twin paradox. Although each twin is under the impression that the other twin's clock is going slower and has to take it into account when, for example, communicating with the other twin, the clocks with the two twins are actually keeping exactly the same time within their own reference frames respectively. Thus, one twin cannot actually age slower than the other twin. My conclusion is that when they meet up one day they will find that they still have the same age.

You do not understand the twin pardox.
What causes the traveling twin to age differently is the acceleration.

Starting with Paul Langevin in 1911, there have been numerous explanations of this paradox, many based upon there being no contradiction because there is no symmetry—only one twin has undergone acceleration and deceleration, thus differentiating the two cases. Max von Laue argued in 1913 that since the traveling twin must be in two separate inertial frames, one on the way out and another on the way back, this frame switch is the reason for the aging difference, not the acceleration per se.[1] Explanations put forth by Albert Einstein and Max Born invoked gravitational time dilation to explain the aging as a direct effect of acceleration.[2]
The twin paradox has been verified experimentally by precise measurements of atomic clocks flown in airplanes and satellites. For example, gravitational time dilation and special relativity together have been used to explain the Hafele–Keating experiment.[3]

You are saying that time dilation never happens because all clocks measured in their own frame keep the same time in that frame. That is ignorant because time dilation is that clocks in different inertial frames keep different times. This has been experimentally verified.
What is the experimental basis of Special Relativity?

 

[GlennB]

This thread is like watching Sumo wrestling.

You don't really understand what it's all about, but it's magnificent anyway

 

[Reality Check]

You missed out my new favourite insult from johanfprins: apparently I am a "yelping doggie" for pointing out undergraduate physics (the derivation of length contraction in SR) to him!

Whoops - just noticed that you include that insult from johanfprins in your list .

 

[Farsight]

Sorry I haven't posted for a few days guys. Friday night was the wife's work Christmas do, on Saturday morning I had my orders to go get a Christmas tree, and it all went downhill from there. Bah humbug!

 

[johanfprins]

Originally Posted by johanfprins:
Then do it.

Zig: That's not the deal, johan.

I'll do half the calculation, IF you promise to do the other half. You even get to pick which half, <V> or <p2>/2m.

It is not a question of not being able to do any of the two calculations, but a refusal on my part to calculate parameters which have no physics-meaning whatsoever.

Or admit that you can't, and I'll do the whole thing.

I know that you can do the whole thing and is willing to do so. Why not: You have posted so many comments which are not physics. Go ahead and calculate two parameters which are also not physics. But I refuse to play a role in calculating paranormal physics.

JFP: Well we know that we need a photon of approximately 13.7 eV to break up a hydrogen atom.

Zig:But you can't predict this number, can you? You can't calculate this energy. You can't tell me why it's 13.7, not 13.8, or 26.4, or any number, can you?

Amazingly I can calculate this number to an excellent approximation by using the Schroedinger equation and inserting the classical potential field between an "electron-particle" and a "proton particle". Why this is possible nobody knows since the Schroedinger equation has been a guess.

After solving the equation, the solution is, however, a standing wave with zero momentum and zero kinetic energy, having an energy of approximately 13.7 eV below the vacuum level. This is the same as the mass-energy that has been radiated away: And therefore the hydrogen atom has a mass of M_H=m_P+m_e-MINUS-13.7(eV)/c² without any "internal" kinetic energy.

JFP: Now go ahead and calculate the masses (delta)mp and (delta)me as you claim that you can do.

Zig: Not until you agree to my terms above. They're quite fair.

They are not fair since you want me to calculate parameters which are not physically relevant. As a realistic physicist I protest.

JFP:
The solution derived from Schroedinger's equation for Sol's famous standing waves in a box

Zig: This is the hydrogen atom we're talking about. The solution isn't the wave Sol gave for the particle in a box.

It does not matter which solution you take: The parameters you want me to calculate, has no physical meaning at all.

JFP: gives the energies En=(hbar*kn)2/me)-V, which must be interpreted as an approximation for (delta)mP+(delta)me: So where is your kinetic energy?


Zig: It's +V, not -V (though V may be negative).

According to the text books and according to you I can choose to measure the potential energy from any level? Or do you agree with me that in the case of the Schroedinger equation this must not be done since the only valid reference level is the vacuum level: If the latter is correct, and it must be correct, then the potential well within which an electron becomes trapped when it bonds to a nucleus, or within a material for that matter MUST always be negative.

And there's no k anywhere in the hydrogen atom solution.

Any harmonic wave, and a Schroedinger wave is a harmonic wave, "lives" within both position space and k space (reciprocal space) simultaneously. To go from one space to the other one uses a Fourier transform. Thus if you have the wave-function within position space, you can always calculate <k> and <k²> for any Schroedinger wave-function. There are then three possibilities:

1. <k>=k and <k²>=0: In this case you have a travelling coherent wave moving past with a momentum p which relates to k by de Broglie's relationship p=(hbar)*k.

2. <k>=0 and <k²> NOT EQUAL TO 0. In this case you have a standing wave without any momentum and without any kinetic energy. Although you can still use (hbar) to write p=(hbar)*k this expression is now physically meaningless since it cannot be a momentum.

3. <k>=k₀, with <(k-k₀)²> NOT EQUAL TO zero. In this case you have a wave packet that moves with a group velocity which cannot be derived from the “momentum” (hbar)*k0. Thus also in this case the expression p=(hbar)*k₀ is meaningless since it does not relate to an actual momentum.

JFP: What calculation do you want me to do? Zig: Either <p2>/2m or <V> for the ground state of hydrogen, take your pick. I'll even do the other one first, if you agree to do one of them.

I told you that I refuse to calculate parameters which have no physics-meaning. You want to argue that the first expression proves that there is kinetic energy. According to Ehrenfest’s theorem you have that: <F>=(d/dt)<p>: Thus for a distance ds one must have that: <F>ds=(ds/dt)*d<p>=<v>*d<p>= (1/m)*<p>*d<p> If you now integrate to get the kinetic energy you get <p>²/2m AND NOT <p²>/2m. Similarly, once quantum mechanical bonding has occurred, the classical potential energy plays no role anymore. Thus to calculate <V> is paranormal nonsense. The physics in which you, Sol, Ben and your choir specialise in.

 

[sol invictus]

Amazingly I can calculate this number to an excellent approximation by using the Schroedinger equation and inserting the classical potential field between an "electron-particle" and a "proton particle". Why this is possible nobody knows since the Schroedinger equation has been a guess.

Yes, it's amazing that a guess works just by coincidence, isn't it? Not only does it calculate the ground state energy of hydrogen correctly, it calculates all the energy levels, tiny corrections like fine and hyperfine shifts, selection rules, and the structure of the periodic table. It predicted nuclear fission and fusion, it predicts the results of accelerator experiments to 10 significant figures, it made possible all modern chemistry and electronics.

All of that just blind luck according to you! Wow!!

According to us, the Schrodinger equation is not a guess at all (it too is derivable - but you can't do that derivation either, and you didn't know that), and we understand quite explicitly where the -13.6eV comes from. And according to us, the electron has a non-zero, positive kinetic energy that's very simply related to that number. Can you tell us how it's related? No, because you haven't a clue how to do any of these computations.