Superconductivity at room temperature??

[Ziggurat]

Wrong. Spheres always appear circular in cross-section, regardless of the observer's velocity or acceleration.

http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2049244

This relates to the common error of confusing what one sees and what one observes. Lorenz contraction is what we observe, not what we see. johan made this same mistake earlier when he mentioned straight lines becoming curved.

 

[ben m]

Thus, Hafele and Keating, and those who followed in their footsteps must have fudged their results to obtain what they wanted to obtain.

On second thought, I want to react more strongly to this.

I submit that the thread is over. Nothing Johan says can be critiqued by comparing it to experiments, because Johan can reject any such comparison by saying someone "must have fudged" their results. It doesn't help if the experiment was replicated, because it's not just a dishonest scientist or two but also "... those who followed in their footsteps".

Johan, I was willing to argue with you in order to figure out what your theory said in order to compare it to experiments. That's what one does with a scientific theory, and your theory (as amateurish as it was) appeared to at least belong to the domain of science. I was wrong---your theory belongs to the domain of delusion. Looking for scientific errors in your theory is like looking for continuity errors in "Finnegan's Wake"---it's just the wrong thing to do. (Of course you find them, if you have the patience to plow through the obscurities and treat it like it's supposed to make sense., but it's not the point.)

I think the best thing to say is, "Uh-huh. Sure. That's interesting, buddy. Say, I see someone I know at the other end of the bar. Good luck with your, uh, stuff. Bye."

 

[D'rok]

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. We have just heard that there are "signs of" the (non-existent) Higgs-boson at CERN. I even doubt Rubbia's claims on the "vector bosons".

In short:

Your intuition trumps theory.
Your intuition trumps mathematics.
Your intuition trumps experiment.

All must give way to Prinsics

And rather than admit error, you accuse other scientists of fraud.

What a piece of work.

 

[ben m]

You know that it looks flattened because you've changed. And you've changed in that you're "smeared out". You're length-stretched rather than length-contracted, and as a result everything you see outside your spaceship looks length-contracted.

No, that's utterly not the right way to think about it. It's rather defeating the point of relativity to say "the rest of the Universe is a special frame, where things have special shapes, and I think they're distorted because I am moving and therefore the wrong shape." The point is that any observer can get a perfectly-valid picture by treating himself as at rest.

Secondly, no, it doesn't actually work. Here's a counterexample. We have you, a dog, at rest w/r/t you, which you measure to be 1m, and a cow (moving towards you at 0.866c) which you also measure to be 1m long. (The cow, in its rest frame, thinks it is 2m long.) Now you accelerate to 0.866c w/r/t your previous frame. In your picture, it sounds like you expect the dog to be measured as 0.5m long ("because you measured it with an 2x-expanded ruler") and the cow as 0.5m long (likewise)

But the actual Lorentz transformation tells that, after having accelerated, you measure the dog to be 0.5m long and the cow to be 0.28m long. Therefore the length-changes cannot be ascribed to "stretches" induced during acceleration of the observer. Your attempt at a clever-but-nonstandard way to explain relativity is simply wrong; it does NOT give you a good basis for understanding why lengths contract when viewed by a moving observer.

(Those decimals can all be made exact---make the speed sqrt(3)/2, and the 0.5 is 1/2, the 0.28 is 2/7.)

ETA: if you want even easier algebra, add a horse. In the original setup, the horse is moving towards you, in the opposite direction as the cow, at 0.877c, and appears 1m long. After you've accelerated, the horse is at rest w/r/t you, and it now appears 2m long. That makes no sense if you suppose that you and your ruler "have expanded"---how would an expanding-ruler simultaneously make the dog appear shorter, the cow appear much shorter, and the horse appear longer? It does make sense in normal relativity.

 

[Farsight]

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

Lorentz contraction is what you see, sol. So the Lorentz transformation correctly predicts it. But when I see you as length contracted, and you see me as length contracted, we know that we aren't both shorter than each other. That's the pole-and-the barn paradox, which like all the other so-called paradoxes, isn't really a paradox.

Wrong. Spheres always appear circular in cross-section, regardless of the observer's velocity or acceleration.
http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2049244

I know about this, see for example http://en.wikipedia.org/wiki/Terrell_rotation, but that isn't the "objective" view, and IMHO suggests we don't observe length contraction. I think this web page gets it across pretty well.

 

[Reality Check]

Another one of your blatant lies: Derivable from what classical physics? .

Another one of your blatant goalpost moves: sol invictus never stated that the Schrödinger equation can be derived from classical physics.
He stated that it can be derived and it can.

The Schrödinger equation can be derived from the classical total energy of a particle to which you add the quantum bits of

• Einstein's light quanta hypothesis of 1905 and
• The de Broglie hypothesis of 1924.

 

[Ziggurat]

Lorentz contraction is what you see, sol.

No it isn't, Farsight. It's what you observe. And what you observe and what you see are not the same thing. You observe length contraction. But you see Terrell rotation.

 

[Reality Check]

We are both pro Special Relativity but we are not stupid about it. For example, even Einstein stated clearly that the time rate of a moving clock slows down within the reference frame relative to which it is moving. You people are so dumb that you believe that the clock also slows down within the reference frame within which it is stationary.

You seem to be lying - I have seen no posts fromn anyone that state that.

What everyone agrees with is what is in the textbooks. An observer in a different inertial reference frame from a clock will observe that the time rate of a moving clock slows down within the reference frame relative to which it is moving.

Thus, Hafele and Keating, and those who followed in their footsteps must have fudged their results to obtain what they wanted to obtain.

That is insanely insulting to Hafele and Keating and many other scientists.

I will be charitable and hope that you are just ignorant about the Hafele–Keating experiment.
They measured the time dilation that happened for atomic clocks that traveled by plane relative to stationary clocks. This is a test of the twin paradox:
What is the experimental basis of Special Relativity?

Tests of the “Twin Paradox”
The so-called “twin paradox” occurs when two clocks are synchronized, separated, and rejoined. If one clock remains in an inertial frame, then the other must be accelerated sometime during its journey, and it displays less elapsed proper time than the inertial clock. This is a “paradox” only in that it appears to be inconsistent but is not.

• Hafele and Keating, Nature 227 (1970), pg 270 (proposal).Science Vol. 177 pg 166–170 (1972) (experiment).They flew atomic clocks on commercial airliners around the world in both directions, and compared the time elapsed on the airborne clocks with the time elapsed on an earthbound clock (USNO). Their eastbound clock lost 59 ns on the USNO clock; their westbound clock gained 273 ns; these agree with GR predictions to well within their experimental resolution and uncertainties (which total about 25 ns). By using four cesium-beam atomic clocks they greatly reduced their systematic errors due to clock drift.
  Criticised in: A. G. Kelly, “Reliability of Relativistic Effect Tests on Airborne Clocks”, Inst. Engineers Ireland Monograph No. 3 (February 1996),http://www.cartesio-episteme.net/H&KPaper.htm. His criticism does not stand up, as he does not understand the properties of the atomic clocks and the way the four clocks were reduced to a single “paper” clock. The simple averages he advocates are not nearly as accurate as the paper clock used in the final paper—that was the whole point of flying four clocks (they call this “correlated rate change”; this technique is used by all standards organizations today to minimize the deficiencies of atomic clocks).
  Also commented on in Schlegel, AJP 42, pg 183 (1974).He identifies the East–West time difference as the Sagnac effect, notes that this is independent of the clock's velocity relative to the (rotating) Earth, and proposes a coordinate system in which it is treated just like the international date line (for use in highly accurate time transfer around the world); while correct, this has been superceded by the ECI coordinate system of the GPS.
  Here is a brief description of a repetition in the UK (2005): http://www.npl.co.uk/upload/pdf/metromnia_issue18.pdf (Page 2)
• Vessot et al., “A Test of the Equivalence Principle Using a Space-borne Clock”, Gel. Rel. Grav., 10, (1979) 181–204.“Test of Relativistic Gravitation with a Space borne Hydrogen Maser”, Phys. Rev. Lett. 45 2081–2084.They flew a hydrogen maser in a Scout rocket up into space and back (not recovered). Gravitational effects are important, as are the velocity effects of SR. This experiment is also known as “Gravity Probe A”.
• C. Alley, “Proper Time Experiments in Gravitational Fields with Atomic Clocks, Aircraft, and Laser Light Pulses,” in Quantum Optics, Experimental Gravity, and Measurement Theory, eds. Pierre Meystre and Marlan O. Scully, Proceedings Conf. Bad Windsheim 1981, 1983 Plenum Press New York, ISBN 0-306-41354-X, pg 363–427.They flew atomic clocks in airplanes that remained localized over Chesapeake Bay, and also which flew to Greenland and back.
• Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268 (July 28, 1977) pg 301.Bailey et al., Nuclear Physics B 150 pg 1–79 (1979).They stored muons in a storage ring and measured their lifetime. When combined with measurements of the muon lifetime at rest this becomes a highly relativistic twin scenario (v ~0.9994 c), for which the stored muons are the traveling twin and return to a given point in the lab every few microseconds. Muon lifetime at rest:Meyer et al., Physical Review 132, pg 2693;Balandin et al., JETP 40, pg 811 (1974);Bardin et al., Physics Letters 137B, pg 135 (1984). Also a test of the clock hypotheses (below).

 

[Farsight]

It's best to avoid visual language for observations, since spherical bodies still look spherical, but if you s/b 'observe', etc., yes.

Thanks. What does s/b mean?

Not really. The situation is precisely analogous to you taking a pen, holding it horizontally and twirling it in front of yourself, and observing that its horizontal length changes. (Or if you prefer, leave it in place and walk around it.)

I'm not fond of that analogy, Vorpal. A better analogy is looking at a square from the top, then changing your viewpoint to a 45-degree angle. You now see a rectangle. You can make this square the face of a cube if you like. The important point is that your perspective changes, but the square doesn't.

You can describe it in several ways, but all that's "really" going on is that your orientation relative to the pen changes. In spacetime, orientation is velocity and Lorentz transformations are rotations.

Sorry, but what's really going on is that you're changing your state of motion through space, and as a result your measurements of time and space change. Yes you get a different perspective on the world, and we do talk of Lorentz transformations and rotations, but you're just moving faster, that's all.

Try this to understand the smearing thing: draw a vertical line say an inch long upwards on a piece of paper to represent the path of the front of a photon in the parallel-mirror light-clock. Now draw another vertical line on top of the first to represent the path of the next portion of the photon. Repeat this half a dozen times. Now do the whole thing again but this time whilst pulling the paper along horizontally. You now see something like this ///////. Take enough sampling points along the photon and there's no gaps. What was a vertical line is now smeared. It's sheared too, but I hope that isn't an issue.

 

[sol invictus]

Lorentz contraction is what you see, sol.

Nope. It's what we infer from careful measurements, which of course take into account the finite speed of light where necessary.

When you hear a car horn go NEEEEEEEUUUUUUUU as it passes you, you hear a shift in frequency, but you infer that the horn had a fixed frequency, and the shift is due to its relative motion and the finite speed of sound. For a spherical star, you see no deformation of the shape - from which you infer that the shape must in fact be squashed, taking into account the finite speed of light.

Put another way - if the star was a sphere in motion, it would look squashed (or rather, stretched) because of light propagation effects. Therefore since it looks circular, it must actually be squashed. Tricky, but not that complicated.

I know about this, see for example http://en.wikipedia.org/wiki/Terrell_rotation, but that isn't the "objective" view

Do you or do you not agree that a photograph (or your own visual perception) of a spherical (in its own rest frame) star will show a circular disk, regardless of your speed or acceleration relative to it?

If you agree, you need to clarify what you wrote before. If you don't you're sharply disagreeing with relativity, because that's a totally unambiguous experimental prediction it makes.

 

[Vorpal]

I'm not fond of that analogy, Vorpal. A better analogy is looking at a square from the top, then changing your viewpoint to a 45-degree angle. You now see a rectangle. You can make this square the face of a cube if you like. The important point is that your perspective changes, but the square doesn't.

That was only half the point of the analogy. The object's projections to (say) your left/right axis and your back/forward axis are what's analogous to the space and time measurements. So if you want a perspective-based analogy, the "horizontal length" of the pen (or whatever) actually is essential here. You're missing out on how the Lorentz transformation works if cut it out.

Sorry, but what's really going on is that you're changing your state of motion through space, and as a result your measurements of time and space change.

That's exactly what I said: "in spacetime, orientation is velocity...".
You figure out measured durations by projecting to your velocity four-vector. You figure out measured spatial lengths by projection to the subspace perpendicular to it.
That's very analogous to the back/front axis and the horizontal axis in the perspective analogy. What's the problem?

What was a vertical line is now smeared. It's sheared too, but I hope that isn't an issue.

Drawing a spacetime diagram for the bouncing-light clock is very simply, but what this drawing is supposed to do escapes me.

 

[ben m]

Farsight, Sol's terminology is the standard. The distinction is important because we there's a distinction between, "In what order, and from what angles, do photons arrive at your eyes from these events?" and "What coordinates do you ascribe to events after taking light-travel-time into account?" The whole business of Lorentz contraction is strictly discussing the latter.

If you want to know about the former, well, figure out where the events are in a coordinate system, then work through the retarded times. This is such an important (and standard) distinction, that whenever I've set a relativity exam for my sophomores, I've tested the distinction explicitly. "What does Sally's watch read at the time the cannonball reaches her? At what time would she infer the cannon had been fired?" Etc.

 

[Farsight]

No, that's utterly not the right way to think about it. It's rather defeating the point of relativity to say "the rest of the Universe is a special frame, where things have special shapes, and I think they're distorted because I am moving and therefore the wrong shape." The point is that any observer can get a perfectly-valid picture by treating himself as at rest.

No it isn't, because the observer can change his speed, and he's smart enough to know that his observations are changing, not the stars, not the distances between the stars, not the universe, and not reality. The real point of relativity comes out when you take on board the wave nature of matter and appreciate The Other Meaning of Special Relativity. People tend to dismiss this paper, but IMHO they're wrong to do so.

Secondly, no, it doesn't actually work. Here's a counterexample. We have you, a dog, at rest w/r/t you, which you measure to be 1m, and a cow (moving towards you at 0.866c) which you also measure to be 1m long. (The cow, in its rest frame, thinks it is 2m long.) Now you accelerate to 0.866c w/r/t your previous frame. In your picture, it sounds like you expect the dog to be measured as 0.5m long ("because you measured it with an 2x-expanded ruler") and the cow as 0.5m long (likewise)

It does work. I measure the dog to be 0.5 metres long, because I'm now moving relative to it, and gamma = √(1 - v²) where v is my speed as a fraction of c. That’s √(1 - .866²) = √(1 - .75) = √0.25 = 0.5. Now forget the dog. I don't measure the cow's length to be half-reduced again because the velocity addition formula applies. (See my last paragraph below for why). I don't see the cow going by at 1.732c. I see it going by me at V+U/1+VU = 1.732 / 1+.866² = 1.732/1.75 = .9897c. Plug that back into √(1 - v²) and gamma = √(1 - 0.9795) = .14317 so I now measure the cow as 0.28m long. I'm smeared out with respect to the cow, so when I measure it, it comes up short. And the cow's smeared out with respect to me, so when it measures me, I come up short too.

But the actual Lorentz transformation tells that, after having accelerated, you measure the dog to be 0.5m long and the cow to be 0.28m long. Therefore the length-changes cannot be ascribed to "stretches" induced during acceleration of the observer. Your attempt at a clever-but-nonstandard way to explain relativity is simply wrong; it does NOT give you a good basis for understanding why lengths contract when viewed by a moving observer.

It isn't wrong ben. The whole point of all this comes out of the wave nature of matter.

(Those decimals can all be made exact---make the speed sqrt(3)/2, and the 0.5 is 1/2, the 0.28 is 2/7.)

No problem.

ETA: if you want even easier algebra, add a horse. In the original setup, the horse is moving towards you, in the opposite direction as the cow, at 0.877c, and appears 1m long.

No problem. The horse and I have a relative velocity of .866c so I measure it like I measured the cow.

After you've accelerated, the horse is at rest w/r/t you, and it now appears 2m long. That makes no sense if you suppose that you and your ruler "have expanded" ---how would an expanding-ruler simultaneously make the dog appear shorter, the cow appear much shorter, and the horse appear longer? It does make sense in normal relativity.

You're seeing this wrong. The horse appears 2m long, the length it always appears when it has no motion relative to me. If I'm smeared out so's the horse, so we don't look smeared out to one another.

To see this right, imagine yourself to be a circle of light. You define your seconds using the motion of light, such that one turn round the circle is a second. You then apply this to orthogonal light waves coming through the middle of the circle and say they're moving at 299,792,458 m/s. (The metre is based on the circumference of the circle but don't worry about it). Now you move towards those oncoming light waves. Your circle is now a helical section, and for light to travel all the way around it takes longer than it did because the light has further to go. But you aren't aware of this. You think you're still a circle wherein one rotation round the circle is a second. You then re-measure the speed of the oncoming light. Your closing rapidity is higher but your second is bigger, and this cancels out the extra closing rapidity so you still measure 299,792,458 m/s. Even if you travel towards the light waves at a significant fraction of c you still measure them coming at you at 299,792,458 m/s. Your helical section is stretched out so it's almost straight, the closing rapidity is nearly 2c, but that's not how you see it. You measure speed as distance over time, both are calibrated on the motion of light round your circle, and there's nothing else to calibrate them on. And of course, if those light waves coming at you are going round in their own circles because they're a cow, you still don't see them coming past you at more than c.

Don't forget that you don't see light waves as infinitely length contracted, and don't forget that the Lorentz factor γ comes straight out of Pythaogras' theorem as per the Simple inference. There's a reciprocal because length contraction is the opposite sense to time dilation. But anyway, I've spent too long on this. The point I wanted to get across is that as per the mass-deficit of the bound electron, Johan has again come out with something that cuts across the grain, but he's not wrong. You just have to cut across the theories to see it.

 

[ben m]

It does work. I measure the dog to be 0.5 metres long, because I'm now moving relative to it, and gamma = √(1 - v²) where v is my speed as a fraction of c. That’s √(1 - .866²) = √(1 - .75) = √0.25 = 0.5. Now forget the dog. I don't measure the cow's length to be half-reduced again because the velocity addition formula applies. (See my last paragraph below for why). I don't see the cow going by at 1.732c. I see it going by me at V+U/1+VU = 1.732 / 1+.866² = 1.732/1.75 = .9897c. Plug that back into √(1 - v²) and gamma = √(1 - 0.9795) = .14317 so I now measure the cow as 0.28m long.

Right. That's the normal Lorentz transformation. That's not "think of it as 'you are smeared out'." If you are smeared out, how can you be smeared out by a factor of 2 when you're looking at one thing, but smeared out by a factor of 7/2 at the exact same time when you're looking at something different? You can't. Because "think of it as 'you are smeared out'" is nonsense.

 

[ben m]

The point I wanted to get across is that as per the mass-deficit of the bound electron, Johan has again come out with something that cuts across the grain, but he's not wrong. You just have to cut across the theories to see it.

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.

 

[leonAzul]

The laws of quantum mechanics---i.e., KE = (p-hat)^2/2m---will look at this wavefunction and get that fact right. (It'll say p = {left or right}, <p> = 0, <p^2> != 0, KE != 0, etc.)

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.

 

[Vorpal]

p is the momentum operator that acts on physical states (often written with a hat). It determines the possible measurement outcomes, and, together with a given state, the distribution of said outcomes. ‹p› is its expected value, i.e., average over many identical systems. That part is standard probability theory; a common alternative notation in math books is E[...] rather than angle brackets.

 

[Ziggurat]

I don't understand the use of <angle brackets> here.

In case Vorpal's answer merely whets your appetite, you can find more information about the notation here. But that might be more information, and on a more mathematically abstract level, than you want.

 

[W.D.Clinger]

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.

To add to what Vorpal and Ziggurat said above, the angle brackets are a standard notation for the probability-weighted average of the thing they surround.

Hence <p> is the average value of the momentum p, and <p²> is the average value of the momentum-squared (which is important because <p²>/2m is the kinetic energy that johanfprins has been denying).

Mathematicians refer to probability-weighted averages as expected values; physicists tend to call them expectation values. Those are two slightly different names for exactly the same thing.

An ordinary arithmetic mean (or average) is a simple case of expected/expectation values. Consider, for example, the possible outcomes of rolling a single die: 1, 2, 3, 4, 5, or 6. All six of those outcomes are equally likely, so the expected/expectation value for rolling a single die is

(1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5​

Note well that the expected/expectation value for that experiment is not a possible outcome of that experiment.

[size=-1](For almost a year now, johanfprins has been denying standard quantum mechanics. One of his main arguments consists of his personal belief that the expected/expectation value must be the most likely outcome of an experiment.)

I am not making that up:
http://www.internationalskeptics.com/forums/showthread.php?postid=6786825#post6786825
http://www.internationalskeptics.com/forums/showthread.php?postid=6795094#post6795094[/size]​

More generally, the expected/expectation value for a discrete set of numerical outcomes x_i is the sum of the products of those outcomes with their probabilities P(i):

<x> = ∑_i (x_i P(i))​

For the continuous case, which is common in physics, we replace the discrete summation by an integral:

<x> = ∫ (x_i P(i)) di​

johanfprins continues to demonstrate his profound non-understanding of expected/expectation values. For example:

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.

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. 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.

It's worth noting that I had corrected johanfprins on essentially this same point just a few days earlier:

Dr Prins is saying <p>=0 does not hold, which implies <p> is nonzero, which implies p cannot be identically zero, which directly contradicts Dr Prins's claim that "it does not have any momentum".

That's just basic calculus, logic, and the definition of the expected value <p>. If p were identically zero, then the expected/expectation value <p> would also be zero.

johanfprins does not learn from his mistakes. He repeats them.

I submit that the thread is over.

As a serious discussion of science, this thread has been over for a long time.

For the past year, johanfprins has been giving us arguments by intimidation, based on his own authority. Ziggurat, sol invictus, and ben m have shown exemplary patience in rebutting those arguments with real science.

D'rok summed it up well:

In short:

Your intuition trumps theory.
Your intuition trumps mathematics.
Your intuition trumps experiment.

All must give way to Prinsics

And rather than admit error, you accuse other scientists of fraud.

What a piece of work.

 

[johanfprins]

Originally Posted by johanfprins:
Quite a coincidence that the correct energy levels were found since the principle quantum numbers later turned out to have nothing to do with orbital momentum, but with the harmonic numbers of standing waves.

Ben: Not a coincidence at all. Or at least, not if you believe anything about quantum mechanics. So: Johan, ignore this.

The Bohr model has a non-zero angular momentum for the ground state of the electron, while the Schroedinger solution has ZERO angular momentum as it should have since the energy of a stationary wave cannot have a kinetic energy component; which would be required if the angular momentum were not zero. The rest of your post is the usual BS that you Sol and Ziggurat are spouting and I will thus not respond to it.