Room-temperature superconductivity?

[tomclarke]

So in earth FOR the muon clock ticks more slowly and it reaches earth before decaying, whereas in muon FOR clocks on earth would appear to tick slowly and the muon decays before it reaches earth?

No in muon FOR the distance from generation to earth is much smaller than in earth FOR. (Sorry, I should have said this earlier).

What happens if you build a muon trap on earth that rapidly decelerates the muon? In earth FOR, would the muon age rapidly and decay before becoming stationary? In muon FOR, would earth clocks speed up massively, and if so, would it start before the muon decays?

The muon would decay shortly after being decelerated, as measured from earth FOR. In muon FOR earth clocks would speed up as muon is slowed down. If you imagine unrealistic very fast deceleration both effects precede the typical lifetime of the muon.

To compare the times of two clocks they need to be co-located at two different times. If there is relative velocity you then need change in FOR (from acceleration) for the colocation ever to be repeated. This will either be symmetric - in which case no difference - or asymmetric, in which case the FOR change accounts for the time difference.

But if the clocks are not just co-located but stationary relative to each other at both comparison events, doesn't the situation have to be symmetric?

No, because the travelling twin would experience acceleration, change of FOR turnaround acceleration, change of FOR, deceleration, chnage of FOR, wheread the "stationary twin would have no acceleration.

So in the case of the twin paradox, if one twin were to launch from earth and land on earth again, as opposed to flying by at close to c, wouldn't both their clocks have to show the same time?

See above.

[johanfprins]

Concerning light and time stopping in a Bose Einstein condensate (BEC), this author implies that the light does not stop, it is absorbed and it's characteristics are preserved. When the BEC is prodded to release the light ( stored information) it gives the illusion that a specific photon was stoped. No need for stopping time.

Obviously it is "absorbed" (please explain to me how this "absorption" occurs): But to be "absorbed" the speed of the incoming light must become zero.

This also happens when an orbital electron-wave absorbs a light photon. The photon stops within the electron-wave to become mass energy so that the mass-intensity of the electron-wave increases. This causes the electron-wave to morph into another shape and size in order to accomodate this mass-energy. If you are stupid you will claim that this involves a "jump by a particle to a higher energy": It is beyond my comprehension why otherwise rational people want to believe such paranormal metaphysical claptrap!!

The intensity of a stationary electron wave is time independent since the amplitude of the wave is a complex function. Thus the absorbed light energy is now also time-independent mass-energy. Thus, within the mass intensity of the electron-wave, time does not exist. If it did the electron would have exploded owing the the concomitant charge-distribution within it.

Light is stopped in the same manner within a Bose-Einstein condensate that forms by the entangling of matter waves. The incoming light is stopped within the Bose-Einstein condensaate to add to the mass-energy. This same amount of light energy can then again be released to move away at light speed.

[johanfprins]

True, but the statement I replied to did not specify how measured, and was therefore meaningless.

I disagree. Within the context that it was asked it was not meaningless.

[johanfprins]

True, as long as both clocks are always in the same FOR. But the inconsistency you claim is not for two clocks always in same FOR.

Which inconsistency are you now referring to?

There is no gloabal universe time.

Oh so the universe is not uniform and has different ages within different regions? This sounds like madness to me. What is then, according to you, the age of our universe?

You can draw a spacelike surface which joins all clock world-lines with clocks showing identical times. But to say the clocks must show the same time beggars the question of how simultaneity can be defined in different FOR. It cannot.

Each and every clock that is stationary within an inertial refrence frame, no matter how fast these clocks are moving relative to one another, must show exactly the same age for our universe; or else inertial reference frames cannot be equivalent as Einstein's first postulate claims that they must be.

[johanfprins]

MWBR

Dan, you are tiring me out at present. In the beginning you asked relevant questions: Now you just regurgitate what is believed by the mainstream theoretical physicists. I know all these theories and models better than you do. I do not believe they are correct. I still do not have the full solution, and therefore it will be a waste of time to pursue this argument in more detail here.

Suffice to say that single quarks have never been observed yet; and that for "particles" to have kinetic energy, there must first be a space within which they can move. They cannot form "within a singularity" and then spew out.

Furthermore, black-body radiation (also called cavity radiation) consists of standing light waves. The energy spacings between these light waves are determined by the size of these standing waves. Thus, if you want to know what the actual, present size of the universe is (not just the observable size) a measurement of the energy-density of the light waves that form the micro-wave background radiation will give it to you directly.

There is even a possibility that this radiation itself might be "dark matter"; since a standing light wave has no kinetic energy: It only has mass energy. I have discussed all these possibilities in my book.

[Teahive]

No in muon FOR the distance from generation to earth is much smaller than in earth FOR. (Sorry, I should have said this earlier).

Instead of saying it travels a shortened distance, why exactly would we not say it's moving faster (or that earth isn't moving faster towards it)? I'm not sure Einsteins principle of constant c precludes this.

No, because the travelling twin would experience acceleration, change of FOR turnaround acceleration, change of FOR, deceleration, chnage of FOR, wheread the "stationary twin would have no acceleration.

So forces cause aging? The twin on earth could easily make herself experience the same acceleration forces without leaving earth. I don't know how this is supposed to explain a difference in aging progress when they meet each other again.

[johanfprins]

I started to write a lengthy rambling message but my circular arguements only increased my confusion, who knows what it would do to others. Instead I will condense it to a few items.

Thanks.

If a frame of reference is centered on a partical, it will have vero velocity- it will be at rest. No mater how the time is manipulated the answer will always be 0 because the numerator is always 0. But this is a classical view.

Yes it can be called "classical" but it remains valid for as long as you invoke mass. Schroedinger's equation invokes mass, thus the electron it models must have x=0 and p=0 within its own inertial reference frame. This means that there can NEVER be any uncertainty in its position OR its momentum.

With quantum mechanics velocity can never be 0 because you never can measureit to such absolute acuracy.

One can NEVER measure any parameter with absolute accuracy, but this does not mean that this parameter cannot have a value that is NOT in the least bit uncertain. Stop believing Heisenberg's paranormal, metaphysicsl claptrap!

Plank length and Plank time (? if time is quantitized)

Prove this paranormal claptrap please.

In a clasical system, instead use a frame of reference where the velocity of the particle is not zero, but some tiny velocity, like 1 micrometer / second. Now the numerator is not zero. Does this allow for distance compression?

NO!!!

I still do not comprehend how this thread answers the Muon issue

The clock on earth measures a longer lifetime for the muon than the clock travelling with the muon measures: Thus, the clock travelling with the muon does not tick slower.

, or the real measured time differences (elapsed time) of an airplane carried atomic clock compared to a non flight atomic clock.

This experiment is flawed and I think the result that they wanted to measure is what they got. It has happenned many times that you find what you set out to find even though what you set out to find is not true. It has also happened that a flawed model predicts the correct result for the wrong reason: The Aharanov-Bohm prediction is an example.

[tomclarke]

No in muon FOR the distance from generation to earth is much smaller than in earth FOR. (Sorry, I should have said this earlier).

Instead of saying it travels a shortened distance, why exactly would we not say it's moving faster (or that earth isn't moving faster towards it)? I'm not sure Einsteins principle of constant c precludes this.

It travelling at almost c anyway, so faster would be a no-no.

No, because the travelling twin would experience acceleration, change of FOR turnaround acceleration, change of FOR, deceleration, chnage of FOR, wheread the "stationary twin would have no acceleration.

So forces cause aging? The twin on earth could easily make herself experience the same acceleration forces without leaving earth. I don't know how this is supposed to explain a difference in aging progress when they meet each other again.

It is not the forces. It is the change in FOR which makes the trajectory through space-time bent.

[tomclarke]

True, as long as both clocks are always in the same FOR. But the inconsistency you claim is not for two clocks always in same FOR.

Which inconsistency are you now referring to?

There is no gloabal universe time.

Oh so the universe is not uniform and has different ages within different regions? This sounds like madness to me. What is then, according to you, the age of our universe?

You can draw a spacelike surface which joins all clock world-lines with clocks showing identical times. But to say the clocks must show the same time beggars the question of how simultaneity can be defined in different FOR. It cannot.

Each and every clock that is stationary within an inertial refrence frame, no matter how fast these clocks are moving relative to one another, must show exactly the same age for our universe; or else inertial reference frames cannot be equivalent as Einstein's first postulate claims that they must be.

"The age of the universe" is a relative measurement. Relative to the observer who is measuring how long his world-line extended back takes to reach the big bang.

How can you measure simultaneity between diferent observers in different FOR and spatially separated?

[Teahive]

It travelling at almost c anyway, so faster would be a no-no.

Why? Because SR says so?

Thought experiment: A. Senna is preparing to race in his relativistic car. He first marches along the track and measures its length L with a tape measure. Then he gets into his car, starts the engine, and accelerates hard. He has an accelerometer and a clock on board, and after a while he calculates that v=a*t exceeds c. He now coasts and times one lap with his clock. He finds that he travelled the distance of the race track in L/v which is less than L/c. He slows down, parks his car and jumps out, proudly exclaiming "I went faster than the speed of light!"

An observer standing next to the race track replies: "Actually you didn't. You just went a shorter distance in your reference frame."

It is not the forces. It is the change in FOR which makes the trajectory through space-time bent.

They're the same. Forces = acceleration = change in FOR.

[tomclarke]

It travelling at almost c anyway, so faster would be a no-no.

Why? Because SR says so?

Thought experiment: A. Senna is preparing to race in his relativistic car. He first marches along the track and measures its length L with a tape measure. Then he gets into his car, starts the engine, and accelerates hard. He has an accelerometer and a clock on board, and after a while he calculates that v=a*t exceeds c. He now coasts and times one lap with his clock. He finds that he travelled the distance of the race track in L/v which is less than L/c. He slows down, parks his car and jumps out, proudly exclaiming "I went faster than the speed of light!"

An observer standing next to the race track replies: "Actually you didn't. You just went a shorter distance in your reference frame."

It is not the forces. It is the change in FOR which makes the trajectory through space-time bent.

They're the same. Forces = acceleration = change in FOR.

I mean it is not the period of acceleration (which could be short) that makes the clocks do different things. It is the whole (different) spacetime trajectory that does this.

[johanfprins]

So in earth FOR the muon clock ticks more slowly and it reaches earth before decaying, whereas in muon FOR clocks on earth would appear to tick slowly and the muon decays before it reaches earth?

No in muon FOR the distance from generation to earth is much smaller than in earth FOR. (Sorry, I should have said this earlier).

In fact it is the actual distance and not the relativistic -distorted distance as observed from the earth. Look at it in this way. Within the muon's FOR (let me also use this term) the decay time of the muon is its actual decay time (tau) as measured within any laboratory. Furthermore, since the muon is stationary, the earth is approaching it at a speed v: Thus if the muon reaches the earth and then decays, the actual height above earth h at which it has been created must be h=v*(tau).

But when looking from earth, the space and time coordinates are distorted; and to obtain this distortion one must use the Lorentz transformation. When synchronizing the clock on earth with the muon’s clock when the muon reaches earth and decays, one finds that the corresponding space and time coordinates at which the muon has been created relative to earth are -H so that H>h and -(TAU) so that (TAU)>(tau): One thus has that H=v*(TAU). Thus the relativistic effect is that both the decay time and the height are observed to lengthen; as they must do since the speed with which the earth and the muon approach each other is the same. There is no length contraction involved.

To compare the times of two clocks they need to be co-located at two different times. If there is relative velocity you then need change in FOR (from acceleration) for the colocation ever to be repeated. This will either be symmetric - in which case no difference - or asymmetric, in which case the FOR change accounts for the time difference.

But if the clocks are not just co-located but stationary relative to each other at both comparison events, doesn't the situation have to be symmetric?

No, because the travelling twin would experience acceleration, change of FOR turnaround acceleration, change of FOR, deceleration, change of FOR, wheread the "stationary twin would have no acceleration.

Consider the two twins within two space ships that start to move at the instant in time when the clocks of the two twins are synchronized. Each one accelerates at the same rate and then decelerates at the same rate: But since the speed that is generated is a relative speed, the acceleration must also be a relative parameter. The first spaceship accelerates at the same acceleration relative to the other spaceship as the other spaceship accelerates to the first. Thus one can choose either one of the spaceships to be stationary and the other one as accelerating. If after a while each decelerates they will eventually reach a relative speed of v=0. They then accelerate and decelerate to reach other. When they reach each other the relative speed will again be zero. Each twin will have observed the same total change in time on the other’s clock. Thus their clocks must show the same time when they meet up again.

Now let only one space ship accelerate (go in free fall) relative to the other space ship, then decelerate to come to a halt and accelerate back and decelerate to come to a halt at the space ship that supposedly did not accelerate. But as far as each clock is concerned it has been the other clock that accelerated away from it and returned. Thus the clocks must again show the same time. This is also borne out by Einstein’s postulate that an accelerating reference frame is also an inertial reference frame. So now we have a contradiction: This is so since Einstein also postulated that acceleration and gravity is the same.

So which is it? Is an accelerating reference frame an inertial reference frame as it must be for sanity to prevail when it comes to the ages of the two twins; or is acceleration and gravity equivalent? If the latter is the case, then an accelerating reference frame cannot be an inertial reference frame in the sense that it is equivalent to a non-accelerating inertial reference frame.

It is my opinion that the slowing down of time within a gravitational field is not a result of acceleration, but that it relates to the wave nature of matter. This would mean that Schroedinger essentially unified gravity with “quantum” mechanics when he postulated his differential wave equation.

[tomclarke]

Consider the two twins within two space ships that start to move at the instant in time when the clocks of the two twins are synchronized. Each one accelerates at the same rate and then decelerates at the same rate: But since the speed that is generated is a relative speed, the acceleration must also be a relative parameter. The first spaceship accelerates at the same acceleration relative to the other spaceship as the other spaceship accelerates to the first. Thus one can choose either one of the spaceships to be stationary and the other one as accelerating. If after a while each decelerates they will eventually reach a relative speed of v=0. They then accelerate and decelerate to reach other. When they reach each other the relative speed will again be zero. Each twin will have observed the same total change in time on the other’s clock. Thus their clocks must show the same time when they meet up again.

Now let only one space ship accelerate (go in free fall) relative to the other space ship, then decelerate to come to a halt and accelerate back and decelerates to come to a halt at the space ship that supposedly did not accelerate. But as far as each clock is concerned it has been the other clock that accelerated away from it and returned. Thus the clocks must again show the same time. This is also borne out by Einstein’s postulate that an accelerating reference frame is also an inertial reference frame. So now we have a contradiction: This is so since Einstein also postulated that acceleration and gravity is the same.

The equivalence is between different FORs. Therefore velocity is relative. Not acceleration is relative.

Gravitation is equivalent to acceleration, but that does not mean that only relative acceleration matters. Absolute acceleration is phyically measurable (though not distinguishable from a gravitational field if GR is correct).

Thus the two cases are different in acceleration experienced by the two spaceships, and not equivalent.

[johanfprins]

The equivalence is between different FORs. Therefore velocity is relative. Not acceleration is relative.

Well in this case a free falling accelerating reference frame cannot be an inertial reference frame as Einstein has postulated and used to develop his theory of gravity.

Gravitation is equivalent to acceleration,

I agree that when you are within a gravitational field in space you will accelerate; but this acceleration does NOT cause time to slow down within your reference frame.

but that does not mean that only relative acceleration matters.

Well if you free-fall past another object and you are of the opinion that you are stationary as you must within an inertial reference frame, then you must conclude that it is the other object that is accelerating past you.

Absolute acceleration is phyically measurable (though not distinguishable from a gravitational field if GR is correct).

How do you measure it? One can measure a force on an object which is enclosed whthin another object that is accelerating. And although the enclosed object will not be able to discern whether this force is caused by a gravity field outside, this force is NOT the same as gravity within the universe. Time does not actually slow down merely because the clock suffers acceleration. It does slow down within a gravitational field owing to the fact that matter consists purely of waves and not of "particles".

Thus the two cases are different in acceleration experienced by the two spaceships, and not equivalent.

I know this has been the belief for nearly 100 years by now: But it is wrong. I have given you the example of the two spaceships accelerating and decelerating symmetrically, and you just ignored it. Do you agree that the twins must, at least in this case, have the same age when they meet up?

[Teahive]

I mean it is not the period of acceleration (which could be short) that makes the clocks do different things. It is the whole (different) spacetime trajectory that does this.

Ok, let's add something to the experiment: Each twin has an extremely long spaceship. They meet tip to tip to synchronize clocks at both ends of each spaceship (now in the same FOR). Then one twin fires up her trusters and heads for the tail end of the other twin's ship at high speed. She decelerates and comes to a halt, gets out and compares her clock to the one at the tail end of the other ship. She radios her readings to the other twin.

At some point in time the "stationary" twin should see the tail end of the other ship approaching. What time will he read at his end?


I'd very much appreciate it if you could find time to answer my other question.