Superconductivity at room temperature??

[ben m]

Good: Thus, when two inertial observers have the same radio-active material with them and they BOTH measure the decay rate of this material on the samples they have with them, will they get the same answer?

Yes. This is standard stuff, Johan, and everyone except you gets this result---along with the "twins have different ages" and "Hafele-Keating clocks run differently"---when we apply the Lorentz transformations.

Go ahead, Johan. Do the math. Consider this experiment as a sequence of events; write down the 4-coordinates of the events. Then show that the Lorentz transformations give you this result. Can you do it?

 

[johanfprins]

Yes. This is standard stuff, Johan, and everyone except you gets this result---along with the "twins have different ages" and "Hafele-Keating clocks run differently"---when we apply the Lorentz transformations.

Go ahead, Johan. Do the math. Consider this experiment as a sequence of events; write down the 4-coordinates of the events. Then show that the Lorentz transformations give you this result. Can you do it?

That is exactly what I have been doing all along.

Let me ask you a simple question: When you have two inertial reference frames K and K^/ moving with a speed v relative to one another, and you have two meter-sticks at rest within both inertial reference frames, do these meter-sticks have the same length as measured within their respective reference frames? Yes or No? And if they do not have the same length, what is the difference in length? Do the latter calculation if you claim that they do not have the same length.

 

[sol invictus]

That is exactly what I have been doing all along.

Let me ask you a simple question: When you have two inertial reference frames K and K^/ moving with a speed v relative to one another, and you have two meter-sticks at rest within both inertial reference frames, do these meter-sticks have the same length as measured within their respective reference frames? Yes or No? And if they do not have the same length, what is the difference in length? Do the latter calculation if you claim that they do not have the same length.

A meter stick at rest in any inertial frame is a meter long, almost (but not quite) by definition. So yes, they have the same length.

 

[ben m]

That is exactly what I have been doing all along.

No you haven't. I have yet to see you write down a single spacetime coordinate.

Let me ask you a simple question: When you have two inertial reference frames K and K^/ moving with a speed v relative to one another, and you have two meter-sticks at rest within both inertial reference frames, do these meter-sticks have the same length as measured within their respective reference frames? Yes or No? And if they do not have the same length, what is the difference in length? Do the latter calculation if you claim that they do not have the same length.

Yes, Johan, this is standard stuff. Get to the point. Why can't you write down your nonstandard result---'Hafele-Keating should have shown no time offset'--- in terms of events and coordinates?

 

[johanfprins]

A meter stick at rest in any inertial frame is a meter long, almost (but not quite) by definition. So yes, they have the same length.

Thank you.

No you haven't. I have yet to see you write down a single spacetime coordinate.

I do not have to write the equations down to use them. After all the Lorentz equations are really very simple.

Yes, Johan, this is standard stuff.

Exactly! So why can you not follow simple logic?

Get to the point. Why can't you write down your nonstandard result---'Hafele-Keating should have shown no time offset'--- in terms of events and coordinates?

Strangely enough, I have come to the same impasse on another thread. What I am thus going to , in order to save time, take the liberty to copy my answers on the other thread.

Here they are:

I'm starting to think that those involved in this discussion are not answering each other arguments.

Ah! A breath of fresh air after being inundated by the paranormal metaphysics that tomclarke and teahive want to believe at all cost and against all logic.

A little question to both parties.

Ender and Peter Wiggin are brothers. Peter stays on Earth while Ender makes multiple trips between solar systems at near-light speed.

After some years Ender stops in the planet Lusitania. This planet is on our galaxy so its movement relative to Earth is at much slower speed than light.

Statement: When Ender stops his travels Peter have already died of old age, as can be seen in broadcasts from Earth.

This is where you go wrong. If you apply the Lorentz transformation correctly you will find that at this point in time they are both alive.

My question is: Is this statement possible or absurd?

In terms of the Special Theory of Relativity it is absurd. Unless Ender has spent a lot of time near a black hole, there should be very little difference between the ages of the twins. If, for example, Ender stayed within gravity free space while travelling, he will in fact be older than Peter who stayed behind within earth's gravity field. If both Ender and Peter were all the time within gravity free space, they will be exactly the same age as determined by the synchronised clocks they are each carrying.

Only a gravity field affects the rates of clocks at different positions within space, and, although one can equate gravity with acceleration, the slow-down in the clock-rate has nothing to do with acceleration. The reason why time slows down is that both light and matter are solely wave-fields (NO PARTICLES INVOLVED: Thank you Schroedinger!). When a light wave interacts with matter the speed of light slows down: For example, when a light-wave enters a glass block its speed slows down and for this reason refraction occurs.

The bending of starlight around the sun is also solely caused by the slowing down of light speed within matter and within matter’s surrounding gravity field. Since the fourth dimension can be written as -ict, a smaller c mandates a decrease in the rate of time. In fact, it is possible stop light-speed and thus stop time within matter: Lene Hau demonstrated this on a macro-scale by stopping a light pulse within a Bose-Einstein Condensate at low temperature. But, in fact, this also happens, and is happening all the time, at much higher temperatures: For example, when an orbital electron around a nucleus absorbs a photon-wave: In this case the light wave stops within the electron wave and thus adds mass energy to this wave.

That time stops within the electron wave is inherent in Schroedinger waves, since, as we have all known for more than 80 years, the intensity of a stationary Schroedinger wave is time independent. A free-electron wave which is stationary within its own inertial reference frame is in effect a mini “black hole” surrounded by a spherical “event horizon”. Outside this “horizon” the wave has tails which decay exponentially to infinity. In text books on quantum mechanics these tails are incorrectly called “tunnelling tails”. They represent the curvature in space-time around the mass of the wave.

In contrast, it is one of the postulates of the Special Theory of Relativity that light speed is the same at each and every point within each and every inertial reference frame. Thus, in this case, the time rate cannot actually slow down at any point as it does within a gravity field. The “time-dilation” and “length-contraction” effects derived from the Theory of Special Relativity can thus not be used, as Einstein has done, to justify the need for a curvilinear four-space in order to model gravity.
Einstein reached the correct conclusion but for the wrong reasons.

If his reasons were correct, then also in the case of gravity, light speed should have remained the same value at every point within our universe. Einstein was forced to abandon this principle in his so-called “General Theory of Relativity”. He never explained how this abandonment can be extrapolated from the Special Theory of Relativity, but only mentioned it in passing that you must abandon the second postulate of the Special theory of Relativity. This to me is proof that Einstein’s Theory of Gravity has nothing to do with any relativistic effect at all.

Now let us look at tomclarke’s tautology:

Let us try and cut through all the BS tomclarke and teahive is posting:

Let me ask a simple question. If you have two inertial reference frames Kp and K, moving with a speed v relative to one another and within each reference frame there is a meter-stick at rest: Do these two meter-sticks have the SAME lengh within their respective reference frames Kp and K? Yes they have or No they have not?

Since you make the question relative to the frame, the answer is yes - tautologically so.

Now, in order to eliminate any time changes owing to gravity, let us place Ender and Peter into their own spaceships which are at first floating side by side in free space (no gravity field). They both have a meter stick of the finest material which they find are exactly the same length within a femtometer. They also both have perfect clocks which they synchronise before Ender accelerates away exploring space without ever entering a gravitational field. As soon as Ender reaches his top speed (near the speed of light) relative to Peter, he switches on his radio and broadcasts a verbal message to Peter. When Peter receives the message he hears a very low frequency rumble. He rewinds his recorder and after experimenting he finds that when he increases the speed with which he plays back the message, Ender’s voice improves, and at a suitable disc-speed Ender's message can be heard clear as a bell.

Fortunately, Ender and Peter can also correspond by e-mail. Peter sends Ender an e-mail telling him of the slow down in his voice and asking him if he is playing a joke. Ender sends an e-mail back and asks Peter to now broadcast a verbal message to him. Ender now finds in turn that he has to speed up Peter’s message to understand it and e-mail this information to Peter. Since they are twins and know that they will not play tricks on each other they finally agreed that not one of them is slowing the verbal messages down before sending them. But what can then be the cause of this slow down?

So they decided to compare their clock rates: Each one of them will send a light pulse from one end of their respective meter sticks to the other and measure the time interval, and then exchange the measured time intervals by e-mail to compare. Since the meter stick with Ender is 1 meter, and the speed of light is c, Ender measures a time interval (delta)tp=(1/c). Similarly Peter’s meter stick is 1 meter long, and even though he and Ender is moving with a high speed relative to one another, the speed of light within Peter’s reference frame is still c (the same as in Ender’s reference frame): And so the time interval measured by Peter is (delta)t=(1/c); which is exactly the same as Ender has measured. Thus although their voices become retarded when they broadcast, their respective clocks keep exactly the same time; as demanded by the constancy of the speed of light (Einstein's second postulate).

But how do you explain the slow down in the voice broadcasts? Well let us consider Ender and Peter passing each other with the relative speed v and both of them shoot off a flare perpendicular to the direction of the speed. Both Ender and Peter will see their own flares moving perpendicularly to the direction of motion of the other ship: i.e. in both reference frames the position of each flare does not change along the x-axis. But if they look at each other’s flares they will each see motion along their x-axis.

If they are stupid they will conclude that the other flare is actually also moving along the x-axis relative to the spaceship from which it has been launched. But being non-dogmatic thinkers they realise that this is not the case. That to really establish the true path relative to the spaceship from which the flare has been launched, they must use the Lorentz transformation to get the correct path.

Exactly the same happens with the time coordinate: The voice broadcasts becomes slowed down as if the time on the other space ship is slower; but to really know what is happening on the other spaceship, you must use the Lorentz transformation and then you will find that the time on the other spaceship has NOT slowed down at all. Since the clocks with the two twins keep exactly the same time, they must remain exactly be the same age at every instant in time as measured on either one of the clocks; no matter how far apart they are or with what speed they are moving relative to one another.

Thus, as far as the Special Theory of Relativity is concerned, time is everywhere the same: All clocks, no matter where they are or with what speeds they move relative to one another, keep time at exactly the same rate.

I hope this helps.

 

[ben m]

I do not have to write the equations down to use them. After all the Lorentz equations are really very simple.

Except that you're getting a different answer than anyone else on Earth.

a) If they're so simple, how did everyone else (including Einstein) get a different answer than you?

b) Is "every physicist on Earth, except Herb Dingle and pages 1--2 of Einstein's 1905 paper, is wrong" not important enough to merit you doing a Freshman-problem-set-level workup? Do you even know what a Freshman relativity problem set looks like?

This is where you go wrong. If you apply the Lorentz transformation correctly you will find that at this point in time they are both alive.

Nope, you made a mistake in the analysis you didn't actually do. Go ahead and do it, Johan. You say "at this point"---can you define that as an event and give its 4-coordinate according to each of the twins?

 

[Farsight]

Johan, you said The reason why time slows down is that both light and matter are solely wave-fields. Boil yourself down to a single wave-field. Imagine you're just some wave sloshing back and forth in a box, like this: ||, repeatedly. It's moving at c. Now imagine I'm your twin brother, and I travel out to some solar system then back. I'm just a wave sloshing back and forth in a box too. My wave moves at c too. Only my wave moved like this: /\/\/\ → followed by this ← /\/\/\ ←. And the total path length of our two waves are both the same. Your total path length is |||||||||||||| but note that all the lines should be on top of one another. My total path length is /\/\/\/\/\/\. The number of reflections represents the time we have each experienced. I've experienced less than you. And I now have fewer grey hairs than you.

 

[johanfprins]

Except that you're getting a different answer than anyone else on Earth.

a) If they're so simple, how did everyone else (including Einstein) get a different answer than you?

Are you saying it is impossible that most physicists can get something wrong over a long period of time? History is not on your side in this respect: The most obvious misinterpretation was the epicycles in Ptolemy's model of the universe which "stood the test of time" for more than a millenium. This is not an argument one expects from a real scientist. A real scientist always accepts that what we think we know today can be proved to be totally false tomorrow. If you are not willing to contemplate this possibility you should get the hell out of science.

Physics should not be judged on the opinion of the majority but on logic and experimental fact which should combine to give a single self-consistent description. As I have demonstrated above, the clocks of two twins moving with a speed v relative to one another must measure the same time interval within their respective reference frames for a light pulse to travel through a distance of one meter. This means that their clocks must be keeping time at the same rate within their respective reference frames: In fact this is demanded by both of Einstein's postulates that it must be so.

If it is found by a competent experiment that one clock actually runs slower than the other, then Einstein's two postulates must be wrong and invalid. My stance is that unless proved othwerwise within a gravity-free experiment that the two clocks do not keep the same time, I accept that Einstein's two postulates must be correct; and that therefore the two clocks must keep the same time.

b) Is "every physicist on Earth, except Herb Dingle and pages 1--2 of Einstein's 1905 paper, is wrong" not important enough to merit you doing a Freshman-problem-set-level workup? Do you even know what a Freshman relativity problem set looks like?

I taught them when I was still caught within the unscientific, dogmatic way of thinking like you still are. The fact is that time-dilation is purely a relativistic effect when you LOOK INTO another inertial reference frame FROM OUTSIDE. You then measure ON YOUR CLOCK that the clock moving relative to you is going slower; while it is actually keeping the exact same time-rate within its own inertial refrence frame that your clock is keeping within your inertial reference frame: What you see is (delta)t where (delta)t=(gamma)*(delta)t^/. This equation tells you clearly that the actual time interval measured by the moving clock within its own inertial refrence frame (within which it is stationary) is given by (delta)t^/ and it is smaller than the dilated time-interval (delta)t; and therefore the moving clock is keeping time within its own inertial reference frame at the non-dilated rate; just like your clock is doing within your inertial reference frame. To repeat, both clocks keep time at the same standard, universal rate which applies when there is no gravity.

Nope, you made a mistake in the analysis you didn't actually do. Go ahead and do it, Johan.

Where did I make the mistake?

You say "at this point"---can you define that as an event and give its 4-coordinate according to each of the twins?

There is no time coordinate which changes with position when there is no gravity. All that happens according to SR, is that two events which occur a time-interval (delta)t apart, as measured by clocks that is stationary relative to the events, will be mesaured at a standard, global time rate; but when these same events are observed from a passing reference frame the time interval is observed to manifest at a slower time rate so that (delt)t>(delt)t^/. This does not mean that the time rate on the stationary clocks within the reference frame where the two events are occurring, is now (delta)t. The time rate within the reference frame where the events are occurring remains such that (delta)t^/<(delta)t: i.e. it is NOT time-dilated at all.

 

[sol invictus]

So they decided to compare their clock rates: Each one of them will send a light pulse from one end of their respective meter sticks to the other and measure the time interval, and then exchange the measured time intervals by e-mail to compare. Since the meter stick with Ender is 1 meter, and the speed of light is c, Ender measures a time interval (delta)tp=(1/c). Similarly Peter’s meter stick is 1 meter long, and even though he and Ender is moving with a high speed relative to one another, the speed of light within Peter’s reference frame is still c (the same as in Ender’s reference frame): And so the time interval measured by Peter is (delta)t=(1/c); which is exactly the same as Ender has measured.

That part is correct.

Thus although their voices become retarded when they broadcast, their respective clocks keep exactly the same time; as demanded by the constancy of the speed of light (Einstein's second postulate).

That part is wrong, or at best written very poorly.

It's true that Peter and Ender each measure the same time interval for the light to travel from one end of their meter stick to the other. So let's use that and consider what Peter sees as he watches Ender make his measurement. Ender has attached a flashlight to one end of his meter stick. It blinks, and the pulse of light travels down Ender's stick to the other end. We know Ender measures a time (1m)/c for it to reach the other end. What will Peter measure?

Well, according to Peter, Ender's meter stick is moving, but the light is still travelling at speed c. Therefore - all else being equal - the light cannot reach the end after a time (1m)/c. For instance, say the meter stick is moving towards the pulse in Peter's frame. Then according to Peter the light arrives at the other end in a time that must be less than (1m)/c, because the end of the stick moved. But since Ender's clock reads (1m)/c, Peter concludes Ender's clock is running slow. Obviously by symmetry the same goes when you exchange Peter for Ender. (One has to be more careful than this, because Peter also concludes that Ender's meter stick is length contracted, and also disagrees on simultaneity - but in the situation I outlined the qualitative conclusion is correct.)

So each thinks the other's clocks run slow. If Ender turns around and comes back to Peter, Ender's clock must read less than Peter's when he returns, because it's been running slow the whole time. To analyze the situation from Ender's point of view is slightly harder, but only slightly, and of course you get the same conclusion.

 

[johanfprins]

Johan, you said The reason why time slows down is that both light and matter are solely wave-fields. Boil yourself down to a single wave-field. Imagine you're just some wave sloshing back and forth in a box, like this: ||, repeatedly. It's moving at c.

I am constructed from matter waves and this means that I am a standing wave relative to the inertial reference frame that I am stationary in, and therefore I cannot be sloshing around at the speed of c.

Now imagine I'm your twin brother, and I travel out to some solar system then back. I'm just a wave sloshing back and forth in a box too. My wave moves at c too. Only my wave moved like this: /\/\/\ → followed by this ← /\/\/\ ←. And the total path length of our two waves are both the same. Your total path length is |||||||||||||| but note that all the lines should be on top of one another. My total path length is /\/\/\/\/\/\. The number of reflections represents the time we have each experienced. I've experienced less than you. And I now have fewer grey hairs than you.

I suppose that you are talking a bout two light clocks with a light pulse being reflected up and down perpendicular between two mirrors: Then two identical clocks will both move I I I I within their respective inertial refrence frames and will thus keep time at exactly the same rate: However, if you can look into your twin's reference frame you will see the "angular behaviour" and conclude that his clock must be running slower than yours, while it is actually not running slower than your clock within its own inertial reference frame. Obviously when your twin looks into your reference frame he will conclude that your clock is running slower, even though both clocks are keeping the exact same time within their respective reference frames. Thus neither you nor your twin brother ages faster than the other.

 

[sol invictus]

However, if you can look into your twin's reference frame you will see the "angular behaviour" and conclude that his clock must be running slower than yours, while it is actually not running slower than your clock within its own inertial reference frame. Obviously when your twin looks into your reference frame he will conclude that your clock is running slower, even though both clocks are keeping the exact same time within their respective reference frames. Thus neither you nor your twin brother ages faster than then other.

Dingleingleing!! That is precisely Dingle's argument - or rather it was, until even he finally was made to understand why it's wrong.

There is no symmetry between the twins. One stays home (i.e. remains at rest in an inertial frame), while the other accelerates. Your argument fails.

 

[sol invictus]

Johan's argument:

if you can look into your twin's reference frame you will see the "angular behaviour" and conclude that his clock must be running slower than yours, while it is actually not running slower than your clock within its own inertial reference frame. Obviously when your twin looks into your reference frame he will conclude that your clock is running slower, even though both clocks are keeping the exact same time within their respective reference frames. Thus neither you nor your twin brother ages faster than the other.

Dingle's argument (as summarized by Hasok Chang in a paper on Dingle):

since the principle of relativity implies that absolute motion or rest is meaningless, we are free to regard either observer R or observer M to be at rest and the other one to be moving; therefore, there is no inherent asymmetry in the phenomenon that could be responsible for the purported effect of asymmetric ageing.

 

[johanfprins]

Originally Posted by johanfprins
So they decided to compare their clock rates: Each one of them will send a light pulse from one end of their respective meter sticks to the other and measure the time interval, and then exchange the measured time intervals by e-mail to compare. Since the meter stick with Ender is 1 meter, and the speed of light is c, Ender measures a time interval (delta)tp=(1/c). Similarly Peter’s meter stick is 1 meter long, and even though he and Ender is moving with a high speed relative to one another, the speed of light within Peter’s reference frame is still c (the same as in Ender’s reference frame): And so the time interval measured by Peter is (delta)t=(1/c); which is exactly the same as Ender has measured.

That part is correct.

Thanks!

Thus although their voices become retarded when they broadcast, their respective clocks keep exactly the same time; as demanded by the constancy of the speed of light (Einstein's second postulate).

That part is wrong, or at best written very poorly.
It's true that Peter and Ender each measure the same time interval for the light to travel from one end of their meter stick to the other. So let's use that and consider what Peter sees as he watches Ender make his measurement.

This is really irrelevant: What Peter sees from outside within Ender's reference frame or what Ender sees from outside in Peter's reference frame has no significance. The incontestable fact is that the two clocks keep the same time within their respective reference frames: Therefore the two twins age at the same rate within their respective reference frames.

See you again when I have time!

 

[ben m]

Where did I make the mistake?

For starters, you didn't even set up a twin-paradox or Hafele-Keating (or radioactive-shuttle, for that matter) calculation. You basically ignored the meat of that calculation, which is that one of the clocks is not inertial. You are misusing an inertial-frames relativity principle, by misapplying it to a non-inertial situation.

Fail. It's not even an interesting mistake, Johan. Your mistake is the reason it's called the Twin Paradox. The "paradox" about it is precisely "ignoring the inertial-frames requirement makes the age-difference sound contradictory, isn't that funny".

 

[ben m]

Therefore the two twins age at the same rate within their respective reference frames.

Yes, and taking this into account, including all of the actual reference frames and displacements required, the twins are not the same age when they meet up at the end of one's trip.

One twin says he was on (say) a 50-year-long stay-at-home. The other twin says he was on (say) a 15-year cruise, followed by an acceleration, followed by another 15-year cruise. Both of them aged at the same rate within their respective inertial reference frames.

 

[johanfprins]

For starters, you didn't even set up a twin-paradox or Hafele-Keating (or radioactive-shuttle, for that matter) calculation. You basically ignored the meat of that calculation, which is that one of the clocks is not inertial.

Both clocks are inertial within their own inertial reference frames.

You are misusing an inertial-frames relativity principle, by misapplying it to a non-inertial situation.

Where have I misused it? A clock that is stationary within an inertial reference frame IS inertial within that reference frame. Thus each of the two clocks travelling with the two twins respectively IS inertial within its twin's inertial reference frame.

 

[johanfprins]

Yes, and taking this into account, including all of the actual reference frames and displacements required, the twins are not the same age when they meet up at the end of one's trip.

One twin says he was on (say) a 50-year-long stay-at-home. The other twin says he was on (say) a 15-year cruise, followed by an acceleration, followed by another 15-year cruise. Both of them aged at the same rate within their respective inertial reference frames.

Note that the Theory of Special Relativity and also the Lorentz transformation has NO information on what happens when one twin turns around. So let us at first ignore this possibility. So you must agree that if one of the twins do not turn back, they will age at exactly the same rate: This means that after one year on the clock of one twin, the clock of the other twin will also show a year, and therefore they will, at that instant in time as shown by both clocks, have exactly the same age; no matter how far they are from one another and no matter with what speed they are travelling relative to one another.

 

[W.D.Clinger]

The highlighted statements are incorrect and/or meaningless.

Note that the Theory of Special Relativity and also the Lorentz transformation has NO information on what happens when one twin turns around. So let us at first ignore this possibility.

So you must agree that if one of the twins do not turn back, they will age at exactly the same rate: This means that after one year on the clock of one twin, the clock of the other twin will also show a year, and therefore they will, at that instant in time as shown by both clocks, have exactly the same age; no matter how far they are from one another and no matter with what speed they are travelling relative to one another.

johanfprins is repeating his fundamental mistakes.

When johanfprins writes "after one year on the clock of one twin", he neglects to specify the inertial frame in which the passage of that "one year on the clock of one twin" is observed. When he writes "the clock of the other twin will also show a year", he neglects to specify the inertial frame in which that clock is seen to show a year. If the twins are at the same position at time 0 (as measured in both of their inertial frames) and are moving with some positive relative velocity v, there is no time within either of their inertial frames at which both clocks will show exactly one year.

When johanfprins writes "at that instant in time as shown by both clocks", he is making Herbert Dingle's mistake of thinking there is some objective meaning for the Newtonian phrase "instant of time" that can be separated from a spacetime event and its coordinates in various inertial frames. As noted in the previous paragraph, there is no value for the time coordinate in either of the twins' inertial frames at which both clocks show that one year has passed.

This would be obvious if johanfprins would take the trouble to do what ben m has been urging him to do: Write down an unambiguous description of each event that johanfprins believes to be important, and use the Lorentz transformation to calculate the spacetime coordinates of those events in all relevant inertial frames.

Freshmen learn how to do that. johanfprins claims to have taught students how to do that. johanfprins has no excuse for not doing that himself.

If johanfprins needs a reminder of how to do those calculations correctly, he could do worse than to follow the example of these two posts:
http://www.internationalskeptics.com/forums/showthread.php?postid=7863273#post7863273
http://www.internationalskeptics.com/forums/showthread.php?postid=7865834#post7865834

 

[leonAzul]

Note that the Theory of Special Relativity and also the Lorentz transformation has NO information on what happens when one twin turns around. So let us at first ignore this possibility. So you must agree that if one of the twins do not turn back, they will age at exactly the same rate: This means that after one year on the clock of one twin, the clock of the other twin will also show a year, and therefore they will, at that instant in time as shown by both clocks, have exactly the same age; no matter how far they are from one another and no matter with what speed they are travelling relative to one another.

Game, set, match.

While researching another usage of language ["Higgs Field particle" for "Higgs boson"] I came across this article:

The Higgs mechanism is a type of superconductivity which occurs in the vacuum. It occurs when all of space is filled with a sea of particles which are charged, or, in field language, when a charged field has a nonzero vacuum expectation value. Interaction with the quantum fluid filling the space prevents certain forces from propagating over long distances (as it does in a superconducting medium, e.g. in the Ginzburg–Landau theory).

retrieved from http://en.wikipedia.org/wiki/Higgs_mechanism#Landau_Model 31 December 2011

Does this sound familiar, Johan, and would it represent previous art and therefore be ineligible for patent protection?

 

[Olowkow]

Non physicist lurkers (like me) may want to check this out.
http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf

I am anxious to contribute this link to the most comprehensible explanation of relativity (including the twin paradox) that I have seen. It answers a lot of "dumb questions" that people like myself have had since our first exposure to these ideas. It uses simple math, mainly algebra, and very well written prose and diagrams. It works for me, and hopefully might be informative to lot of others who are trying to follow the thread.

This has been an extremely educational thread, and I have followed it since its beginning. Thanks to everyone involved, including of course Johan without whose participation none of this discussion would have ever taken place. Happy new year to all.