Superconductivity at room temperature??

[johanfprins]

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.

As usual a blatant lie and a distortion of the facts. We are clearly talking about two twins which are stationary within their own inertial reference frames where these two inertial refrence frames are moving relative to one another with speed v.

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.

I have not read all that Herbert Dingle has written. He might have made mistakes but some of his statements might have been correct. I find it disgusting that Herbert Dingle is now being used as an ogre to classify opponents instead of using logic. A clear case of an ad hominen attack. I also consider it as a criminal violation of Herbert Dingle's human rights.

Consider the following thought experiment:

Two twins have identical spaceships, of identical lengths, engines etc. floating side by side within free space and they have identical clocks which they synchronise at the instant that both spacehips start to move way from one another. In order to start the journey, each twin simultaneously with the other twin, switches on his/her identical engines for exactly the same pre-arranged length of time as measured on their respective clocks; and then switches off their engines. When switching off their engines they coast at a constant speed v relative to one another.

They then both allow their space ships to coast until at a pre-arranged time on both their clocks they both switch on their engines to de-acceleate until they both come to a halt relative to one another. Thus when they start to de-accelerate the time on their clocks must be identical and also when they come to a halt relative to one another

They then simultaneously, as shown by their clocks, gun their engines to identically accelerate towards one another, until at a pre-arranged time on both their clocks they switch off their engines in order to coast towards one another with the same speed v with which they coasted away from one another. Then again at a pre-arranged time on their respective clocks, they switch on the engines to de-accelerate towards one another so that they end up again being stationary relative to one another.

Now answer the following questions:

1. Have their engines been switched on and off simultaneously during the whole journey?

2.Which twin will be younger than the other?

 

[johanfprins]

Game, set, match.

Thanks!

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

Quote:
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?

I will never patent to protect nonsense!

Yes these are fundamental concepts that are being used in quantum field theory. I know that I am now again opening myself up to be accused that I must be a crackpot to claim that hundreds of thousands of "experts" on superconduction and quantum field theory could have been wrong over a period of more than 50 years. But unfortunately this is the actual state of affairs. They should have called quarks, quacks!

Firstly the Ginsberg-Landau model, like all the subsequent models (BCS etc.) leads to the requirement that one must be able to equate a conservative vector field point-for-point with a circular vector field. It is inteteresting to note that Feynman's path integrals lead to the same requirent, as well as Aharanov and Bohm's model for the experimentally verified Aharanov-Bohm effect. This effect is in reality caused by a diffracting electron wave which actually does move through both slits simultaneously, and therefore its centre-of-charge does move smack-bang through the magnetic field within the solenoid between the two slits. Nonetheless to assume that a conservative vector field can match a circular vector field point-for-point, grossly violates one of the most fundamental rules of elementary vector calculus.

It is absolutely amazing that most of our modern day theoretical physicists have made this mistake and doggedly keep on making this mistake and aggresively defend making this mistake. They think that baptising a differentiable conservative vector field, having more than one dimension, to be circular, they can treat it as circular even though the most elementary rules of vector calculus demand that it cannot be so.

You do not need a Higg's boson to explain mass-energy. Mass energy is simply stationary energy of a stationary (light or matter) wave. It is a natural consequence of Maxwell's equations; just as the Special Theory of Relativity is.

 

[johanfprins]

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.

Thank you: I have downloaded the book and will read it when time permits.

Best for the New Year.

 

[sol invictus]

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.

It's not irrelevant - it's the whole point. Relativity allows you to choose any reference frame you like, and make correct predictions for all experiments. You don't ever have to change frames. That's why it's trivial to determine that the travelling twin is younger - just use the frame of the stay at home twin.

Note that the Theory of Special Relativity and also the Lorentz transformation has NO information on what happens when one twin turns around.

Wrong. SR is adequate to determine the results. For one thing, you can just use the frame of the stay-at-home twin.

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.

Wrong again. The effect you termed "irrelevant" just above directly contradicts this. In the rest frame of twin A (B), twin B's (A's) clock is running slow. That follows directly from the constancy of the speed of light, as I just explained to you.

Game, set, match.

You can't be serious, so I'll ignore this.

 

[sol invictus]

Consider the following thought experiment:

Two twins have identical spaceships, of identical lengths, engines etc. floating side by side within free space and they have identical clocks which they synchronise at the instant that both spacehips start to move way from one another. In order to start the journey, each twin simultaneously with the other twin, switches on his/her identical engines for exactly the same pre-arranged length of time as measured on their respective clocks; and then switches off their engines. When switching off their engines they coast at a constant speed v relative to one another.

They then both allow their space ships to coast until at a pre-arranged time on both their clocks they both switch on their engines to de-acceleate until they both come to a halt relative to one another. Thus when they start to de-accelerate the time on their clocks must be identical and also when they come to a halt relative to one another

They then simultaneously, as shown by their clocks, gun their engines to identically accelerate towards one another, until at a pre-arranged time on both their clocks they switch off their engines in order to coast towards one another with the same speed v with which they coasted away from one another. Then again at a pre-arranged time on their respective clocks, they switch on the engines to de-accelerate towards one another so that they end up again being stationary relative to one another.

Now answer the following questions:

1. Have their engines been switched on and off simultaneously during the whole journey?

Meaningless question. The answer is yes or no, depending on which frame you use. A big part of your problem is that you've failed to comprehend one of the basic facts about relativity - simultaneity is relative. Pick a frame (or specify an experiment - if you were observing this, what precisely would you do to determine the answer, and what trajectory would you follow?), then I can answer the question.

2.Which twin will be younger than the other?

At the end of the journey they will be the same age (obviously, that follows immediately by symmetry). But if they were triplets and the third sibling remained at rest at home, the two travelers would both be younger than the one that remained behind.

 

[Farsight]

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. I suppose that you are talking about two light clocks with a light pulse being reflected up and down perpendicular between two mirrors:

That's where this comes from, but it applies at a deeper level because we're constructed from waves. Don't worry about the distinction between a standing wave and a wave going back and forth at c.

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.

Agreed. I see my light like this || and yours like this /\. You see your light like this || and mine like this /\. We each say that the other's clock is running slower than his own. But there's no mystery to it. We have relative motion, that's all. It's something like the way you look smaller to me and I look smaller to you when we're separated by distance.

Thus neither you nor your twin brother ages faster than the other.

But when I turn round and come back, we both then agree that your light moved like this |||||||||||| and mine moved like this /\/\/\/\/\/\. We agree that the total light-path lengths in our clocks was the same. And we agree that your clocks says 7 years have elapsed and mine says 6 years have elapsed. And we both know that it applies at a deeper level because we're constructed from waves. Time is a cumulative measure of local motion, "in my reference frame". Or yours. That's all it is. And my local motion was reduced because I moved out and back through space. So I have fewer gray hairs than you. Or if I was a muon good for a billion rotations, you've decayed and I'm still going.

 

[johanfprins]

It's not irrelevant - it's the whole point. Relativity allows you to choose any reference frame you like, and make correct predictions for all experiments. You don't ever have to change frames. That's why it's trivial to determine that the travelling twin is younger - just use the frame of the stay at home twin.

Who is the "stay-home" twin? There is no absolute reference frame relative to which one can choose which twin is moving and which one not. Thus BOTH twins are "stay-home twins" and BOTH are also travelling away from the "stay-home twin"!

Wrong. SR is adequate to determine the results. For one thing, you can just use the frame of the stay-at-home twin.

You can use any one of the two reference frames as the "stay-home" twin, and for each reference frame you get the opposite result that you get for the other reference frame: Both results cannot be simultaneously correct.

Wrong again. The effect you termed "irrelevant" just above directly contradicts this. In the rest frame of twin A (B), twin B's (A's) clock is running slow. That follows directly from the constancy of the speed of light, as I just explained to you.

Twin B's clock is NOT running slower within twin B's own inertial reference frame; and it is THIS clock speed that determines twin B's rate of aging, NOT the relativistically transformed clock speed of twin B's clock within twin A's reference frame.

You can't be serious, so I'll ignore this.

He is of course TOTALLY correct. Not that you will ever be able to grasp this simple fact.

 

[johanfprins]

Meaningless question. The answer is yes or no, depending on which frame you use.

This is the typical BS you have posted on this thread. The answer cannot be "yes or no". If both spaceships stop accelerating simultaneously when their clocks read the same time, then this must happen simultaneously across both reference frames.

A big part of your problem is that you've failed to comprehend one of the basic facts about relativity - simultaneity is relative.

Oh my God! now simultaneity is the same as non-simultaneity .

At the end of the journey they will be the same age (obviously, that follows immediately by symmetry). But if they were triplets and the third sibling remained at rest at home, the two travelers would both be younger than the one that remained behind.

Good try, but wrong again! The formulas for SR only applies to two inertial reference frames moving with a speed v relative to one another. When you bring in a third reference frame relative to which the other two move, you are assuming a "preferred" uniquely stationary reference frame relative to which the other two are moving. This cannot exist!

This does not mean that you cannot describe the motion relative to a third inertial reference frame; but then the relativistic equations become horrendous since the speeds of the two twins relative to their triplet, are not +v/2 and -v/2 when the relative speed is v between the two travelling twins. So let us first stick to the Lorentz equations which only model TWO inertial reference frames moving with a SINGLE speed v, RELATIVE to one another.

The fact is that it does not matter how "asymmetrical" the motion is between the two twins, there will always exist an inertial reference frame relative to which this motion will be symmetrical. This must thus surely mean, by your own argument, that the two twins cannot age at different rates.

 

[Ziggurat]

Who is the "stay-home" twin?

The twin who doesn't accelerate.

There is no absolute reference frame relative to which one can choose which twin is moving and which one not.

One can pick an inertial reference frame in which either twin is stationary at any one moment. But there is NO inertial reference frame in which the "traveling" twin twin remains stationary for the entire journey, whereas one can choose such a frame for the "stay-home" twin. In any inertial reference frame, only one twin accelerates, therefore in ALL reference frames, it's easy to distinguish between them and they are not equivalent.

Thus BOTH twins are "stay-home twins" and BOTH are also travelling away from the "stay-home twin"!

No. The two twins are easily distinguished from any inertial reference frame.

You can use any one of the two reference frames as the "stay-home" twin, and for each reference frame you get the opposite result that you get for the other reference frame: Both results cannot be simultaneously correct.

You are wrong. The reference frame of the traveling twin is not inertial. Non-inertial reference frames are not equivalent to inertial reference frames. Not even in Newtonian physics.

Twin B's clock is NOT running slower within twin B's own inertial reference frame

The traveling twin's reference frame isn't inertial, johan. He accelerates. That makes it non-inertial. Epic failure.

 

[johanfprins]

But when I turn round and come back, we both then agree that your light moved like this |||||||||||| and mine moved like this /\/\/\/\/\/\.

No we do not: From your perspective it will be your light that moved like ||||||||||||, and my light that moved like \/\/\/\/\/\.

And we agree that your clocks says 7 years have elapsed and mine says 6 years have elapsed.

No we do not agree: According to me your clock says 7 years while my clock says 6 years.

And we both know that it applies at a deeper level because we're constructed from waves.

I must admit that I am not following your argument here.

Time is a cumulative measure of local motion, "in my reference frame". Or yours. That's all it is. .

OK.

And my local motion was reduced because I moved out and back through space.

We both moved away from ecah other and back towards one another.

So I have fewer gray hairs than you.

No you do not.

Or if I was a muon good for a billion rotations, you've decayed and I'm still going.

No you would have decayed after reaching the same age on your clock that I have reached on my clock when I decayed.

Within your reference frame you will conclude that I have outlived you, just as I will conclude in my reference frame that you have outlived me. It is as if we are looking at each other through a thick pane of glass within which light is slowed down by an incredible amount. When observers on your side see me die, I have already died long ago, while when observers on my side see you die, you have already died long ago. I remember a delightful story where a person had such a glass pane to his outside garden and sat watching his deceased children playing outside under a tree when they still lived.

 

[W.D.Clinger]

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

That looks like a nice exposition of special relativity. As the author says, it's a "rigorous but non-technical" book for a general audience. Thank you for bringing it to our attention.

It is as if we are looking at each other through a thick pane of glass within which light is slowed down by an incredible amount. When observers on your side see me die, I have already died long ago, while when observers on my side see you die, you have already died long ago. I remember a delightful story where a person had such a glass pane to his outside garden and sat watching his deceased children playing outside under a tree when they still lived.

That would be one of Bob Shaw's "slow glass" stories. The only one I've read is "Light of Other Days", which was nominated for a Hugo Award in 1967. It has nothing to do with relativity.

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.

As usual a blatant lie and a distortion of the facts. We are clearly talking about two twins which are stationary within their own inertial reference frames where these two inertial refrence frames are moving relative to one another with speed v.

It was a statement of fact, not a distortion of fact. The event itself is unambiguous, but you did in fact neglect to mention that the other twin observes that event at a time other than one year (in that other twin's inertial frame). The fact that the inertial frames of the two twins assign different times to that event is crucial to understanding special relativity, but you're trying to obscure that point.

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.

I have not read all that Herbert Dingle has written. He might have made mistakes but some of his statements might have been correct. I find it disgusting that Herbert Dingle is now being used as an ogre to classify opponents instead of using logic. A clear case of an ad hominen attack.

No, it was another statement of fact. When you wrote "at that instant in time as shown by both clocks", you did indeed make the same mistake as Herbert Dingle. Like you, he forgot that Einstein's theory of special relativity says there is no observer-independent "instant of time" that can be attached to two distinct events. When two events are separated in time and/or space, the times at which they occur depend on the observer's inertial frame.

As Einstein, sol invictus, Ziggurat, ben m, and thousands of others have explained, simultaneity is relative.

The relativity of simultaneity (hence non-simultaneity) is covered in chapter 7 of Professor Styer's book that Olowkow recommended to us. Time dilation is covered in chapter 4, length contraction in chapter 6, and the twin "paradox" in chapter 13.

I also consider it as a criminal violation of Herbert Dingle's human rights.

After what you've said about lawyers, I'm surprised to see you impersonate one on the Internet.

 

[johanfprins]

Originally Posted by johanfprins
Who is the "stay-home" twin?

The twin who doesn't accelerate.

You have agreed that if the motion is “symmetric” as I have described where both twins accelerate and decelerate identically, they will not age at different rates. Correct?

This can only mean that during the whole journey the clocks of the two twins must have kept synchronous time, or else the age of the twins would have had to be different.. And this synchronous time-keeping also happened during the time that both twins coasted, even though according to special relativity there is now a time dilation when observing each clock from its opposite inertial reference frame: For some wondrous reason, this time dilation does not “bite” across the two reference frames when the motion is “symmetric” but supposedly only when the motion is “not symmetric”?

Quote:
There is no absolute reference frame relative to which one can choose which twin is moving and which one not.

One can pick an inertial reference frame in which either twin is stationary at any one moment. But there is NO inertial reference frame in which the "traveling" twin twin remains stationary for the entire journey, whereas one can choose such a frame for the "stay-home" twin. In any inertial reference frame, only one twin accelerates, therefore in ALL reference frames, it's easy to distinguish between them and they are not equivalent.

Not true: In my “symmetric” example both accelerate at the same time. I assume that you have deliberately chosen the “asymmetric” condition to try and make another one of your irrelevant points?

Quote:
Thus BOTH twins are "stay-home twins" and BOTH are also travelling away from the "stay-home twin"!

No. The two twins are easily distinguished from any inertial reference frame.

Only when you choose the asymmetrical condition where you claim that only one of the twins accelerate on the round trip. And then reach the experimentally unproven conclusion that the twins will have different ages according to the "equations of special relativity".

Quote:
You can use any one of the two reference frames as the "stay-home" twin, and for each reference frame you get the opposite result that you get for the other reference frame: Both results cannot be simultaneously correct.

You are wrong. The reference frame of the traveling twin is not inertial.

Only while it is accelerating you can venture to make such a claim: Consider again the two twins with their identical space ships and, to please you, only one of them accelerates away for two days until he reaches his top speed v relative to his “stay-home” brother. He then coasts at that speed for 10 years, spends four days to turn around and coasts back for another 10 years, then decelerates for two days and stops next to his brother. Thus the trip took 20 years and eight days. How do you interpret the relative ages of the twins solely in terms of the Special Theory of Relativity?

Firstly, Special Relativity tells us nothing about what happens to clocks during acceleration. The Lorentz equations are only valid for a symmetric unidirectional motion with a single relative speed v. Furthermore, in the thought experiment I have just given, the twins are for twenty years coasting symmetrically relative to one another. Thus, since this twenty years of coasting is symmetric, the twins should, by your own reasoning, not age relative to one another during this time: Thus the twins cannot age differentially owing to the time-dilation formula of the Special Theory of Relativity. The only possibility is that, if there is an age difference, this must be solely caused by the 8 days of acceleration and deceleration of the one twin.

The latter age difference can obviously NOT be explained in terms of the Special Theory of Relativity. The effect of acceleration was only considered by Einstein many years (about 10?) later. By that time there had already been many laughable arguments about the twin paradox. These arguments were clearly premature, since the Lorentz transformation only applies to a symmetrical situation where the reference frames move unidirectional relative to one another. And as we have already agreed above, this must mean that clocks within both reference frames (which move relative to one another) must then keep synchronous time when the time is measured within their own inertial reference frames respectively: i.e. if the “stay-home” clock shows that 8 years have passed during coasting, the “travelling clock” will also show that 8 years have passed during coasting. There is no time-dilation caused by the Theory of Special Relatvity between these clocks at all.

Quote:
Twin B's clock is NOT running slower within twin B's own inertial reference frame

The traveling twin's reference frame isn't inertial, johan. He accelerates. That makes it non-inertial. Epic failure.

So you are claiming that when the travelling twin only accelerates for 8 days of a twenty year journey his reference frame is permanently inertial, also while he is coasting? I do not believe that Newton’s mechanics claim this. Epic! Epic! Epic! failure.

 

[johanfprins]

After what you've said about lawyers, I'm surprised to see you impersonate one on the Internet.

History has shown that most lawyers are very willing indeed to toe the line and apply laws which violate human rights. So as usual you are posting nonsense.

 

[Farsight]

Johan, I'm not telling you porkies here. You're right about a lot of things, but not this one, and it's harming your case. Sit down with a piece of paper and draw lines on it like this ǁ x 7, and out and back like this /\/\/\→ followed by this ←/\/\/\. Once I'm back motionless with respect to you, I adopt your view of how my light moved, even though while I was travelling it appeared to me to be moving like this ǁ. The spacetime interval between the beginning and end of my trip is given by the lightpath length. Your lightpath length is the same as mine. The elapsed time measured by you is 7, and the elapsed time measured by me is 6. But there isn't any actual time passing, all we're measuring is the number of bounces. That's all there is to it. Time is a cumulative measure of local motion. I measured less local motion because I was the one moving through the universe. I know I was the one moving because I felt the accelerations.

 

[johanfprins]

Johan, I'm not telling you porkies here. You're right about a lot of things, but not this one, and it's harming your case. Sit down with a piece of paper and draw lines on it like this ǁ x 7, and out and back like this /\/\/\→ followed by this ←/\/\/\. Once I'm back motionless with respect to you, I adopt your view of how my light moved, even though while I was travelling it appeared to me to be moving like this ǁ. The spacetime interval between the beginning and end of my trip is given by the lightpath length. Your lightpath length is the same as mine. The elapsed time measured by you is 7, and the elapsed time measured by me is 6. But there isn't any actual time passing, all we're measuring is the number of bounces. That's all there is to it. Time is a cumulative measure of local motion. I measured less local motion because I was the one moving through the universe. I know I was the one moving because I felt the accelerations.

I am really sorry to disappoint you Farsight, but on this issue I just cannot agree since what you are doing (what has been done for over 100 years) is to misunderstand what a coordinate transformation really means.

Let me try and illustrate it with an example: When I look through a microscope at a flea I transform the coordinates so that within my frame of reference (FOR) the flea is larger than it is within its own FOR.

Now consider the case where I look at a flea under a scanning electron microscope (SEM) so that its picture can be seen on the screen. On the screen the flea is 5 cm long. A person walking in who does not know a flea and an SEM might after asking what it is, respond: "So a flea is 5 cm long". I will then respond: "No, what you are seeing is a coordinate transformation of the flea which, within the electron microscope is only a few microns long. It is magnified by this coordinate transformation so that its actual size d is (gamma) times larger; and therefore you see the size D (equal to 5 cm) is given by D=(gamma)*d. The flea actually remains equal to d within its own FOR.

The Lorentz transformation is still a coordinate transformation which now includes time as an extra coordinate; and it is found that for a time-interval (delta)tp on a moving clock within its own FOR the TRANSFORMED time interval within our FOR is (delta)t=(gamma)*(delta)tp. Even though within our FOR (delta)t is larger than (delta)tp, (delta)t is NOT the time interval on the moving clock within its own FOR. In fact, if the moving clock could be brought to be within our own FOR, the time interval (delta)tp on this clock will correspond exactly with the same time interval on our clock within our FOR. Thus, the moving clock is keeping the exact same time within its own FOR than our stationary clock is keeping within our FOR. It is similar to the flea within the SEM being exactly the same size as a flea within our FOR even though the coordinate transformation increases the size of the flea within the SEM to be larger within our FOR.

Thus, the time-dilation formula does not imply that clocks that are stationary within different inertial reference frames are not keeping the exact same time. All stationary clocks keep the same time no matter within which inertial refrence frame each one of them is stationary. As I have pointed out before: If this is not the case, Einstein's first postulate on which he based Special Relativity would be wrong. A physics experiment involving time will then not give the same answer within any inertial refrence frame as is required by Einstein's first postulate.

 

[sol invictus]

Who is the "stay-home" twin?

As Zig told you, the one that remains at rest in an inertial frame. In the standard twin setup, one does and one doesn't. In the twin scenario you described above, neither does and both follow identical trajectories, hence they are the same age - but a stay-at-home third triplet that doesn't accelerate at all will be older than both travelers when they all meet at the end.

You can use any one of the two reference frames as the "stay-home" twin, and for each reference frame you get the opposite result that you get for the other reference frame: Both results cannot be simultaneously correct.
Twin B's clock is NOT running slower within twin B's own inertial reference frame; and it is THIS clock speed that determines twin B's rate of aging, NOT the relativistically transformed clock speed of twin B's clock within twin A's reference frame.

Wrong. The travelling twin does not remain at rest in any inertial frame.

This is the typical BS you have posted on this thread. The answer cannot be "yes or no".

Is the Eiffel tower on the left or on the right? The answer depends on the observer, and so does simultaneity. But I'm not going to discuss this with you, because you're failing to understand even more basic issues.

The fact is that it does not matter how "asymmetrical" the motion is between the two twins, there will always exist an inertial reference frame relative to which this motion will be symmetrical.

Oh really? Very well, take the standard twin setup, where one twin never accelerates, while the other one accelerates away, coasts for a while, accelerates to turn around and come back, coasts for a while, and then stops at the first twin. What's the inertial frame in which that's symmetric, Johan? There obviously isn't one, since the degree of acceleration is independent of choice of inertial frame.

 

[Ziggurat]

You have agreed that if the motion is “symmetric” as I have described where both twins accelerate and decelerate identically, they will not age at different rates. Correct?

They will be the same age when they rejoin each other. More than that we cannot say without specifying the reference frame.

This can only mean that during the whole journey the clocks of the two twins must have kept synchronous time, or else the age of the twins would have had to be different..

It means nothing of the sort. This is what their worldlines will look like:

I drew this for a triplet problem, but since you're using two traveling twins, we can ignore the central red line and just look at the green and blue lines (the black lines are light cones). In reference frame 1, their clocks will remain synchronous for the entire journey. But in reference frames 2 and 3, their clocks are NOT synchronous. Time dilation will apply to each of the twins for only part of their journey, but which part (the first or second leg) is different between them. Clocks synchronized at the start of the journey will not remain synchronous during the journey, but will BECOME synchronous again at the endpoint. In fact, you can see this quite clearly in frames 2 and 3 because the turning points are NOT at the same time. Yet the paths will still maintain a sort of symmetry which simple inspection tells us (and calculations back up) both have the same proper time.

In other words, what you assumed (but did not show) was a logical requirement is nothing of the sort. I have shown that it is not. It's been a while since I drew those up so I forget exactly what speeds I used, but those diagrams are properly Lorentz-transformed world lines of the same scenario in each panel.

Not true: In my “symmetric” example both accelerate at the same time.

Only in reference frame 1, NOT in reference frame 2 or 3. Epic fail.

Only while it is accelerating you can venture to make such a claim

And that is enough. The traveling frame is not inertial. And two DIFFERENT inertial reference frames (if you want to pick the outbound and inbound frames) are not equivalent to a single frame either.

Consider again the two twins with their identical space ships and, to please you, only one of them accelerates away for two days until he reaches his top speed v relative to his “stay-home” brother. He then coasts at that speed for 10 years, spends four days to turn around and coasts back for another 10 years, then decelerates for two days and stops next to his brother. Thus the trip took 20 years and eight days. How do you interpret the relative ages of the twins solely in terms of the Special Theory of Relativity?

Firstly, Special Relativity tells us nothing about what happens to clocks during acceleration.

Not true.

The Lorentz equations are only valid for a symmetric unidirectional motion with a single relative speed v.

It's funny how many things become "only valid" for situations which don't contradict you. Too bad you're wrong.

But it's easier to use the space-time metric ds² = dr² - (ct)² for such a problem, not the Lorentz transformations in their canonical form. For your trajectory, we simply integrate ds along the path. Quite simple. Your claim is mathematically exactly equivalent to saying that the Pythagorean theorem can't tell us the length of a curve, so we can't calculate it using Euclidean geometry. But of course, we can.

Furthermore, in the thought experiment I have just given, the twins are for twenty years coasting symmetrically relative to one another.

So what? You still can't ignore the acceleration in the middle.

Thus, since this twenty years of coasting is symmetric, the twins should, by your own reasoning, not age relative to one another during this time: Thus the twins cannot age differentially owing to the time-dilation formula of the Special Theory of Relativity. The only possibility is that, if there is an age difference, this must be solely caused by the 8 days of acceleration and deceleration of the one twin.

If you insist on doing this from the non-inertial reference frame of the traveling twin, then YES. Non-inertial reference frames are funny that way: the passage of time at a distance can go amazingly fast, or even backwards, but that's an artifact of being non-inertial. If you want to keep it inertial, then the twins are NEVER symmetric, and even in a frame where the traveling twin is at rest for part of his journey, that just means he's moving even faster for the other part, and thus has an even higher time dilation. And once you're no longer in the "stay-at-home" twin's frame, those two legs don't even constitute half the journey. Which you can see from my above diagrams.

Seriously, try it. Try actually DOING the Lorentz transformations on this journey, from any reference frame you choose.

So you are claiming that when the travelling twin only accelerates for 8 days of a twenty year journey his reference frame is permanently inertial, also while he is coasting? I do not believe that Newton’s mechanics claim this. Epic! Epic! Epic! failure.

An accelerated frame is not inertial, regardless of how much of the time that frame is accelerating. And yes, Newtonian mechanics claims this rather explicitly. How much the frame accelerates and for how long can make a difference in terms of how much it deviates from a truly inertial one, but you have to actually crunch the numbers to find out. Which you haven't done.

 

[sol invictus]

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.

Incidentally, Johan, if you accept this you should also accept that the traveling twin is younger. The traveller necessarily experiences acceleration, which according to the basic principle behind general relativity is indistinguishable from a gravitational field. The stay-behind twin experiences no acceleration.

Therefore the traveller spends part of his time in a gravity field, while the stay-behind twin doesn't - and therefore the traveller is younger.

 

[ben m]

Another 10 posts by Johan, another ten posts with no Lorentz transformations.

Seriously, Johan, it's about six lines of algebra. Name the events that set up your scenario, find their coordinates in a particular reference frame, then apply the Lorentz transformations. Then add up the proper times along each worldline. The result---every time I've done this---has been that all observers agree, with no contradictions, that the traveling/accelerating twin has aged less.

If you could do this, and if that math agreed with you it would be a slam-dunk, wouldn't it? Why avoid it? Is it because you can't? Or because you know it'll prove you wrong? It's like someone claiming that their pendulum-based perpetual motion machine "must work", and talking through an "impeccable proof" that it works, but refusing to apply Newton's Laws.

 

[johanfprins]

Oh really? Very well, take the standard twin setup, where one twin never accelerates, while the other one accelerates away, coasts for a while, accelerates to turn around and come back, coasts for a while, and then stops at the first twin. What's the inertial frame in which that's symmetric, Johan? There obviously isn't one, since the degree of acceleration is independent of choice of inertial frame.

I should have known that when stating this it will be well above your ability to grasp. I admit that I made a mistake to include acceleration in the symmetry discussion, since this requires Einstein's GR to justify. I just do not have the time to do so and lead us into another topic. Thus, as I have described above, I am asking you to only consider the time of 20 years when the two twins coasted symmetrically relative to one another: During this time as you all admitted, the two twins cannot age relative to one another.

You claim that it requires acceleration of the one twin to ensure that this twin is younger. This cannot be proved by using the two postulates of Einstein and the Lorentz transformation. But OK, if acceleration is required, it must mean that the differential ageing only occurs during acceleration, and can thus not be due to the time-dilation formula obtained from Special Relativity.

Good night for now.