I have found many new interpretations (I mean different physical phenomena to explain the same results) for well known experiments rather than the relativistic explanations!: http://www.geocities.com/anewlightinphysics/new_interpretations/Summary_of_new_interpretations.htm
Someone will find the proper explanation on Gravity Probe B results at its time.

What is important now is that Relativity Theory is THEORETICALLY wrong as demonstrated in: http://www.geocities.com/anewlightinphysics/sections/Section1-1_Considerations_against_Relativity.htm

A new theory is NEEDED!

A new theory is NEEDED!

Needed for what?

Hi martillo,

What is important now is that Relativity Theory is THEORETICALLY wrong as demonstrated in: http://www.geocities.com/anewlightinphysics/sections/Section1-1_Considerations_against_Relativity.htm
It doesn't look like your arguments have changed since the last time I saw them (except for one which seems to have disappeared). But anyway. I'll deal with your point (C) about gyroscopes now because it's fairly trivial:

Another consideration against Relativity is about the Foucault pendulum and the gyroscope behavior that is related to the conservation of the angular momentum of the bodies. If there is no special absolute referential then we may wonder: related to what referential are the directions determined by these apparatus fixed?
The answer is: relative to any inertial reference frame. Relativity denies the existence of a single, unique absolute frame, but it doesn't claim the opposite extreme: that all reference frames are indistinguishable and equally valid. The fact that things behave differently in rotating frames is obvious to anyone who's ever been on a merry-go-round or in a car turning a sharp corner, so you don't need to bring up pendulums, gyroscopes, or the Sagnac effect to prove this point.

STR claims that the laws of physics are invariant with respect to transformations in the Poincaré group (http://en.wikipedia.org/wiki/Poincar%C3%A9_group). This includes boosts, translations, and fixed rotations (as opposed to rotating frames, where the angle of rotation is not constant), so you can see relativity as proposing a preferred set of reference frames if you like.

As for (A), I don't see why this:

We must also note that the rate of aging is different as seen in the mother-ship (velocity v) than seen by the twins (velocity w).
should be a problem for you (I'd expect ageing rates to be relative).

As for this:

This means that for each one the other twin is getting younger than himself.
It may seem counter-intuitive, but it isn't actually a contradiction. In general, \frac{\part a}{\part b} doesn't necessarily equal \left( \frac{\part b}{\part a} \right)^{-1} in mathematics.

There's nothing magical about reciprocity, so if you like I could come up with a though experiment that might make it seem more reasonable to you.

Hi przyk,

It doesn't look like your arguments have changed since the last time I saw them (except for one which seems to have disappeared).
Yes, thanks to some discussions in this forum I have realized that Relativity can be developed with constant mass. It's only necessary to redefine the momentum as p=γmv where γ=1/(root(1-v2/c2)). So the argument that mass is constant although considered true is not any proof that Relativity is wrong.

Relativity denies the existence of a single, unique absolute frame, but it doesn't claim the opposite extreme: that all reference frames are indistinguishable and equally valid.
Yes it does, General Relativity does. It even include any accelerated frame!

STR claims that the laws of physics are invariant with respect to transformations in the Poincaré group. This includes boosts, translations, and fixed rotations
Boosts, translations and fixed rotations relative to what??? You must consider a first basic reference frame to say that and which is it? That frame must not be accelerated to be an inertial frame but not accelerated relative to what?
It must exist a basic frame or a set of frames which we could call them to be at rest! Then you after can state that any frame having a boost, translation or constant rotation relative to them is an inertial frame.
You must consider that any frame obtained by a boost, translation or constant rotation from an accelerated frame will also be an accelerated frame.
Then once we accept that "rest" frames in the Universe exist just one of them selected by some property of the universe like possible symmetry would be the "privileged" absolute frame of the Universe.
Iknow it would not be easy to determine but I believe some day it will be.

As for (A), I don't see why this:

“ We must also note that the rate of aging is different as seen in the mother-ship (velocity v) than seen by the twins (velocity w). ”

should be a problem for you (I'd expect ageing rates to be relative).

As for this:

“ This means that for each one the other twin is getting younger than himself. ”

It may seem counter-intuitive, but it isn't actually a contradiction. In general, doesn't necessarily equal in mathematics.

There's nothing magical about reciprocity, so if you like I could come up with a though experiment that might make it seem more reasonable to you.

The problem is that ages are not relative! Age is directly related to all physiological phenomena that have haened to an individual in his history. You cannot say something has happened if one frame of observation is selected but that thing hasn't if another one is selected!
Suppose that in one frame of observation twin1 aged more than twin2 and so twin1 has a long bear while twin2 has not. That situation cannot change just changing the frame of observation. The fact that the twin has or not has bear cannot depend in the reference frame.
A consistent theory will give the same observation of that kind of things (which can be called Intrinsic properties) in any frame of observation. Otherwise is an inconsistent and a wrong theory!

(Q),

“ Originally Posted by martillo
A new theory is NEEDED! ”

Needed for what?
Well I think nobody wants to have to believe in a wrong theory. It could give wrong predictions and could make us think or even do something wrong.

(Q),

Well I think nobody wants to have to believe in a wrong theory. It could give wrong predictions and could make us think or even do something wrong.

It hasn't yet. So, what's your point?

It is just a theory.

Yes it does, General Relativity does. It even include any accelerated frame!
And it accounts for the pseudo-forces in those frames with the existence of a gravitational potential gradient and space-time curvature. I don't think GR handles rotating frames, though.

Boosts, translations and fixed rotations relative to what??? You must consider a first basic reference frame to say that and which is it? That frame must not be accelerated to be an inertial frame but not accelerated relative to what?
I suppose the short answer is to say that there's an absolute state of acceleration. Once you've detected an inertial reference frame (not very difficult), the Poincaré group gives you all the other reference frames with the same properties.

It must exist a basic frame or a set of frames which we could call them to be at rest! Then you after can state that any frame having a boost, translation or constant rotation relative to them is an inertial frame.
You must consider that any frame obtained by a boost, translation or constant rotation from an accelerated frame will also be an accelerated frame.
This gives absolute acceleration. Relativity denies absolute velocity. This, by the way, is exactly the way it was with Galilean relativity.

Just in case you didn't know, a "boost" is a transformation into a frame in relative motion. For example:

t' = \gamma (t - \frac{v}{c^2}x)

x' = \gamma (x - vt)
is a Lorentz boost along the x-axis.

Then once we accept that "rest" frames in the Universe exist just one of them selected by some property of the universe like possible symmetry would be the "privileged" absolute frame of the Universe.
If the laws of physics are the same in an entire set of reference frames, you cannot attribute special properties to just one.

The problem is that ages are not relative! Age is directly related to all physiological phenomena that have haened to an individual in his history.
All this tells me is that relativity is incompatible with your own worldview. A theory only contradicts itself if it predicts (in two different ways) that one observer will make two contradictory observations.

In any case, the success of relativity is due to its history of making accurate predictions. If you like, you can imagine that your mother ship is in an absolute reference frame, claim that time dilation and length contraction occur relative to that frame only, and still show that the twins will observe exactly what relativity predicts they will. In this sense, you can consider relativity to be an illusion if you want.

You cannot say something has happened if one frame of observation is selected but that thing hasn't if another one is selected!
If an event occurs in one reference frame, it occurs in all reference frames. It's only a question of when.

Suppose that in one frame of observation twin1 aged more than twin2 and so twin1 has a long bear while twin2 has not. That situation cannot change just changing the frame of observation. The fact that the twin has or not has bear cannot depend in the reference frame.
The problem here is simultaneity. Two events that occur simultaneously in one frame do not occur simultaneously in all frames. Look at the equation for the Lorentz boost I posted above. Notice that t' is also a function of x. Again, you can call relativity of simultaneity "real" or "apparent", but it is possible to show that either one of the moving twins will "naturally" map out a reference frame related to your mother ship frame by a Lorentz transformation.

Imagine one of the twins moving in the +x direction at velocity v. Suppose he wants to place a clock in front of him, and another behind him. From the point of view of the mother ship, the clock he pushes forward will move faster than v, and so will dilate more than the twin. The clock he pushes back will move slower than v, and so will experience less time dilation. If the clocks were synchronized before our twin moved them, they won't be synchronized afterwards. But he'll think they're still in sync because of the difference in time it takes light to reach him from each of the clocks.

I'll address point (B) here.

Suppose there's a particle traveling at velocity v respect to a referential R. R’ is a referential fixed to the particle. In R De Broglie formula is λ = h/mv where m and v are the velocity and the mass of the particle as seen from R.

Now we consider a change to R’. λ as a wavelength is a distance. If we apply the Lorentz transforms to λ, at t=0 for simplicity, λ is enlarged by the denominator s = (1-v2/c2)1/2. We will have in R’: λ’ = λ/s which gives us some finite value.

But what happens if we apply De Broglie equation in the same form in R’? The particle is at rest in R’, this means v’ = 0 and we would have λ’ = infinite which is in discrepancy with the result above.
The transformation of a wavelength (and distances in general) is not that simple. The expression of a wave (assuming unit amplitude and no phase) looks like this:

\cos(kx + \omega t) (wavelength \lambda = \frac{2 \pi}{k}, period T = \frac{2 \pi}{\omega})

The inverse Lorentz boost along the x axis is:

t = \gamma (t' + \frac{v}{c^2}x')

x = \gamma (x' + vt')

Substituting into the expression kx + \omega t gives:

\gamma k (x' + vt') + \gamma \omega (t' + \frac{v}{c^2}x')
This can be rearranged to give:

\gamma (k + \frac{v}{c^2}\omega) x' + \gamma (\omega + vk) t'
So we can rewrite our wave as:

\cos(k' x' + \omega' t')
where:

k' = \gamma (k + \frac{v}{c^2}\omega)

\omega' = \gamma (\omega + vk)

You can see that it is perfectly possible for the wavenumber k to be zero (corresponding to an infinite wavelength) in one frame and non-zero in another, provided \omega \neq 0. Where \omega = 0, k' = \gamma k, and we get the familiar length contraction formula for the wavelength:

\lambda' = \frac{1}{\gamma}\lambda

(Q),

It hasn't yet. So, what's your point?
Well, you haven't but I believe there are some contradictions and inconsistencies and is what I'm presenting here.

(Q),

Well, you haven't but I believe there are some contradictions and inconsistencies and is what I'm presenting here.

So, it is YOU personally who needs a new theory. Any particular reason why?

przyk,

I don't think GR handles rotating frames, though.

It does but the rotaton is treated the same as classically. If you have an object with a composed motion of linear velocity relative to an observer plus a rotation you just need to apply Lorentz Transform in the time varying plane of both, the observer and the object (that which also has the direction of the rotation) plus a classical rotation. It is a composed transform.


I suppose the short answer is to say that there's an absolute state of acceleration. Once you've detected an inertial reference frame (not very difficult), the Poincaré group gives you all the other reference frames with the same properties.

But this way you are accepting there is a special set of frames and is not what General Relativity says. When I'm referring to the frame that can be determined by pendulums and gyroscopes I'm referring to GR statement of no privileged frames of reference at all.


Relativity denies absolute velocity.
Yes and I believe it exist. We can just choose a frame with directions determined by gyroscopes far away from massive objects to ensure no gravitatonal effects and the center of the frame to be the center of the Universe which I believe must exist (although difficult to determine).
These way we can define rest frames in the Universe. One of them would be the Absolute Frame of the Universe which could be determined by some property of the Universe like symmetries in it.


If the laws of physics are the same in an entire set of reference frames, you cannot attribute special properties to just one.

I'm talking about special properties of the Universe and not of the laws of the Universe.


All this tells me is that relativity is incompatible with your own worldview. A theory only contradicts itself if it predicts (in two different ways) that one observer will make two contradictory observations.

A theory also contradicts itself if two different observers make contradictory observations what is present in the presented problem.



The problem here is simultaneity. Two events that occur simultaneously in one frame do not occur simultaneously in all frames.
The crossing event of the twins after the symmetrical travel gives only one event to be measured by all the frames in the instant when the center of all the frames coincide. This avoids the problem of the relativity of the simultaneity. This is one of the important features of the problem.

(Q),

So, it is YOU personally who needs a new theory. Any particular reason why?
Yes, I just want the truth.

przyk,

...and we get the familiar length contraction formula for the wavelength:
λ’ = (1/γ)λ

(copy/paste didn't work for the formula)
First note that the right equation is λ’ = γλ since the lenght observed by the frame "at rest" (λ) must be smaller than the lenght observed by the moving frame: the contraction must be λ < λ’ .
Second, you just arrived at the same formula as me so where is the problem???

(copy/paste didn't work for the formula)
Hit "quote" to reply. By the way, I'm using \TeX (added to the forum a few months ago) to typeset formulae. There's a thread about it here, in case you find it useful.

First note that the right equation is λ’ = γλ since the lenght observed by the frame "at rest" (λ) must be smaller than the lenght observed by the moving frame: the contraction must be λ < λ’ .
Second, you just arrived at the same formula as me so where is the problem???
The more general formula is:

\frac{1}{\lambda'} = \gamma \left( \frac{1}{\lambda} + \frac{v}{c^2} \, \frac{1}{T} \right)
where T is the period of the wave.

\lambda' = \frac{1}{\gamma} \lambda is true only if the period T (in the unprimed frame) is infinite.

\lambda' = \gamma \lambda is true only if the period T' (in the primed frame) is infinite.

The general relation between T and T' is:

\frac{1}{T'} = \gamma \left( \frac{1}{T} + v \frac{1}{\lambda} \right)

I'll respond to your other post when I have time.

(Q),

Yes, I just want the truth.

That would be your own, personal, self-gratifying truth?

Will that so-called truth be of any use to you? What do you expect to gain?

(Q),

That would be your own, personal, self-gratifying truth?

Yes, it is self-gratificating to find at least part of a truth.


Will that so-called truth be of any use to you?
I don't have a practical use for now but it is all under development and something can surge...


What do you expect to gain?
I used to dream to gain some things, it was a good incentive while having a hard work taking much of my time... but now I don't expect anything. I will receive just what others really would want to give, it all depends on the value it would have for otherones not for me.

That would be your own, personal, self-gratifying truth?

Will that so-called truth be of any use to you? What do you expect to gain?

i am sure this exactly what the greatest scientist were told by the orthodox community, so stop being narrow minded.

I don't have a practical use for now but it is all under development and something can surge...

What is under development? What will surge?


I used to dream to gain some things, it was a good incentive while having a hard work taking much of my time... but now I don't expect anything. I will receive just what others really would want to give, it all depends on the value it would have for otherones not for me.

Others already have found value in the existing theory. In fact, it works very well, with solid experimental results. What makes you think you will provide any more value?

(Q),

What is under development? What will surge?
You should take a look at the main page: http://www.geocities.com/anewlightinphysics


Others already have found value in the existing theory. In fact, it works very well, with solid experimental results. What makes you think you will provide any more value?
Well, a right theory has more value than a wrong theory.

(Q),

You should take a look at the main page: http://www.geocities.com/anewlightinphysics

A lot of nonsense. An electrical engineer who doesn't understand physics. He claims relativity is wrong and photons have mass. He is a confused individual.


Well, a right theory has more value than a wrong theory.

If it were wrong, yes. But, that is not the case. The theory is used and shows correct results every time. How is that wrong?

(Q),

The theory is used and shows correct results every time. How is that wrong?
And that is why is still considered true after 100 years I know. Nevertheless I have found theoretical inconsistencies and I also have found a totally new theory that seems to be right. It is under development and many things remains to be proven, still much work remains to be done, but is a good one...


A lot of nonsense. An electrical engineer who doesn't understand physics. He claims relativity is wrong and photons have mass. He is a confused individual.

I'm not confused, I have made some theoretical discoveries...

It does but the rotaton is treated the same as classically. If you have an object with a composed motion of linear velocity relative to an observer plus a rotation you just need to apply Lorentz Transform in the time varying plane of both, the observer and the object (that which also has the direction of the rotation) plus a classical rotation. It is a composed transform.
If GR doesn't treat rotating reference frames any better than they were treated classically, then it isn't "handling" them in the sense I was talking about (ie. that the same laws are valid in rotating and non-rotating frames).

But this way you are accepting there is a special set of frames and is not what General Relativity says. When I'm referring to the frame that can be determined by pendulums and gyroscopes I'm referring to GR statement of no privileged frames of reference at all.
If GR doesn't claim that the laws of physics are invariant with respect to transformations into rotating frames, it certainly cannot claim all reference frames are equally valid.

Yes and I believe it exist. We can just choose a frame with directions determined by gyroscopes far away from massive objects to ensure no gravitatonal effects and the center of the frame to be the center of the Universe which I believe must exist (although difficult to determine).
Do you have anything more than your own personal belief to offer? Because every indication is that this belief is wrong. Gyroscopes may detect an absolute state of rotation, but that is all they detect.

These way we can define rest frames in the Universe. One of them would be the Absolute Frame of the Universe which could be determined by some property of the Universe like symmetries in it.

I'm talking about special properties of the Universe and not of the laws of the Universe. For example the way the axis of rotation of Earth is determined.
The "absolute" frame that would contradict special relativity is the unique reference frame in which the laws of physics would take their simplest form, which requires variance of the laws of physics with respect to transformations. If you're working on a different definition of "absolute" (eg. considering the CMB or centre of mass of the universe), fine, but it no longer has anything to do with relativity.

A theory also contradicts itself if two different observers make contradictory observations what is present in the presented problem.
So what makes two different observations made by two different observers a contradiction in and of itself? I'd say it just made whatever was being observed frame (or observer) dependent.

The crossing event of the twins after the symmetrical travel gives only one event to be measured by all the frames in the instant when the center of all the frames coincide. This avoids the problem of the relativity of the simultaneity. This is one of the important features of the problem.
You can't measure a rate of change with only one data point.

Nevertheless I have found theoretical inconsistencies and I also have found a totally new theory that seems to be right. It is under development and many things remains to be proven, still much work remains to be done, but is a good one...

Good luck with that. Although, I'll have to assume those so-called 'theoretical inconsistencies' are merely assertions based on misunderstandings.


I'm not confused, I have made some theoretical discoveries...

You mean to tell me that is YOUR website, that it is YOU? YOU are the confused electrical engineer who wrote all that nonsense?

przyk,

If GR doesn't treat rotating reference frames any better than they were treated classically, then it isn't "handling" them in the sense I was talking about (ie. that the same laws are valid in rotating and non-rotating frames).

Can you give an example?


Gyroscopes may detect an absolute state of rotation, but that is all they detect.

They "detect an absolute state of rotation" and you think this is irrelevant?
They determine very special frames of reference!


The "absolute" frame that would contradict special relativity is the unique reference frame in which the laws of physics would take their simplest form, which requires variance of the laws of physics with respect to transformations. If you're working on a different definition of "absolute" (eg. considering the CMB or centre of mass of the universe), fine, but it no longer has anything to do with relativity.

When I say that an absolute frame can be determined by gyroscopes directions and the center at the center of the Universe I'mnot defining "absolute frames" but just giving a way to determine them. Of course I propose that some laws are non-invariant. I present DE Broglie law as an example in part B of page cited at the head post.


So what makes two different observations made by two different observers a contradiction in and of itself? I'd say it just made whatever was being observed frame (or observer) dependent.

There are some things that are not frame dependent. If I has a long bear I will have it in any frame of reference that I could be observed. Age is an intrinsic property of living individuals that cannot be frame dependent anyway. You cannot observe that something have happened in a frame of observation while not happening just changing the frame of observation.


You can't measure a rate of change with only one data point.
Of course I can, just look at Lorentz transformation for time!

Hi martillo,

Can you give an example?
Of what? A law that is variant with respect to a transformation into a rotating frame? Just take either classical or relativistic mechanics. An object with no external forces acting upon it doesn't even obey Newton's first law in a rotating frame (see Coriolis effect (http://en.wikipedia.org/wiki/Coriolis_effect)).

They "detect an absolute state of rotation" and you think this is irrelevant?
As far as special relativity is concerned, yes (I don't really feel like defending a theory like GR that I don't understand).

They determine very special frames of reference!
They don't contradict Lorentz invariance, and therefore don't contradict special relativity. Relativity only denies absolute linear velocity. It says nothing about absolute angular velocity. The two are completely independent.

When I say that an absolute frame can be determined by gyroscopes directions and the center at the center of the Universe I'mnot defining "absolute frames" but just giving a way to determine them.
You have to show that the "absolute frame" you are determining is the same type of "absolute frame" relativity claims is undetectable. Otherwise, you don't have a case against relativity.

Of course I propose that some laws are non-invariant. I present DE Broglie law as an example in part B of page cited at the head post.
The de Broglie relations (http://en.wikipedia.org/wiki/De_Broglie_hypothesis#The_de_Broglie_relations) can be expressed as:

\vec{p} = \hbar \vec{k}
where \vec{p} and \vec{k} are the four-momentum (http://en.wikipedia.org/wiki/Four-momentum) and the wavenumber, respectively (the timelike component of \vec{k} is related to the frequency of the wave, which the above equation links to the particle's energy). Since \vec{k} is proportional to \vec{p}, it, like the four-momentum, is a four-vector (http://en.wikipedia.org/wiki/Four-vector) (meaning it transforms by the Lorentz transformation).

The equation of a (real) wave can be expressed as:

\Psi = A \cos(\vec{k} \cdot \vec{x} + \varphi)
where \vec{x} is the four-position, and \vec{k} \cdot \vec{x} = - k^0 x^0 + k^1 x^1 + k^2 x^2 + k^3 x^3 is the Minkowski inner product (http://en.wikipedia.org/wiki/Minkowski_space#The_Minkowski_inner_product) of \vec{k} and \vec{x}. The Minkowski inner product of any two four-vectors is a Lorentz scalar (http://en.wikipedia.org/wiki/Lorentz_scalar) (ie. invariant). To anyone familiar with the Minkowski formalism, this ends the discussion on the Lorentz invariance of the relativistic de Broglie relations.

I posted the correct, general transformations for the wavelength (\lambda' = \gamma \lambda is only a special case) and period of a wave here, following a derivation here. I suggest you reread these posts - then we can discuss any specific points you don't find convincing.

There are some things that are not frame dependent. If I has a long bear I will have it in any frame of reference that I could be observed. Age is an intrinsic property of living individuals that cannot be frame dependent anyway.
If I have a theory that claims that the existence of your long beard is frame dependent, my theory clearly would not fit observation and experience, but it could well be internally consistent.

You cannot observe that something have happened in a frame of observation while not happening just changing the frame of observation.
This is the way things happen to be and what human intuition has evolved to expect. It is not a logical necessity.

Of course I can, just look at Lorentz transformation for time!
Staring at an equation and deriving something based on physical measurements are not the same thing. How would your two twins measure their relative ageing rates?

przyk,

A law that is variant with respect to a transformation into a rotating frame? Just take either classical or relativistic mechanics. An object with no external forces acting upon it doesn't even obey Newton's first law in a rotating frame (see Coriolis effect).

From your link:

The Coriolis effect is caused by the Coriolis force, which appears in the equation of motion in a rotating frame of reference. Sometimes this force is called a fictitious force (or pseudo force), because it does not appear when the motion is expressed in an inertial frame of reference. Regardless of the chosen frame of reference, the motion is the same. In an inertial frame of reference, the real impressed forces, together with inertia, are sufficient to explain the motion. In a rotating frame, the Coriolis and centrifugal forces are needed in the equation to correctly describe the motion.

A rotational frame is an accelerated frame (non inertial) and of course "fictitious" forces are needed and used even in Classical Physics! In an accelerated frame every object must be considered as under the effect of the "fictitious" forces. The same happens in frames under gravitational fields.


You have to show that the "absolute frame" you are determining is the same type of "absolute frame" relativity claims is undetectable. Otherwise, you don't have a case against relativity.

Of course it is. Is obvious.


I posted the correct, general transformations for the wavelength ( λ’ = γλ is only a special case) and period of a wave here, following a derivation here. I suggest you reread these posts - then we can discuss any specific points you don't find convincing.

I repeat: you arrive at the same formula as me so where is the problem?


If I have a theory that claims that the existence of your long beard is frame dependent, my theory clearly would not fit observation and experience, but it could well be internally consistent.

I don't say Relativity is "internally" inconsistent, I say it is physically inconsistent.



You cannot observe that something have happened in a frame of observation while not happening just changing the frame of observation.

This is the way things happen to be and what human intuition has evolved to expect. It is not a logical necessity.

"Human intuition has evolved to expect???
It is a physical necessity. It's about reality in the real world. The reality in some place (as the representation of all phenomena happening in some place of the Universe at some time) is only one, is unique!

From your link:

A rotational frame is an accelerated frame (non inertial) and of course "fictitious" forces are needed and used even in Classical Physics! In an accelerated frame every object must be considered as under the effect of the "fictitious" forces. The same happens in frames under gravitational fields.
So...?

Of course it is. Is obvious.
So you've found Lorentz variant laws?

I repeat: you arrive at the same formula as me so where is the problem?
You did not arrive at:

\frac{1}{\lambda'} = \gamma \left( \frac{1}{\lambda} + \frac{v}{c^2} \, \frac{1}{T} \right)
This equation allows the wavelength to be finite in one frame and infinite in another. If you don't see a problem then you have either not read my posts or you have changed your mind about the de Broglie wavelength violating Lorentz invariance.

I don't say Relativity is "internally" inconsistent, I say it is physically inconsistent.
So you can cite experimental results that are inconsistent with relativity?

"Human intuition has evolved to expect???
Yes. Humans have evolved minds designed to cope with the world we live in as we see it. We did not directly experience curved space, reciprocity, wave-particle duality and so on in everyday life in the history of our species, so it's not surprising that we never evolved to consider them natural or obvious. They seem impossible, but in reality they just contradict a worldview you take for granted.

It is a physical necessity.
What makes it a necessity? Necessity for what, anyway?

It's about reality in the real world. The reality in some place (as the representation of all phenomena happening in some place of the Universe at some time) is only one, is unique!
Relativity doesn't contradict this. The descriptions of a phenomenon gets transformed from one reference frame to another, but the phenomenon remains the same.

przyk,

So...?

So there is no problem with rotational frames in GR nor even in Classical Physics as you have said.


So you've found Lorentz variant laws?
Don't you read properly the posts?
Part B shows De Broglie law non-invariant in a Lorentz transform.


This equation allows the wavelength to be finite in one frame and infinite in another.
My equation is the same as that you called "simplified equation" and is derived directly from the Lorentz transform of distance (lenght). This is the right way to analize the variance or non-invariance of De Broglie law and it appears non-invariant.


So you can cite experimental results that are inconsistent with relativity?
Not for the moment. I made a theoretical work and have found Relativity THEORETICALLY inconsistent.


They seem impossible, but in reality they just contradict a worldview you take for granted.

That the way Relativity defenders have to think. I don't believe it. I believe our intuitive perception of nature is right and Relativity is wrong.


What makes it a necessity? Necessity for what, anyway?

I have said: "You cannot observe that something have happened in a frame of observation while not happening just changing the frame of observation."
This is a so basic principle that is not discussed!



It's about reality in the real world. The reality in some place (as the representation of all phenomena happening in some place of the Universe at some time) is only one, is unique!
Relativity doesn't contradict this. The descriptions of a phenomenon gets transformed from one reference frame to another, but the phenomenon remains the same.
There are some descriptions or observations that are frame dependent (like trajectory) but there are others like age that are not. As I said, for example if I have a bear at some time I will have it independently of the referential chosed to observe me.
Relativity is inconsistent with the intrinsic property of age of living beings.

So there is no problem with rotational frames in GR nor even in Classical Physics as you have said.
I never said that you couldn't transform the laws of physics into a rotating reference frame. In general, you can do this for any arbitrarily twisted coordinate system (you can even use the Galilean transformation if you really want) - it just boils down to variable substitutions and a bit of algebra. The point is that the transformed laws generally don't take the same form in the new coordinate system as they do in the original inertial frame. The paragraph you quoted from Wiki supports this in the case of rotating frames.

Don't you read properly the posts?
Part B shows De Broglie law non-invariant in a Lorentz transform.
I do, and no, it doesn't.

My equation is the same as that you called "simplified equation" and is derived directly from the Lorentz transform of distance (lenght). This is the right way to analize the variance or non-invariance of De Broglie law and it appears non-invariant.
No it's not. If it was, you'd be able to point out specific flaws in my arguments here and here and here, rather than resorting to reflexive denial and contradiction. I, on the other hand, can point out the specific flaw in your reasoning: you are applying a formula of limited applicability where it isn't applicable. Unless you can find a flaw in the derivation I provided, the correct transformation of the wavelength is this one:

\frac{1}{\lambda'} = \gamma \left( \frac{1}{\lambda} + \frac{v}{c^2} \, \frac{1}{T} \right)
This only reduces to \lambda' = \frac{1}{\gamma} \lambda if T = \infty. Were this the case, your point would be valid. But T isn't infinite. It's related to the particle's energy by:

T = \frac{h}{E}
(this is a rearrangement of the equation E = hf)

Not for the moment. I made a theoretical work and have found Relativity THEORETICALLY inconsistent.
Relativity is an internally consistent and tested theory. It is only inconsistent with your own preconceptions. You aren't trying to observe and explain the laws of nature. You are trying to dictate what you think they should be.

That the way Relativity defenders have to think. I don't believe it. I believe our intuitive perception of nature is right
There's no reason for our intuition to be naturally tuned to understanding nature on scales and in situations we don't encounter in everyday life.

I have said: "You cannot observe that something have happened in a frame of observation while not happening just changing the frame of observation."
This is a so basic principle that is not discussed!
I'm not claiming this is right or wrong. I am claiming it cannot be shown that the universe must obey this principle, not that we don't observe that it does. There's a difference.

Relativity is inconsistent with the intrinsic property of age of living beings.
Therefore, your notion that age is an "intrinsic property" of living beings is inconsistent with reality.

przyk,
I will answer the point related to the wavelenght now since I think is the most important now.
In my manuscript I derived the Lorentz transformation of the wavelenght directly as a lenght and I think is right.
Your formula is also right but you have found the Lorentz transformation of the equation of the wave and derived which condition the wavelenght must satisfy.
In principle both derivation should be the same but they aren't which show some inconsistenct but I will not treat this point. I find more important to say that your formula is equivalent in showing the same type of inconsistency of Relativity Theory I have found with "my" formula!
You must consider that the De Broglie wave has a velocity of propagation equal to c2/v (See: http://www.davis-inc.com/physics/broglie/broglie.shtml) what gives a finite value of the wavelenght in your formula since there is a relationship between λ and T: λf = λ/T = velocity of propagation = c2/v. This value is the value of the wavelenght in the frame of the object moving at velocity v and as I said is finite but if we apply the De Broglie formula in the same form (invariantly) in this frame we obtain a wavelenght equal to infinity since the velocity there is zero (λ=h/mv).
So your formula also demonstrates that the De Broglie formula is not invariant under a change of referentials!

De Broglie law is non-invariant and so the first principle of Relativity is false. It says that all laws in Physics must be invariant under a change of frames while in practice we have found one that is not.

martillo:

In my manuscript I derived the Lorentz transformation of the wavelenght directly as a lenght and I think is right.
No it's not. The length contraction formula only works if whatever you're measuring the length of is at rest in one of the two frames you are considering. As you just worked out for yourself, the phase velocity of a de Broglie wave is never less than c in any frame.

Your formula is also right but you have found the Lorentz transformation of the equation of the wave and derived which condition the wavelenght must satisfy.
Which condition would that be?

In principle both derivation should be the same but they aren't which show some inconsistenct but I will not treat this point. I find more important to say that your formula is equivalent in showing the same type of inconsistency of Relativity Theory I have found with "my" formula!
You must consider that the De Broglie wave has a velocity of propagation equal to c2/v (See: http://www.davis-inc.com/physics/broglie/broglie.shtml) what gives a finite value of the wavelenght in your formula since there is a relationship between λ and T: λf = λ/T = velocity of propagation = c2/v. This value is the value of the wavelenght in the frame of the object moving at velocity v and as I said is finite but if we apply the De Broglie formula in the same form (invariantly) in this frame we obtain a wavelenght equal to infinity since the velocity there is zero (λ=h/mv).
So your formula also demonstrates that the De Broglie formula is not invariant under a change of referentials!
Er, how can \frac{c^2}{v} be finite and \frac{h}{\gamma m v} be infinite for the same v?

przyk,

Er, how can be finite and be infinite for the same ?
Is not the same.
v is the velocity of the object as seen by the "fixed" frame only.
You must remember we are considering at first time the wave of an object of mass m travelling at velocity v relative to the "fixed" frame and at second time we are trying to see the wave from the moving frame at the moving object. The velocity of the object in the moving frame is zero and so the De Broglie wavelenght of it as seen by the moving frame is infinity (its velocity is also infinity but it doesn't matter now).
We use the velocity v to Lorentz transform the wavelenght from that seen by the "fixed" frame to the moving frame.
The objective is to compare the results and they appear different even with your formula and so there is a contradiction. If De Broglie law would be invariant the results would be the same but they aren't.
Then De Broglie is not invariant under a change of referentials.
Is a non invariant law in contradiction with the first principle of Relativity. Then the principle is not valid.

martillo:

No it's not. The length contraction formula only works if whatever you're measuring the length of is at rest in one of the two frames you are considering. As you just worked out for yourself, the phase velocity of a de Broglie wave is never less than c in any frame.

Which condition would that be?

Er, how can \frac{c^2}{v} be finite and \frac{h}{\gamma m v} be infinite for the same v?

Can u please shed some light on how length contraction was know first time, or any links, please.

Singularity,
Please read the thread from the head post.

Singularity,
Please read the thread from the head post.

I am not professional , i dont understand your HiFi language, i am reading this thread from the top. Hope someone answers the question to the point, i have been asking this question for many months now, and none of the experts could utter a word on it.:rolleyes:

Singularity,
Lorentz (before Einstein) was the first to point that lenght contraction would be necessary to explain some electrodynamics of the electrons. I don't remember which exactly were those experiments.

Is not the same.
v is the velocity of the object as seen by the "fixed" frame only.
Look at your own formula:

\frac{\lambda}{T} = \frac{c^2}{v}
\lambda and T are both frame dependent quantities, therefore so is v. v, \lambda, and T are all as observed in the same reference frame.

You must remember we are considering at first time the wave of an object of mass m travelling at velocity v relative to the "fixed" frame and at second time we are trying to see the wave from the moving frame at the moving object. The velocity of the object in the moving frame is zero and so the De Broglie wavelenght of it as seen by the moving frame is infinity
And \lambda' = \frac{c^2}{v'} T is also infinite if v' = 0, so where's the problem?

To generalize for a particle of velocity u:

\lambda = \frac{c^2}{u} T = \frac{c^2}{u} \frac{h}{E} = \frac{c^2}{u} \frac{h}{\gamma(u) m c^2} = \frac{h}{\gamma(u) m u} = \frac{h}{p}
so \lambda = \frac{c^2}{u} T is the de Broglie wavelength. This shouldn't come as a surprise considering it's the two de Brogie relations that are used to derive \frac{\lambda}{T} = \frac{c^2}{u} in the first place!

(its velocity is also infinity but it doesn't matter now).
See this (http://en.wikipedia.org/wiki/Phase_velocity#Matter_wave_phase) and this (http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#5). The latter gives a whole list of trivial ways in which you can get "something" to travel faster than light in a way that isn't a problem for relativity.

We use the velocity v to Lorentz transform the wavelenght from that seen by the "fixed" frame to the moving frame.
The objective is to compare the results and they appear different even with your formula and so there is a contradiction. If De Broglie law would be invariant the results would be the same but they aren't.
Then De Broglie is not invariant under a change of referentials.
Is a non invariant law in contradiction with the first principle of Relativity. Then the principle is not valid.
Given that a particle's velocity is u and mass is m (and p = \gamma(u) m u, E = \gamma(u) m c^2) in a reference frame K, there are two ways of calculating its wavelength in a frame K' travelling at velocity v with respect to K.

The first method is to find the particle's velocity u' and momentum p' in K', and apply the de Broglie relation there:

\lambda' = \frac{h}{p'}

The second approach is to use the de Broglie relations in K to substitute the particle's momentum and energy into the wavelength transformation formula I posted earlier, giving:

\frac{1}{\lambda'} = \gamma(v) \left( \frac{p}{h} - \frac{v}{c^2} \frac{E}{h} \right)

If these formulae agree, the de Broglie wavelength formula is Lorentz invariant. You can see quite easily that both equations will yield \lambda' = \infty for u = v.

NB: The appearance of the minus sign in the second equation for \lambda' is worth explaining: a Lorentz invariant wave would be dependent on the Minkowski product \vec{k} \cdot \vec{r} = - k^0 r^0 + k^1 r^1 + k^2 r^2 + k^3 r^3 (\vec{r} is a four-position vector), with k^0 proportional to the \omega related to the energy. When deriving the wavelength transformation formula, I was working with k x + \omega t, with the opposite sign in front of the products of the timelike coefficients of both vectors. Apologies if this leads to any confusion - I'm not exactly copying all this out of a textbook, and I was initially trying to avoid bringing details and the Minkowski formalism into this.

Can u please shed some light on how length contraction was know first time, or any links, please.
See: http://en.wikipedia.org/wiki/Length_contraction

przyk,

I accept the change of the minus sign in the equation. You would have it in that way if you had started with the equation cos(kx-wt) for the wave.
I must recognize you have found a very good point with your reasoning.
This will make me think more about.

I still wonder: if the wavelenght is a lenght (a distance) in the x axis why it does not transform as a normal lenght contraction λ=λ'/γ? I think that in principle it should do.

Returning to other subject:

I never said that you couldn't transform the laws of physics into a rotating reference frame. In general, you can do this for any arbitrarily twisted coordinate system (you can even use the Galilean transformation if you really want) - it just boils down to variable substitutions and a bit of algebra. The point is that the transformed laws generally don't take the same form in the new coordinate system as they do in the original inertial frame. The paragraph you quoted from Wiki supports this in the case of rotating frames.

But the concept of the "fictitius"forces enables for exmple the motion law: F=dp/dt have the same form. Is just that the forces are different when seen from different frames.
This is why I asked for an example. In which cases a law would take different form?
Note that if this is the case you are pointing to a possible inconsistency in Relativity Theory.

martillo,

I accept the change of the minus sign in the equation. You would have it in that way if you had started with the equation cos(kx-wt) for the wave.
I must recognize you have found a very good point with your reasoning.
This will make me think more about.
Happy thinking :)

I'd just like to point out something: The Minkowski product of any two four-vectors is invariant (this is why you'll sometimes hear Lorentz boosts called "rotations" in space-time), so you can see immediately that a relation like \Psi = e^{i \vec{k} \cdot \vec{r}} is Lorentz invariant. This is why Minkowski's formalism is so popular - it allows laws to be expressed in a form that emphasizes their invariance using terms that are Lorentz scalars. There are also invariant equivalents to the classical gradient, divergence, and laplacian operators.

I still wonder: if the wavelenght is a lenght (a distance) in the x axis why it does not transform as a normal lenght contraction λ=λ'/γ? I think that in principle it should do.
You can derive the length contraction formula from the Lorentz transformation by considering the worldlines of two objects stationary in one frame.

If you have an object at rest at the origin in K' (x_1' = 0) and another at x_2' = L', then for object #1:

x_1' = \gamma ( x_1 - v t ) = 0


\Rightarrow x_1 = v t

For #2:

x_2' = \gamma ( x_2 - v t ) = L'


\Rightarrow x_2 = vt + \frac{L'}{\gamma}

The difference L = x_2 - x_1 gives:

L' = \gamma L

I'm sure you can extend this reasoning to see what happens to the relationship between L and L' if objects #1 and #2 are not stationary in K'.

Returning to other subject:

But the concept of the "fictitius"forces enables for exmple the motion law: F=dp/dt have the same form. Is just that the forces are different when seen from different frames.
This is why I asked for an example. In which cases a law would take different form?
Note that if this is the case you are pointing to a possible inconsistency in Relativity Theory.
On its own \vec{F} = \frac{d \vec{p}}{d \tau} has no predictive power. To make real predictions you need another law that determines \vec{F} (electromagnetism, gravity, etc.). While it's true that you can introduce centrifugal forces and the like to keep \vec{F} = \frac{d \vec{p}}{d \tau} invariant, this comes at the expense of making the laws determining the force variant.

przyk,

Thanks to our discussion I got deeper in De Broglie wave features and I have found that I had a different concept of the matter-wave.
I used to think it was a real wave but now I see it is not!
I realize now that the De Broglie waves seen by different frames are not geometrically the same wave in the space! The limit case is that while in the "fixed" frame is a periodic wave with λ=h/p in the "moving" frame it is a straight line (λ=infinite) moving at infinite velocity (actually not a wave)!
So it is stated by current Physics that for any massive object exist a wave associated to it but it is frame dependent! Different frames see different waves!
This is too strange...

martillo,

Thanks to our discussion I got deeper in De Broglie wave features and I have found that I had a different concept of the matter-wave.
I used to think it was a real wave but now I see it is not!
Ironically, I thought this in your "Classical Physics is coming back, RELOADED!!!" thread, and have since realized that it is the same wave in both frames!

I realize now that the De Broglie waves seen by different frames are not geometrically the same wave in the space! The limit case is that while in the "fixed" frame is a periodic wave with λ=h/p in the "moving" frame it is a straight line (λ=infinite) moving at infinite velocity (actually not a wave)!
It's the same wave, in the sense that any two observers will agree on the value of \Psi at any given point in space and time. The strangeness is all in the phase velocity being greater than c and the Lorentz transformation (specifically, relativity of simultaneity). An infinite wavelength corresponds to an infinite phase velocity, and the Lorentz transformation allows a velocity to be infinite in one frame and finite in another. Velocities transform according to:

u' = \frac{u - v}{1 - \frac{u v}{c^2}}
u' can be infinite if the denominator 1 - \frac{u v}{c^2} = 0, or rearranged:

u = \frac{c^2}{v}
This is the phase velocity you derived earlier!

przyk,

It's the same wave, in the sense that any two observers will agree on the value of at any given point in space and time. The strangeness is all in the phase velocity being greater than and the Lorentz transformation (specifically, relativity of simultaneity). An infinite wavelength corresponds to an infinite phase velocity, and the Lorentz transformation allows a velocity to be infinite in one frame and finite in another.
Yes, the variation of the time in the different referentials introduce strange features...

I must rethink what I have wrotten.
It is seeming De Broglie law is invariant within Relativity statements.

I give the point for you in this part but not yet in the other two. They involve conceptual subjects (not mathematical ones) in which we probably will not agree:


“ Relativity is inconsistent with the intrinsic property of age of living beings. ”

Therefore, your notion that age is an "intrinsic property" of living beings is inconsistent with reality.

I think that the "intrinsic property" of the age is consistent with reality while Relativity is inconsistent with reality.


“ They "detect an absolute state of rotation" and you think this is irrelevant? ”

As far as special relativity is concerned, yes (I don't really feel like defending a theory like GR that I don't understand).
I'm referring to GR here and its statement that there are no privileged frames.

What do u think will happen to Magnetic Induction if the conductor and the magnetic fields pass by each other at near light speed compared to earth but in opposite directions.


http://en.wikipedia.org/wiki/Electromagnetic_induction

Please correct the question if u understood, i am not much aware of what i am quoting here.

Singularity,
Your question is out of the topic of this thread. Please open a new thread for it.

I think that the "intrinsic property" of the age is consistent with reality while Relativity is inconsistent with reality.
We could go in circles arguing about what nature can and can't do, but in this case I don't think it's necessary. At a first glance, reciprocity (the "each twin ageing slower than the other" claim) seems impossible, but unlike quantum mechanics, relativity doesn't require you to abandon your worldview, provided you are willing to accept contraction and slowing of physical processes relative to an absolute frame. Here's a scenario I think you should consider carefully:

Imagine we have an observer A at rest (let's say at the origin) in an absolute frame, and B in motion (velocity +v to the right) in this absolute frame. B's motion results in him ageing slower than A in accordance with the time dilation formula. While A ages \Delta t seconds, B ages \Delta t' = \frac{1}{\gamma} \Delta t seconds.

The equations of motion for the two observers are:

x_A = 0


x_B = v t + X (with X > 0)

Now suppose A sends two pulses of light out toward B: the first at t = 0 and the second at t = T. The equations for these are then:

x_{\gamma 1} = c t


x_{\gamma 2} = c ( t - T ) (with T > 0)

The first pulse reaches B when x_{\gamma 1} = x_B:

c t_1 = v t_1 + X \Rightarrow t_1 = \frac{X}{c - v}

The second pulse reaches B when x_{\gamma 2} = x_B:

c ( t_2 - T ) = v t_2 + X \Rightarrow t_2 = \frac{X + cT}{c - v}

then:

\Delta t = t_2 - t_1 = \frac{c}{c - v}T

During this time, taking B's slow ageing rate into account, B will age:

\Delta t' = \frac{1}{\gamma} \frac{1}{1 - \frac{v}{c}}T = \frac{1}{\gamma} \gamma^2 \left( 1 + \frac{v}{c} \right) T

so:

\Delta t' = \gamma \left( 1 + \frac{v}{c} \right) T

Now we repeat the same scenario, but this time it is B who sends the pulses at A.

B ages T' seconds between sending out the two pulses (first one at t = 0). In the absolute frame, the second pulse is emitted at t = T = \gamma T'. The equations for these two pulses are:

x_{\gamma 1} = -ct + X


x_{\gamma 2} = -ct + X + ( c + v ) T

(the second having to satisfy x_{\gamma 2} = x_B at t = T)

The first pulse reaches A when x_{\gamma 1} = 0:

- c t_1 + X = 0 \Rightarrow t_1 = \frac{X}{c}

The second pulse reaches A when x_{\gamma 2} = 0:

- c t_2 + X + ( c + v ) T = 0 \Rightarrow t_2 = \frac{X + ( c + v ) T}{c}

So the time A ages between receiving both pulses from B is:

\Delta t = t_2 - t_1 = \left( 1 + \frac{v}{c} \right) T = \gamma \left( 1 + \frac{v}{c} \right) T'

so:

\Delta t = \gamma \left( 1 + \frac{v}{c} \right) T'

There you go: reciprocity, and we never had to consider B's rest frame!

Hope that wasn't too long.

I'm referring to GR here and its statement that there are no privileged frames.
I don't know GR, so I don't know how justified it would be in claiming all frames are completely equivalent.

What do u think will happen to Magnetic Induction if the conductor and the magnetic fields pass by each other at near light speed compared to earth but in opposite directions.


http://en.wikipedia.org/wiki/Electromagnetic_induction

Please correct the question if u understood, i am not much aware of what i am quoting here.
I'm not too familiar with electromagnetism, so you'd be better off starting a new thread in the Physics subforum or asking a moderator to split this one.

What's prompting this question, anyway?

przyk,

I understand your point but that reciprocity does not applly in my problem.

At the final cross point where the three observers coincide there is only only one state possible for each of the twins. As I said in the page each twin can take photographs of themselves and send them to the others to everybody see what really happen to them at that moment. But note that the problem now is not to choose or determine which prediction would be the right one, the problem is that Relativity gives three different predictions in each frame which are not compatible with each other. They are contradictory. You cannot observe a twin aged with a long bear in one frame and see a half bear or not bear at all in the other frames!
Relativity is inconsistent in this problem.

At the final cross point where the three observers coincide there is only only one state possible for each of the twins. As I said in the page each twin can take photographs of themselves and send them to the others to everybody see what really happen to them at that moment. But note that the problem now is not to choose or determine which prediction would be the right one, the problem is that Relativity gives three different predictions in each frame which are not compatible with each other. They are contradictory. You cannot observe a twin aged with a long bear in one frame and see a half bear or not bear at all in the other frames!
Hang on, is it the ages or the ageing rates that are bothering you?

All three observers will agree on their relative ages. At the cross point, the two twins are both the same age, and both are younger than the observer on the mothership. This is true in all reference frames.

przyk,

is it the ages or the ageing rates that are bothering you?

One is consequence of the other.


All three observers will agree on their relative ages. At the cross point, the two twins are both the same age, and both are younger than the observer on the mothership. This is true in all reference frames.
Not at all. You must "sit" with each observer to see what they see. You must observe the phenomenon from each of the three frames and all the observations are different.

Not at all. You must "sit" with each observer to see what they see. You must observe the phenomenon from each of the three frames and all the observations are different.
The observer on the mothership sees the two twins age slowly as they move away and as they come back, such that they are both the same age as each other and both younger than the mothership observer when they get back.

Each twin sees the mother ship observer and the other twin age slowly while he's moving away from the mothership, age rapidly while he's turning his ship around (accelerating) and sees them age slowly on the way back, such that he sees he's the same age as the other twin and younger than the mothership observer when he gets back.

At the crossover point, they all agree on who has the longest beard.

przyk,

The observer on the mothership sees the two twins age slowly as they move away and as they come back, such that they are both the same age as each other and both younger than the mothership observer when they get back.

Each twin sees the mother ship observer and the other twin age slowly while he's moving away from the mothership, age rapidly while he's turning his ship around (accelerating) and sees them age slowly on the way back, such that he sees he's the same age as the other twin and younger than the mothership observer when he gets back.

At the crossover point, they all agree on who has the longest beard.
I'm sorry but this is wrong.
One twin could see the other aging faster while going away and slowler while coming but the age accumulates and at the crossover he will always see the other younger. But exactly the same happens to the other twin seeing the first younger!

The unique way they could all agree in time at the crossover point is if they brake and stop the ships at the mothership's place making v=0 but this is not what they do in the proposed problem. The problem is thought with the twins continuing travelling at their constant velocity and interchanging photographs at the crossover point to all see what really happen to all of them.

I'm sorry but this is wrong.
One twin could see the other aging faster while going away and slowler while coming but the age accumulates and at the crossover he will always see the other younger. But exactly the same happens to the other twin seeing the first younger!
Each twin ages slower in the other's rest frame when the other is not accelerating. What happens during the acceleration periods (both twins have to decelerate, turn around, and accelerate back toward the mothership) isn't so simple. Just thinking about it, while each twin is accelerating, the other must be ageing much faster than him in his rest (accelerating) frame if they're going to be the same age at the cross-over point. This is General Relativity's gravitational time dilation. You can derive this effect by extending the Lorentz transformation to include accelerating frames, by considering all the rest frames an accelerating object passes through.

The unique way they could all agree in time at the crossover point is if they brake and stop the ships at the mothership's place making v=0 but this is not what they do in the proposed problem. The problem is thought with the twins continuing travelling at their constant velocity and interchanging photographs at the crossover point to all see what really happen to all of them.
This accelerated/gravitational time dilation effect increases with distance. Braking a short distance from the mother ship changes very little.

przyk,
Acceleration has nothing to do here. If you look at Lorentz Transform it depends on velocity only, not in the acceleration.
Einstein showed the timing problem in a relativistic train without even mentioning the necessary acceleration to move it at a relativistic velocity.
One twin age less than the other due to its relative velocity, if the velocity returns to zero again the time difference disappear and their clocks are synchronized again.

Acceleration has nothing to do here. If you look at Lorentz Transform it depends on velocity only, not in the acceleration.
Both twins accelerate: they move away from the mothership then turn around to get back. If you want to find the relative ages of all the observers as seen by one of the twins, you can't ignore this.

Einstein showed the timing problem in a relativistic train without even mentioning the necessary acceleration to move it at a relativistic velocity.
Do you have a link? I'm not sure which thought experiment you are referring to.

if the velocity returns to zero again the time difference disappear and their clocks are synchronized again.
Who told you this?

przyk,

“ Originally Posted by martillo
Acceleration has nothing to do here. If you look at Lorentz Transform it depends on velocity only, not in the acceleration. ”

Both twins accelerate: they move away from the mothership then turn around to get back. If you want to find the relative ages of all the observers as seen by one of the twins, you can't ignore this.

I assume acceleration in neglihible time to have travels with constant velocity.
The time dilation is due to their relative velocity not their acceleration. Look at Lorentz equations.


Do you have a link? I'm not sure which thought experiment you are referring to.

I have an Spanish translation of Einstein's book "Sobre la Teoria de la Relatividad Especial y General".
Here he explain Special Relativity with a relativistic train.


“ if the velocity returns to zero again the time difference disappear and their clocks are synchronized again. ”

Who told you this?
I deduce this from the Lorentz Transform. There is a time dilation while a relative velocity and distance exist between the two frames. When they coincide again in space with zero velocity there is no more time difference. Just look at the equations!

By the way I have updated the manuscript in the site taking away the argument of the De Broglie law, rewriting a little of the twins problem and some other minor edition corrections.

I assume acceleration in neglihible time to have travels with constant velocity.
The time dilation is due to their relative velocity not their acceleration. Look at Lorentz equations.
Lets assume the acceleration is instantaneous for the moment. If we call the twins "A" and "B", and the mothership observer "S", then twin A's trajectory in S's frame might look something like this:

x_s = \left{ \begin{matrix} - v t_s & \qquad t_s < T \\ v ( t_s - 2 T) & \qquad t_s \geq T \end{matrix} \right.
where T is the time twin A turns around, as measured by S.

To analyze this from A's point of view we need to consider two reference frames in addition to S. The first, A1, is A's rest frame up until time T, while A is moving away from S. The transformation from S to A1 is:

\begin{eqnarray}
t_{a_1} = & \gamma \left( t_s + \frac{v}{c^2} x_s \right) \\
\\
x_{a_1} = & \gamma \left( x_s + v t_s \right) \end{eqnarray}

The second, A2, is A's rest frame after T, when A is moving toward S. It is related to S by:

\begin{eqnarray}
t_{a_2} = & \gamma \left( t_s - \frac{v}{c^2} x_s - 2 \frac{v^2}{c^2} T \right) \\
\\
x_{a_2} = & \gamma \left( x_s - v t_s + 2 v T) \end{eqnarray}
(the extra terms ensure x_s = v ( t_s - 2T ) for x_{a_2} = 0, and t_{a_1} = t_{a_2} for t_s = T and x_s = - v T)

When twin A returns to S, S has aged 2 T. The second set of transformations (for x_s = 0 and t_s = 2T) gives t_{a_2} = \frac{2T}{\gamma}, which is A's age at the crossover point. This is true for all observers.

The interesting thing is what happens to S's age from A's point of view as A turns around. Rearranging the equations for t_{a_1} and t_{a_2} for x_s = 0 yields:

t_s = \frac{1}{\gamma} t_{a_1} \\
\\
t_s = \frac{1}{\gamma} t_{a_2} + 2 \frac{v^2}{c^2} T

At t_{a_1} = t_{a_2} = \frac{1}{\gamma} T, the difference between these two is:

\Delta t_s = 2 \frac{v^2}{c^2} T

So twin A's infinite acceleration results in S ageing by 2 \frac{v^2}{c^2} T (infinitely fast) as seen by A.

I expect considerations like these are what led to General Relativity's gravitational time dilation.

I have an Spanish translation of Einstein's book "Sobre la Teoria de la Relatividad Especial y General".
Here he explain Special Relativity with a relativistic train.
Could you could briefly explain it and Einstein's conclusions, if you still think it's relevant?

przyk,

The first part of your calculations when you deduce the age of the twin as seen by the mothership seems right to me but the second does not. Observing from the twin's frame the mothership makes a completely symmetric travel! I mean the mothership goes away at velocity V and at time t=T turns back with velocity v and so the calculations should be exactly the same.
The motherships sees the twin exactly the same way the twin sees the mothership. Here reciprocity applies and so the results should be the same but at the inverse: themothership observer younger than the twin.


Could you could briefly explain it and Einstein's conclusions, if you still think it's relevant?
Is a very interesting book for those interested in Relativity. I recommend you to find one copy. You can see in it the real thoughts Eintein had about all Relativity. Some of them are very abstract and need much attention.

He begins talking about the classical concepts of space and time always with the example of frames in a train and in the railways.After he introduce the principle of the relativity as it is applied in Special Relativity and talks about the relativity of distance and simultaneity (much conceptual talk and no math).
Then he introduce Lorentz Transform. He doesn't do all the math. He describe well the problem and finally says that the Lorentz formula is the right answer. He present two frames of reference, one with coordinates x', y', z', t' in the train and the other x, y, z, t in the railways and he states that for them to measure the velocity of light exactly the same the change of coordinates must verify Lorentz equations. After he talks about how that transformation implies in lenght contraction and time dilation and how the experiment of Fizeau agree with the relativistic addition of velocities under some aproximation.
He also talks that the energy of a massive object is E=mc2 (without deriving it) and how an object that absorbs energy E0 by radiation increases its mass in the value E0/c2.
He very briefly mention Minkowsky space-time and after he enters in concepts about General Relativity for which he dedicates more than half of a small book with 142 pages.

I have found this links:
http://www.bartleby.com/173/
http://www.marxists.org/reference/archive/einstein/works/1910s/relative/index.htm

Einstein also wrote another book with a title like "On the electrodynamic of moving bodies".
Links:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

The first part of your calculations when you deduce the age of the twin as seen by the mothership seems right to me but the second does not. Observing from the twin's frame the mothership makes a completely symmetric travel! I mean the mothership goes away at velocity V and at time t=T turns back with velocity v and so the calculations should be exactly the same.
The kinematics may be the same, but the mechanics isn't. You only get reciprocity between inertial frames, and the rest frames of both twins are not inertial throughout their entire trips. Both twins will feel (and could even be knocked out or killed by) pseudo g-forces as they accelerate to get back, while the observer on the mothership feels nothing. This is the short answer to the twin paradox. I took it a bit further by giving you an indication of what the accelerating twins will observe.

Note that you do have reciprocity on the two segments of the trip where twin A is leaving S, and A is returning to S at constant velocity:

t_s = \frac{1}{\gamma} t_{a_1} \\
\\
t_s = \frac{1}{\gamma} t_{a_2} + 2 \frac{v^2}{c^2} T
For every second increase in t_{a_1} or t_{a_2}, t_s increases by \frac{1}{\gamma} seconds - ie. S ages slower than A for most of the trip, as seen by A. It's only while A is accelerating that you lose reciprocity.

Is a very interesting book for those interested in Relativity. I recommend you to find one copy. You can see in it the real thoughts Eintein had about all Relativity. Some of them are very abstract and need much attention.

He begins talking about the classical concepts of space and time always with the example of frames in a train and in the railways.After he introduce the principle of the relativity as it is applied in Special Relativity and talks about the relativity of distance and simultaneity (much conceptual talk and no math).
Then he introduce Lorentz Transform. He doesn't do all the math. He describe well the problem and finally says that the Lorentz formula is the right answer. He present two frames of reference, one with coordinates x', y', z', t' in the train and the other x, y, z, t in the railways and he states that for them to measure the velocity of light exactly the same the change of coordinates must verify Lorentz equations. After he talks about how that transformation implies in lenght contraction and time dilation and how the experiment of Fizeau agree with the relativistic addition of velocities under some aproximation.
He also talks that the energy of a massive object is E=mc2 (without deriving it) and how an object that absorbs energy E0 by radiation increases its mass in the value E0/c2.
He very briefly mention Minkowsky space-time and after he enters in concepts about General Relativity for which he dedicates more than half of a small book with 142 pages.

I have found this link: http://www.bartleby.com/173/

Einstein also wrote another book with a title like "On the electrodynamic of moving bodies".
Link: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Thanks, but I wasn't asking for a book report. You brought up a specific though experiment as if it justified a point you were making, and spoke of a "timing problem". This is what I was asking you to elaborate on.

przyk,

It's only while A is accelerating that you lose reciprocity.
It has been proven, at particle accelerators etc., that acceleration does not affect clock rates. The 'clock postulate' is well known. Only motion through spacetime, or changes in gravitational potential, affect clocks beat rates. The old 'reciprocity between inertial frames' of Special Theory has long ago been shown to be a false premise. Yes, there is a preferred frame of reference, though difficult to define exactly. It is most often defined by the rest frame of the CMB, the frame in which the CMB exibits no Doppler shifts in any vector.

It has been proven, at particle accelerators etc., that acceleration does not affect clock rates. The 'clock postulate' is well known. Only motion through spacetime, or changes in gravitational potential, affect clocks beat rates.
Where did I say anything that contradicts any of this?

The old 'reciprocity between inertial frames' of Special Theory has long ago been shown to be a false premise.
It's a correct premise of limited applicability. It is only exhibited between inertial frames.

Yes, there is a preferred frame of reference, though difficult to define exactly. It is most often defined by the rest frame of the CMB, the frame in which the CMB exibits no Doppler shifts in any vector.
The rest frame of the CMB is the rest frame of the CMB. Nothing more.

przyk,
I think the problem you have in your calculations is about this:

The interesting thing is what happens to S's age from A's point of view as A turns around. Rearranging the equations for ta1 and ta2 for xs=0 yields:
You "rearranged" the equation so you are still considering the same frames while you should consider now two new frames, a "fixed" frame on the twin and a "moving" frame on the mothership as seen by the twin. This will give you the same equations but with the variables interchanged. This way the results will be the same (but with inversion in the meaning of who aged less) as they must be.

martillo,

See for yourself what happens when you rearrange:

\begin{eqnarray}
t' = & \gamma \left( t - \frac{v}{c^2} x \right)\\
\\
x' = & \gamma \left( x - v t \right)
\end{eqnarray}
to get t and x in terms of t' and x'. Inverting a transformation in this fashion is how you see what the unprimed frame looks like with respect to the primed frame.

przyk,

Inverting a transformation in this fashion is how you see what the unprimed frame looks like with respect to the primed frame.
But this is not what must be done!
The problem is to analyze the phenomenon as seen from the three observers and compare the results. This means to consider three different "fixed" frames. One time the "fixed" frame on the mothership, other time other "fixed" frame in one twin and the third time the "fixed" frame on the other twin.
This is the unique way to obtain the three different observations of the same phenomenon as seen by each one of the three observers.

The equations must not be "rearranged" but new equations must be setted up in each situation.

But this is not what must be done!
The problem is to analyze the phenomenon as seen from the three observers and compare the results. This means to consider three different "fixed" frames. One time the "fixed" frame on the mothership, other time other "fixed" frame in one twin and the third time the "fixed" frame on the other twin.
This is the unique way to obtain the three different observations of the same phenomenon as seen by each one of the three observers.
To analyze this from all points of view, we need five inertial reference frames. You can't simply attach a "fixed" frame to each twin and apply the same rules as you did from the mothership frame. Those rules are only valid in an inertial frame, and both twins can detect that they've accelerated at one point. The correct way to view this is to say that each twin's rest frame changes. They leave one inertial frame and enter another, if you like.

The equations must not be "rearranged" but new equations must be setted up in each situation.
Did you try rearranging the Lorentz transformation as I suggested? If rearranging the transform gave a different result than reciprocity would imply, relativity would be in serious trouble.

przyk,

Where did I say anything that contradicts any of this?
What you posted:

You only get reciprocity between inertial frames, and the rest frames of both twins are not inertial throughout their entire trips. Both twins will feel (and could even be knocked out or killed by) pseudo g-forces as they accelerate to get back, while the observer on the mothership feels nothing.
As I stated, acceleration does not affect clock rates, only velocity through spacetime. The twin in motion will experience different clock rates as his velocity through spacetime (the CMB rest frame) changes, but only because of changes in velocity, not acceleration. If inertial frames are 'reciprocal', the non-inertial frames can be broken down into 'inertial slices' which would retain reciprocity throughout the trip. But they are not. Only a clock in motion relative the preferred frame, the CMB rest frame, will slow. This also applies to the 'mothership' if it is in motion relative to the CMB. Almost all clocks in the Milky Way galaxy are in motion relative to the CMB rest frame, because the Milky Way is moving relative to the CMB rest frame, and are slightly slowed because of this fact. You can also use the International Celestial Reference Frame because it is based on the rest frame of the CMB.

As I stated, acceleration does not affect clock rates, only velocity through spacetime. The twin in motion will experience different clock rates as his velocity through spacetime (the CMB rest frame) changes, but only because of changes in velocity, not acceleration.
The "time dilation affected by acceleration" I was talking about was the effect the twin's acceleration had on the mothership observer's ageing rate in the twin's accelerating frame.

If inertial frames are 'reciprocal', the non-inertial frames can be broken down into 'inertial slices' which would retain reciprocity throughout the trip. But they are not.
I imagine compounding all an accelerating observer's rest frames (for motion along the x axis) will yield a transformation that looks something like this:

t' = \displaystyle\int_0^t \frac{ d \tau }{ \gamma ( \dot{u} ( \tau ) ) } \, - \, \gamma ( \dot{u} ) \frac{ \dot{u} }{c^2} ( x - u )


x' = \gamma ( \dot{u} ) ( x - u )

Here, u \equiv u(t) is the location of the primed frame's origin (an arbitrary function of time) in the unprimed frame, \dot{u} \equiv \dot{u}(t) \equiv \frac{d}{dt} u(t), and \gamma ( s ) \equiv \frac{1}{ \sqrt{ 1 - \frac{s^2}{c^2} }.

This transformation will not exhibit reciprocity in general.

Only a clock in motion relative the preferred frame, the CMB rest frame, will slow. This also applies to the 'mothership' if it is in motion relative to the CMB. Almost all clocks in the Milky Way galaxy are in motion relative to the CMB rest frame, because the Milky Way is moving relative to the CMB rest frame, and are slightly slowed because of this fact. You can also use the International Celestial Reference Frame because it is based on the rest frame of the CMB.
Clocks in motion with respect to the CMB are slowed as measured in the CMB frame. Clocks at rest in the CMB frame are slowed as measured in an inertial reference frame in motion with respect to the CMB.

przyk,

Clocks at rest in the CMB frame are slowed as measured in an inertial reference frame in motion with respect to the CMB.
I realize Special Theory predicts this, but that is where the theory does not reflect reality. Clocks at rest in the CMB frame will, in reality, be beating fast when measured in an inertial frame in motion wrt the CMB.

I have several ways to prove this, but let's start with a simple example keeping in tune with the premis of this thread.

Let's send one twin away from the Earth on a relativistic voyage that measures one year in duration when measured by an Earth clock, start to finish back on Earth. The Earth will rotate 365 times in that year by an Earth clock. Assume the travelling twin's clock measures a duration of six months in the moving frame. How many times will the Earth rotate in the travelling twin's reference frame? Keep in mind the Earth clock will beat very slightly slower than a CMB rest frame clock, but not enough to make any difference when measured in days.

I realize Special Theory predicts this, but that is where the theory does not reflect reality. Clocks at rest in the CMB frame will, in reality, be beating fast when measured in an inertial frame in motion wrt the CMB.
This would be consistent with a transformation like:

t' = \frac{1}{\gamma} t


x' = x - v t (for example)

Inverting the first equation would indeed give t = \gamma t' - ie. if B ages slow in A's frame, A ages fast in B's frame.

The laws of physics, however, are Lorentz invariant, and the Lorentz transformation is:

t' = \gamma \left( t - \frac{v}{c^2} x \right)


x' = \gamma \left( x - v t \right)

I'll also point you to a thought experiment I posted here if you have the time and patience for it.

How many times will the Earth rotate in the travelling twin's reference frame?
366* times. At some point, the travelling twin will have to accelerate toward Earth to get back. During this time, the Earth will rotate faster than once per day and clocks on it will speed up from the accelerating twin's perspective. You should be able to derive this from the transformation I gave in my last post. General Relativity calls this gravitational time dilation (http://en.wikipedia.org/wiki/Gravitational_time_dilation#Definition), attributed to the Earth's higher gravitational potential in the twin's accelerating frame.

*This is an off-topic detail: the Earth rotates just over 366 times every year. The number of times any planet orbiting a star rotates will always differ from its number of solar days by one per year.

przyk,

366* times. At some point, the travelling twin will have to accelerate toward Earth to get back. During this time, the Earth will rotate faster than once per day and clocks on it will speed up from the accelerating twin's perspective.
Acceleration has no effect on clock rates, remember the experiments and the clock postulate? Also, the Earth never rotates faster than once per day, that conjecture is due to measurements made with the spaceship's on-board clock, which is beating slow. I do agree the clocks on Earth will beat faster than spaceship clocks during the entire trip, even when the spaceship is travelling orthogonal to the Earth during the turn-around phase, except if the spaceship were to come to rest wrt the rest frame of the CMB or Earth. Again, acceleration has no effect on clock rates, only speed through spacetime. Of course, gravitational potential does affect clock rates, but gravitational time dilation is irrelavent in this gedankin. We could also introduce a distant pulsar into the gedankin to illustrate that the direction the spaceship's travel in the accelerating frames has no effect on its clocks. I could explain what 'time' and 'time dilation' really is, but that topic is outside the scope of this thread.



*This is an off-topic detail: the Earth rotates just over 366 times every year. The number of times any planet orbiting a star rotates will always differ from its number of solar days by one per year.
Yes, I am very much aware of Sidereal time, but when I spoke of an Earth clock, I was speaking of UTC time, our standard of time measurement on Earth. Sidereal time would be a more accurate method to state Earth rotations, especially as seen from the spaceship frame.

Thanks, I will take a look at your link when I have the time to do so.

przyk,

Did you try rearranging the Lorentz transformation as I suggested? If rearranging the transform gave a different result than reciprocity would imply, relativity would be in serious trouble.
If you just rearrange the equations you are getting the results from the same frame you considered at first time, that is the mothership's frame not the twin's frame so you are not getting the twin's view of the phenomenon.


The correct way to view this is to say that each twin's rest frame changes. They leave one inertial frame and enter another, if you like.

Right and you MUST choose the new frames to get the right results for the twin's point of view, this means new equations.


To analyze this from all points of view, we need five inertial reference frames. You can't simply attach a "fixed" frame to each twin and apply the same rules as you did from the mothership frame. Those rules are only valid in an inertial frame, and both twins can detect that they've accelerated at one point.
As I said before the acceleration was in a neglihible at time and as you said the twins entered in a new inertial frame in the travel.
Of course I can simply attach a "fixed" frame to each twin and apply the same rules. This is what MUST be done. If you don't you are not understanding the problem properly.

Acceleration has no effect on clock rates, remember the experiments and the clock postulate?
The clock postulate states that an accelerating clock's rate is \frac{1}{\gamma} as seen by an inertial observer. On its own it says nothing about the rate of an inertial clock as seen by an accelerating observer. This is just mathematics: in general, \frac{ \part a }{ \part b } alone tells you nothing about \frac{ \part b }{ \part a }. The Lorentz transformation is a case in point.

Also, the Earth never rotates faster than once per day, that conjecture is due to measurements made with the spaceship's on-board clock, which is beating slow. I do agree the clocks on Earth will beat faster than spaceship clocks during the entire trip, even when the spaceship is travelling orthogonal to the Earth during the turn-around phase
GR predicts that the Earth will rotate faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin only while he's accelerating toward Earth.

except if the spaceship were to come to rest wrt the rest frame of the CMB or Earth.
Only if the twin's acceleration is also zero at this point.

Of course, gravitational potential does affect clock rates, but gravitational time dilation is irrelavent in this gedankin.
Apply the equivalence principle. A gravitational field exists in the accelerating frame, and the Earth is at a higher gravitational potential if the acceleration is toward Earth, so Earth ticks/rotates/ages faster. How do you think this effect was first derived?

If you just rearrange the equations you are getting the results from the same frame you considered at first time, that is the mothership's frame not the twin's frame so you are not getting the twin's view of the phenomenon.

Right and you MUST choose the new frames to get the right results for the twin's point of view, this means new equations.
No. The relation between two frames is given by only one set of equations. There can only be one relationship between the coordinates of an event in two frames. Since you don't seem to have inverted the Lorentz transform as I suggested, I'll do it algebraically here (you can also do it via matrix inversion). We start with:

t' = \gamma \left( t - \frac{v}{c^2} x \right) (1)


x' = \gamma \left( x - v t \right) (2)

Rearranging (1) and (2) give, respectively:

t = \frac{1}{\gamma} t' + \frac{v}{c^2} x (3)


x = \frac{1}{\gamma} x' + v t (4)

Substituting (4) into (1) and (3) into (2) give, respectively:

t' = \gamma \left( t - \frac{v}{c^2} \frac{1}{\gamma} x' - \frac{v^2}{c^2} t \right) = \gamma \left( 1 - \frac{v^2}{c^2} \right) t - \frac{v}{c^2} x' = \frac{1}{\gamma} t - \frac{v}{c^2} x' (5)


x' = \gamma \left( x - v \frac{1}{\gamma} t' - \frac{v^2}{c^2} x \right) = \gamma \left( 1 - \frac{v^2}{c^2} \right) x - v t' = \frac{1}{\gamma} x - v t' (6)

Rearranging (5) and (6) gives the inverse Lorentz transform:

t = \gamma \left( t' + \frac{v}{c^2} x' \right)


x = \gamma \left( x' + v t' \right)

which tells us what the unprimed frame looks like from the perspective of the primed frame. If the result of inverting the Lorentz transformation like this were incompatible with reciprocity, relativity wouldn't have lasted five minutes - let alone a century.

As I said before the acceleration was in a neglihible at time and as you said the twins entered in a new inertial frame in the travel.
The acceleration is never negligible. The only way to shorten the period of acceleration necessary is to increase the acceleration, which increases its effect. It does not tend to zero as the acceleration approaches infinity.

The total increase in the twin's age as seen by the mothership observer can be broken down into three parts:

\Delta T' = \Delta T_1' + \Delta T_a' + \Delta T_2'

Here, \Delta T_1' and \Delta T_2' are the twin's age increases during the first part and second parts of his journey. \Delta T_a' is his age increase during his acceleration period. Special relativity allows \Delta T_1' and \Delta T_2' to be calculated directly, and we can apply the clock postulate 2inquisitive brought up to calculate \Delta T_a' - which is negligible in this case. This was a condition I applied when working out the transformation between S and A2.

We can do the same thing for the mothership observer's age increase as seen by the travelling twin:

\Delta T = \Delta T_1 + \Delta T_a + \Delta T_2
Here, \Delta T_1 and \Delta T_2 are the age increases of the mother ship observer during the first and second periods of the trip (ie. before and after the twin's acceleration). \Delta T_a is the mothership observer's age increase as seen by the twin during the twin's acceleration. Again, we can use STR to calculate \Delta T_1 and \Delta T_2. \Delta T_a is the problem here. Because \Delta T_a is measured in an accelerating frame, STR does not explicitly provide a rule for calculating it.

Of course I can simply attach a "fixed" frame to each twin and apply the same rules. This is what MUST be done. If you don't you are not understanding the problem properly.
If you do this, you are misapplying relativity and knocking down a strawman.

przyk,

“ Of course I can simply attach a "fixed" frame to each twin and apply the same rules. This is what MUST be done. If you don't you are not understanding the problem properly. ”

If you do this, you are misapplying relativity and knocking down a strawman.

Well, we are not going to reach an agreement this time.

martillo: you mean you want to keep this rule you've inserted into STR just so you can disprove your home-made version of the theory? Why?

przyk,

366* times. At some point, the travelling twin will have to accelerate toward Earth to get back. During this time, the Earth will rotate faster than once per day and clocks on it will speed up from the accelerating twin's perspective.

GR predicts that the Earth will rotate faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin only while he's accelerating toward Earth.
Why do you believe this supports your viewpoint? The Earth also rotates faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin during the travelling twin's inertial segments. It supports the clock postulate that states acceleration has no effect on clocks. Again, if the travelling twin measures the Earth as rotating faster than once every 24 hours by his local clock, it means his clock is counting time at a slower rate than an Earth clock.

przyk,

you mean you want to keep this rule you've inserted into STR just so you can disprove your home-made version of the theory? Why?
I haven't inserted any rule to STR. I have a developed a tought experiment where the basic problem is to compare the different views of the different observers and for that new frames must be considered each time.
You refuse to do that so you are not understanding the problem properly.

The Earth also rotates faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin during the travelling twin's inertial segments.
How do you know this? The example you brought up only tells you what the average rotation rate of the Earth must be as seen by the travelling twin, and my point so far has been that relativity is perfectly compatible with this. Of course, there's still the seperate issue of how good a model of reality relativity is.

It supports the clock postulate that states acceleration has no effect on clocks.
What supports this?

Again, if the travelling twin measures the Earth as rotating faster than once every 24 hours by his local clock, it means his clock is counting time at a slower rate than an Earth clock.
Again, there's no reason for the rate in one frame (K) of a process at rest in another frame (K') to be the reciprocal of the rate of a different process at rest in K as seen by K'. The relationship depends on the coordinate transformation between K and K'. So far you haven't proposed a coordinate system or justified its use.

I haven't inserted any rule to STR.
Yes you have: that recprocity applies between all frames and not just inertial frames.

When analysing the ageing rate of the mothership observer as seen by one of the travelling twins, you keep imposing that what I call T_a and T_a' in post #75 are the same (or at least that the former is "negligible"). This is your own rule, not relativity's. Special relativity doesn't directly tell you what T_a is.

There isn't a single Lorentz transformation that relates the travelling twin's and mothership observer's rest frames for the duration of the trip. If you use two transformations for the two inertial segments (before and after the acceleration), fine, but you can't simply assume whatever you like about what the accelerating twin sees as he switches from one inertial rest frame to another, then claim you've disproved a theory that doesn't make this assumption.

As far as General Relativity is concerned, your view is quite explicitly contradicted:

Gravitational time dilation is manifested in accelerated frames of reference or, by virtue of the equivalence principle, in the gravitational field of massive objects. In more simple terms, clocks which are far from massive bodies (or at higher gravitational potentials) run faster, and clocks close to massive bodies (or at lower gravitational potentials) run slower. (http://en.wikipedia.org/wiki/Gravitational_time_dilation)
If you're in an accelerating reference frame, anything you are accelerating toward is at a higher gravitational potential than you.

“ Originally Posted by 2inquisitive
The Earth also rotates faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin during the travelling twin's inertial segments. ”

przyk,

How do you know this? The example you brought up only tells you what the average rotation rate of the Earth must be as seen by the travelling twin, and my point so far has been that relativity is perfectly compatible with this. Of course, there's still the seperate issue of how good a model of reality relativity is.
I can deduce my statement from experimental evidence + logic. First, experimental evidence confirms the fact that acceleration does not alter clock rates in excess of what velocity through spacetime predicts in the inertial observer's frame of reference. Your statements were as follows, quote:

On its own it says nothing about the rate of an inertial clock as seen by an accelerating observer.

GR predicts that the Earth will rotate faster than once per day (24 of the travelling twin's hours) as seen by the travelling twin only while he's accelerating toward Earth.
GR also predicts that the Earth will rotate slower than once per day as seen by the travelling twin while he is accelerating away from the Earth. These accelerating frames (away from and toward the Earth) should offset each other during the course of the entire trip, leaving the inertial frames as the frames where clock dilation would occur. Do you disagree with that assumption?

GR also predicts that the Earth will rotate slower than once per day as seen by the travelling twin while he is accelerating away from the Earth. These accelerating frames (away from and toward the Earth) should offset each other during the course of the entire trip, leaving the inertial frames as the frames where clock dilation would occur. Do you disagree with that assumption?
Yes. The gravitational time dilation is related to a gravitational potential difference. This means that, apart from the field strength, the effect will also increase with distance. The effect while the twin is turning around far from Earth will be much more significant than when he's taking off or landing.

przyk,

Whatever the effect on the intervals of accelerations of the twins can be you must consider:
1) The twins make a symmetrical travel and so the effect would be the same and will give exactly the same "Ta" for both and so this doesn't affects the problem.
2) I can choose the travel of the twins going so far as I want and the twins ageing so much as I want and the interval of acceleration so small as I want and so the effect of the acceleration can always be made neglihible.

We will ever find the same contradictions in the different observations seen from the different frames.

1) The twins make a symmetrical travel and so the effect would be the same and will give exactly the same "Ta" for both and so this doesn't affects the problem.
And the T_a's each twin measures for the other couldn't possibly be such that the twins are the same age when they cross over?

2) I can choose the travel of the twins going so far as I want and the twins ageing so much as I want and the interval of acceleration so small as I want and so the effect of the acceleration can always be made neglihible.
If the gravitational time dilation effect increases with gravitational field strength (ie. acceleration) and distance, why should an infinite acceleration for an infinitely short duration be negligible? You keep claiming it is, but you can never justify this. A time interval approaching zero multiplied by an average ageing rate approaching infinity could quite easily yield a T_a that approaches a finite, non-zero value.

przyk,

And the 's each twin measures for the other couldn't possibly be such that the twins are the same age when they cross over?

No, because of the symmetric travel the same "Ta" would be added to the relativistic predictions for both twins and the difference of the the different ages would remain the same.
Also note that "Ta" we can vary the time of the travel by varying arbitrary the travelling distance and so "Ta" would compensate the difference of age in a very particular case only not in the general case.


If the gravitational time dilation effect increases with gravitational field strength (ie. acceleration) and distance, why should an infinite acceleration for an infinitely short duration be negligible? You keep claiming it is, but you can never justify this. A time interval approaching zero multiplied by an average ageing rate approaching infinity could quite easily yield a that approaches a finite, non-zero value.
I agree that is my assumption that the travel could be made such a way that the accelerations in both twins would have neglihible effects but may be some calculations must be made to ensure if this is possible or not. Even if not we fall in the consideration 1) that both twins would have the same effect and the difference between the relativistic predictions between them remains the same.

So, even if a considerable "Ta" exist the thought experiment will give contradictory observations from the different frames.

No, because of the symmetric travel the same "Ta" would be added to the relativistic predictions for both twins and the difference of the the different ages would remain the same.
If we look at thing from, say, twin A's frame, then A's age by the time they meet up again is:

\text{A's age increase before accelerating} \; + \; \text{A's age increase after accelerating}

B's age increase, as seen in A's frame is:

\text{B's age increase before A's acceleration} \; + \; T_a \; + \; \text{B's age increase after A's acceleration}

With T_a such that the twins are the same age when they cross over. The analysis will be completely symmetrical in B's frame.

Also note that "Ta" we can vary the time of the travel by varying arbitrary the travelling distance and so "Ta" would compensate the difference of age in a very particular case only not in the general case.
No, it would just mean that T_a is a function of distance. The further an object is, the greater its jump in age for the same acceleration toward it.

I agree that is my assumption that the travel could be made such a way that the accelerations in both twins would have neglihible effects but may be some calculations must be made to ensure if this is possible or not.
Well STR only really gives you one way of calculating all these T_a's: work out all the other terms, and there will only be one T_a such that everyone agrees on all ages.

przyk,

That's the way what you believe it could work for both twins have the same age at the crossover point.
I don't think it works, I don't think it really happen that way.
I ask you now, if that were true why the traditional twin's paradox gives the travelling twin getting younger? Why your "Ta" is not considered in the traditional paradox? (The travelling twin has to accelerate to go away and to come back)
Following your reasoning there would be a "Ta" which make these both twins have the same age while the well known result is that the travelling one gets younger!
Sorry but your reasoning doesn't work.

I believe that acceleration effect could not be relevant to the final results of the problem.

Why your "Ta" is not considered in the traditional paradox? (The travelling twin has to accelerate to go away and to come back)
Because "STR doesn't apply in accelerating frames" is a much quicker rebuttal, also valid against your own version of the paradox. Rather than responding this way, though, I looked at the paradox in a bit more detail. (For the traditional paradox, your mothership observer is the equivalent of the Earth twin.)

Following your reasoning there would be a "Ta" which make these both twins have the same age while the well known result is that the travelling one gets younger!
The Earthbound observer doesn't accelerate in the traditional twin paradox. There's only one T_a: the jump in the stationary twin's age as seen by the accelerating twin. In your scenario each twin measures such an age jump for the other twin, as both accelerate.

I believe that acceleration effect could not be relevant to the final results of the problem.
Then relativity contradicts your personal belief - not itself.

If relativity theory is anything then try answerin this ;) http://sciforums.com/showthread.php?p=1387441#post1387441

przyk,

I will not discuss anymore.
I can only say that for me, as personal belief, your reasoning about the "Ta" has never been presented in any discussion about twin's paradoxes in forums and does not apply.

your reasoning about the "Ta" has never been presented in any discussion about twin's paradoxes in forums and does not apply.
Why? Because I demonstrated it to be a consequence of the Lorentz transformation? Because it shows how STR can be internally consistent? Honestly I don't know why you keep choosing to ignore something that is a prediction of General Relativity anyway.

Relativity is correct. Even if Einstein re-incarnates tommorrw (** I AM NOT DISCUSSING REINCARNATION PPL! ;) ****) and says its incorrect i wont blv him.

Rick