Relativity Theory is THEORETICALLY wrong!

[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 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.

 

[2inquisitive]

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.

 

[przyk]

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.

 

[martillo]

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.

 

[przyk]

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.

 

[martillo]

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.

 

[przyk]

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.

 

[2inquisitive]

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.

 

[przyk]

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.

 

[2inquisitive]

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.

 

[przyk]

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 [POST=1375494]here[/POST] if you have the time and patience for it.

How many times will the Earth rotate in the travelling twin's reference frame?

366[sup]*[/sup] 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, 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.

 

[2inquisitive]

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.

 

[martillo]

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.

 

[przyk]

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?

 

[przyk]

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 A[sub]2[/sub].

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.

 

[martillo]

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.

 

[przyk]

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?

 

[2inquisitive]

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.

 

[martillo]

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.

 

[przyk]

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.