# Alternative Twin’s Experiment

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### #1 jakuta

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Posted 19 October 2010 - 07:09 PM

Alternative Twin’s Experiment

The traditional twins experiment is generally solved such that all agree the traveling twin is younger because of non-symmetric acceleration. However, this twin’s paradox will produce symmetric acceleration and force SR to exclusively explain reciprocal time dilation up against the SR clock synchronization method which is based on the universal truth of t = d/c. In short, SR fails.

Setup
O and O' are observers with clocks in the same frame in the vacuum of space.

Procedure
1. Both set their clocks to 0 and O instantly acquires v relative to O'.
2. O and O' are in relative motion for some agreed up time t' on the clock of O'.
3. After time t', O' will acquire v in precisely the same way as O in precisely the same direction.
4. At the same instant O' acquires v, O' enters the frame of O and O' sends a light pulse to O and records this as the end of the experiment at time t'.
5. O' receives the light pulse and records the time as te.
6. Since, O and O' are again in the same frame, O performs with O' the round trip speed of light calculation using a time trial to calculate the distance between the two. Let D be that distance. O then subtracts D/c from te and determines its proper for when O' entered the frame to determine the correct end of the experiment which will match the end for O'. Let this time be t.
7. Finally, O and O' perform the SR clock synchronization method to determine the ordinality of the two clocks. To do this, O sends its time t1O to O'. O' immediately sends its time tO' back to the clock of O. O receives this light at t2O. In particular, Einstein concludes the two clocks are in sync if tO' = ½ (t2O - t1O). Thus, the age ordinality is established as tO' - ½ (t2O - t1O).
http://www.fourmilab...in/specrel/www/

***Note, the above can be replaced with equal agreed upon proper burn times and proper accelerations and will produce the same results according to the SR uniform acceleration equations.

The steps above represent an effective procedure for deciding the age ordinality of O and O'. More specifically, step 7 determines the clocks of O and O' as one of the following.
· O is younger than O'.
· O is older than O'.
· O and O' are the same age.

Since each step of the procedure above is well defined and produces a unique outcome with the output at step 7 producing only one value, then this problem of determining the age ordinality of O and O' is recursive and thus logically decidable.

Now, LT will be applied to solve the problem above.

From O as stationary, its proper time is t. Hence, O concludes O' elapses t/λ and therefore concludes O' is younger.

From O' as stationary, its proper time is t'. Hence, O' concludes O elapses t'/λ and therefore concludes O is younger.

Now, Einstein’s clock synchronization method by his own admission is “free from contradictions”.
http://www.fourmilab...in/specrel/www/

Therefore, we may conclude the effective procedure listed above produces one correct answer to the age ordinality of O and O' otherwise, the SR clock synchronization method does not work and hence LT fails.

On the other hand, LT produces contradictory results and thus provides an answer that is not logically decidable.

Hence, under SR, the problem above is logically decidable and not logically decidable depending on the method used.

Worse, at least one of the solutions provided by LT must contradict the results of the clock synch and therefore, LT must calculate at least one incorrect result. Clearly, this proves LT is not a reliable tool for mapping proper times between frames since at least one of its solutions must be wrong above.

### #2 coldcreation

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Posted 20 October 2010 - 01:25 AM

The traditional twins experiment is generally solved such that all agree the traveling twin is younger because of non-symmetric acceleration. However, this twin’s paradox will produce symmetric acceleration and force SR to exclusively explain reciprocal time dilation up against the SR clock synchronization method which is based on the universal truth of t = d/c. In short, SR fails.

Actually the situation is non-symmetric, not symmetric as you imply. Indeed, according to special relativity all observers are not equivalent. Only observers at rest in inertial frames of reference are equivalent. The asymmetry arises in the frame of the twin inside a space ship when he turns around (switching from one frame to another). Whereas, the twin at home remains in the same inertial frame during the entire experiment. The homebound twin experiences no acceleration or deceleration. The twin inside the rocket experiences two different frames (one on the way out, and another coming home). Switching frames is the reason asymmetry arises. But the same asymmetry can be attributed to a gravitational time dilation (due to the acceleration and deceleration experienced by the rocket-bound twin), something the home-bound twin does not experience.

CC

### #3 jakuta

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Posted 20 October 2010 - 05:49 PM

Actually the situation is non-symmetric, not symmetric as you imply. Indeed, according to special relativity all observers are not equivalent. Only observers at rest in inertial frames of reference are equivalent. The asymmetry arises in the frame of the twin inside a space ship when he turns around (switching from one frame to another). Whereas, the twin at home remains in the same inertial frame during the entire experiment. The homebound twin experiences no acceleration or deceleration. The twin inside the rocket experiences two different frames (one on the way out, and another coming home). Switching frames is the reason asymmetry arises. But the same asymmetry can be attributed to a gravitational time dilation (due to the acceleration and deceleration experienced by the rocket-bound twin), something the home-bound twin does not experience.

CC

No, there is no turn around.

Two ships are co-located and times synched.

One ship takes off at some uniform acceleration for some time.

There is a period of relative motiuon.

Then, the other ship takes off at the same uniform acceleration for the same time.

They end up in the same frame after all this. Hence, all acceleration is symmetric.

This point of the OP is to show the light postulate is incompatible with the relativity postulate..

That means the SR clock sync (light postulate) is forced up against reciprocal time dilation (relativity postulate) and the clock sync contradicts reciprocal time dilation.

### #4 CraigD

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Posted 20 October 2010 - 07:48 PM

Procedure
1. Both set their clocks to 0 and O instantly acquires v relative to O'.
...

I don’t think I’m following your instructions well enough to complete step 7, jakuta.

If you could provide an specific example from the point of view of a distant 3rd observer, I believe you’re example would be clear, and you would no longer find a failure or special relativity.

Calculation of this kind are often made clear by writing specific examples. Here’s one, where $v= \frac{\sqrt{3}}{2} \dot = \, 0.866 \,\mbox{c}$, a good example value because then $\sqrt{1-\left(\frac{v}{c}\right)^2} = \frac{1}{2}$.
X and V are the position and velocity of O as measured by the distant observer.
T is the time that appears on O’s clock at that position.
X’, V’, and T’ are the position, velocity, and time observed on the clock for O’.
X_s and V_s are the position and velocity of the light signal between O and O’.
S is the content of that signal.
T_0 is the time that appears on the distant observer’s clock.
X	V	T	X'	V'	T'	X_s	V_s	S	T_0	note
0	0	-1	0	0	-1				-1
0	.866	0	0	0	0				0
.866	.866	.5	0	0	1				1
1.732	.866	1	0	.866	2	0	1	"2"	2
2.598	.866	1.5	.866	.866	2.5	1	1	"2"	2
...
12.93	.866	7.465	11.20	.866	8.465	12.93	1	"2"	14.93	te=2
12.93	.866	7.465	11.20	.866	8.465	12.93	-1	"7.465"	14.93	t10=7.465
13.73	.866	7.929	12.00	.866	8.929	12.00	-1	"7.465"	15.87
13.73	.866	7.929	12.00	.866	8.929	12.00	1	"8.929"	15.87
24.94	.866	14.40	23.21	.866	15.40	24.94	1	"8.929"	28.80	t20=14.40

Hopefully, you can see the trouble I'm having following your steps, and correct it in a specific example of your own, using my symbols and table layout.

### #5 jakuta

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Posted 20 October 2010 - 08:25 PM

I don’t think I’m following your instructions well enough to complete step 7, jakuta.

If you could provide an specific example from the point of view of a distant 3rd observer, I believe you’re example would be clear, and you would no longer find a failure or special relativity.

Calculation of this kind are often made clear by writing specific examples. Here’s one, where $v= \frac{\sqrt{3}}{2} \dot = \, 0.866 \,\mbox{c}$, a good example value because then $\sqrt{1-\left(\frac{v}{c}\right)^2} = \frac{1}{2}$.
X and V are the position and velocity of O as measured by the distant observer.
T is the time that appears on O’s clock at that position.
X’, V’, and T’ are the position, velocity, and time observed on the clock for O’.
X_s and V_s are the position and velocity of the light signal between O and O’.
S is the content of that signal.
T_0 is the time that appears on the distant observer’s clock.

X    V    T    X'    V'    T'    X_s    V_s    S    T_0    note
0    0    -1    0    0    -1                -1
0    .866    0    0    0    0                0
.866    .866    .5    0    0    1                1
1.732    .866    1    0    .866    2    0    1    "2"    2
2.598    .866    1.5    .866    .866    2.5    1    1    "2"    2
...
12.93    .866    7.465    11.20    .866    8.465    12.93    1    "2"    14.93    te=2
12.93    .866    7.465    11.20    .866    8.465    12.93    -1    "7.465"    14.93    t10=7.465
13.73    .866    7.929    12.00    .866    8.929    12.00    -1    "7.465"    15.87
13.73    .866    7.929    12.00    .866    8.929    12.00    1    "8.929"    15.87
24.94    .866    14.40    23.21    .866    15.40    24.94    1    "8.929"    28.80    t20=14.40

Hopefully, you can see the trouble I'm having following your steps, and correct it in a specific example of your own, using my symbols and table layout.

I have no need for a third observer. What makes you think one is needed?

All information is self contained within the two observers. If I needed to select a third observer to make calculations, that would imply these two 2 are inferior and some preferred observer is needed.
That would immediately refute SR.

### #6 jakuta

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Posted 20 October 2010 - 08:40 PM

I don’t think I’m following your instructions well enough to complete step 7, jakuta.

If you could provide an specific example from the point of view of a distant 3rd observer, I believe you’re example would be clear, and you would no longer find a failure or special relativity.

Calculation of this kind are often made clear by writing specific examples. Here’s one, where $v= \frac{\sqrt{3}}{2} \dot = \, 0.866 \,\mbox{c}$, a good example value because then $\sqrt{1-\left(\frac{v}{c}\right)^2} = \frac{1}{2}$.
X and V are the position and velocity of O as measured by the distant observer.
T is the time that appears on O’s clock at that position.
X’, V’, and T’ are the position, velocity, and time observed on the clock for O’.
X_s and V_s are the position and velocity of the light signal between O and O’.
S is the content of that signal.
T_0 is the time that appears on the distant observer’s clock.

X    V    T    X'    V'    T'    X_s    V_s    S    T_0    note
0    0    -1    0    0    -1                -1
0    .866    0    0    0    0                0
.866    .866    .5    0    0    1                1
1.732    .866    1    0    .866    2    0    1    "2"    2
2.598    .866    1.5    .866    .866    2.5    1    1    "2"    2
...
12.93    .866    7.465    11.20    .866    8.465    12.93    1    "2"    14.93    te=2
12.93    .866    7.465    11.20    .866    8.465    12.93    -1    "7.465"    14.93    t10=7.465
13.73    .866    7.929    12.00    .866    8.929    12.00    -1    "7.465"    15.87
13.73    .866    7.929    12.00    .866    8.929    12.00    1    "8.929"    15.87
24.94    .866    14.40    23.21    .866    15.40    24.94    1    "8.929"    28.80    t20=14.40

Hopefully, you can see the trouble I'm having following your steps, and correct it in a specific example of your own, using my symbols and table layout.

I don’t think I’m following your instructions well enough to complete step 7, jakuta.

Sorry, I have the two in the same frame at the start. Each applied the same acceleration from the original frame.

When the second observer enters the frame, it sends a light pulse. I don't know the distance at this point but it could be calculated with the acceleration equations of SR. But, I am using instant acceleration.

So, the first observer waits to receive a light signal from O2.

When it comes in, the 1st observer records the time. Now, what is not known is the time from the view of observer 1 when observer 2 entered the frame. O1 only knows when it received the signal.

So, O1 does a SR clock sync to determine the distance between the observers. Once that is known as d, O1 uses the time it received the original signal from O2 and subtracts d/c.
O1 then knows the time I2 entered the frame closing the loop.

The logic is now recursive meaning logically decidable

### #7 coldcreation

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Posted 21 October 2010 - 07:31 AM

I have no need for a third observer. What makes you think one is needed?

All information is self contained within the two observers. If I needed to select a third observer to make calculations, [...]

CraigD asked you for the point of view of a distant 3rd observer to help anyone interested in understanding your thought experiment see how exactly each twin is supposed to be moving and/or accelerating. The third observer needs not make calculations. The third observer (you?) needs to tell us what he/she sees, in order to make your thought experiment intelligible.

You write in your number 3. "After time t', O' will acquire v in precisely the same way as O in precisely the same direction." Then you claim in number 4. "At the same instant O' acquires v, O' enters the frame of O..."

It is unclear is how O' can be in the same frame as O at the same instant he acquires v. In order for this to occur, acceleration of the two cannot be the same, yet you say they are the same. Do you mean they become stationary relative to each other, while accelerating relative to a hypothetical third observer?

CC

### #8 jakuta

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Posted 21 October 2010 - 07:46 PM

CraigD asked you for the point of view of a distant 3rd observer to help anyone interested in understanding your thought experiment see how exactly each twin is supposed to be moving and/or accelerating. The third observer needs not make calculations. The third observer (you?) needs to tell us what he/she sees, in order to make your thought experiment intelligible.

You write in your number 3. "After time t', O' will acquire v in precisely the same way as O in precisely the same direction." Then you claim in number 4. "At the same instant O' acquires v, O' enters the frame of O..."

It is unclear is how O' can be in the same frame as O at the same instant he acquires v. In order for this to occur, acceleration of the two cannot be the same, yet you say they are the same. Do you mean they become stationary relative to each other, while accelerating relative to a hypothetical third observer?

CC

Oh, I am not following the need for a 3rd observer.

This after all is the scientific method.

If SR is a consistent theory, then if two elements are in the same frame and each perform exactly the same actions, ie uniform acceleration, then they should end up in the same frame after all this.

Is this false or am I not understanding how I am miscommunicating.

### #9 CraigD

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Posted 22 October 2010 - 12:22 AM

If I needed to select a third observer to make calculations, that would imply these two 2 are inferior and some preferred observer is needed.
That would immediately refute SR.

A distant third observer is not a privileged, because it’s velocity and precise distance is arbitrary, other than it should be about on a line perpendicular to the velocities of O and O', and much more distant than the distance traveled by O and O'.

A privileged or preferred, observer is one for whom the laws of physics are different than some other. In the calculations I made in post #4, there’s nothing special about the third observer – it’s just a different one, which conveniently doesn’t accelerate at any time in the thought experiment.

Oh, I am not following the need for a 3rd observer.

A distant third observer provides a convenient computational technique for calculating the various measurements made by O and O'. Because of its position (described above), simultaneity differences, which make calculations like these difficult, are made arbitrarily small, simplifying the calculations.

This after all is the scientific method.

Since we’re not actually making predictions and testing them with real physical experiments, this isn’t strictly “the scientific method” – observe, theorize, predict, test – but a sub-process of the theorize step, usually called a “thought experiment” – a means to better understand, and possibly expand or correct, a theory.

If SR is a consistent theory, then if two elements are in the same frame and each perform exactly the same actions, ie uniform acceleration, then they should end up in the same frame after all this.

But O and O’ do not perform exactly the same actions, even though they undergo the same change in velocity (instant, in the thought experiment) because they undergo it at different instants, and travel different distances, relative to each other or a distant observer.

After step 4, O and O’ have the same velocity, so are in the same inertial frame. Their clocks, however, don’t agree. In my specific example (distance in light-seconds, speed in c, and clock times in seconds), they disagree by 1 second, as observed by the distant observer, or by each other when each compensated for their known separation (of 3.46 light-seconds)

My suggestion to you, jakuta, is that you perform the actual calculations of your thought experiment, using specific speeds and times, using a table notation like I did. In my experience, though experiments such as yours are much easier to understand in terms of precise numbers than in more philosophical terms like “frames”, “consistency”, and “refuted”. Theoretical predictions must be precise, measurable, and, in this case, numeric.

I can post a program (actual computer code, or pseudo code) for these calculations, if that would be useful to you. I did the calculations in post #4 by hand, but could quickly write a program to do them.

### #10 Kharakov

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Posted 22 October 2010 - 01:29 AM

Just think of it this way: matter has a "total speed" at which it always moves relative to other matter:

[imath]c = \sqrt{v_{3dspace}^2+v_{orthogonal}^2}[/imath]

The 3dspace component of the velocity is that which is relative to other matter in 3dspace (3 dimensions of space). The other component is the matter's orthogonal velocity component, which is how fast it 'moves forwards' in time (time being the dimension orthogonal to 3dspace in spacetime) relative to other matter.

If you have a higher 3dspace velocity component (this is only relative to other matter) you have a lower orthogonal velocity component (once again only relative to other matter): you are moving sssslslllllloooooowowwwwwweeerrrrr through the 4th dimension (time) compared to the other matter.

[imath]v_{orthogonal} = \sqrt{c^2-v_{3dspace}^2}[/imath]

I didn't do the math, but I think you end up with a time factor of:

[imath]time= time *\frac{v_{orthogonal}}{c}[/imath]

Aight, checked it.. yup. Just set c to one unit per second, do the good ole sqrt(1-sin(60)^2) and you got your orthogonal velocity component of 1/2 units per second.

### #11 jakuta

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Posted 22 October 2010 - 06:41 PM

A distant third observer is not a privileged, because it’s velocity and precise distance is arbitrary, other than it should be about on a line perpendicular to the velocities of O and O', and much more distant than the distance traveled by O and O'.

A privileged or preferred, observer is one for whom the laws of physics are different than some other. In the calculations I made in post #4, there’s nothing special about the third observer – it’s just a different one, which conveniently doesn’t accelerate at any time in the thought experiment.

A distant third observer provides a convenient computational technique for calculating the various measurements made by O and O'. Because of its position (described above), simultaneity differences, which make calculations like these difficult, are made arbitrarily small, simplifying the calculations.

Since we’re not actually making predictions and testing them with real physical experiments, this isn’t strictly “the scientific method” – observe, theorize, predict, test – but a sub-process of the theorize step, usually called a “thought experiment” – a means to better understand, and possibly expand or correct, a theory.

But O and O’ do not perform exactly the same actions, even though they undergo the same change in velocity (instant, in the thought experiment) because they undergo it at different instants, and travel different distances, relative to each other or a distant observer.

After step 4, O and O’ have the same velocity, so are in the same inertial frame. Their clocks, however, don’t agree. In my specific example (distance in light-seconds, speed in c, and clock times in seconds), they disagree by 1 second, as observed by the distant observer, or by each other when each compensated for their known separation (of 3.46 light-seconds)

My suggestion to you, jakuta, is that you perform the actual calculations of your thought experiment, using specific speeds and times, using a table notation like I did. In my experience, though experiments such as yours are much easier to understand in terms of precise numbers than in more philosophical terms like “frames”, “consistency”, and “refuted”. Theoretical predictions must be precise, measurable, and, in this case, numeric.

I can post a program (actual computer code, or pseudo code) for these calculations, if that would be useful to you. I did the calculations in post #4 by hand, but could quickly write a program to do them.

I do not need a third observer because under SR uniform acceleration, the calculations are absolute. You can see here, all I need do is perform a burn time and some acceleration and the twins will end of up in the frame after both burn but at a distance.

http://math.ucr.edu/.../SR/rocket.html

I have the actual acceleration equations but I dod not think anyone would want to fool with it. Here is a longer version with the math you requested.

Alternative Twins Experiment

The traditional twins paradox is solved such that all agree the traveling twin is younger because of non-symmetric acceleration. So, this twins paradox will produce symmetric acceleration with no turn-around and force SR to exclusively explain reciprocal time dilation. In particular, this experiment will force SR to calculate two different answers for the age of the twins while at the same time use the SR clock synchronization method to prove there can only be one correct answer. Clearly, a valid theory cannot assert there exists only one correct answer for a problem and at the same time assert there are two correct contradictory answers for that same problem.

O and O' are physicists testing special relativity.

These physicists are going to test the aging of two twins.

O, O', twin1 and twin2 are in the same frame and sync their clocks with Einstein's clock synchronization method.

They all agree to burn for BT and agree on the number.

Next, all set their clocks to 0 and reside at the same origin for a common start point. Then, twin1 takes off from the frame in a spaceship at proper acceleration a and for burn time BT in the proper time of the accelerating frame such that they attain a relative motion v compared to the prior frame.

Then, for a long period of time, they remain in relative motion to O' and twin2.

So, O and twin1 see O' and twin2 moving away at relative motion v and O' and twin2 see O and twin1 moving away at relative motion v.

After some long period of t, O' and twin2 take off in an identical spaceship in exactly the same direction as O and twin1 did at the same acceleration a and for the same burn time BT.

After the burn is complete, all are back in the same frame but at a distance between them and O' immediately sends out a light pulse. O records the time it receives the light pulse. Then, O performs the round trip speed of light calculation to decide the distance between the two ships. Once the distance D between the two ships is known, O subtracts D/c from the time it received the light pulse from O'. This gives O the exact time in its own proper time when O' stopped accelerating and entered the frame of O. In addition, this gives a correct endpoint to the experiment that matches the endpoint of O'. Thus, O and O' share a common start point and end point to the experiment. After this is complete, both twins perform the clock synchronization method to determine the ordinality of the clocks.

Let BT be the burn time in the accelerating frame. Let a = acceleration and v = the relative speed attained by the acceleration.

O and O' calculate the proper time of the twins as follows.

Calculations of O for the twins

Elapsed proper time calculation for twin 1

twin1's acceleration phase
BT

twin1's relative motion phase
t'

twin2's acceleration phase
c/a sinh(aBT/c) Total elapsed proper time calculation of O for twin1 BT + t' + c/a sinh(aBT/c)

Please note above, O marked an endpoint to the experiment after receiving a light pulse from O', call that time, Te. Then,

Te = BT + t' + c/a sinh(aBT/c) + D/c

where D is the distance between the two ships in the resulting frame as determined above.

Thus,

t' = Te - BT - c/a sinh(aBT/c) - D/c.

Elapsed proper time calculation for twin 2

twin1's acceleration phase
c/a sinh(aBT/c)

twin2's relative motion phase
t'/ λ

twin2's acceleration phase
BT

Total elapsed proper time calculation of O for twin2 c/a sinh(aBT/c) + t'/λ + BT

Conclusion of O, twin1 is older.

Calculations of O' for the twins

Elapsed proper time calculation for twin 1

twin1's acceleration phase
BT

twin1's relative motion phase
t/λ

twin2's acceleration phase
c/a sinh(aBT/c)

Total elapsed proper time calculation of O' for twin1 BT + t/λ + c/a sinh(aBT/c)

Elapsed proper time calculation for twin 2

twin1's acceleration phase
c/a sinh(aBT/c)

twin2's relative motion phase
t

twin2's acceleration phase
BT

Total elapsed proper time calculation of O' for twin2 c/a sinh(aBT/c) + t + BT

Conclusion of O', twin1 is younger.

Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks. This is how it is done.

Twin1 sends its time ttwin1 to twin2. Twin2 immediately sends its time ttwin2 back to the clock of twin1. Twin1 receives this light at t'twin1

Einstein concludes the two clocks are in sync if ttwin2 = ½ ( t'twin1 - ttwin1). Thus, the age ordinality is established as ttwin2 - ½ ( t'twin1 - ttwin1)

http://www.fourmilab...in/specrel/www/

SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame.

However, both O' and O cannot be correct with their SR conclusions when presented with the results of the clock sync.

So, when using the thought experiment above, the clock sync at the end of the experiment produces only one correct answer. The following are the possibilities.

1) Twin1 is older than twin2
2) Twin2 is older than twin1
3) Both twins are the same age.

Only one of these results can be true from the result of the clock sync. On the other hand, LT calculates two different contradictory answers for the same thought experiment.

Therefore one path of logic in SR leads to strictly one answer.

Hence, SR is logically inconsistent.

### #12 jakuta

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Posted 22 October 2010 - 06:49 PM

Just think of it this way: matter has a "total speed" at which it always moves relative to other matter:

[imath]c = \sqrt{v_{3dspace}^2+v_{orthogonal}^2}[/imath]

The 3dspace component of the velocity is that which is relative to other matter in 3dspace (3 dimensions of space). The other component is the matter's orthogonal velocity component, which is how fast it 'moves forwards' in time (time being the dimension orthogonal to 3dspace in spacetime) relative to other matter.

If you have a higher 3dspace velocity component (this is only relative to other matter) you have a lower orthogonal velocity component (once again only relative to other matter): you are moving sssslslllllloooooowowwwwwweeerrrrr through the 4th dimension (time) compared to the other matter.

[imath]v_{orthogonal} = \sqrt{c^2-v_{3dspace}^2}[/imath]

I didn't do the math, but I think you end up with a time factor of:

[imath]time= time *\frac{v_{orthogonal}}{c}[/imath]

Aight, checked it.. yup. Just set c to one unit per second, do the good ole sqrt(1-sin(60)^2) and you got your orthogonal velocity component of 1/2 units per second.

The rockets move in exactly the same direction during acceleration and so I do not understand why orthogonal plays a role. So, normally under SR, one needs to be concerned with errors in making preferences, the SR uniform acceleration equations are absolurte motion and distance, time differentials etc are all absolutely mathematically decidable.

Anyway here are several links on the SR uniform acceleration equations.

http://www.ias.ac.in...b252007/416.pdf

http://www.ejournal....21/RMF52110.pdf

http://users.telenet...celeration.html

http://math.ucr.edu/.../SR/rocket.html

http://arxiv.org/PS_...1/0411233v1.pdf

### #13 modest

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Posted 23 October 2010 - 02:56 AM

Are you willing to work through your thought experiment?

Let's set v=.6c

Procedure
1. Both set their clocks to 0 and O instantly acquires v relative to O'.
2. O and O' are in relative motion for some agreed up time t' on the clock of O'.

we'll set the duration of relative motion as t' = 2.5.

From the perspective of O' when t' = 2.5, t = 2. From the perspective of O, when t' = 2.5, t = 3.125 (we have to keep in mind that simultaneity is relative, so "when" means something different to the two people)

3. After time t', O' will acquire v in precisely the same way as O in precisely the same direction.

From the perspective of O', t jumps from 2 to 3.125 (because the plane of simultaneity shifts) and now advances more quickly than it had before (t' still equals 2.5). From the perspective of O, t' = 2.5 (it does not jump because O did not accelerate or change frames of reference) and t = 3.125. t' now advances more quickly than it had before from the perspective of O. In fact, both clocks now advance the same rate from either perspective.

4. At the same instant O' acquires v, O' enters the frame of O and O' sends a light pulse to O and records this as the end of the experiment at time t'.

From the perspective of O' the time of emission is t' = 2.5 and t = 3.125. From the perspective of O the time of emission is t' = 2.5 and t = 3.125.

Being in the same frame, you might now notice that each agrees on the time of the other's clock.

5. O' receives the light pulse and records the time as te.

From both observer's perspective, the pulse takes 1.875 seconds to travel the distance of 1.875 light-seconds. From the perspective of O' the time of detection is t' = 4.375 and t = 5. From the perspective of O the time of detection is t' = 4.375 and t = 5.

Again, they both agree on the time of the other's clock.

There is not a single detection time, te, but rather two times because there are two clocks which now run at the same rate but display different times.

6. Since, O and O' are again in the same frame, O performs with O' the round trip speed of light calculation using a time trial to calculate the distance between the two. Let D be that distance. O then subtracts D/c from te and determines its proper for when O' entered the frame to determine the correct end of the experiment which will match the end for O'. Let this time be t.

During the light's trip, both observer's agree that t' changes from 2.5 to 4.375 and t changes from 3.125 to 5. Both clocks changed by 1.875 seconds. Both observer's agree that D is 1.875 light-seconds.

7. Finally, O and O' perform the SR clock synchronization method to determine the ordinality of the two clocks. To do this, O sends its time t1O to O'. O' immediately sends its time tO' back to the clock of O. O receives this light at t2O. In particular, Einstein concludes the two clocks are in sync if tO' = ½ (t2O - t1O). Thus, the age ordinality is established as tO' - ½ (t2O - t1O).

tO' is 0.625 seconds less than 1/2(t2O - t1O) because t' is .625 seconds less than t. In other words, to synchronize their clocks, O must set his clock back .625 seconds or O' must set his clock forward .625 seconds.

The difference in their time at the end of the thought experiment will be t'(ɣ - 1) where t' is from step 2.

The steps above represent an effective procedure for deciding the age ordinality of O and O'. More specifically, step 7 determines the clocks of O and O' as one of the following.
· O is younger than O'.
· O is older than O'.
· O and O' are the same age.

Step 7 would determine that O is older than O'.

Now, LT will be applied to solve the problem above.

From O as stationary, its proper time is t. Hence, O concludes O' elapses t/λ and therefore concludes O' is younger.

From O' as stationary, its proper time is t'. Hence, O' concludes O elapses t'/λ and therefore concludes O is younger.

No, that is mistaken.

There is no problem in considering either O or O' as stationary during the relative velocity part of the thought experiment. You'll see in my description above that step 2 has O thinking the clock of O' runs slow and O' thinking the clock of O runs slow because each thinks the other has a velocity and a time dilated clock relative to themselves. That part is symmetric.

But, I agree with Coldcreation and Craig, the thought experiment is not symmetrical.

The important difference in considering things from the perspective of O or O' is that O' accelerates (or changes frames of reference) while there is a distance between O and O'. When O accelerates, there is no distance between O and O'. This is a non-symmetrical difference between the two. The reason this matters is given on wiki's relativity of simultaneity page:

If two events happen at the same time in the frame of the first observer, they will have identical values of the t-coordinate. However, if they have different values of the x-coordinate (different positions in the x-direction), we see that they will have different values of the t' coordinate; they will happen at different times in that frame. The term that accounts for the failure of absolute simultaneity is that vx/c^2

For the acceleration of O they do NOT have different values of the x coordinate, for the acceleration of O' they do have different values of the x coordinate.

In other words, changing ones velocity changes the time of an event in a manner that depends on the distance to that event. If I accelerate and there is a clock right next to me then its time is not shifted from what it was, but if the clock is a good distance from me it will be shifted (either forward or backward depending on if I accelerate toward or away from it).

Let me know if anything is unclear or if making a spacetime diagram would help.

~modest

EDIT:

here is that spacetime diagram in case it will help:

Messy, I know, sorry

### #14 jakuta

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Posted 23 October 2010 - 09:00 AM

Are you willing to work through your thought experiment?

Let's set v=.6c

we'll set the duration of acceleration as t' = 2.5.

From the perspective of O' when t' = 2.5, t = 2. From the perspective of O, when t' = 2.5, t = 3.125 (we have to keep in mind that simultaneity is relative, so "when" means something different to the two people)

This is not correct. Under acceleratrion, this is not relativity. All parties agree on the timing differentials.

So, assume from the context of O', the accelerating twin (BT) elapses as the burn time.

Then, t = (c/a) sinh(a(BT)/c) us the elapsed proper time of the stay at home twin.

These values are absolute.

For example, we can solve for BT,

t = (c/a) sinh(a(BT)/c)

at/c = sinh(a(BT)/c)

sinh-1(at/c) = a(BT)/c

c/a sinh-1(at/c) = (BT)

You will find this at mathpages,

http://math.ucr.edu/.../SR/rocket.html

From the perspective of O', t jumps from 2 to 3.125 (because the plane of simultaneity shifts) and now advances more quickly than it had before (t' still equals 2.5). From the perspective of O, t' = 2.5 (it does not jump because O did not accelerate or change frames of reference) and t = 3.125. t' now advances more quickly than it had before from the perspective of O. In fact, both clocks now advance the same rate from either perspective.

Where do you get "the plane of simultaneity shifts" when dealing with acceleration?

If you look here, you will find the derivation of the SR acceleration equations. It does based itself on 3 observers though and you will find this technique used pretty much everywhere when dealing with acceleration.
Basically, it starts with the velocirty addition equations.

Then, the 3 observers are the at rest frame, the accelerating frame and and an instaneous co-moving frame that is at rest with the accelerating frame at any time T in the context of the accelerating frame.
http://users.telenet...celeration.html

Oh, "the plane of simultaneity shift" is a myth.
If you set
x = vtγ /(1+γ)
x' = - vtγ /(1+γ)

and run this through LT, you will find t' = t.

From the perspective of O' the time of emission is t' = 2.5 and t = 3.125. From the perspective of O the time of emission is t' = 2.5 and t = 3.125.

Being in the same frame, you might now notice that each agrees on the time of the other's clock.

Step 7 would determine that O is older than O'.

No, that is mistaken.

Agreed.

There is no problem in considering either O or O' as stationary during the relative velocity part of the thought experiment. You'll see in my description above that step 2 has O thinking the clock of O' runs slow and O' thinking the clock of O runs slow because each thinks the other has a velocity and a time dilated clock relative to themselves. That part is symmetric.

Agreed

But, I agree with Coldcreation and Craig, the thought experiment is not symmetrical.

The important difference in considering things from the perspective of O or O' is that O' accelerates (or changes frames of reference) while there is a distance between O and O'. When O accelerates, there is no distance between O and O'. This is a non-symmetrical difference between the two. The reason this matters is given on wiki's relativity of simultaneity page:

This I cannot agree on. The SR velocity addition equation is the basis for the integral describing the timing differentials. This equation is not sensitive to distance and is valid regardless of the distance.

However, I did consider this a problem originally by looking into the Born rigidity problem. But, what I find in all solutions is the answer only depends on the front ship pulling the string and thus the string breaks. You can see that here.

http://www.mathpages...22/kmath422.htm

It never considers the fact, that the ship on the other end is pushing the string from the other end toward the front of the ship. So, until I see a Born solution with the back end considered, I do not find them valid.

However, I came up with an easy solution to this possibility. Let's assume that is true that we must accelerate the front ship less than the rear to maintain symmetry. Again, I cannot agree with this because it falsifies the SR acceleration equations.
But, anyway, let that be a1 and a for the two accelerations. Then, the symmetry is solved and the refutation of SR still holds.

I can also test they are in fact in the same frame by pulsling pico second flashes from one to the other. If the receiver ship measures something other than pico second flashes, then they are not in the same frame and a recalculation of acceleration a1 needs to be done.

If I can't perform the math for some reason for a1 though the Born solution provides the acceleration differential, I find an a1 that causes the frequency to be too high. Then I find one too low. Then I simply perform a divide and conquer algorithm which is a Cauchy sequence and hence convergent to the correct acceleration. So, the entire process above remains recursive and thus logically decidable.

Hopefully, I have show symmetry is achievable regardless.

### #15 modest

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Posted 23 October 2010 - 12:48 PM

Where do you get "the plane of simultaneity shifts" when dealing with acceleration?

The special theory of relativity developed by Einstein in 1905. In particular, section 2 of On The Electrodynamics of Moving Bodies -- 1905 -- A. Einstein which concludes:

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

To accelerate is to move from one system of coordinates to another. There are two frames of reference for O'. In the first, t' = 2.5 is simultaneous with t = 2. In the second, t' = 2.5 is simultaneous with t = 3.125. Because your thought experiment relies on "when" t' = 2.5 (a matter of simultaneity) it is important not to neglect this detail.

A cynical person would think you designed this thought experiment to obfuscate the relativity of simultaneity.

Oh, "the plane of simultaneity shift" is a myth.

If that is your position, then I can only conclude that you are intentionally solving the problem in a manner inconsistent with relativity.

The "plane of simultaneity shift" is explained on wiki's relativity of simultaneity page. It is not a myth. It is explained in the context of the twin paradox like so:

In special relativity there is no concept of absolute present. A present is defined as a set of events that are simultaneous from the point of view of a given observer. The notion of simultaneity depends on the frame of reference (see relativity of simultaneity), so switching between frames requires an adjustment in the definition of the present. If one imagines a present as a (three-dimensional) simultaneity plane in Minkowski space, then switching frames results in changing the inclination of the plane.
Minkowski diagram of the twin paradox

In the spacetime diagram on the right, drawn for the reference frame of the stay-at-home twin, that twin's world line coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. Just before turnaround, the traveling twin calculates the age of the resting twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnaround, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin.

I will mark the shift in simultaneity on the spacetime diagram I made:

When O' turns to the right (at the red dot) the plane of simultaneity shifts from the horizontal green line to the tilted green line. Before the shift the present instant intersects t = 2 (the black dot in the blue coordinate system), after the shift it intersects t = 3.125 (the green dot in the blue coordinate system). If the acceleration is 'instant' then the shift from t = 2 to t = 3.125 from the perspective of O' is instant.

Unless you want to invoke a pseudo gravitational field in general relativity, or a third reference frame, the thought experiment must be done this way. If you refuse to solve the problem this way then you are simply insisting on doing it wrong. You can't blame relativity if you don't follow it.

This is not correct. Under acceleratrion, this is not relativity. All parties agree on the timing differentials.

So, assume from the context of O', the accelerating twin (BT) elapses as the burn time.

Then, t = (c/a) sinh(a(BT)/c) us the elapsed proper time of the stay at home twin.

These values are absolute.

For example, we can solve for BT,

t = (c/a) sinh(a(BT)/c)

at/c = sinh(a(BT)/c)

sinh-1(at/c) = a(BT)/c

c/a sinh-1(at/c) = (BT)

You will find this at mathpages,

http://math.ucr.edu/.../SR/rocket.html

The burn time is not absolute. You found the burn time relative to t. You could also find the burn time relative to t'. There are two equations given in that link:

t = (c/a) sh(at'/c)

t' = (c/a) sh^-1(at/c)

You can substitute t = BT or t' = BT. The two observers disagree on the burn time, but that is not at issue. The burn time is zero in the thought experiment—we are switching frames instantly.

If you look here, you will find the derivation of the SR acceleration equations. It does based itself on 3 observers though and you will find this technique used pretty much everywhere when dealing with acceleration.
Basically, it starts with the velocirty addition equations.

Then, the 3 observers are the at rest frame, the accelerating frame and and an instaneous co-moving frame that is at rest with the accelerating frame at any time T in the context of the accelerating frame.
http://users.telenet...celeration.html

The amount of time it takes to accelerate is not at issue.

This I cannot agree on. The SR velocity addition equation is the basis for the integral describing the timing differentials. This equation is not sensitive to distance and is valid regardless of the distance.

The velocity addition equation is not at issue. Nowhere in this thought experiment are we adding velocities.

However, I did consider this a problem originally by looking into the Born rigidity problem.

Rigidity is not at issue here. The thought experiment would work for point particles.

Hopefully, I have show symmetry is achievable regardless.

Symmetry is is not achievable because the situation is non-symmetric. When O accelerates, O' is co-located with O. When O' accelerates they are separated by vt'. Clearly O does not experience the same thing as O'.

You can, however, achieve a solution in which both observers are 'at rest' through the duration of the thought experiment, which is, I think, what you are really looking for. To do this you must invoke general relativity as Einstein explains here:

http://en.wikisource...y_of_relativity

~modest

### #16 HydrogenBond

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Posted 23 October 2010 - 02:37 PM

Special relativity has three equations one for mass, one for distance and one for time, since these all three will change with velocity. To get past the confusing reference special effects that can occur using only distance and time, I prefer to do an energy balance by looking at mass/relativistic mass changes. Because of the conservation of energy, relativistic mass changes can not be reference dependent but need to be absolute. If it was reference dependent, we could use a reference special effects, to create or take away energy, making an universal energy balance impossible since it would be relative. We could also do perpetual motion tricks.

You simply compare SR energy/mass changes, for both references, the one with most energy, will time dilates the most.

### #17 jakuta

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Posted 23 October 2010 - 02:41 PM

The special theory of relativity developed by Einstein in 1905. In particular, section 2 of On The Electrodynamics of Moving Bodies -- 1905 -- A. Einstein which concludes:

Yea, I understand that part.

But, since the velocity add equations are used to derive the the acceleration equations based on everything I read from you, you did agree with the SR acceleration equations.

It that true?

To accelerate is to move from one system of coordinates to another. There are two frames of reference for O'. In the first, t' = 2.5 is simultaneous with t = 2. In the second, t' = 2.5 is simultaneous with t = 3.125. Because your thought experiment relies on "when" t' = 2.5 (a matter of simultaneity) it is important not to neglect this detail.
A cynical person would think you designed this thought experiment to obfuscate the relativity of simultaneity.

May I see your equations to prove this?

And my thought epxeriment relies on the acceleration equations. I doubt you disagree with the relative motion part.

And, further given any acceleration phase "when" can be determined since that phase is absolute.

It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames.
In special relativity accelerating frames are different from inertial frames. Velocities are relative but acceleration is treated as absolute.

http://math.ucr.edu/...celeration.html

So, you are trying to inject the relativity of simultaneity, but it is not applicable to acceleration under SR. Both frames will agree on the time differential between the two frames between the start of the acceleration and the end.

If that is your position, then I can only conclude that you are intentionally solving the problem in a manner inconsistent with relativity.

No, I am not using that tool to operate on this. But, can you explain the simultaneity shift when t' = t for the x and x' I gave. This is a completely issue from the twins issue.

The "plane of simultaneity shift" is explained on wiki's relativity of simultaneity page. It is not a myth. It is explained in the context of the twin paradox like so:

I will mark the shift in simultaneity on the spacetime diagram I made:

I do not have a turn around in this thought experiment, so this is not applicable.

The burn time is not absolute. You found the burn time relative to t. You could also find the burn time relative to t'. There are two equations given in that link:

t = (c/a) sh(at'/c)

t' = (c/a) sh^-1(at/c)

You can substitute t = BT or t' = BT. The two observers disagree on the burn time, but that is not at issue. The burn time is zero in the thought experiment—we are switching frames instantly.

The amount of time it takes to accelerate is not at issue.

I think I see our disagreement. The equations are based on BT in the context of the accelerating frame. So, BT is the known value. Since this is controlled in the context of the accelerating, then we can solve for t in the context of the twin that remained behind.

t = (c/a) sh(a(BT)/c)

If you disagree with this, then you disagree with the SR equations.

Therefore, if the accel frame elapses BT duing the acceleration period, then the stay at home frame elapses (c/a) sh(a(BT)/c). The frames will not disagree on this.

Since the frames do not disagree on the timing differential, then instantly acquiring v is not a problem, Einstein used this same tool in the explanation of time dilation.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other
http://www.fourmilab...in/specrel/www/

The velocity addition equation is not at issue. Nowhere in this thought experiment are we adding velocities.

It is used to derive the acceleration equations as I said.

You know that SR says that
"When an object P has a relative velocity u w.r.t.
to an inertial frame Q , of which the observer itself
has a relative velocity v w.r.t. an inertial observer R ,
then the object P has a relative velocity
w = (u+v)/(1+u v)
w.r.t. the observer R."

http://users.telenet...celeration.html

Rigidity is not at issue here. The thought experiment would work for point particles.

I tend to agree, but others would disagree.

Symmetry is is not achievable because the situation is non-symmetric. When O accelerates, O' is co-located with O. When O' accelerates they are separated by vt'. Clearly O does not experience the same thing as O'.

You can, however, achieve a solution in which both observers are 'at rest' through the duration of the thought experiment, which is, I think, what you are really looking for. To do this you must invoke general relativity as Einstein explains here:

http://en.wikisource...y_of_relativity

No it is not what I am looking for.

I did exactly what I wanted to do.

Here is the velocity attained by the acceleration
v = c th(a(BT)/c)

So, you have no basis to claim just because they are separated by a distance each will arrive at a different v relative to the original co-located position.

This implies the rules of physics are different basd on distance.

So, since v = c th(a(BT)/c) for both, then they are in the same frame after each accelerates because that is a v relative to the same frame that each had been a member.