# Twin paradox (Pete and MacM)

Discussion in 'Physics & Math' started by Pete, Sep 6, 2004.

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1. ### PeteIt's not rocket surgeryRegistered Senior Member

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Who says that the spaceship stopwatch is travelling? The spaceship stopwatch doesn't think it's travelling - it thinks that the Earth stopwatch is travelling.

I think I see where your confusion lies now.

Are you thinking that:

1) The time observed from Earth to pass on the spaceship clocks during the laser beam transit

is the same as

2) The time passing on spaceship clocks during the laser beam transit as observed from the spaceship

?

Time (1) is 1 second
Time (2) is 4 seconds

3. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
Here's the bottom line, I think:

[highlight]If two events occur in the same place in some reference frame, the time between those events will be longer in any other reference frame.[/highlight]

This can be thought of as:
The time between two consecutive clock ticks will be shortest in the clock's rest frame (the frame in which the two ticks occur in the same place).

Or:
Moving clocks run slower.

Note that the first way of expressing time dilation is the most precise. It is the one that should be used in ambiguous situations.

Applied to this case, it becomes clear what relativity predicts:
The laser beam's firing and return happens in the same place in only one frame: Earth's.
This means that the transit time of the laser beam should be shortest in Earth's frame, and dilated in all other frames.

Essentially this means that we're timing an Earth clock in this situation. The laser is an Earth clock, and it runs slower (time between ticks is longer) in all other frames.

Last edited: Nov 3, 2004

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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Thoughts on the clock topic...

Imagine the laser apparatus fired a new beam immediately on reception of the last one, and there were a series of spaceship passing at a distance of 1 light-second.

Now, watch the laser, and count the times that it fires.
Can you use that count to measure time?
If you watched someone drive around a 1 kilometre track while the laser fired 30 times, could you use that information to determine their speed?

7. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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The travel time of the laser LIGHT has to be 'c' by any clock. You are not looking at a
clock attached to a laser on Earth, it doesn't matter what one observer 'sees' on the
clock in the other frame. On the Earth's stopwatch, it takes two seconds for the laser
beam to travel to the reflector on the spaceship located 299,792,458 meters and back
to Earth. In the spaceships frame of reference, the laser beam travels a longer path
because SR puts the Earth and laser in motion and the spaceships stopwatch will record 4 seconds. The spaceship stopwatch HAS to record the laser beam as traveling
299,792,458 meters per second or constancy of the speed of light is violated. THAT is
why I used the reflected laser beam in the first place.

8. ### MacMRegistered Senior Member

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10,104
It would be appropriate for you to answer his question and not to be putting words in my mouth as to what I think. You clearly do not know.

9. ### MacMRegistered Senior Member

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10,104
Thanks. I appreciate your contribution. You seem to understand the issue.

You will note that Janus, James R nor Pete responded to your inquiry of the relavistically calibrated clocks and timers being used in such a test.

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi Mac,
geistkiesel is on my ignore list.
Even if he wasn't, you'll notice I stopped responding to posts on that experiment some time ago - I've said all I want to (more than once).

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12. ### Quantum QuackLife's a tease...Valued Senior Member

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2inquisitive,
I have started a thread that may be of interest for you titled 'light clock revisited". I think it takes up the issue about perpendicular to vector length contractions using a multiple light clock.

13. ### MacMRegistered Senior Member

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I'm only curious at this point as to why I too have not been put on your list.

14. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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My point is very simple. Special Relativity predicts moving clocks run FAST.
JamesR, you don't believe me? Prove it. My simple paradox is posted.

15. ### geistkieselValued Senior Member

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I am on the "ignore" list, so sad I am. Anyway what could they possibly say that would have any relevance?

16. ### geistkieselValued Senior Member

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Your thesis is incorrect. if the frames, Vm, Vn and Vo all accelerate from the earth frame Ve, assmed zero, and Vm > Vn > Vo and Vn is the one with the clock ticks to measure, then relativity theory must predict the Vm rate < the Mn rate and the Vo rate > than the Vn rate and the Ve rate the slowest of all.

17. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's a simplification of what SR says, which you are applying incorrectly and drawing incorrect conclusions.

Here is a more precise statement:
[highlight]If two events occur in the same place in some reference frame, the time between those events will be longer in any other reference frame.[/highlight]

Last edited: Nov 4, 2004
18. ### PeteIt's not rocket surgeryRegistered Senior Member

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Because your range of posts is not limited to a single debate which I feel has been exhaustively discussed already (in more than one arena).

19. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi 2inquisitive,
I've just noticed your new thread for this issue, and am responding there.

20. ### Paul TRegistered Senior Member

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460
Pete,

Yes, those times that I computed were times in the spaceship. They are indirectly connected to the events of laser leaving and returning to the earth. I used word indirectly because I had introduced three points (a, b, c), where ab and bc are equal distance (0.866c second according to earth observer), instead of using the original point (on earth) where the laser was fired.

To summarized, I computed times at 'a', 'b' and 'c' respectively 1.5sec, 2sec and 2.5sec. If you compute the position of events corresponding to those three times, you will get for all three x' = -2*0.866c second. It means that those three events (when the spaceship arrives at points 'a', 'b' and 'c') occur on the same position in the spaceship reference frame. We can compare those times and obtain their interval, for instance 2.5sec - 1.5sec = 1.0 sec which is the time interval in the spaceship corresponding to the 2.0 seconds time interval on earth reference frame, in this case, the time interval between the arrival of spaceship at point 'a' and 'c'.

I disagree with this. The reason is, the time t'=0 and t'=4sec are not from the same position on the spaceship reference frame, as shown by your own calculation (x'=0 and x'=4*0.866c seconds). You cannot compute the time interval for this two events that occur at two difference location in x' reference frame (a separation of 4*0.866c second). Your conclusion that the 4 seconds time interval in the spaceship reference frame corresponds to 2 seconds time interval in earth reference frame is therefore incorrect.

I am sure my computation is appropriate. If you still think that it is not, please let me know.

In addition, we can also compute t' and x' for following two events:

1) Laser leaves earth (t=0, x=0, z=-1 light second)
t' = 0
x' = 0​
2) Laser returns to earth (t=1sec, x=0, z=0)....sorry, laser reaches the spaceship!
t' = 2*(1-(0.866c/c)*0) = 2 seconds
x' = 2*(0-0.866c*1) = -2*0.866c seconds​
As shown, those two events occur respective at x' = 0 and x' = -2*0.866c seconds. However, it is incorrect to say that while the clock on earth ticks 1 second, the clock on the spaceship ticks 2 seconds. The time interval of 1 second in x frame is the correct time interval as it is obtained from time in the same x, but the time interval of 2 seconds calculated this way is not....

Last edited: Nov 4, 2004
21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi Paul,
Here's a spacetime diagram in Earth's frame.
<img src="/attachment.php?attachmentid=3404&stc=1">

A: Beam leaves Earth
B: Beam reflected
C: Beam returns to Earth
D: Spaceship event simultaneous (in Earth's frame) with A
E: Spaceship event simultaneous (in Earth's frame) with C

The task as I understand it is to determine the transit time of the beam in the spaceship frame.
To me, this means determining the time difference between points A and B.
[highlight]Correction - should be between points A and C[/highlight]

Yes, these points are in different places in the spaceship frame - and that's OK. (see note below)

Your interpretation of the problem is that it means determining the time interval separating points D and E, right?

The reason I don't like that is that in the spaceship frame, point D isn't simultaneous with the beam leaving Earth, and point E isn't simultaenous with the beam returning... so to me it doesn't seem meaningful to consider that interval to be the spaceship's observation of the transit time.

Note:
I don't have a problem with computing a time separation between events in different places.
In the Earth frame, for example, it seems quite meaningful to say "The reflection occurred one second after the beam was fired".
Similarly, it seems meaningful to me to say that in the spaceship frame "The beam returned to Earth four seconds after it was fired."
Isn't that what events have a time coordinate for?

Last edited: Nov 5, 2004
22. ### PeteIt's not rocket surgeryRegistered Senior Member

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Minor correction: t = 2, leading to t' = 4, x' = -3.464
(that's same transform I did earlier)
It's true in the spaceship frame, isn't it?

23. ### Paul TRegistered Senior Member

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460
Sorry, the second event that I wanted to compute was laser reaches the spaceship at t=1 according to earth reference frame.