Twin paradox (Pete and MacM)

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You explanation of your new scenario is correct - but it is not the same as 2inquisitive's scenario.

In 2inquisitive's scenario, there is only one location of interest in Earth's frame - not three.
The are three events of interest in Earth's frame - but they all happen at the same place.

In your scenario, you have three different locations in Earth's frame, hence it is a different problem.

This is the core of the problem in 2inquisitive's scenario - in Earth's frame, the end points of the interval of interest are in the same place.
In your alternative scenarios, you are choosing events which are not in the same place in Earth's frame, which is why you get a different result.

If you want to change the scenario, make sure that the interval end points are in the same place in Earth's frame.


Please,
What problem are your answering in 2inquisitive's scenario?
Can you state the problem that you are answering without reference to the laser events?
 
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Pete said:
In 2inquisitive's scenario, there is only one location of interest in Earth's frame - not three.
The are three events of interest in Earth's frame - but they all happen at the same place.

In your scenario, you have three different locations in Earth's frame, hence it is a different problem.
Click to expand...
In my calculation, elapsed time of 4 sec is for two events occur at A. It is therefore the same as 2inquisitive's scenario. You should answer these questions concerning 2inquisitive's scenario:

1) Does point at x=-0.866c sec mean anything? If yes, what?
2) Does point at x=0 mean anything? If yes, what?
3) Does point at x=0.866c sec mean anything? If yes, what?

Pete said:
This is the core of the problem in 2inquisitive's scenario - in Earth's frame, the end points of the interval of interest are in the same place.
In your alternative scenarios, you are choosing events which are not in the same place in Earth's frame, which is why you get a different result.
Click to expand...

Which different result were referring to? I obtained that 4 seconds elapsed time too. In fact, you can compute for any points (B or C), so long you use the the same x you will get 4 second elapsed time. So, it is not true that my three points scenario is essencially not the same as 2inquisitive's scenario. You just did not recognize point A and C in 2inquisitive's scenario although these two points indeed exist and crucial.

Pete said:
If you want to change the scenario, make sure that the interval end points are in the same place in Earth's frame.
Click to expand...
When you do that, you do not compute meaningful elapsed time in the ship frame. Your 4 second elapsed time is not for time at one particular location in the ship (as I have pointed out much ealier, it was obtained by comparing time at two separate locations in the ship frame), so it is not "pure time", unlike 1 second elapsed time that I suggested.
 
I'm sorry, Paul... I read your previous post too fast, and drew incorrect conclusions.
Give me a minute, and I'll reply properly!
 
Paul T, there is one item of interest that you may have forgotten. I used
colored laser beams so they could be seen in transit, the path of the beam
could be seen by both frames. Impossible? Well, is it possible to see only the
points on Earth where the beams were emitted? There is no length contraction in this scenario because of the 90 degree angle perpendicular
to the spaceship trajectory, from the Earth based laser intersect the path
of the spaceship/reflecter. As per Special Relativity, you have to put the
spaceship in the rest frame and the Earth/laser in the moving frame when
using the spaceship frame of reference. The spaceship does not care WHEN
the laser was fired, only its apparent flight path, which is LONGER in the
resting spaceship/moving laser frame of reference. What is the speed of
light in any frame? Light has to travel 299,792,458 of that frame's meters
in one of that frame's seconds to remain constant. One second to travel
this longer path dictated by Special Relativity? Relativity of simultaneity
does not enter into the picture because we are calculating the DURATION
of an event, not WHEN it happened in each frame of reference.
 
Hi 2inquisitive,
Look at the space-time diagrams, and let me know when you understand them.
 
Pete said:
Hi 2inquisitive,
Look at the space-time diagrams, and let me know when you understand them.
Click to expand...

Pete, As the title of this thread reflects you started this thread as a challenge to my views. I have been setting back while you and Paul T debated 2inquisitive's question.

You are now challenging 2inquisitive and I feel it is time for me to reassert some reality.

[post=711958]Please reply[/post]
 
Hi Mac,
You might be interested in trying to understand the space-time diagrams yourself.
 
Pete said:
Hi Mac,
You might be interested in trying to understand the space-time diagrams yourself.
Click to expand...

I understand them fine. Why can't you answer the question. You made the challenge, I'm challenging you to provide a viable answer.
 
Hi Paul,
It looks like our only difference is over which two events we should use to determine the elapsed time in the spaceship frame.

I really think we're answering different questions.

You are answering:
"What time elapses on ship clocks during a two second time period in Earth's frame?"


I am answering:
"How long does it take in the ship's frame for two seconds to elapse on Earth clocks?"


We both agree that "1 second" is the answer to the first question, and "4 seconds" is the answer to the second, right?
 
Pete said:
Prove it.

I choose not to.
Our SR discussions were [post=697274]exhausted[/post] in this thread.
Click to expand...

I suggest that your refusal to answer the question means perhaps I understand things better than yourself and that your challenge of me has ended in you failing in that challenge.

You know you don't have the answer. No relativists does. You fail to respond with a correct solution you lose, I win.
 
MacM said:
You know you don't have the answer. No relativists does. You fail to respond with a correct solution you lose, I win.
Click to expand...
So not only does he not remember you showing him the answer... he's proud of it.
 
Paul,
Some things I should have covered in my last post:
1) Does point at x=-0.866c sec mean anything? If yes, what?
Click to expand...
In Earth's frame, the laser is fired when the ship is at this point.
In the Ship's frame, this event (the ship at that point) means nothing more than any other event on the ship's worldine.
2) Does point at x=0 mean anything? If yes, what?
Click to expand...
In Earth's frame, the laser strikes the ship at this point.
This is also a meaningful event in the ship's frame.
Does point at x=0.866c sec mean anything? If yes, what?
Click to expand...
In Earth's frame, the laser is received when the ship is at this point.
In the Ship's frame, this event (the ship at that point) means nothing more than any other event on the ship's worldine.

...it was obtained by comparing time at two separate locations in the ship frame), so it is not "pure time", unlike 1 second elapsed time that I suggested.
Click to expand...
Do you have a reference for your concept of "pure time"?
I suggest that a difference in times at different locations is not more nor less meaningful than a difference in times at the same location.

Consider the muons from cosmic rays experiment.
Is the Earth-frame time difference between muon birth and muon reception meaningful?
 
2inquisitive,

2inquisitive said:
Paul T, there is one item of interest that you may have forgotten. I used
colored laser beams so they could be seen in transit, the path of the beam
could be seen by both frames. Impossible? Well, is it possible to see only the
points on Earth where the beams were emitted? There is no length contraction in this scenario because of the 90 degree angle perpendicular
to the spaceship trajectory, from the Earth based laser intersect the path
of the spaceship/reflecter. As per Special Relativity, you have to put the
spaceship in the rest frame and the Earth/laser in the moving frame when
using the spaceship frame of reference. The spaceship does not care WHEN
the laser was fired, only its apparent flight path, which is LONGER in the
resting spaceship/moving laser frame of reference. What is the speed of
light in any frame? Light has to travel 299,792,458 of that frame's meters
in one of that frame's seconds to remain constant. One second to travel
this longer path dictated by Special Relativity? Relativity of simultaneity
does not enter into the picture because we are calculating the DURATION
of an event, not WHEN it happened in each frame of reference.
Click to expand...

Actually, this is not possible for a couple of reasons:

  • You see laser only if it hit something like wall, smog, dast, etc. You don't see laser passing through the empty space. Science fiction movie might had misled you.
  • You know that laser moves at the speed of light. So does image like the laser firing process or the "red laser" on the way to your spaceship. You don't see the incoming laser before it hit your ship. The moment you see the laser, you see the image of the laser firing too (assuming you have the appropriate equipment for that).
 
Pete,

Pete said:
“ 1) Does point at x=-0.866c sec mean anything? If yes, what? ”

In Earth's frame, the laser is fired when the ship is at this point.
In the Ship's frame, this event (the ship at that point) means nothing more than any other event on the ship's worldine.
Click to expand...

That's true. But, it does mean something for the ship observer. What is the elapsed time between the event of the ship passing this point and the ship hit by the laser, according to the ship observer? It is 0.5 second, right?

Pete said:
“ 2) Does point at x=0 mean anything? If yes, what? ”

In Earth's frame, the laser strikes the ship at this point.
This is also a meaningful event in the ship's frame.
Click to expand...

Yes.

Pete said:
“ Does point at x=0.866c sec mean anything? If yes, what? ”

In Earth's frame, the laser is received when the ship is at this point.
In the Ship's frame, this event (the ship at that point) means nothing more than any other event on the ship's worldine.
Click to expand...

Again, in the ship frame, this event coincide with the ship clock record another 0.5 sec time since it hit by the laser.

Pete said:
“ ...it was obtained by comparing time at two separate locations in the ship frame), so it is not "pure time", unlike 1 second elapsed time that I suggested. ”

Do you have a reference for your concept of "pure time"?
I suggest that a difference in times at different locations is not more nor less meaningful than a difference in times at the same location.
Click to expand...

Honestly, I am not aware of any reference to that concept. What I meant was "the elapsed time as recorded by one clock (or more generally clock at the same location)". When the time that you compared influence by space, then -- in my opinion -- it's no longer "pure time". That 4 seconds elapsed time that we were taling about all this time is one of that "not pure time".

Pete said:
Consider the muons from cosmic rays experiment.
Is the Earth-frame time difference between muon birth and muon reception meaningful?
Click to expand...

In this case, there is another reference frame, the moun frame. You have two sets of time differences, one on moun frame and one on earth frame. And there are two events, moun birth and moun reception. They both meaningful in their respective frame.
 
In this case, there is another reference frame, the moun frame. You have two sets of time differences, one on moun frame and one on earth frame. And there are two events, moun birth and moun reception. They both meaningful in their respective frame.
Click to expand...
But in the Earth frame, muon birth and moun reception are in different places.
Does that mean that the muon's extended life in Earth's frame isn't meaningful, because its life is not measured in "pure time"?
 
That's true. But, it does mean something for the ship observer. What is the elapsed time between the event of the ship passing this point and the ship hit by the laser, according to the ship observer? It is 0.5 second, right?
Click to expand...
Right.
But how does that make the event meaningful?
 
Persol said:
So not only does he not remember you showing him the answer... he's proud of it.
Click to expand...

FYI: There is indeed a differance in "AN" answer and a "Correct" answer. He, YOU, nor anyone here has actually provided a valid answer to these questions.

Can you do anything but bump your gums and pretend to know everything
 
Paul T said:
2inquisitive,



Actually, this is not possible for a couple of reasons:

  • You see laser only if it hit something like wall, smog, dast, etc. You don't see laser passing through the empty space. Science fiction movie might had misled you.
  • You know that laser moves at the speed of light. So does image like the laser firing process or the "red laser" on the way to your spaceship. You don't see the incoming laser before it hit your ship. The moment you see the laser, you see the image of the laser firing too (assuming you have the appropriate equipment for that).
Click to expand...

Of course I agree that the laser beam cannot be seen any more than a muon
in flight. My stupid mistake was thinking of the laser pointing device on my
telescope. The green beam 'seems' to go all way into space, but I now realize
the glow is only in the atmosphere.
 
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