1. part of the explanation is in your quote "PERCEIVE".
perception is not reality.
2. back to first principles: the speed of light is independent of its origin.
it would require 4 yrs from that location.
Who are all these people that think that Special Relativity is only some kind of illusion? Who is spreading this garb? If you perceive something, it becomes a part of your reality.

JesseM
windscar, you're presenting as fact a lot of totally false information about how relativity works--have you read the Physics Forums Global Guidelines policy on "overly speculative posts", or the IMPORTANT! Read before posting message at the top of this forum? For example:
Actually according to the twin paradox, both observes would see each other as a black hole at the same time. So being sucked into a singularity would be relative to how far the object's passed by each other. They then wouldn't know wich one was the singularity untill on accelerated to the speed of the other object. And then this acceleration would allow them to observe themselves being sucked into their own singularity, since acceleration is equal to gravitation.
This is nonsense--right now we are travelling at some extremely large fraction of light speed in some frame, so why don't we see ourselves being sucked into a singularity? In any case, if you actually calculate the answer in general relativity instead of just making baseless speculations, the answer you get is that a high relativistic mass due to large velocity does not create a black hole if the object's mass in its own frame is not enough to create one. It is certainly not true that "according to the twin paradox" each twin sees the other one become a black hole!
windscar said:
Then again, anyone trying to approach the speed of light would have to accelerate and this would seperate them from observers traveling at a constant speed, hence allowing them to distinctivly observe themselves becomeing a singularity.
Again, there are perfectly valid frames in which any object you like--including the Earth--is already moving arbitrarily close to light speed, no acceleration necessary.

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Pervect and phyti, thank you for your insights. I think I got what I wanted on my original question. JesseM gave me the clue I needed – I didn’t realize that according to my calculations the clocks on AC are actually ahead of similar calculations done on Earth; and that the reason that I see the clocks rolling fast during my journey is due to Doppler effect.

But now, here’s still the mind bending twist. If you can reconcile the following, things would be a lot clearer. I’m going to try to remove the Doppler effect from the picture, by setting up clocks all along my journey to AC and I’ll be “looking” at these clocks as I go by them.

So, AC and Earth are in the same frame. I synchronize all the clocks in this frame and I set a whole bunch of them on the way from Earth to AC. It’s the year 2000 in this frame, it takes light 4 years to reach Earth from AC. I fly by Earth at almost c. You stand on Earth. Both you and I see the image of the clock from AC showing 1996. According to your calculations it’s 2000 on AC, according to my calculations it’s year 2003.9 on AC. So, by the time I reach AC in 0.1 years, the image of the clock I see from AC will go from 1996 to 2004, in 0.1 years of my time. That is due to Doppler effect, as you assert. That’s good; but now, as I fly, instead of looking at the image of the clock from AC, I will be looking at the clocks set up all the way from Earth to AC. They will go from 2000 to 2004 in 0.1 years of my time. THAT is not due to Doppler effect! So, according to SR, the clocks in the AC-Earth frame should be ticking slower, time should dilate. But as I watch these clocks on my journey, they roll a lot faster than mine. Again, these clocks are all synchronized, they show real time of the frame, and since I look at them at very close proximity, there is no need to adjust for delay to receive their images. So, if you ask me, time does not dilate here, it runs faster realtive to my clock. What’s the deal here?

Thanks,

Pavel

I was using the twin paradox to draw a new conclusion and if the first conconlusion (traveling close to light create's a singularity) was correct than this one would also be correct. Have you read my forum about Delta T > To? I think I understand the basis of relativity and I can show how your change in time just happens to be the inverse of Einsteins origianal equation. Also, we are moveing close to the speed of light relative to other object's in the universe. But, this is due to the expansion of the universe itself. We do not feel any affect from the acceleration due to this expansion.

The expanion of the universe can be described as being on the surface of a balloon that is being blown up. You can draw different points on the balloon and blow it up and watch them get farther and farther away from them. But they all stay on the same points of the balloon. They are moving away from each other with a constant speed in a higher dimension. It is as if they are in constant free fall, traveling a straight line in a dimension, which reduces the force of acceleration felt to zero. That is why we don't feel any forces on ourselves or on the other object with relative acceleration. When looking at the expansion from this stand point, all object's with relative motion due to the expansion of the universe are moveing at a constant velocity relative to each other through a higher dimension. All of these objects are not moveing relative to space-time itself and are at rest relative to the universe itself. (all the point's stay on the same place on the balloon). But, if an object traveling close to the speed of light relative to us in a far away galaxy, started to approach us and their velocity was close to the speed of light relative to the velocity of the expansion of the universe in our area, the object would undergo all the affect's of relativity. Therefore, an object traveling at high velocity due to the expansion of space-time, does not prove anything about the conditions of object's traveling close to the speed of light. Furthermore, the "event horizon" of the universe turns out to be the distance away from us that the expansion of the universe comes out to be the speed of light. As we do not see any objects that are farther out than this.

JesseM
So, AC and Earth are in the same frame. I synchronize all the clocks in this frame and I set a whole bunch of them on the way from Earth to AC. It’s the year 2000 in this frame, it takes light 4 years to reach Earth from AC. I fly by Earth at almost c. You stand on Earth. Both you and I see the image of the clock from AC showing 1996. According to your calculations it’s 2000 on AC, according to my calculations it’s year 2003.9 on AC.
OK, in this case you must be traveling at 0.975c.
Pavel said:
So, by the time I reach AC in 0.1 years, the image of the clock I see from AC will go from 1996 to 2004, in 0.1 years of my time.
Not exactly. If you're going at 0.975c relative to Earth and Alpha Centauri, then the distance between Earth and Alpha Centauri is shrunk from 4 light years to $$4 * \sqrt{1 - 0.975^2}$$ = 4 * 0.2222 = 0.8888 light years, and since Alpha Centauri is moving towards you at 0.975c, it takes you 0.9116 years to reach it in your frame. However, in your frame Alpha Centauri's clock is also slowed down by a factor of 0.2222, so it only advances forward by 0.9116 years * 0.2222 = 0.2026 years, so it will read 2003.9 + 0.2026 = 2004.1 when you arrive. This matches the prediction of the Earth frame, where you travel 4 light years at 0.975c and therefore take 4.1 years to reach Alpha Centauri, so if you leave in 2000 you'll arrive in 2004.1
Pavel said:
That is due to Doppler effect, as you assert.
Well, the Doppler effect accounts for the fact that if I look through a telescope I'll see the clock at Alpha Centauri ticking fast, going from 1996 to 2004.1 in only 0.9116 years according to my own clock. But the fact that Alpha Centauri's clock reads 2004.1 when I arrive can also be deduced by ignoring what I see and just calculating what is true in my frame, as I did above, where Alpha Centauri's clock was ticking slower than mine throughout the journey but it started off at 2003.9 as I was passing Earth.
Pavel said:
That’s good; but now, as I fly, instead of looking at the image of the clock from AC, I will be looking at the clocks set up all the way from Earth to AC. They will go from 2000 to 2004 in 0.1 years of my time. THAT is not due to Doppler effect!

You might like to look at my thread An illustration of relativity with rulers and clocks, where I posted some diagrams of a situation much like this, where two rulers with clocks placed along them at regular intervals were passing alongside each other, and you can see how it's consistent for each ruler to measure the other ruler's clocks as both slowed down closer together relative to its own, as measured in the ruler's own rest frame.

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Awesome reply, Jesse! This is exactly what I was looking for - a detailed and consistent account of what is going on. I need to go now disect and chew on it for a little until it's all digested in my head. I appreciate your time and effort illustrating the math too.

Pavel.

a slight correction

quote from jesseM post #30
"However, in your frame Alpha Centauri's clock is also slowed down by a factor of 0.2222, so it only advances forward by 0.9116 years * 0.2222 = 0.2026 years, so it will read 2003.9 + 0.2026 = 2004.1 when you arrive."

Let 'E' be the earth observer and 'S' be the ship observer.
The trip takes 4/.975=4.10 yrs earth time.
The S clock reads (4.1)*sqrt(1-.975^2)=.88 yr elapsed time to AC.
S sees the elapsed time of the AC clock from 1996 to 2004.1 as 8.1 yrs.
E calculates the doppler factor as (8.1)/(4.1)=1.975 (=1+v/c)
With time dilation, S calculates the doppler factor as (8.1)/(.88)=9.20.
S perceives the AC clock to be running faster, not slower.
Reasoning if S's clock has slowed, a unit of time on the ship is longer
than the same unit on earth, and will contain more events.
For S world events outside the ship are happening at a faster rate.
Time dilation depends on your speed relative to light.
The doppler effect depends on your speed and direction relative to a light
signal.

JesseM
quote from jesseM post #30
"However, in your frame Alpha Centauri's clock is also slowed down by a factor of 0.2222, so it only advances forward by 0.9116 years * 0.2222 = 0.2026 years, so it will read 2003.9 + 0.2026 = 2004.1 when you arrive."

Let 'E' be the earth observer and 'S' be the ship observer.
The trip takes 4/.975=4.10 yrs earth time.
The S clock reads (4.1)*sqrt(1-.975^2)=.88 yr elapsed time to AC.
S sees the elapsed time of the AC clock from 1996 to 2004.1 as 8.1 yrs.
E calculates the doppler factor as (8.1)/(4.1)=1.975 (=1+v/c)
With time dilation, S calculates the doppler factor as (8.1)/(.88)=9.20.
S perceives the AC clock to be running faster, not slower.
Yes, but in the sentence you quoted I was referring to how fast Alpha Centauri's clock was ticking in the ship observer's rest frame, which is different from how fast the ship observer sees Alpha Centauri's clock ticking if he's watching it through a telescope. They are two different concepts--what happens in your frame is what you calculate after you have factored out delays due to the finite speed of light (for example, in the Earth observer's frame he sees AC's clock reading 1996 when his own clock reads 2000, but he knows the light took 4 years to reach him, so he concludes that the events he is currently seeing happened at the same time-coordinate of his own clock reading 1996, 4 years in the past).

"Time dilation" in relativity is understood in terms of how fast a clock is calculated to be ticking in your own rest frame--if the clock is moving this is always slower, never faster than your own clock--not based on how fast you see the clock ticking.

not convinced

S sees the AC clock at 1996 as he leaves earth.
S sees the AC clock at 2004.1 when he arrives at AC.
S sees the elaplsed time on his (S) clock as .88 yrs.
S sees nothing wrong with his clock.
It is running slower only to the other observers but not to him.
Even if S subtracts 4 yrs for the light to reach earth,
there are 4.1 yrs elapsed on the AC clock, which he sees during his trip of .88 yrs. His time dilation increases the doppler effect by a signicant amount.
If he looked back at earth, he would see events moving very slowly because of exteme doppler and time dilation.
Look at the numbers, they tell the story.

Hurkyl
Staff Emeritus
Gold Member
S sees the AC clock at 1996 as he leaves earth.
S sees the AC clock at 2004.1 when he arrives at AC.
S sees the elaplsed time on his (S) clock as .88 yrs.
S sees nothing wrong with his clock.
It is running slower only to the other observers but not to him.
Even if S subtracts 4 yrs for the light to reach earth,
there are 4.1 yrs elapsed on the AC clock, which he sees during his trip of .88 yrs. His time dilation increases the doppler effect by a signicant amount.
If he looked back at earth, he would see events moving very slowly because of exteme doppler and time dilation.
Look at the numbers, they tell the story.
If you do all of the calculations in Earth's rest frame, you will get exactly what Earth's rest frame describes.

Have you tried doing any of the calculations in S's rest frame?

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JesseM
S sees the AC clock at 1996 as he leaves earth.
S sees the AC clock at 2004.1 when he arrives at AC.
S sees the elaplsed time on his (S) clock as .88 yrs.
S sees nothing wrong with his clock.
It is running slower only to the other observers but not to him.
Even if S subtracts 4 yrs for the light to reach earth
S wouldn't subtract 4 yrs. In S's frame, the distance between Earth and AC is not 4 light years, it's shorter by a factor of $$\sqrt{1 - v^2/c^2}$$ due to Lorentz contraction. Also, in S's frame AC is moving towards him, so its distance when the light was emitted is different from its distance when the light actually reaches him.

To make the calculations a little easier, let's say S is traveling at 0.8c relative to Earth and Alpha Centauri, rather than 0.975c as in my previous calculations. This means that in S's frame, the distance between E and AC is not 4 light years but $$4 * \sqrt{1 - 0.8^2}$$ = 4*0.6 = 2.4 light years. So how far away would AC have had to have been when it emitted the light, in order for it to be 2.4 light years away at the moment the light reaches S? If the distance is x, and the time the light took to reach S is t, then we have:

x - ct = 0

and in this time AC moved closer a distance of 0.8c * t:

x - 0.8ct = 2.4

Solving for t:

0.8ct + 2.4 = ct
2.4 = 0.2ct
t = 12 years

Which means x = 12 light years. So in S's frame, if he sees Alpha Centauri's clock read 1996 as he passes Earth, he concludes this event actually happened 12 years earlier in his frame. And S reaches AC 2.4 ly/0.8c = 3 years after passing Earth in his frame, at which point AC's clock reads 2005. So, S concludes that it took AC's clock 12+3 = 15 years to advance forward 9 years from 1996 to 2005. This means that in S's frame, AC's clock is slowed down by a factor of 9/15 = 0.6, which is exactly the time dilation factor $$\sqrt{1 - v^2/c^2}$$

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Which means x = 12 light years. So in S's frame, if he sees Alpha Centauri's clock read 1996 as he passes Earth, he concludes this event actually happened 12 years earlier in his frame. And S reaches AC 2.4 ly/0.8c = 3 years after passing Earth in his frame, at which point AC's clock reads 2005. So, S concludes that it took AC's clock 12+3 = 15 years to advance forward 9 years from 1996 to 2005. This means that in S's frame, AC's clock is slowed down by a factor of 9/15 = 0.6, which is exactly the time dilation factor $$\sqrt{1 - v^2/c^2}$$

Here you are adding undilated time (12) to dilated time (3).
It appears you have both S and A moving at .8c.

You state S is moving at .8c therefore E and A are at rest.
E is still 4 ly distant from A .
In E-time A's signal took 4 yrs to arrive at 2000.
In E-time S moved (.8)*4=3.2 ly.
S would be 4+3.2=7.2 ly from A according to E. (4.32 ly for S).
To calculate time or distance for S, multiply E's values by .6, the time dilation
factor. It's simpler and more consistent.
E and S synchronize at 2000.
Subtract (.6)*4=2.4 yrs to get the S-time for the origin of A's signal, 1997.6
In E-time S takes 4/(.8)=5 yrs to reach A, 2005.
The S-time would be (.6)*5=3 yrs, 2003.
For S, (4.32 ly)/(.8)=5.4 yrs for A to meet him.
Everyone is in the right place at the right time for each viewpoint.

Once again S sees 9 yrs on A happen during his 5.4 yrs.,
i.e. he perceives A's clock running faster than his.

JesseM
Here you are adding undilated time (12) to dilated time (3).
No I'm not, they're both in S's frame. In S's frame, the Earth and Alpha Centauri are 2.4 light years apart, and both are approaching S at a constant speed of 0.8c. At t=1988 in S's frame, Alpha Centauri is 12 light years away, and the clock at Alpha Centauri reads a time of 1996. 12 years later at t=2000 in S's frame, the light signal is just reaching S, and Alpha Centauri has gotten closer by (12 years)*(0.8c) = 9.6 light years, so it's now at a distance of 12-9.6=2.4 light years away (which means S must be passing Earth at that moment, since Earth and Alpha Centauri are 2.4 light years apart in S's frame). Then 3 years later at t=2003 in S's frame, Alpha Centauri has gotten closer by (3 years)*(0.8c) = 2.4 light years, so this must be when S passes Alpha Centauri. So, 12+3=15 years is the time it takes in S's frame for Alpha Centauri to get from a distance of 12 light years away (at the moment it emits a light signal which reaches S as S is passing Earth in 2000) to a distance of 0 from S.
phyti said:
It appears you have both S and A moving at .8c.
In S's rest frame, S is of course at rest, and Earth and Alpha Centauri are approaching it at 0.8c. In the Earth/Alpha Centauri rest frame, Earth and Alpha Centauri are at rest, and S is moving towards Alpha Centauri at 0.8c.
phyti said:
You state S is moving at .8c therefore E and A are at rest.
Only in the Earth/Alpha Centauri rest frame, not in S's frame.
phyti said:
E is still 4 ly distant from A .
Only in the E/AC frame. But my above calculations, showing 15 years between the time AC's clock reads 1996 and the time S passes AC, were done from the point of view of S's frame. In S's frame the distance from Earth to Alpha Centauri is shorter due to length contraction, they're just 2.4 light years apart.
phyti said:
In E-time A's signal took 4 yrs to arrive at 2000.
Yes, that's true in the Earth's frame, but not in S's frame.
phyti said:
In E-time S moved (.8)*4=3.2 ly.
Sure, in the E/AC rest frame. But this frame defines simultaneity differently than the S frame, so they disagree on where S was on his journey "at the same time" that AC's clock was reading 1996 and sending a signal in the direction of Earth. In S's frame, this event was simultaneous with the event of S's own clock reading a date of 1988; but in the E/AC frame, this event was simultaneous with the event of S's own clock reading a date of 1997.6 (so that 4 years later in the E/AC frame, S's clock only advanced forward by 2.4 years due to time dilation, leading to the correct prediction that S's clock reads 1997.6+2.4=2000 at the moment S and the signal are both reachign Earth).
phyti said:
S would be 4+3.2=7.2 ly from A according to E. (4.32 ly for S).
No, it doesn't work that way--you're forgetting about the relativity of simultaneity. It's true that in E's frame, S was 7.2 ly from AC at the moment that S's clock read 1997.6, and according to E's definition of simultaneity, the event of S's clock reading 1997.6 happened "at the same time" as the event of AC's clock reading 1996 and sending a signal towards Earth. And in S's frame, it is also true that at the moment his own clock read 1997.6, he was a distance of 4.32 ly from Earth. But it is not true that in S's frame the event of his own clock reading 1997.6 is simultaneous with the event of AC's clock reading 1996--you understand that different frames always disagree about the simultaneity of distant events, right? In S's frame, the event of AC's clock reading 1996 is simultaneous with the event of his own clock reading 1988, at which point he is 12 light years away from AC.
phyti said:
To calculate time or distance for S, multiply E's values by .6, the time dilation
factor. It's simpler and more consistent.
Again, you're forgetting the relativity of simultaneity. You can only use the time dilation factor this way if you have the time interval between two events which happened at the same position in E's frame, at different times (like two different readings of a clock at rest on Earth), and you want to know the time interval between these same two events in S's frame. But if you're talking about the time interval in E's frame between events that do not have the same spatial coordinates, you can't just multiply by the time dilation factor to find the time between them in S's frame.
phyti said:
E and S synchronize at 2000.
It's simpler to say that S's clock was presynchronized in just the right way so it reads a date of 2000 at the moment it passes Earth and Earth's clock also reads 2000; that way you can talk about the time on S's clock at earlier times, like the fact that in S's frame, S's clock reads 1988 "at the same time" that AC's clock reads 1996, or that in E's frame, S's clock reads 1997.6 "at the same time" that AC's clock reads 1996.
phyti said:
Subtract (.6)*4=2.4 yrs to get the S-time for the origin of A's signal, 1997.6
In E-time S takes 4/(.8)=5 yrs to reach A, 2005.
The S-time would be (.6)*5=3 yrs, 2003.
For S, (4.32 ly)/(.8)=5.4 yrs for A to meet him.
Everyone is in the right place at the right time for each viewpoint.
Once again, you must remember the relativity of simultaneity. A simple way of showing your numbers must be wrong is that they'd imply that in S's frame, the speed of light is not c! If you think that in S's frame his own clock read 1997.6 at the moment the signal was sent, then since in S's rest frame the Earth is moving towards him at 0.8c, the Earth must have moved a distance of 2.4*0.8=1.92 light years between 1997.6 and 2000 in S's frame. Since the distance between E and AC is 2.4 light years in S's frame, then if the Earth was 1.92 light years away at the moment the signal was sent, AC must have been 1.92 + 2.4 = 4.32 light years away at the moment the signal was sent, in S's frame. And since the signal reached S in 2000, this would mean that in the 2.4 years between 1997.6 and 2000, the signal would have covered a distance of 4.32 light years, moving at 4.32/2.4 = 1.8c! Obviously this cannot be correct. With my numbers there is no such problem--the signal was sent in 1988 in S's frame, at which point AC was 12 light-years away, and then 12 years later in 2000, the signal reached S, moving at 12/12 = 1c.

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OK Jesse:
After using a space-time diagram to construct the sequence of events you describe for S,the difference of opinion is obvious. I agree with your results.
You are calculating the values for the simultaneous (it was mentioned but I missed it) timesand locations for an observer,
whereas I'm calculating the values for the experiences of the observers.

"Time dilation" in relativity is understood in terms of how fast a clock is calculated to be ticking in your own rest frame--if the clock is moving this is always slower, never faster than your own clock--not based on how fast you see the clock ticking.
Not if they are approaching each other. IF nothing else you have to consider the doppler effect.

Although I agree with your calculations, I do not agree with your conclusions.
Based on calculations you conclude that (2003-1988) 15 yrs for S pass
while (2005-1996) 9 yrs pass for A, therefore A's clock runs slower than S's clock.
1.
Both experience the same amount of time, but the rate of time depends on their motion.
If S's clock cycle was stretched by a factor of 1/(.6), then converting it to non-dilated time is just (.6)*15=9 yrs. the same as A.
We know by logic and reason the motion of S cannot alter the
rate of distant clocks, so it all reduces to perception. The predictions of the theory have to match perception/experience.
2.
Knowledge is historical/after the fact. S can in 1988 calculate and assume A sends a signal in 1996, but he will not know it happened until 2000!
S only experiences A's 9 year span, from 2000 to 2003 on his clock, a 3 to 1 ratio.
S sees 9 A yrs in his 3 yrs. Since both have uniform motion, the ratio is constant while S and A appoach each other.
His experience does not support his prediction.
How does he explain this?

A's clock appears to run faster not slower while S moves towards A,
and slower as S moves away from A.

You can calculate simultaneous positions but you can't experience them.

JesseM
Not if they are approaching each other. IF nothing else you have to consider the doppler effect.
Not if you are talking about what is true in each observer's frame, rather than what they see with their eyes. Do you understand that frames are a very basic concept in SR, and that most calculations about the dynamics of objects are done from the perspective of a particular reference frame?
phyti said:
Although I agree with your calculations, I do not agree with your conclusions.
Based on calculations you conclude that (2003-1988) 15 yrs for S pass
while (2005-1996) 9 yrs pass for A, therefore A's clock runs slower than S's clock.
That statement does not accurately describe my position. What I would say is that in S's frame, 15 years pass between the event of A's clock reading 1996, and the event of S passing next to A; while in A's frame, 9 years pass between the event of A's clock reading 1996 and S passing next to A. This would not lead me to conclude that "therefore A's clock runs slower than S's clock", however--in relativity all questions of which clock runs slower are frame-dependent, at least as long as both clocks move inertially. After all, it is equally true that in A's frame, 15 years pass between the event of S's clock reading 1994, and the event of S passing A; while in S's frame, only 9 years pass between those same two events. Would this lead you to conclude that S's clock runs slower than A's clock?

What is true is that if two events take place at the same position-coordinate in one frame (like two events on the worldline of a given observer, as in both the examples above), then the time between the events in that frame will always be smaller than the time between the same two events in any other frame.
phyti said:
Both experience the same amount of time
Experience the same amount of time between what and what?
phyti said:
We know by logic and reason the motion of S cannot alter the
rate of distant clocks, so it all reduces to perception. The predictions of the theory have to match perception/experience.
Not sure what you mean. Time dilation is not some type of illusion--if I traveled away from the Earth at 0.8c for 4.5 years according to my own clock, then rapidly turned around and traveled back towards Earth at 0.8c for another 4.5 years, when I returned to Earth I would have aged by 9 years, while everyone on Earth would have aged 15 years.
phyti said:
Knowledge is historical/after the fact. S can in 1988 calculate and assume A sends a signal in 1996, but he will not know it happened until 2000!
Sure, but this is just how reference frames work in relativity, of course you can't assign a time-coordinate to an event until you actually know about it.
phyti said:
S only experiences A's 9 year span, from 2000 to 2003 on his clock, a 3 to 1 ratio.
S sees 9 A yrs in his 3 yrs. Since both have uniform motion, the ratio is constant while S and A appoach each other.
Why do you think it contradicts his calculations? His calculations were about when events actually happened in his frame, not when he saw them. In S's frame, A was 12 light-years away when its clock read 1996 (indeed, if S was sitting at one end of a measuring-rod 12 light years long, then in 2000 when he looks through his telescope and sees A's clock reading 1996, he will also see A passing the other end of his measuring-rod at the same moment). So of course if light moves at a speed of c in his frame, he will not see A's clock read 1996 until 12 years later!
phyti said:
His experience does not support his prediction.
How does he explain this?
The prediction is not about when he sees events, so you are inventing a strawman here. If you'd like the "prediction" to be a bit more physical, then again imagine S is sitting on one end of a rod 12 light-years long, and further imagine that at the other end of the rod is a clock which has been synchronized with his own using the Einstein clock synchronization convention (which means that if a flash was set off at the center of the rod, his clock and the clock at the other end would read the same time at the moment the light from the flash reached them). In this case, in 2000 when he looks through his telescope and sees Alpha Centauri's clock reading 1996, he will also see Alpha Centauri passing next to the other end of the rod, and he will see that the clock sitting at that end reads 1988.
phyti said:
A's clock appears to run faster not slower while S moves towards A,
and slower as S moves away from A.
If you define "appears" in terms of what is seen, you're correct. However, it is much more useful to calculate things from the perspective of a frame of reference in SR (just as it is in Newtonian mechanics), almost all the equations you see in an SR textbook are written under the assumption that position and time coordinates are those of some inertial frame (again, just like Newtonian mechanics). And if you like, you're free to think of these coordinates in terms of the local physical readings on rulers at rest relative to you and clocks at different points on the rulers which are synchronized with your own using the Einstein synchronization convention, as in my example above where a synchronized clock on a ruler 12 light-years long reads a time of 1988 at the moment it is passing next to A and A's clock reads 1996.