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## Speed of the light and dilation of time

 Quote by uniqueland So then basically I would see the train station going by at what the amount of revolutions would be if I were traveling at just under light speed for 30 years, even though my watch onboard my light speed train would only show that I had been on the train for a few hours so it would seem to me as if the train were traveling a much higher speed than just light speed, figuring that it would take me one whole light second to make say around 6 revolutions around the planet.
No, it won't seem to you that you are traveling faster than light speed. As I said in post #13, you're overlooking length contration. And HallsofIvy explained this way back in post #4:
 Quote by HallsofIvy Any integer result, such as the number of times you have gone around the earth is invariant. Both you and son would calculate the circumference of the earth, in your respective frames of reference, he dividing by your speed relative to the earth, you dividing by the speed of the earth relative to you, to find the number of rotations you have made. But, comparing your calculations, you would have used the contracted distance (relative to his distance) and the dilated time (compared to his time). Since the same Lorenz contraction factor is used in both distance and time, that will cancel giving you both the same result.
In other words, you would determine that the distance around the earth was much shorter than your son would determine it to be.
 Quote by uniqueland but I would "see" (putting aside for my example any consideration that it would be impossible for me to actually make out what I would be seeing even without my view outside being accelerated for me) that train station passing by thousands or millions of times greater per second than just 6 because everything outside would be accelerated proportionate to the degree to which time slowed down for me. Conversely, if my son at the train station were looking at a monitor hooked up to a video camera focused on me onboard the train, it would seem as if I was frozen in place since it might take me a month just to blink my eyes from his perspective. Is that correct?
Yes.

 Then my son at the train station viewing an webcam onboard my light speed train would see me as all but frozen with it taking maybe a month for me to make a smile whereas me viewing a webcam of the train station would show people moving around in a complete blur because I would see them moving speeded up by thousands of times the actual earth speed. Right?
 Recognitions: Gold Member Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.

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 Quote by uniqueland T 08:28 AM uniqueland This Message is Moderated If I am on this train traveling at 99.99% light speed around the circumference of the earth for 30 earth years. My watch, which is synchronized to count each revolution my train makes around the earth, will show those revolutions blurring by at a much faster rate than the earth based revolution counter display outside the train station because it will count 30 years worth of revolutions in just say one day to me
As observed by whom? To you your watch moves normally, not "blurring by". But, looking out the window, you would see the earth based counter moving very very slowly.

 When I get off the train, my watch will match the number of revolutions shown on the revolution counter outside the train station but the time on my watch might say that only a day has passed when 30 years have passed outside my train. If I was skyping with my wife, my "viewing" would be one big blur, because I would be seeing her over 30 years but speeded up to all be shown in one day to me and she would see me as practically a frozen image because I would be going by in only one day to her 30 years. It might take a year for her to see my mouth move. Correct?

So then, if we could ever figure out how to practically travel just under the speed of light, we would be able to theoretically travel to any spot in the universe we chose with the time it took to get to that spot no longer being a factor, since, while it may take thousands or even millions of years to get there, we could do it in the same generation since we would not age practically at all, at least in theory anyway. Correct?

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 Quote by ghwellsjr Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.
This is false, time dilation is symmetrical, each observer sees the other in slow motion.

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 Quote by ghwellsjr Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.
This is false, time dilation is symmetrical, each observer sees the other in slow motion.
Apparently you did not read the first post.

 Quote by ghwellsjr Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.
 Quote by ghwellsjr Apparently you did not read the first post.
What does it have to do with your erroneous claim? Each observer sees the other observer moving slow(ly), time dilation is mutual (symmetrical).

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 Quote by GAsahi What does it have to do with your erroneous claim? Each observer sees the other observe moving slow(ly), time dilation is mutual (symmetrical).
It's not my claim and it's not erroneous. Einstein made the claim in his 1905 paper introducing Special Relativity near the end of §4 that two clocks starting together and one of them taking a circular path arriving back at the other clock will accumulate less time:
 If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be ½tv²/c² second slow.

 Quote by ghwellsjr It's not my claim and it's not erroneous. Einstein made the claim in his 1905 paper introducing Special Relativity near the end of §4 that two clocks starting together and one of them taking a circular path arriving back at the other clock will accumulate less time:
You are confused, Einstein's claim is about elapsed time and requires that the observers are reunited, you made your claim about two observers in uniform relative motion observing each other's motion. Do you understand the difference between the two?

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 Quote by GAsahi You are confused, Einstein's claim is about elapsed time and requires that the observers are reunited, you made your claim about two observers in uniform relative motion observing each other's motion. Do you understand the difference between the two?
Of course I understand the difference but you are not reading the posts in this thread. I was not commenting about two observers in uniform relative motion but rather it was about the scenario in the first post regarding a high speed train circling around the earth many times and passing through a train station.

Here's the question I was answering from post #19:
 Quote by uniqueland Then my son at the train station viewing an webcam onboard my light speed train would see me as all but frozen with it taking maybe a month for me to make a smile whereas me viewing a webcam of the train station would show people moving around in a complete blur because I would see them moving speeded up by thousands of times the actual earth speed. Right?
And here's my answer from post #20:
 Quote by ghwellsjr Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.

 Quote by ghwellsjr Of course I understand the difference but you are not reading the posts in this thread. I was not commenting about two observers in uniform relative motion but rather it was about the scenario in the first post regarding a high speed train circling around the earth many times and passing through a train station. Here's the question I was answering from post #19: And here's my answer from post #20:
Same difference, you are mixing up mutual time dilation (the way the two observers see each other moving) with the calculation of total elapsed time (the circular variant of the twins paradox).

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 Quote by GAsahi Same difference, you are mixing up mutual time dilation (the way the two observers see each other moving) with the calculation of total elapsed time (the circular variant of the twins paradox).
You are mixing up time dilation (which is frame dependent and arbitrary) with what observers see (which is relativistic Doppler and not dependent on any arbitrarily selected frame).

Einstein went on to describe:
 Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.
Einstein was talking about the slow transport of a clock but in this thread it is a fast transport so the only difference is that the clock zipping around the equator will be ticking much more slowly than one at the equator.

Now, if each clock could actually see the other one (without the curvature of the earth getting in the way), then the zipping one would see the one at the pole as ticking much more quickly all the time and the one at the pole would see the one zipping around as ticking much more slowly all the time. In the earth's inertial frame, all the time dilation occurs for the zipping clock and it is constant. You can pick a non-inertial frame in which the zipping clock is at rest and the clock at the pole is running faster, not slower, and it is constant. Since the distance between the pole clock and the zipping clock is constant, the relativistic Doppler and the "time dilation" can be made the same.

But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".

I believe uniqueland wanted to avoid all these complications, especially of not being able to see the other one during the entire orbit and so he introduced a couple webcams. Now it will depend on where the mutual antenna is located as to how much fluctuation would be seen by each observer. I mentally put this antenna above the earth's pole to eliminate any fluctuation but to be more general, I allowed for the antenna or antennas to be located anywhere and so I included "on average".

I'm really sorry that you had to make me go into all these gory details as they have nothing to do with what uniqueland is asking about and I hope it doesn't undo all the work I have been doing in trying to help him understand the answers to his questions.

 Quote by ghwellsjr You are mixing up time dilation (which is frame dependent and arbitrary) with what observers see (which is relativistic Doppler and not dependent on any arbitrarily selected frame).
1.The two observers are in motion wrt each other, therefore, the observe mutual time dilation (i.e. a slowdown in measured clock rate) when, according to you, they "look at each other's webcam".
2. It is when they get reunited that they notice the discrepancy on total elapsed time. The observer that had the longest spacetime trip has the lowest elapsed time.
3. You are freely mixing the two different effects.

 Quote by ghwellsjr But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".
This is further compounding the confusion, the Doppler effect on frequency follows a different set of rules in accelerated frames. The revolving observer is continuously accelerating , so you cannot extrapolate from the Doppler effect in inertial frames. The only thing that you got right is the fact that, quantitatively, the observers notice a mutual blueshift when they approach each other. When they are separating from each other, they are experiencing a mutual redshift.
In both cases the effect is mutual i.e. you cannot have:

 Quote by ghwellsjr Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual earth speed", just thousands of times your speed.

 Recognitions: Gold Member Do you agree with Einstein's claim that if we had one inertial clock and a second non-inertial clock always equidistant from the first one but traveling at some speed in a circle, then the inertial clock will see the non-inertial clock ticking slower than it is ticking and the non-inertial clock will see the inertial clock ticking faster than it is ticking?