Takes longer on the surface of the earth than it takes in geosynch orbit. Takes longer in a stationary frame than it takes in a jet plane.
If I count 1-mississippi exactly the same every time it is always the same. If I count different that changes everything. If a constant velocity shaft is rotating 100 revolutions per second, does it change rotational velocity if I get further form the Earth's core?
Motor Daddy, there is something more fundamental going on with, say, Cesium-133 than your watch. Your watch can break, be set incorrectly, or be generally inaccurate, while the rate of local Cesium oscillations are theoretically unchangeable (which is why it makes a convenient reference for defining a "second").
If you're near the shaft as it is moved, no. If you stay on the Earth while the shaft is moved away, YES! That's the whole point of Relativity. What you are thinking in your head as duration only has local significance in terms of being an absolute.
Oh, so you mean my watch can change rates and is generally not accurate, and cesium-133 is more accurate at staying consistent? So consistency is of high importance in maintaining the standard?
Correct. If we could replace the Cesium with a "perfect watch" it would be easier for people to understand. Take two of these perfect watches and put one on the top floor of a tall building while the other remains on the ground floor...we will discover that they become out of sync, even though neither one is broken! Also, the wearer of each watch would report that time passed normally for him, even though they were passing at differing rates.
So the rotational velocity of a constant velocity shaft is dependent on me to maintain its constant velocity? So is there any measurable force that makes the shaft change rotational velocity(ie, torque) when It travels away and I stay at the Earth's surface? If I then travel to it does it then retain its original rotational velocity?
How do you mean sync, like NY and Florida are sync'd? If it's 12:00:00 in NY is it not 12:00:00 in Florida?
If the shaft is moving relative to you, you will not measure the same velocity as you would if the shaft were stationary relative to you. From the shafts point of view, nothing changes.
If a constant velocity shaft moves away from me, what right do I have to say it changed its constant velocity, just because it is getting further away, and the light is becoming redshifted? Does the shaft experience a torque?
Take two identical watches. Set them to exactly the same time, when they are stationary relative to each other. (that's what is meant by sync) Take one watch on a fast moving jet and fly it around the world. When you then compare the two watches, they will have recorded a different amount of time.
No, it's not "really" changing. That's what is so confusing about Relativity. If your twin brother moved with the shaft and came back later he would tell you the velocity remained the same the whole time. You would be saying to him "no way bro, I could see you up there in space and it was moving FASTER!!" I mean sync'd like when Doc put his dog Einstein in the DeLorean at the shopping mall with a clock around his neck and it was "sync'd" with his clock. When the DeLorean pops back into existence all covered with ice the clocks are no longer sync'd.
It's not a question of right, it's a question of measurement. If the constant velocity shaft moves away from you, you will measure it as having a different velocity than you would if it were stationary relative to you. The distance it is from you is NOT a factor. It's relative velocity is a factor.
If the watches were in sync, and one was then flown and became out of sync, one watch's rate was altered. Two watches can't become out of sync unless the rate was different.
Yes the rate is different. That's the whole point. The watches haven't changed, they're still identical, and both are operating without a flaw. But the rate at which time passes is different.
Again, the light is becoming redshifted, do I not realize that and take that into consideration when doing my calculations? If one watch is not reading the same as the other, one watch 's rate was altered, hence, all bets are off.