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Old 09-November-2008, 12:32 AM
Digix Digix is offline
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Default Twin paradox (relativity)

I ran into some problems on this famous paradox.
if you move fast time is supposed to slow down.

so lets say what will happen in this case:

I have one stationary clock and another one is sent some to distance away and it returns back to same point.
later we compare them and do they show different time ?

Some Time ago I expected that they should show different time, but now I think that they will show exactly same time like no relativity ever existed.
because all effects will cancel if you return to the same point.
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Old 09-November-2008, 02:05 AM
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Some Time ago I expected that they should show different time, but now I think that they will show exactly same time like no relativity ever existed.
because all effects will cancel if you return to the same point.
I'm not sure why you'd think that. The effect, as you've described it, is special relativity, and Einstein described the effect in his first paper in 1905. They will not show the same time, at least at the level of the effect--and experiments have agreed.
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Old 09-November-2008, 02:31 AM
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Default The Same Point

The Same Point is the point here. You are considering a return to the same point. But it hasn't, it is at a different point in the four dimensional sense. Both clocks have left that original point "somewhere back there" on their individual time lines. The usual set up is for the clock that is "stationary" to be assigned coordinate time because it will have the longest proper time possible for the twins scenario you describe. Whilst the traveled clock will have less time elapsed meaning it was observed to be time-dilated by the "stationary clock" Also SR is largely about speed, a scalar, so the direction relative to you doesn't matter, so a difference is not cancelled out.
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Old 09-November-2008, 02:33 AM
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The one that have changed reference frames(the one that moved) will be behind the one that stayed behind.

If you have clock a and b, and they start in the same reference frame(same speed and position), one accelerates at 1g for a week, does nothing for a week, then decelerates for a week. Now that clock will be behind the one left behind..

Now if the OTHER clock does the same thing, then when they meet up they will show the same time. But if the one that traveled first travels back, it will still be behind.
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Old 09-November-2008, 03:06 AM
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Originally Posted by Digix View Post
I ran into some problems on this famous paradox.
if you move fast time is supposed to slow down.

so lets say what will happen in this case:

I have one stationary clock and another one is sent some to distance away and it returns back to same point.
later we compare them and do they show different time ?

Some Time ago I expected that they should show different time, but now I think that they will show exactly same time like no relativity ever existed.
because all effects will cancel if you return to the same point.
Why would you think that? As another member has already pointed out speed is a scalar... It has nothing to do with location, rather it's all about speed. Perhaps, you should read up on special relativity.
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Old 09-November-2008, 06:11 AM
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Lets say two clocks are set to exactly the same time. Neither will wear down with time or anything. One clock is left on Earth, while the other is sent out into the cosmos at some non-relativistic speed, say 20k mph. The clock going out into space spends 2 billion years going out, and then 2 billion years returning to Earth.

Should the clocks still show the same time?
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Old 09-November-2008, 11:47 AM
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Why would you think that? As another member has already pointed out speed is a scalar... It has nothing to do with location, rather it's all about speed. Perhaps, you should read up on special relativity.
Speed i usually vector so I don't see a reason why it should be scalar here
and if we think in that way then time will slow down when distance increases, and it will speed up if it increases. all that because of simple Dopler shift. which can explain everything on all relativity cases.

in case if we assume that speed is scalar, I don't see any way to solve problem with invariance. neither observer can tell who is moving and who is not.
acceleration cant be used for that, because we can accelerate for few minutes and then fly free for many years. And relativity does not mention acceleration anywhere too.

in case if all this happens because of acceleration, then just putting clock into centrifuge should slow down it indefinitely, since acceleration is constant for unlimited time.

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I'm not sure why you'd think that. The effect, as you've described it, is special relativity, and Einstein described the effect in his first paper in 1905. They will not show the same time, at least at the level of the effect--and experiments have agreed.
actually I don't know any experiment where one of 2 clocks travels somewhere and returns, it would be nice to see what happens.
experiments where clock trawled only one direction doest count since effect will not cancel in that way
since to read data from clock you must catch your clock so all experiments will be performed on decreasing distance or else if clock is running away you cant catch and measure it.

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Old 09-November-2008, 11:58 AM
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The one that have changed reference frames(the one that moved) will be behind the one that stayed behind.

If you have clock a and b, and they start in the same reference frame(same speed and position), one accelerates at 1g for a week, does nothing for a week, then decelerates for a week. Now that clock will be behind the one left behind..

Now if the OTHER clock does the same thing, then when they meet up they will show the same time. But if the one that traveled first travels back, it will still be behind.
again, how that acceleration incorporates into relativity?
none of the formulas mentions acceleration, and acceleration is also non constant but only temporary so after acceleration process ended all observers are equal and are free to think that they are stationary and another observer is moving.
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Old 09-November-2008, 01:09 PM
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Originally Posted by Kebsis View Post
Lets say two clocks are set to exactly the same time. Neither will wear down with time or anything. One clock is left on Earth, while the other is sent out into the cosmos at some non-relativistic speed, say 20k mph. The clock going out into space spends 2 billion years going out, and then 2 billion years returning to Earth.

Should the clocks still show the same time?
Valuable question because it points out that the Lorentz is always there even at non-relativistic speeds although the effect is very small.

The clock that traveled 4Gyears relative to an observer that is stationary WRT to a clock on earth will have had less time elapse.
The equation being 4GyearsX Squareroot(1-(8940.8^2/c^2))= not quite 4Gyears. It comes down to the precision of the calculator used but 8940^2/c^2 is very small. Nevertheless 4Gyears should be plenty enough time for a small dilation at non-relativistic speeds to be seen; especially given the clocks would be 12hr or 24hr

Edit: Managed to get3.999999998 billion years, approx 73days (Edit correction 730 days being two years, sorry )shorter time on the clock that travelled not counting leap years
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Old 09-November-2008, 01:29 PM
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No Digix. It has to be scalar. Your scenario would have two people waiting on two train stations with synchronised watches seeing two different effects for a relativistic train travelling between those two stations. But observers stationary relative to one another must agree on what they observe.
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Old 09-November-2008, 01:50 PM
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No Digix. It has to be scalar. Your scenario would have two people waiting on two train stations with synchronised watches seeing two different effects for a relativistic train travelling between those two stations. But observers stationary relative to one another must agree on what they observe.
no that is incorrect approximation because if you have synchronized watches then you are only measuring distance decrease mode. in that case it will look like non stationary clock became slower.

but the whole idea is if I not use any clock synchronization. just 2 watches, (not 3) and you do all measuring in the starting point.

alternative question would be if I put some clock into centrifuge and rotate it there for some time, will it slow down?
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Old 09-November-2008, 02:56 PM
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Speed i usually vector so I don't see a reason why it should be scalar here
Digix, speed is a scalar. Velocity is a vector. Vectors have a direction and a magnatude. In the case of velocity, that magnitude is the scalar, speed.
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Old 09-November-2008, 03:00 PM
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Valuable question because it points out that the Lorentz is always there even at non-relativistic speeds although the effect is very small.

The clock that traveled 4Gyears relative to an observer that is stationary WRT to a clock on earth will have had less time elapse.
The equation being 4GyearsX Squareroot(1-(8940.8^2/c^2))= not quite 4Gyears. It comes down to the precision of the calculator used but 8940^2/c^2 is very small. Nevertheless 4Gyears should be plenty enough time for a small dilation at non-relativistic speeds to be seen; especially given the clocks would be 12hr or 24hr

Edit: Managed to get3.999999998 billion years, approx 73days shorter time on the clock that travelled not counting leap years
I atempted to calculate effect again and it really does not cancel
so seems that moving clock will lag afterall.

but again how to solve invariance problem here speed just cant have any effect.
so it means that all is because of acceleration.
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Old 09-November-2008, 03:05 PM
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Digix, speed is a scalar. Velocity is a vector. Vectors have a direction and a magnatude. In the case of velocity, that magnitude is the scalar, speed.
In any case it impossible to run in place so speed must be vector, unless we invent tome strange concept with no meaning.
how can you be moving with no direction?
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Old 09-November-2008, 03:05 PM
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again, how that acceleration incorporates into relativity?
none of the formulas mentions acceleration, and acceleration is also non constant but only temporary
Special Relativity, as originally written by Einstein, doesn't incorporate acceleration. That was one of the reasons for his efforts to produce General Relativity (GR). GR, does incorporate acceleration, which is why most people would say the acceleration is covered under GR. However, Special Relativity can handle acceleration. All you have to do is calculate the acceleration during each time slice, then integrate all those time slices over the the total acceleration time.

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so after acceleration process ended all observers are equal and are free to think that they are stationary and another observer is moving.
While the observers are now stationary, they took different world lines to get to that place from the original starting point. The observer whose world line included acceleration is the one whose time passed slower.
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Old 09-November-2008, 03:17 PM
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In any case it impossible to run in place
Treadmill?
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Old 09-November-2008, 03:30 PM
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Special Relativity, as originally written by Einstein, doesn't incorporate acceleration. That was one of the reasons for his efforts to produce General Relativity (GR). GR, does incorporate acceleration, which is why most people would say the acceleration is covered under GR. However, Special Relativity can handle acceleration. All you have to do is calculate the acceleration during each time slice, then integrate all those time slices over the the total acceleration time.
Ok, thank you. Now I finally managed to understand all that nonsense
so basically SR is just some approximation and real effects are because of that unmentioned acceleration
Ok, now everything is almost fine.
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Old 09-November-2008, 03:31 PM
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Treadmill?
verry funny...
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Old 09-November-2008, 03:36 PM
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In any case it impossible to run in place so speed must be vector, unless we invent tome strange concept with no meaning.
how can you be moving with no direction?
No, check any begining physics book. The definition of a scalar is a one component quantity (a magnitude) that is invariant under rotations. Speed fits this definition as the quantity will stay the same no matter which direction you go. A vector has two components. A magnitude and a direction. In the case of velocity, it has two components. It has speed and a direction.
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Old 09-November-2008, 03:47 PM
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Ok, thank you. Now I finally managed to understand all that nonsense
so basically SR is just some approximation and real effects are because of that unmentioned acceleration
Ok, now everything is almost fine.
Be careful here. SR is not just an approximation. In the case of an observer going out and back, you have to calculate five different time effects. The original acceleration, the first coasting, the acceleration to turn around, the second coasting, and the final acceleration to arrive at the final point. All of those combine to determine how fast the observer aged. The acceleration will just tell you which of the observer's ages the least. The time during coasting parts can be calculated as relative motion using only the original SR equations.
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Old 09-November-2008, 03:50 PM
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No, check any begining physics book. The definition of a scalar is a one component quantity (a magnitude) that is invariant under rotations. Speed fits this definition as the quantity will stay the same no matter which direction you go. A vector has two components. A magnitude and a direction. In the case of velocity, it has two components. It has speed and a direction.
well yes, logically this is correct but practically I cant imagine such situation
it is like voltage only can be between 2 points and it cant be in single point.

also back to the relativity
Do I understand correctly now.

if you accelerate in one direction then time slows down and it speed up if you accelerate in another direction, if there is no acceleration time speed does not change.
basically time speed is just matter of energetic potential. like in gravity, if you move up time accelerates for you if you move down it slows down.
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Old 09-November-2008, 03:54 PM
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Be careful here. SR is not just an approximation. In the case of an observer going out and back, you have to calculate five different time effects. The original acceleration, the first coasting, the acceleration to turn around, the second coasting, and the final acceleration to arrive at the final point. All of those combine to determine how fast the observer aged. The acceleration will just tell you which of the observer's ages the least. The time during coasting parts can be calculated as relative motion using only the original SR equations.
I want to use only most fundamental theories and get rid of all redundancy. Since such thing as SR really mess up everything they give good results but they destroy logic.
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Old 09-November-2008, 04:10 PM
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I want to use only most fundamental theories and get rid of all redundancy. Since such thing as SR really mess up everything they give good results but they destroy logic.
But, there is a reason for using them. Have you ever seen the math required for GR? Here is the first page of the equations used in two spatial dimensions (feel free to click to see the other two pages). The full three dimensional equations are even worse. Now, you must admit, if you can use the more simpler SR equations (a good high school math student would be able to), why not use them.
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Old 09-November-2008, 04:19 PM
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Do I understand correctly now.

if you accelerate in one direction then time slows down and it speed up if you accelerate in another direction, if there is no acceleration time speed does not change.
basically time speed is just matter of energetic potential. like in gravity, if you move up time accelerates for you if you move down it slows down.
Not quite. Any acceleration, in any direction, will slow down time. The greater the acceleration, the slower the time. This fits with the greater gravity slowing time more. In addition,relative motion, between inertial observers, also slows time.
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Old 09-November-2008, 04:25 PM
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Digix,

It is quite easy to set up the twin "paradox" in a way that completely
avoids acceleration. The clocks are already in relative motion and are
compared at the instant they pass. Acceleration is not essential to
understanding time dilation. It is pure special relativity.

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Old 09-November-2008, 04:40 PM
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Special Relativity, as originally written by Einstein, doesn't incorporate acceleration.
OK
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That was one of the reasons for his efforts to produce General Relativity (GR). GR, does incorporate acceleration, which is why most people would say the acceleration is covered under GR. However, Special Relativity can handle acceleration. All you have to do is calculate the acceleration during each time slice, then integrate all those time slices over the the total acceleration time.
Which is pretty much how Einstein presented it in his first paper on special relativity.

But we've talked about this before.
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Old 09-November-2008, 04:47 PM
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OKWhich is pretty much how Einstein presented it in his first paper on special relativity.

But we've talked about this before.
Yabut, how many people actually know this?
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Old 09-November-2008, 05:05 PM
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Quote:
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It is quite easy to set up the twin "paradox" in a way that completely
avoids acceleration. The clocks are already in relative motion and are
compared at the instant they pass. Acceleration is not essential to
understanding time dilation.
But acceleration is essential for understanding the twin paradox, which involves not just time dilation. That's the point of it-- most people "get" time dilation at some level, but they think it's a "paradox" because it is symmetric with both observers-- both observers think the other is younger. But in the twin "paradox", one is definitely younger, because they have been reunited. This is why the resolution of the paradox absolutely does require acceleration, or gravity.
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Old 09-November-2008, 06:15 PM
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In that case, can we go back in time before I posted that, please? I must
have had some other "paradox" in mind, if acceleration is essential to the
"twin paradox". Thinking about it for a few seconds, I don't know how to
get two clocks to tick at identical rates if they are in relative motion but
one is not accelerated, so, uh, yeah... Sorry about that. Just go back
two hours, that's all.

... However... You could eliminate the turnaround at the far end of the
trip by having a third clock already in motion. That reduces the number
of acceleration events from three to two. If that has any value...

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Old 09-November-2008, 06:20 PM
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It's a useful mistake that underscores the difference between the twin paradox and time dilation.
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