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Relativity questions

Ashles said:
I thougt relativity was accepted fact.
His equations hold true for everything we have ever observed such as the exact location of GPS satellites, differences in atomic clocks sent into orbit etc.
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Best definition I could find quickly (of course, in Wikipedia):
In various sciences, a theory is a logically self-consistent model or framework for describing the behavior of a certain natural or social phenomenon, thus either originating from observable facts or supported by them (see scientific method). In this sense, a theory is a systematic and formalized expression of all previous observations made that is predictive, logical, testable, and has never been falsified.
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Most notably, theories (like the Theory of Relativity) have to be falsifiable.

Also, to quote good ol' Albert: "E=m c^2, but this could be a local phenomenon." That's why I pointed out that I am not aware of experimental verification of time dilation in distant star systems.
 
epepke said:
That's the hard way of doing it.

But you can still ignore it and have no acceleration at all. Just have Alice hold her watch up to the window. When the inbound train passes, Agnes in the inbound train looks at Alice's watch and synchronizes it to Alices. When Agnes gets back to Bob, her watch will read less elapsed time of exactly the same amount.
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But then it's no longer a twin paradox -- but that of triplets, no?

Anyway, what is suppose to happen is that the original two get back together from where they originally separated with one being noticably older than the other. This way of avoiding the accelerations does indeed show time dilations and yields the same results (as far as how much their respective clocks differ) but it allows for one to always counter with "It's not exactly the same conditions -- they do not start off from (nor return to) the same inertial frame."

Also, when you transpose the observers' points of view, you're going to have to adjust for the velocities of the two opposing trains as not being so easily additive. This too can get confusing, even though it's all Special Relativity.
 
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I note that when we get into discussions like this the phrase "time for x runs slower" or some variant, is often used.

In slightly different discussions- for example concerning spacetime curvature, a phrase like this might be used- "X experiences less time than Y".
These two ways of viewing the relation between time and a moving object seem rather different, but are they mathematically equivalent?

Does "X experiences less time" mean the same as "For X, time moves more slowly"?
 
Anytime i see an explanation for the twin paradox it always involves the outbound twin coming back to meet the stationary twin.
Does this mean there would be no effect if he didn't come back?
For example, The outbound twin flys off at near the speed of light and then stops. Both twins then compare the time (or length of beard whatever) - obviously taking into account the amount of time it takes to get the info to each other.
So, in this scenario, are both twins the same age?
 
Ashles said:
I assume that our solar system is travelling massively fast compared to some of the other stars that we can see in the sky (maybe even a fair proportion of the speed of light), so are we experiencing time dilation with regard to them?
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Actually, the sun is moving around the galactic center at ~220km/s which, although fast, is much less than the speed of light. If you work out v^2/c^2 for this velocity you'll see that the amount of time dilation is about 1 part in 10^7! The relative velocity between the sun and other stars is much less, ~ tens of km/s.
 
monoman said:
Anytime i see an explanation for the twin paradox it always involves the outbound twin coming back to meet the stationary twin.
Does this mean there would be no effect if he didn't come back?
For example, The outbound twin flys off at near the speed of light and then stops. Both twins then compare the time (or length of beard whatever) - obviously taking into account the amount of time it takes to get the info to each other.
So, in this scenario, are both twins the same age?
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If he didn't come back, then there would be no way to measure their relative ages. The question has no answer and no meaning.

One of the most important things to "get" about relativity is that there is no universal frame of reference (another way to look at it is that God doesn't have his own clock). This implies several things, one of which is that there is no such thing as universal simultaneity. Simultaneity -- the observance of two things happening "at once," or more broadly, the observance of the order of things occurring, first one, then the other -- depends on one's frame of reference. In order to compare two frames of reference, they need to be close enough for a third observer to see or observe both of them at once, from his frame of reference, or they need to pass by each other close enough for each of the two moving observers to observe the clock of the other.

Information cannot be transmitted across great distances instantaneously, as would have to occur to compare two relativistic clocks from a great distance ("great distance" here means one large enough so that relativistic effects can be detected and appreciated, without being negligible).

AS
 
It's not a meaningless question, AS. They could arrange to send a message to an observer located at the midpoint between the final destinations at a particular time, as read by their own clocks. Since the twin that travelled has experienced time dilation relative to the twin that didn't travel, the observer at the midpoint will receive that twin's message first, and the amount of dilation can be calculated from the time difference between the messages arriving.
 
SpaceFluffer said:
It's not a meaningless question, AS. They could arrange to send a message to an observer located at the midpoint between the final destinations at a particular time, as read by their own clocks. Since the twin that travelled has experienced time dilation relative to the twin that didn't travel, the observer at the midpoint will receive that twin's message first, and the amount of dilation can be calculated from the time difference between the messages arriving.
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Sorry. You're right.

AS
 
AmateurScientist said:
Simultaneity -- the observance of two things happening "at once," or more broadly, the observance of the order of things occurring, first one, then the other -- depends on one's frame of reference. In order to compare two frames of reference, they need to be close enough for a third observer to see or observe both of them at once, from his frame of reference, or they need to pass by each other close enough for each of the two moving observers to observe the clock of the other.
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The first sentence is true, but the second seems to show a misunderstanding of the term "frame of reference."

"Frame of reference" refers to how fast you're moving, not to where you are. So it's meaningless to talk about two frames of reference being close to each other or far from each other. Two observers are in the same frame of reference if they're not moving relative to each other, regardless of how far apart they are.

If a bunch of observers are all at rest relative to each other, they can consistently measure the order of events, and in particular simultaneity, in a way that they can all agree about, even if the observers are spread out over an arbitrarily large distance. It's just that if there's a different bunch of observers, at rest relative to each other, but moving relative to the first bunch, then the two bunches won't agree. So, I guess it's somewhat a matter of taste how much reality to ascribe to any one bunch's notion of simultaneity.

This is how special relativity works, anyway (i.e., no gravity). General relativity is ... harder.

(By "frame of reference", I mean "inertial frame of reference", i.e., the observers just float along and feel no acceleration.)
 
SpaceFluffer said:
It's not a meaningless question, AS. They could arrange to send a message to an observer located at the midpoint between the final destinations at a particular time, as read by their own clocks. Since the twin that travelled has experienced time dilation relative to the twin that didn't travel, the observer at the midpoint will receive that twin's message first, and the amount of dilation can be calculated from the time difference between the messages arriving.
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Is that third person at the halfway point even really needed?

After all, isn't he in the same inertial frame of reference as the twin that remains home? If the twins synchronize their clocks just before the second twin's departure, and the first twin could somehow watch or time when the second twin arrives at his destination and notes the time on his (the at-rest twin's) clock for that moment, it should display the same dilation to the time the traveling twin claims is on his clock upon his arrival in a signal sent back to the first twin. Of course, the first twin will have to wait some time for that signal -- but comparrisons of the two times should match that of a third observer doing the same thing at the midpoint.

ETA: OK, just watching wouldn't do it -- it must be timed as predicted.
 
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I don't understand why the need for the middle man either. What if the twin went to a distance plant that was stationary relative to the earth. Once there he sends a message to say what his time is. The twin on earth receives the message, knows how far the distant planet is away and calcuates the other twins time.

eg: the other planet is 20 light years away, so he deducts 20 years from the outbound twins time and compares
 
monoman said:
I don't understand why the need for the middle man either. What if the twin went to a distance plant that was stationary relative to the earth. Once there he sends a message to say what his time is. The twin on earth receives the message, knows how far the distant planet is away and calcuates the other twins time.

eg: the other planet is 20 light years away, so he deducts 20 years from the outbound twins time and compares
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I don't think he (the twin at home) should have to deduct any time at all. Here's why ...

Let's assume that the traveling twin can accelerate to 0.99999 c almost instantaneously and survive (just like on Star Trek -- but without the warp speed nonsense). To him (the traveling twin) it takes just one day to get to the planet 20 light years away. But to the twin at home, 20 years have passed. The twin at home predicts the traveling twin's arrival at time = 20 years. True, it will take another 20 years for the signal to get back home, or if he uses a method of observation to determine the traveling twin's arrival, but the at home twin should not be timing that. His clock should stop when the traveling twin arrives as predicted, not when his signal gets back. I guess if you have the at home twin's clock always running, then you must do as you say, so it depends on what he does with his clock and what he actually is timing.
 
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Just thinking said:
I don't think he (the twin at home) should have to deduct any time at all. Here's why ...

Let's assume that the traveling twin can accelerate to 0.99999 c almost instantaneously and survive (just like on Star Trek -- but without the warp speed nonsense). To him (the traveling twin) it takes just one day to get to the planet 20 light years away. But to the twin at home, 20 years have passed. The twin at home predicts the traveling twin's arrival at time = 20 years. True, it will take another 20 years for the signal to get back home, or if he uses a method of observation to determine the traveling twin's arrival, but the at home twin should not be timing that. His clock should stop when the traveling twin arrives as predicted, not when his signal gets back. I guess if you have the at home twin's clock always running, then you must do as you say, so it depends on what he does with his clock and what he actually is timing.
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Good point - i was just covering all bases!
So any experts out there who can tell me if this scenario won't result in time dialation? If you can explain why (the need to turn round and come home) i may finally fully understand the paradox.
 
monoman said:
Good point - i was just covering all bases!
So any experts out there who can tell me if this scenario won't result in time dialation? If you can explain why (the need to turn round and come home) i may finally fully understand the paradox.
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Oh, it will (not that I'm an expert) -- it's just that the twin paradox does not involve any signal sending as originally described. The only way for them to see a difference in ages is for them getting back together. I guess it just makes for more of an impact that way.
 
That's good to know. Simplifies it a little. It just surprises me that every explanation i see involves turning round and coming back, as if that was an integral part of the paradox.
 
I just thought of something -- I'm not 100% sure of it, but I suspect I'm not too far off.

Let's assume that somehow the at-home twin can watch through a telescope the traveling twin's journey. Let's also assume that the traveling twin is headed for a destination 20 light years distant and will send back a signal (at the speed of light) upon arrival. Finally, let's give him the ability to accelerate to 0.9999 c and back to rest almost instantaneously.

The observer at home will now watch the journey -- but how long will it be (for him) to note his twin's arrival? 20 years? No -- 40! Because it will take light twice as long to get to the destination (the speed of the spacecraft) and then travel back again (his seeing it happen) as compared to the predicted time he is expected to get there. But what will be even more interesting is that the at home twin will get the traveling twin's signal of arrival when the at home twin actually views it happening. It will seem as if the traveling twin sent a signal at infinite velocity!
 
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monoman said:
Good point - i was just covering all bases!
So any experts out there who can tell me if this scenario won't result in time dialation? If you can explain why (the need to turn round and come home) i may finally fully understand the paradox.
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An easier way to think of it is - from the time the star first saw the twin approaching, how long did it take him to get there?

Roughly the same time the travelling twin experienced ('roughly' due to interstellar gas interfering with the travel of light - anyway...)

Another way to think of what is going on is a 'spacetime triangle'. Distance doesn't matter here, just picture a graph with a perfect triangle on it - 60 degree angles at all sides. We can pick one point as the origin of the twins.

The 60 degree angle represents them moving apart at about .84 of c. This ends up with a lorentz contraction factor of 2. That is, twin A experiences 2 seconds for every second he sees twin B experience, and twin B experiences 2 seconds for every second he sees twin A experience.

So, let's say at t = 15 seconds (from the POV of the origin), when they are 13 light-seconds apart, twin A immediately changes course and returns at .84 of c to twin B, completing our spacetime triangle.

If you draw parallel lines at each of twin B's seconds (29 lines for 30 bars), you will notice that the relative lines on twin A's side are twice as long. Prior to twin A's return, however, who was undergoing contraction would be a matter of dispute - up until twin A switched inertial frames.

I don't know if that makes any more sense :-/
 
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