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Time Dilation/Proper Time Question 
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#1
Oct1008, 05:15 PM

P: 159

(This isn't a homework problem, I'm struggling with this concept) My professor today was talking about time dilation. He stated that a clock that does not move in the observers point of view (a stationary clock) is, in that observer's viewpoint: a proper time. Thus, an observer on Earth will see a spaceship's clock tick slower, and thus less time will have elapsed on the ship than on Earth. So far so good for me. Here is what is tripping me up. We struggled with this for about five minutes, but felt we should just stop and think about it.
Let's say it took 10 years in the Earth frame and 6 years in the spaceship frame (the ship was travelling 80% light speed.) However, in the spaceship frame, the Earth is also moving away at 80% of light speed. If the spaceship sees his clock ticking at a normal rate (6 years), the Earth clock will be ticking slower (as all moving clocks do.) This will mean that the time for the Earth is less than six years, and not ten? And this could go on ad infinitum until both observers take zero time to go anywhere. Obviously something has gone wrong with my thinking. But I can't quite pin it down. Both clock in their respective stationary frames represent proper times. Applying the time dilation formula I get these results. I can't figure this one out... 


#2
Oct1008, 05:44 PM

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P: 8,470

The relativity of simultaneity has to do with the fact that clocks at different positions at rest in the same frame are supposed to be "synchronized" in that frame using the assumption that light moves at the same speed in all directions in that frame. If I'm on a ship with clocks at either end, I can synchronize the clocks in the ship's rest frame by setting off a flash at the midpoint of the two clocks, and setting them to read the same time when the light reaches them. But if the ship is moving forward in your frame, then you'll observe the back clock moving towards the position where the flash was set off while the front clock is moving away from that position, so if you assume light moves at the same speed in both directions in your own frame, naturally you'll say the light reaches the back clock before the front one, so the two clocks are outofsync in your frame. There's a nice illustration of a similar thoughtexperiment here: http://www.youtube.com/watch?v=wteiuxyqtoM Einstein also discusses the relativity of simultaneity in chapters VIII and IX of this book. 


#3
Oct1108, 08:59 AM

P: 1,544

Hence when you look at http://en.wikipedia.org/wiki/Proper_velocity explained it surly is confusing and even shows apparent observations by a traveler of FTL. It is important to remember that is only an apparent observation – nothing that can be directly observed at any point during travel. You describe a traveler ending his trip 8 light years from earth in 10 years earth time but in only 6 years on the traveler clock. Obviously the will seem like FTL to the traveler, but that is only a “Proper Velocity” not a speed the traveler could ever directly observe for real any time during the trip. The wiki info has some additional links that might help. 


#4
Oct1108, 09:25 AM

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Time Dilation/Proper Time Question



#5
Oct1108, 09:54 AM

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What do you call the apparent speed his traveler seems to see after ending the OP trip after 6 years traveler time 8 Lightyears from the start. 


#6
Oct1108, 12:54 PM

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#7
Oct1108, 01:11 PM

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#8
Oct1108, 01:51 PM

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#9
Oct1108, 02:33 PM

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Lucretius posted
The reason the spaceship's clock appears slower when both clocks are compared at the end of any journey is that the spaceship has moved from an inertial (non accelerating reference frame) to an accelerated frame whereas earth has remained in it's inertial reference frame. 


#10
Oct1108, 03:42 PM

P: 1,544

Just as scientists do when they need to use the “proper” terms. No Naty 1; accelerating has nothing to do with it The OP traveler was not in an accelerated frame The trip was made in an inertial frame of a different speed (0.8c) 


#11
Oct1108, 03:53 PM

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#12
Oct1208, 01:10 PM

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But if you use Proper Time you need to use it correctly with other proper terms like Proper Distance, as it means you are defining measures wrt to some other ref frame. The rate of Proper Time wrt to the Time Rate in the other frame varies dependent on the speed differences between the two frames, as does any measure of Proper Distance. And yes of course if one of the two frames involved is a noninertial frame then it can be as confusing and easy to make a mistake as any other approach. That it does not make the acceleration case simple and easy is why some choose to ignore the “Propers” I guess. But that is well beyond the issue for this OP. The OP confusion with SR here is rather simple. IMO a proper understanding (pun intended) of the “Proper terms” approach, does a better job of making clear of how traveler can apparently travel FTL in their Proper Time experience. This more directly addresses how the OP could be confused by miss reading the wrt conditions between two frames. And that miss reading is what lead to the OP thinking “this could go on ad infinitum until both observers take zero time to go anywhere.” 


#13
Oct1208, 02:24 PM

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#14
Oct1208, 05:17 PM

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The scenario you're describing is just the standard "twin paradox" and there are lots of threads about it. See e.g. posts #142 and #3 in this thread for a pretty complete resolution. 


#15
Oct1308, 08:42 AM

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I'm sure the OP has enough  I'm done here.



#16
Oct1308, 09:16 AM

P: 56

The twin paradox involves a set of twins, or rather a set of two watches set to an equal time and traveling at different velocities. After some interval they are supposed to read different times.
Now the time measured by a watch is its proper time, whatever route it has traveled; there is no difficulty there. But to compare the watches, they must be brought together again, or else some standard of simultaneity must be imposed in order to read them at different locations but at the same time. Any inertial system imposes a standard of simultaneity, i.e. a set of parallel hyperplanes in spacetime, each hyperplane corresponding to a certain point in time. However, if the watches are traveling at different velocities (which they must, if they do not remain together), they inhabit different inertial frames. The hyperplanes corresponding to one watch are slanted relative to the hyperplanes corresponding to the other one. If one watch is traveling to the right, its hyperplanes slant futurewards to the right and pastwards to the left (just the opposite of what one would expect in a spatial rotation). (If it travels at lightspeed, it remains in the same hyperplane during its travel.) If the other watch travels to the left, its hyperplanes slant the other way. Now let both watches start out together, and travel for one hour each in its own proper time. Draw their worldlines; the diagram should look like a letter V. Now draw the hyperplanes through the two points where they end up (the upper corners of the V); each hyperplane will intersect the worldline of the other watch at some earlier point (it should look like a shallow X intersecting the V). so, seen by each watch, the other watch is slow. If the watches are brought together again, at least on of them must change its velocity. There are now three inertial systems, each with its own standard of simultaneity. Two of these pairs can be made to match (for example: departure and arrival back home). But the third pair can then not be made to match, and one of the watches will have to be reset to reflect this (for example, when starting the return journey). So the watches (or the twins) will in general not have experienced the same amount of proper time. As a rule of thumb, watches are slowed down by the same amount in any inertial frame where they have the same speed (absolute value v of velocity v). In such a frame, if the hyperplanes are horizontal, the worldlines of the two watches make the same angles with these hyperplanes. 


#17
Oct1308, 06:44 PM

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Al 


#18
May1810, 07:51 PM

P: 14

This problem is easily remedied:
You start off with the earth at 10 yrs, but it views the spaceship as six years (due to the spaceship moving 0.8c). Relative to the spaceship, it is the earth moving at 0.8c. Thus the spaceship measures 10 yrs, and sees the earth as six years. 


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