# Time Dilation/Proper Time Question

1. Oct 10, 2008

### Lucretius

(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. Oct 10, 2008

### JesseM

Different frames define simultaneity differently--i.e. they disagree about whether two events happen at the same time-coordinate or different time-coordinates (see relativity of simultaneity). In the Earth rest frame, the event of the Earth clock reading 10 years is simultaneous with the event of the spaceship clock reading 6 years. In the ship rest frame, the event of the ship clock reading 6 years is simultaneous with the event of the Earth clock reading 3.6 years.

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 out-of-sync in your frame. There's a nice illustration of a similar thought-experiment here:

Einstein also discusses the relativity of simultaneity in chapters VIII and IX of this book.

Last edited by a moderator: Sep 25, 2014
3. Oct 11, 2008

### RandallB

The problem you are having comes from thinking the “real time” observed by the traveler is the same as “proper time”. Proper time, proper distances and proper speeds can only be defined with respect to (wrt) to some reference. When you say “Applying the time dilation formula I get these results” you were figuring proper time wrt the Earth frame. If you had used another frame (say something traveling the other way at 0.8c) wrt that frame you would get a different “proper time” result, because you would be basing your measurements from a different inertial reference frame.

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. Oct 11, 2008

### JesseM

No, http://www.iep.utm.edu/ancillaries/Proper-Time.htm#Proper%20Time [Broken] refers to the time along a given worldline as measured by an actual clock moving along that worldline--it's frame-invariant, unlike coordinate time. "Proper velocity" is a term that almost no one ever uses, I've never seen it in any textbook, although you're correct that it depends on picking a frame to use. "Proper distance" doesn't give you a choice of reference frames, because it always refers to the distance between events in the frame where those events are simultaneous.

Last edited by a moderator: May 3, 2017
5. Oct 11, 2008

### RandallB

Really - Wikipedia must need fixing then; Looking forward to seeing your updates to "Proper velocity" there.

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

Last edited by a moderator: May 3, 2017
6. Oct 11, 2008

### JesseM

What would need updating? I just said it's rarely used, nothing in the article is incorrect. For example, introductory textbooks like Special Relativity by A.P. French and Spacetime Physics by Taylor and Wheeler include proper time and proper length/distance in the index, but not proper velocity. In fact one of the papers linked to by the wikipedia article backs up my point that it's a notion sometimes used in the literature that's not presently taught in introductions to relativity, although the author argues it should be:
That "apparent speed" would of course be the proper velocity in the rest frame of the destination, but again, most textbooks would just talk about coordinate speed, time dilation, proper time, etc.

7. Oct 11, 2008

### RandallB

“the rest frame of the destination” is the Earth frame – so exactly what is it as seen from the earth frame (from anywhere in that frame, including the stationary wrt earth destination) has an “apparent speed” speed of 1.33c ??

8. Oct 11, 2008

### JesseM

If the destination and the Earth were 8 light years apart in their rest frame, and the traveler took 6 years of his own time to travel between them, that's where the "apparent speed" of 1.33c comes from (the astronaut would have a natural reason to use this frame if it was his own original rest frame before he started the journey). In any other frame, the distance between the Earth and the destination would be different than 8 light years, so if you calculated (distance)/(traveler's time) in this frame you'd get something different than 1.33c.

Last edited: Oct 11, 2008
9. Oct 11, 2008

### Naty1

Lucretius posted

Yes, but keep in mind there is relative velocity between earth and spaceship....each thinks the other is moving...both are correct! There is no absolute velocity between two entities, its all relative. Each observes the other's clock is slower...!!!!!! Both are right! Each observe's their own clock as "normal"....again both are right!!!

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.

Last edited: Oct 11, 2008
10. Oct 11, 2008

### RandallB

Now you are using Proper Velocity (and proper time) exactly as I did in post 3.
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. Oct 11, 2008

### JesseM

I never denied your definition of proper velocity was correct, I just said the term is not widely-used. And you said that proper time is defined relative to a particular reference frame, that's incorrect, it refers to the actual clock time along a given wordline; for an object moving inertially, proper time is the same as the coordinate time in the object's rest frame, but in the case of the worldline of a non-inertial object, the proper time won't match the coordinate time of any inertial frame.

12. Oct 12, 2008

### RandallB

But you cannot apply "proper time" to just any distance as if it were the "absolute real time" or against coordinate distances for that you use coordinate time. Just because Coordinate time and Proper time are the same time does not mean you are using Proper Time when making a coordinate system calculation – you are using coordinate time. As you have already said; you don’t use the system of “Proper” terms – your choice, although lots do – scientists can pick own systems I’m not stopping you.

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 non-inertial 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. Oct 12, 2008

### JesseM

It's not clear what you mean by "apply 'proper time. to just any distance". Proper time has nothing to do with distance, it's a measure of time elapsed on a physical clock. Are you trying to talk about the fact that proper velocity is defined as coordinate distance/proper time? If so, there is nothing in the definition of proper velocity that specifies which frame's distance you should use--you are free to use "just any distance". This is one of the reasons I don't like the term "proper velocity". Proper time and proper distance don't leave you any choice of what reference frame to use, so I understand the use of the term "proper" to have something to do with the fact that they are frame-invariant. On the other hand, "proper velocity" does leave you a choice of what frame to use.
Again, I don't really know what you're talking about here. When did I say you used proper time when making a coordinate system calculation? All I said was that in the case of an object moving inertially, proper time has the same value as coordinate time in the object's rest frame.
When did I say I "don't use the system of proper terms"? I use proper time all the time, and I wouldn't have any objection to using proper distance or proper velocity if the situation called for it, I just don't find them useful very often. I think it's also a misconception to view these terms as a "system"--they are just three terms that happen to have the word "proper" in them, but just because you use one doesn't mean you have to use the others. Proper time is extremely useful and I imagine pretty much every physicist working in relativity often calculates the proper time along various worldlines, proper distance seems less commonly-used to me, and I think proper velocity is not used at all by most physicists (this is supported by the fact that if you search google scholar for 'proper time' and 'relativity' you get 12,000 results; if you search for 'proper distance' and 'relativity' you get 2,010; and if you search for 'proper velocity' and 'relativity' you get only 409). This isn't because any of the terms are invalid, just because there are a lot more situations where proper time would actually be relevant to the sorts of problems physicists (or physics students) would be trying to solve.
Again, just because you use one doesn't require you to use the others--proper time is generally more useful than proper distance or proper velocity, and the terms were never meant to be part of a single "system", they are just different terms that have the word "proper" in them--the wikipedia article suggests the term 'proper velocity' was not invented until around the late 1960s, whereas you can see here the term 'proper time' has been in use at least since 1919.
Defining what measures wrt to some other frame? Proper time itself is not defined in terms of any frame at all, it's a frame-invariant quantity measured along worldlines of physical objects.
What "two frames"? What was the original frame that you are comparing the "other" frame with?
I never said anything about non-inertial frames, I said that the proper time along the worldline of an object which is moving non-inertially will not coincide with the coordinate time of any inertial frame (whereas the proper time of an object moving inertially will coincide with the coordinate time of the inertial frame where the object is at rest). But again, proper time isn't based on any frame at all, it's analogous to the length of a curve in 2D space. If your curve in 2D space is a straight line, then if you use a cartesian coordinate system and the line starts at point A at the origin and runs along the y-axis, then of course the length of the line from point A to some other point B will just be equal to the y-coordinate of point B; this is analogous to the fact that the proper time along an inertial worldline coincides with the coordinate time in the object's inertial rest frame. But if your curve in 2D space is not a straight line, then there is no way to orient a cartesian coordinate system such that the length along the curve from the start at point A to some other point B on the curve is always equal to the y-coordinate of point B, and this is analogous to the fact that the proper time along the worldline of an object which is moving non-inertially will not coincide with the coordinate time in any inertial coordinate system.

Last edited: Oct 12, 2008
14. Oct 12, 2008

### Fredrik

Staff Emeritus
"Proper time" is defined as an integral along the curve that represents the clock's motion, and since we're talking about two different curves, the results can of course be different.

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. Oct 13, 2008

### RandallB

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

16. Oct 13, 2008

### Almanzo

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. Oct 13, 2008

### Al68

The clocks only read the proper time locally, if viewed from a different frame. So, assuming we define an event at 8 ly from earth in earth's frame, and v=0.8c, the event will occur simultaneously with earth's clock reading 10 yrs in earth's frame and the ship's clock reading 6 yrs in the ship's frame. This is local proper time for the event, and is frame invariant. Sure, in the ship's frame, earth's clock will read 3.6 yrs simultaneously with the ship's clock reading 6 yrs, but that 3.6 yrs does not represent proper time in earth's frame non-local to earth. Just like the 6 yrs the ship's clock reads in earth's frame when the earth's clock reads 10 yrs would not represent proper time in the ship's frame non-local to the ship. A clock in motion relative to the observer only represents proper time local to the clock.

Al

Last edited by a moderator: Oct 14, 2008
18. May 18, 2010

### YangMills

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.