Is this a paradox in Special Relativity?

[Shackleford]

We covered SR last semester in Modern Physics I and also covered early Quantum Theory and started Modern Quantum Theory. I've been speculating a bit on SR recently since I started reading The Universe and Dr. Einstein a guy at work gave me.

If we have the Earth as a stationary reference frame and a rocket ship as a moving frame at some appreciable value of c with respect to Earth, according to Earth's observations, humans on the rocket ship will age more slowly, the physiological processes will slow down and so forth.

However, it's equally as valid to reverse the situation, i.e. the stationary frame is the rocket ship and the moving frame is Earth. The people on Earth from the rocket ship's observations will appear to age more slowly and the proper time will then be on the rocket ship.

Which is actually physically true? Because both cannot simultaneously be true. They cannot both age more slowly and age more rapidly. I'm wondering if this is a conventionally-valid paradox I conjured up. I'm sure my textbook covered this situation, but I just haven't opened it up to refresh my memory on such an event.

 

[russ_watters]

This is a variant of the so-called "twins paradox", which is only an apparent paradox due to a misunderstanding of the theory and poor definition of the case. The problem with your case is not understanding that you can't have a clear "who is older" unless both clocks are sitting next to each other, stationary, in the same reference frame. And in order to do that, you have to break the symetry of your scenario.

 

[Shackleford]

This is a variant of the so-called "twins paradox", which is only an apparent paradox due to a misunderstanding of the theory and poor definition of the case. The problem with your case is not understanding that you can't have a clear "who is older" unless both clocks are sitting next to each other, stationary, in the same reference frame. And in order to do that, you have to break the symetry of your scenario.

Well, of course to show which is "older" you would have to explicitly compare the rates of time in the same reference frame, but that would destroy the apparent relativistic effect. However, you can still compare the situations and their logical conclusions according to the theory because they both happen actually physically concurrently. They are just viewed through different prisms and yield different results depending on your frame of reference.

 

[russ_watters]

Well, of course to show which is "older" you would have to explicitly compare the rates of time in the same reference frame, but that would destroy the apparent relativistic effect.

No, it wouldn't. If you compare them in the same reference frame, one really would be ahead of the other*.

However, you can still compare the situations and their logical conclusions according to the theory because they both happen actually physically concurrently. They are just viewed through different prisms and yield different results depending on your frame of reference.

If you make a truly mirror-image scenario, then either really could be considered older/younger, but making a truly mirror-image scenario is tougher than you may think and your scenario isn't specific enough to know if it is truly symmetrical.

For example: where did the spaceship come from? If it came from earth, their situations are not symmetrical because the spaceship had to fire its engines to move away from earth. *So to define the situation explicitly and relate it to the above requirement:

Start: Two clocks sitting next to each other on Earth are synchronized.
Next: One clock is put into a rocket and launched on a high-speed trip to Alpha Centuari and back.
Finish: The two clocks are sitting next to each other again on earth.

Which is ahead and which is behind at the finish?

 

[Shackleford]

No, it wouldn't. If you compare them in the same reference frame, one really would be ahead of the other*. If you make a truly mirror-image scenario, then either really could be considered older/younger, but making a truly mirror-image scenario is tougher than you may think and your scenario isn't specific enough to know if it is truly symmetrical.

For example: where did the spaceship come from? If it came from earth, their situations are not symmetrical because the spaceship had to fire its engines to move away from earth. *So to define the situation explicitly and relate it to the above requirement:

Start: Two clocks sitting next to each other on Earth are synchronized.
Next: One clock is put into a rocket and launched on a high-speed trip to Alpha Centuari and back.
Finish: The two clocks are sitting next to each other again on earth.

Which is ahead and which is behind at the finish?

I remember reading in the textbook the situation you defined. The rocket ship undergoes acceleration leaving Earth, and when approaching Alpha Centauri, slowing down and turning back towards Earth. In this scenario, they mapped out the light-signal intervals and so forth over the span of years of the trip. The longest amount (interval) of time is always the proper time which is measured by a clock at rest with respect to the stationary frame. So, in this case the rocket ship would be slightly behind.

I must admit I'm not familiar with the symmetry requirement in elucidating the application of SR. Why is it not valid to consider a rocket ship randomly passing Earth with some constant velocity?

 

[JustinLevy]

I must admit I'm not familiar with the symmetry requirement in elucidating the application of SR. Why is it not valid to consider a rocket ship randomly passing Earth with some constant velocity?

This problem usually comes about when people think of "time dilation" and "length contraction" as the defining results from which everything is derived. Instead, "time dilation" is actually referring to a very specific case of measurements.

Consider two inertial frames with their spatial origins moving with respect to each other. If there are two events which occur at the same location according to frame1, then frame2 will claim there was less coordinate time between between the two events than frame1.

Now let's look at it in terms of clocks. For the coordinate time in an inertial frame to match that of an actual clock, the clock has to be at rest according to the inertial frame. Therefore, in the previous experiment, you need three clocks. One in the frame1, and two in frame2 (at different spatial locations).

The overarching point is: time dilation is NOT a relationship between two clocks moving with respect to each other. It is a relationship between coordinate time between two events according to two inertial frames, and only in a very specific case (when the events are at the same location according to one inertial frame).

It is just hard to discuss SR accurately without using math. So many "popularist introductions" to SR unfortunatly cause a lot of confusion.

 

[Fredrik]

Which is actually physically true? Because both cannot simultaneously be true. They cannot both age more slowly and age more rapidly.

They're both equally true, and neither is objectively true (i.e. true in a coordinate independent way). These statements are just saying that if these guys assign coordinates (t,x,y,z) to events in spacetime using the standard procedure, the difference between the t coordinates that either of them is assigning to two arbitrary events where the other guy is present will be greater (by a factor of gamma) than the difference between his ages at the two events.

There's no contradiction here. What you can learn from this is that the time coordinate assigned by the standard procedure isn't a good measure of a person's age. (The correct measure of his age, according to both SR and GR, is the proper time of the curve in spacetime that represents his motion).

Why is it not valid to consider a rocket ship randomly passing Earth with some constant velocity?

It's certainly valid, but it's not useful if you're trying to find a logical inconsistency. There's nothing inconsistent about the fact that their coordinate assignments disagree with the other guy's age.

 

[Shackleford]

Okay, so there's a difference between coordinate time and actual clock time, or rather rate of time. Feel free to use math. I'm not sure if you read the first sentence in my original post. Your explanations make sense. However, the overarching point seems to imply merely a difference in coordinate time, not necessarily a difference in rate of time, which is my primary concern - actual rate of time in different inertial frames. The textbook seemed to make the connection between time dilation and a slowing of biological processes and so forth in that moving inertial frame.

I fully understand assigning arbitrary units and measures of time in different inertial frames and relating them using the Lorentz transformations. I'm concerned with the conclusions and experimental verifications of SR that claim the biological processes slow down in that moving inertial frame, i.e. the rate of time slows down. Let's assume we have a mirror-image, perfectly symmetrical situation involving two moving inertial frames. The statement was made that either could be considered older or younger. This of course cannot actually be true. So, as was stated, to determine the correct age of someone or something you must look at the object's proper time which is of course the longest interval of time. Is the claim that biological processes slow down misleading?

 

[ZikZak]

Let's assume we have a mirror-image, perfectly symmetrical situation involving two moving inertial frames. The statement was made that either could be considered older or younger. This of course cannot actually be true.

There is no "actually." If you have two astronauts freely floating past each other, then each observes the other's clock to run slow. There is no answer to which clock is faster or slower in an absolute "actually" sense. There is no universal clock. There is no absolute rest frame with the "real," "actual" clock. There are only clocks that move with the observers. The answer depends on reference frame. That is why it is called "Relativity."

Is the claim that biological processes slow down misleading?

No, all processes, biological or not, occurring in a moving laboratory will be observed to run slow. "Moving," that is, relative to the person doing the observing. Each astronaut observes the other's biology to run slow.

 

[Shackleford]

There is no "actually." If you have two astronauts freely floating past each other, then each observes the other's clock to run slow. There is no answer to which clock is faster or slower in an absolute "actually" sense. There is no universal clock. There is no absolute rest frame with the "real," "actual" clock. There are only clocks that move with the observers. The answer depends on reference frame. That is why it is called "Relativity."
No, all processes, biological or not, occurring in a moving laboratory will be observed to run slow. "Moving," that is, relative to the person doing the observing. Each astronaut observes the other's biology to run slow.

So then if I observe someone's clock to be running more slowly, they're not really aging more slowly? There doesn't have to be an "absolute" sense but an actual sense, i.e. the person and their proper time, etc.

 

[JustinLevy]

So, as was stated, to determine the correct age of someone or something you must look at the object's proper time which is of course the longest interval of time.

Alright! Proper time. This concept makes things much easier.

First though, your concept of proper time seems to be based on coordinate time in inertial frames. Let's make it as general as possible: the proper time is the length of the wordline of whatever object/observer/hypothetical we are talking about.

So if two people want to compare their aging, using proper time, they need to meet up at two different times. Indeed, one's worldline can be longer than the others between these two "meeting events". So yes, one can actually age more (or less) than the other.

Since proper time is coordinate system invariant, regardless of what coordinate system you use to describe the travels, you will get the same answer for who aged more.

Is the claim that biological processes slow down misleading?

It depends on what you are extracting from that claim. Locally, neither astronaut will "feel" like their biology is going slow. In fact they can perform any experiment they want and they shouldn't be able to tell that they are moving in any absolute sense. Yet if they look out and try to describe the physics of another astronaut flying by, they will measure that astronaut's clocks running slow -- any clocks, biological or otherwise. And when they meet up, their ages will be different just as any inertial observer (an inertial frame used consistently for the entire experiment) would have expected from time dilation.

 

[ZikZak]

So then if I observe someone's clock to be running more slowly, they're not really aging more slowly? There doesn't have to be an "absolute" sense but an actual sense, i.e. the person and their proper time, etc.

What do you mean by "really" and "actual" other than "that which is observable?"

 

[matheinste]

What do you mean by "really" and "actual" other than "that which is observable?"

I suspect, as you do, that Shackleford is still thinking of absolute time which of course does not exist in SR. This absolute time, or Newtonian time was thought to have an existence distinct from our measurement (observation) of it.

Matheinste.

 

[ZikZak]

I suspect, as you do, that Shackleford is still thinking of absolute time which of course does not exist in SR. This absolute time, or Newtonian time was thought to have an existence distinct from our measurement (observation) of it.

Matheinste.

Yes, well, the purpose of asking the question was to get him to think about that ;)

 

[matheinste]

Yes, well, the purpose of asking the question was to get him to think about that ;)

Sorry for any misunderstanding. I was really aiming the comment at Skackleford but your question seemed appropriate to quote.

Matheinste.

 

[Fredrik]

So then if I observe someone's clock to be running more slowly, they're not really aging more slowly?

The question actually doesn't make sense, because "really" indicates that you're talking about something coordinate independent, and the word "slowly" is a reference to a coordinate system.

 

[Shackleford]

Well, I'm not talking about an absolute time. I'm well aware of that historical notion being discarded in SR.

The statement was made that if you have a mirror-image, perfectly symmetrical situation, you could compare the clocks of the reference frame and moving frame to see which one aged more or less. This of course is a relative difference, not absolute, because you can always find some arbitrary frame to modify the perspective and observations. This implies to me that it's not simply an observation based on coordinates and equations, but an actual change in the rate of time in space the object experiences. That has to be the case if you bring back together your inertial frames to compare the clocks at rest in the stationary frame and they yield different times as russ said.

No, it wouldn't. If you compare them in the same reference frame, one really would be ahead of the other*.

 

[dulrich]

...to see which one aged more or less...

Realize that in order to measure this, you must have the two parties at the same place twice: once at the beginning and once at the end. In order to do this one must accelerate out of his initial inertial reference frame. This change not only breaks the symmetry of the situation but also destroys the synchronization of the clocks (relativity of simultaneity).

If you want a truly symmetric situation, then have BOTH parties accelerate toward one another in the same way. If you do that, then they will meet with the same biological age.

 

[Shackleford]

Okay, if it's possible for someone to age more or less relative to someone else, and measure this accurately, why does the rate of time vary at appreciable values of c? What happens along that journey through the space-time continuum?

 

[Shackleford]

I'm also a little confused as to why my use of words actually and physically are controversial. In reference to the muon decay experiment, the decay of muons traveling near c is physically less than that predicted by an Earth-frame observer because the muons actually experience less time and so lost less mass through nuclear decay. Right? Or am I just mistaken in thinking something actually, physically changes affecting the matter and the disparities are simply positional and observational?

 

[Fredrik]

The statement was made that if you have a mirror-image, perfectly symmetrical situation, you could compare the clocks of the reference frame and moving frame to see which one aged more or less.

But you're not comparing the clocks in that scenario. You're comparing the time displayed by clock B at some event on its world line, to the time coordinate assigned to that event by the inertial coordinate system associated with the motion of clock A.

Suppose that you're moving with clock A and read the time it displays at some event P on its world line. How do you know which event on the world line of clock B that you should compare it with? We can imagine that you have an assistant moving with clock B and taking readings all the time, but which one of them should you compare to yours?

This implies to me that it's not simply an observation based on coordinates and equations, but an actual change in the rate of time in space the object experiences. That has to be the case if you bring back together your inertial frames to compare the clocks at rest in the stationary frame and they yield different times as russ said.

Actual=coordinate independent
Experiences=coordinate dependent
...so the statement doesn't quite make sense.

Suppose you set both clocks to 0 at an event P where they're both present, and then have them move differently for a while, until they meet again at an event Q. They will usually display different "times" at Q because what they're displaying isn't a time coordinate, it's a coordinate independent property (proper time) of the curves that describe their motion.

 

[Shackleford]

But you're not comparing the clocks in that scenario. You're comparing the time displayed by clock B at some event on its world line, to the time coordinate assigned to that event by the inertial coordinate system associated with the motion of clock A.


Actual=coordinate independent
Experiences=coordinate dependent
...so the statement doesn't quite make sense.

Suppose you set both clocks to 0 at an event P where they're both present, and then have them move differently for a while, and then meet again at an event Q. They will usually display different "times" at Q because what they're displaying isn't a time coordinate, it's a coordinate independent property (proper time) of the curves that describe their motion.

Sorry for all my confusion, guys. I think this is what I'm a bit fuzzy on. I'll have to go back and read the SR chapter in my textbook. The proper time, a coordinate independent property, accurately describes the instantaneous physical "experience" of matter. Is this correct?

 

[jtbell]

If we have the Earth as a stationary reference frame and a rocket ship as a moving frame at some appreciable value of c with respect to Earth, according to Earth's observations, humans on the rocket ship will age more slowly, the physiological processes will slow down and so forth.

However, it's equally as valid to reverse the situation, i.e. the stationary frame is the rocket ship and the moving frame is Earth. The people on Earth from the rocket ship's observations will appear to age more slowly and the proper time will then be on the rocket ship.

Which is actually physically true? Because both cannot simultaneously be true.

Yes, they can both "simultaneously" be true. To see this better, instead of a single Earth and a single rocket ship, imagine two sets of rocket ships arranged along a line. One set is stationary (in the Earth's reference frame, say), and the other set is moving at the same speed in the same direction in a sort of "parade." Of course, in the frame in which the second set of ships is stationary, the first set is moving backwards with the same speed.

The first set of ships synchronizes their clocks so they all read the same time in their "rest frame" (the frame in which they are stationary). The second set of ships synchronizes their clocks so they all read the same time in their rest frame.

In the first set of ships' rest frame, the second set of ships' clocks run at a slower rate than the first set of ships' clocks. In the second set of ships' rest frame, the first set of ships' clocks run at a slower rate than the second set of ships' clocks.

Yet, whenever two ships (one from each set) pass each other, everyone agrees on the time readings of those two ships' clocks at that moment! (We assume that they pass right next to each other, closely enough that they can read each other's clocks with a high-speed camera or something, without significant error caused by light's travel time between the two ships.)

How is this possible? Because of relativity of simultaneity, the first set of ships' clocks are not synchronized in the second set's rest frame. Nor are the second set of ships' clocks synchronized in the first set's rest frame.

I'm sure there are diagrams illustrating this setup somewhere on the web, but I can't find one at the moment. I first saw it many years ago in an article in the American Journal of Physics, I think by N. David Mermin. I'd make up a set of diagrams myself, and attach them, but I'm on the road and don't have the tools or the time to do this properly. If nobody's found something like this by the time I get home in about a week, I'll do it then.

 

[Fredrik]

When we talk about an inertial observer's "experience" we're really talking about a description in terms of the inertial coordinate system associated with his motion. His world line is taken as the time axis of that coordinate system. The time axis is labeled by the times displayed by his clock at each event on it. The assignment of coordinates to events not on his world line is done using a specific synchronization procedure. For example, if he emits a light pulse at (t=-T,x=0), and the light is reflected once and returns to him at (t=T,x=0), the reflection event is assigned coordinates (t=0,x=T).

 

[Shackleford]

Okay, I busted out the textbook and perused the SR chapter a bit, particularly the muon experiment. This site also helped too by explicitly putting the situation in both reference frames.

In the Earth Frame, I am at rest with respect to my ruler, so I observe no length contraction. My clock is not at rest with respect to the muon, so I do observe time dilation.

In the Muon Frame, I am not at rest with respect to the ruler (Earth), so there is length contraction. My clock is at rest with respect to me, so I observe no time dilation.

Now for the Twin Paradox, the conclusion is the Moving twin is younger because of the asymmetrical journey, i.e. the acceleration undergone and leaving the initial inertial frame. I also of course understand the shortest time-interval being at rest with respect to the inertial frame and that moving clocks undergo time dilation and yield a larger time interval (magnitude). However, it is said they run slowly and in the case of the Twin Paradox the Moving twin is younger. Why is it said it runs slowly when it yields a larger time interval? I guess you could say the proper time finishes more quickly and the moving frame takes more time to end.

 

[yuiop]

In the Earth Frame, I am at rest with respect to my ruler, so I observe no length contraction. My clock is not at rest with respect to the muon, so I do observe time dilation.

In the Earth frame the muon takes about 30 ms to reach sea level. (Longer time interval)

In the Muon Frame, I am not at rest with respect to the ruler (Earth), so there is length contraction. My clock is at rest with respect to me, so I observe no time dilation.

In the muon frame the trip takes about 6 ms. (Shorter time interval)

The Earth frame considers the interval measured in the muon frame to be time dilated (shorter) than the interval measured in the Earth frame.

THe Earth/muon experiment can not be directly equated with the twin paradox, because if we ignore gravitational time dilation then both the Earth frame and the muon frame are essentially inertial while in the twin experiment one of the twins is non-inertial.

However, it is said they run slowly and in the case of the Twin Paradox the Moving twin is younger. Why is it said it runs slowly when it yields a larger time interval?

This is not correct. The interval measured by the moving twin is shorter.

 

[Shackleford]

I'm quoting the book,

In all cases, the actual time interval on a moving clock is greater than the proper time as measured on a clock at rest.

In the muon decay, the event would be the muon's journey down the arbitrary ruler and so you would assign the muon frame the proper time (shorter interval). The moving clock would be on the 0.98c Earth and would yield the larger interval.

In the Twin Paradox you would identify the event as being the journey of the spaceship and assign it the proper time or the shorter interval and the moving Earth would again get the larger interval of time since the Moving twin does actually experience the accelerations thus breaking the symmetry.

 

[yuiop]

... However, it is said they run slowly and in the case of the Twin Paradox the Moving twin is younger. Why is it said it runs slowly when it yields a larger time interval? ...

I take it from your last post that you now agree that the clock of the the moving twin runs slower and yields a shorter time interval?

In the Twin Paradox you would identify the event as being the journey of the spaceship and assign it the proper time or the shorter interval and the moving Earth would again get the larger interval ...

 

[Shackleford]

I take it from your last post that you now agree that the clock of the the moving twin runs slower and yields a shorter time interval?

If I properly identified the correct event in the right reference frame, yes.

 

[matheinste]

Hello Shackleford.

In an earlier post you describe the spaceship journey as an event, which, in normal language usage it is. However in Relativity the word event has a specific meaning. To identify the location of an object at a certain place at a certian time relative to the three spatial and one coordinate axes of sapcetime requires four numbers. This spacetime location at a certain place at a certian time determined by this set of numbers is an event. It has no spatial or temporal extension. A journey is a path in spacetime which is made up of a series of events which together constitute the wordline of an object, be it a spaceship, clock etc. The reading on a clock traveling along this worldline is a measure of the spacetime path, necessarily timelike, traversed and is called proper time. This may seem a bit long-winded but it is important for the answer to your question.

If we take the traveling twins path, as is usual in the twin scenario, as cosisting of two inertial paths out and back with instantaneous acceleration at the turn around (ignoring initial and final stages of the journey), during these two inertial phases each twin will observe (not see) the others clock running slow compared to his own. But at the turnaround, the traveller, and only the traveller, chages to a different inertial frame. This inertial frame's line of simultaneity meets the stay at home twin's coordinate time axis at a later coordinate time than the line of simultanety of the frame of the traveller before turnaround. So the traveller is now in a frame where the stay at home has aged somewhat more than before the turnaround. The rest of the journey home (ignoring the final deceleration) is again symmetrical inasmuch as each observes the other's clock to be running relatively slower than his own.

Now if we want it to be more realistic and not have an instantaneous turnaroud and also to take into account the initial and final accelerations at parting and reuniting we will have some parts of the journey following a curved path in spacetime due to accelerated, non inertial, motion. To measure proper time along this path it is necessary to add all the infinitessimally small inertial frames that the ship is at rest in during this time. With the traveller at rest in any of these frames the situation would again be symmetrical as regards clock rates but the line of simultaneity of the traveller meets the stay at home's time axis at continually later times as he prgresses from one such small inertial portion of his worldline to the next. The addition of these small spacetime path lengths, proper time, as the path length of the individual inertial portions tend to zero, is achieved by integration of proper time along the world line.

Now all this window dressing can be avoided by simply stating that the clock accompanying the twin who traverses the longest spacetime path length records the shorter proper time. The inertial spacetime path between two events is always the shortest spacetime path and a clock traveling along it will record more proper time than a clock traversing any other spacetime path. The stay at home twin follows such an inertial spacetime path. No matter how you pose the scenario, both twins traveling etc. the shortest spacetime path will give the longest proper time. Clock rates can be, in some cases, just an unnecessary complication and the cause of much confusion.

As regards biological ageing, it is only another physical process subject, as any other, to the laws of nature and so what can be said about clocks can be said about biological mechanisms.

There are many finer points and loose ends to be tied up, as always, but the above, as I see it, are the basics.

Matheinste.

 

[jtbell]

Okay, I busted out the textbook and perused the SR chapter a bit, particularly the muon experiment.

Does your textbook say anything about relativity of simultaneity?

In the Earth frame the muon takes about 30 ms to reach sea level. (Longer time interval)

In the muon frame the trip takes about 6 ms. (Shorter time interval)

Let's apply relativity of simultaneity here. Suppose there are two clocks at rest in the Earth's reference frame, one at the muon's starting point and one on the ground; and a clock that rides along with the muon (and a hypothetical observer who rides along with them). We set the clocks so that in the Earth's reference frame, all three read zero when the muon passes its starting point.

In the Earth's reference frame, the muon takes 30 ms to reach the ground, so both the clock at the starting point and the clock on the ground read 30 ms when the muon reaches the ground. Because of time dilation, the clock on the muon reads 6 ms when it passes the clock on the ground.

In the muon's reference frame, 6 ms elapse on its clock during the trip, of course. In this frame the other two clocks are time-dilated, so only about [STRIKE]0.17 ms[/STRIKE] 1.2 ms elapse on those clocks. (The time-dilation ratio is the same in both cases: 1/5 = 6/30 = [STRIKE]0.17/6[/STRIKE] 1.2/6).

In the muon's reference frame, the two "Earth frame clocks" are also not synchronized. At the start of the trip, the clock at the starting point reads 0, and the clock on the ground reads about [STRIKE]21.83 ms[/STRIKE] 20.8 ms. During the trip, [STRIKE]0.17 ms [/STRIKE] 1.2 ms elapse on both of these clocks. At the end of the trip, the clock at the starting point reads [STRIKE]0.17 ms[/STRIKE] 1.2 ms, and the clock on the ground reads 22 ms.

So both an "Earth observer" and the "muon observer" agree that when the trip starts, the clock at the starting point and the clock on the muon both read 0; and when the trip ends, the clock on the ground reads 22 ms and the clock on the muon reads 6 ms. However, they disagree about whether the clock at the starting point and the clock on the ground are synchronized.

 

[Shackleford]

Hello Shackleford.

In an earlier post you describe the spaceship journey as an event, which, in normal language usage it is. However in Relativity the word event has a specific meaning. To identify the location of an object at a certain place at a certian time relative to the three spatial and one coordinate axes of sapcetime requires four numbers. This spacetime location at a certain place at a certian time determined by this set of numbers is an event. It has no spatial or temporal extension. A journey is a path in spacetime which is made up of a series of events which together constitute the wordline of an object, be it a spaceship, clock etc. The reading on a clock traveling along this worldline is a measure of the spacetime path, necessarily timelike, traversed and is called proper time. This may seem a bit long-winded but it is important for the answer to your question.

Thanks for clearing up the vernacular. An event is basically a well-defined four-"dimensional" point. A journey is a path comprised of all these space-time "points" from one event to the next.

If we take the traveling twins path, as is usual in the twin scenario, as cosisting of two inertial paths out and back with instantaneous acceleration at the turn around (ignoring initial and final stages of the journey), during these two inertial phases each twin will observe (not see) the others clock running slow compared to his own. But at the turnaround, the traveller, and only the traveller, chages to a different inertial frame. This inertial frame's line of simultaneity meets the stay at home twin's coordinate time axis at a later coordinate time than the line of simultanety of the frame of the traveller before turnaround. So the traveller is now in a frame where the stay at home has aged somewhat more than before the turnaround. The rest of the journey home (ignoring the final deceleration) is again symmetrical inasmuch as each observes the other's clock to be running relatively slower than his own.

So, my original assertion was partly correct insofar as I was looking only at the comparison of the inertial frames (inertial phase). However, my analysis was incomplete. I didn't take into account the acceleration, or non-inertial phase, of the moving frame which breaks any symmetry in the simplified case. Why does the line of simultaneity change when the traveler changes to another inertial frame? Is this where the diagram is helpful?

Now if we want it to be more realistic and not have an instantaneous turnaroud and also to take into account the initial and final accelerations at parting and reuniting we will have some parts of the journey following a curved path in spacetime due to accelerated, non inertial, motion. To measure proper time along this path it is necessary to add all the infinitessimally small inertial frames that the ship is at rest in during this time. With the traveller at rest in any of these frames the situation would again be symmetrical as regards clock rates but the line of simultaneity of the traveller meets the stay at home's time axis at continually later times as he prgresses from one such small inertial portion of his worldline to the next. The addition of these small spacetime path lengths, proper time, as the path length of the individual inertial portions tend to zero, is achieved by integration of proper time along the world line.

Now all this window dressing can be avoided by simply stating that the clock accompanying the twin who traverses the longest spacetime path length records the shorter proper time. The inertial spacetime path between two events is always the shortest spacetime path and a clock traveling along it will record more proper time than a clock traversing any other spacetime path. The stay at home twin follows such an inertial spacetime path. No matter how you pose the scenario, both twins traveling etc. the shortest spacetime path will give the longest proper time. Clock rates can be, in some cases, just an unnecessary complication and the cause of much confusion.

As regards biological ageing, it is only another physical process subject, as any other, to the laws of nature and so what can be said about clocks can be said about biological mechanisms.

There are many finer points and loose ends to be tied up, as always, but the above, as I see it, are the basics.

Matheinste.

That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.

Does this even make sense? Let's say you're comparing a stationary frame and a moving frame. A slower-moving object will "accumulate" more time at each event along a space-time path, but a faster-moving object will "accumulate" less time at each event along a space-time path.

 

[yuiop]

That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.

You are right to think that it seems counter-intuitive, but the statement by matheinst is correct. Consider two cars moving traveling from A to B. Both start at A at the same time and both arrive at B at the same time, but one takes the straight path and the other takes a long winding path. The car that takes the long winding route has to travel faster and as a consequaence experiences more time dilation and less proper time. This is not not an exact analogue. The cars in this example are traveling a path through 3D space rather than 4D spacetime, but the idea is similar.

In the muon's reference frame, 6 ms elapse on its clock during the trip, of course. In this frame the other two clocks are time-dilated, so only about 0.17 ms elapse on those clocks. (The time-dilation ratio is the same in both cases: 1/5 = 6/30 = 0.17/6).

Did you mean 1.2 ms?

 

[Shackleford]

You are right to think that it seems counter-intuitive, but the statement by matheinst is correct. Consider two cars moving traveling from A to B. Both start at A at the same time and both arrive at B at the same time, but one takes the straight path and the other takes a long winding path. The car that takes the long winding route has to travel faster and as a consequaence experiences more time dilation and less proper time.

Hmm. So, is my "accumulation" analogy kind of close?

 

[Fredrik]

Thanks for clearing up the vernacular. An event is basically a well-defined four-"dimensional" point. A journey is a path comprised of all these space-time "points" from one event to the next.

I don't see the point of using the word "journey" like this. I would call that curve the object's "world line", and use words like "journey" only when I'm talking about the corresponding sequence of events in the real world.

Why does the line of simultaneity change when the traveler changes to another inertial frame? Is this where the diagram is helpful?

It changes in that particular way only when we have chosen to use the comoving inertial frames. (That's not the only possible choice, but we don't need to get into that now). The comoving inertial frame is constructed by applying the synchronization procedure that I described briefly in my previous post, to the tangent of your world line at some point on it. How to do this is one of the first things you learn when you study spacetime diagrams, which by the way is by far the best and easiest way to learn SR.

See this post for my standard reply (including a spacetime diagram) to questions about the twin paradox. There are good contributions from other posters as well in that thread.

That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.

Don't forget that the "length of the spacetime path" is defined as the proper time of the curve. It's actually a useless concept. It's better to just use the term proper time.

Edit: Now I see that you actually said "shorter space-time path" = "larger proper time". I guess what you (and matheinsteine) meant by "length of the spacetime path" is just the length of the curve in a diagram, using the Euclidean concept of distance and length. (I recommend that you don't use such terminology. People like me tend to assume that we're talking about Lorentzian geometry, not Euclidean, when "spacetime" is mentioned).

Does this even make sense? [...] A slower-moving object will "accumulate" more time at each event along a space-time path, but a faster-moving object will "accumulate" less time at each event along a space-time path.

That's not the prettiest way of saying it, but yeah, it makes sense, because expressed in the coordinates of an inrtial frame, the proper time is the integral of \sqrt{dt^2-dx^2-dy^2-dz^2} along the curve. A line parallel to the time axis (which represents zero velocity) has dx=dy=dz=0 everywhere, so every other curve between two points on that line that's parallel with the time axis has to have a shorter proper time, because of the negative contribution from every little movement in space. (Higher speed=bigger dx,dy,dz for a given dt).

 

[Shackleford]

So, it's critical to make the observation that the observer in the Fixed Frame never leaves his respective inertial frame, but the observer in the Moving frame does in fact change inertial frames. Because of those non-inertial frames/phases, it's valid to put the relativistic effect/corrections on the Moving frame and look at it from the "fixed" Fixed frame, i.e. the Moving frame is the moving clock which runs slowly and yields the shorter time interval. You then also have to make corrections for length contraction.

I don't see the point of using the word "journey" like this. I would call that curve the object's "world line", and use words like "journey" only when I'm talking about the corresponding sequence of events in the real world.

Is that not what a journey is? That's how I'm looking at it - a series of infinitesimally-small events that constitute the curve. Any point on the world line or curve is an event at a specific point in space and time. There's just a bunch of them depending on how you sub-divide the time interval along that curve.

 

[Fredrik]

So, it's critical to make the observation that the observer in the Fixed Frame never leaves his respective inertial frame, but the observer in the Moving frame does in fact change inertial frames.

I'm not a fan of phrases like "leaves his inertial frame". I recommend that you never say that an object is "in" an inertial frame or that it "changes" inertial frames. Instead, say that it's stationary in an inertial frame and that it changes velocity in that same inertial frame. If it changes velocity, it now has a different comoving inertial frame.

Inertial frames are just global coordinate systems. A global coordinate system is a function from spacetime into \mathbb R^4. So the inertial frames are just functions that assign coordinates to events. You can't be "in" one, but you have a velocity in all of them.

Is that not what a journey is? That's how I'm looking at it - a series of infinitesimally-small events that constitute the curve.

I'm just saying that if you travel to South Africa, that would be a journey (unless you happen to live there). I don't see a reason to use the word "journey" for the mathematical representation of that journey when the we already have a standard term for it: "world line".

 

[jtbell]

Did you mean 1.2 ms?

Ugh, you're right. I don't remember how I could have possibly come up with 0.17 ms this morning. I should have waited until after I had my quota of coffee. I've corrected it and the other related numbers. Thanks for catching that.

 

[GRDixon]

Start: Two clocks sitting next to each other on Earth are synchronized.
Next: One clock is put into a rocket and launched on a high-speed trip to Alpha Centuari and back.
Finish: The two clocks are sitting next to each other again on earth.

Which is ahead and which is behind at the finish?

The rocket clock is behind (less time has elapsed on it) upon its return. On the way out to Alpha Centauri the rocket occupant can measure the rate of the Earth clock (using the clocks distributed in his own rest frame) and will deduce that the Earth clock runs more slowly than those at rest in his own rest frame. But when he changes rest frames upon reaching Alpha Centauri, he will admit that the clocks in his initial rest frame are actually not synchronized with one another, etc., and that it is actually his clock that has been running slowly during the trip. The same situation repeats on his way back to Earth. By way of a real example, short half-life mesons last much longer (don't decay) when shot around the loop of an accelerator.