# Confused about time dilation. (Just got introduced to relativity))

1. Dec 19, 2011

### PrincePhoenix

I just got my first official introduction to relativity in my textbook (I'm equivalent to a high school senior student). It is part of a chapter on modern physics.It first mentions the two basic postulates and then briefly introduces what length contraction, time dilation, increase in mass, energy mass equivalence (all with an equation each) and speed of light as a limit are.
I am a little confused with time dilation.

According to the textbook, the clock at rest in a reference frame appears to tick slower to an observer moving with respect to that frame of reference.The time interval they measure is t' = t/√(1-v2/c2) , where t is the time interval as measured at rest in the frame of reference of the clock.

1-)To me it seems that this means that the moving observer will measure a longer time interval, meaning the time in the reference frame of the clock has slowed down from his point of view. But that's exactly opposite to what I've read about the decay of pions where time seems to slow down for the fast moving particle as viewed by an observer in the lab's frame of reference.

2-) Also in the topic just before relativity, the textbook mentions that motion and rest are relative. So to an observer at rest in the reference frame with the clock, the other observer (moving w.r.t the frame with the clock) seems to be moving. But then to the moving observer (in his own reference frame) this clock and its frame of reference seems to be moving. Shouldn't each observer see the other's clock ticking slower? What would that mean???

2. Dec 19, 2011

### Staff: Mentor

1) the moving observer measures a second as a second, the stationary observer measures a second as a second too. The stationary observer looking at the moving observer's clock says it is moving slower so that to his mind time has dilated / slowed down for the moving observer.

How did they determine a second transpired? By watching a light beam travel from the floor to ceiling and back again. The moving observer sees it go up and come down, one second. The stationary observer uses the same method with a light clock in his frame.

Both observers agree that the speed of light is the same. Hence the stationary observer concludes that it took longer for the light on the moving observer's train to bounce up and down because it had to travel a greater distance due to the forward motion of the train. Hence the stationary observer concludes that time is slower for the moving observer. (By similar reasoning, the moving observer says No! time has slowed for the stationary observer because to the moving observer the stationary dude is the real moving dude).

Back to the pion, it lasts for a short amount of time but if it is traveling at near light speed then to us it is aging more slowly and thus hangs around for a longer period of time. (The pion is the moving observer).

I guess if I were pion, (yada dadada dadadada... fiddler on the roof...) then I would see the world as slowing down but that's another story.

Your question is similar to the twin paradox. Which twin ages more? the moving one or the stationary one. SR is ambiguous on it as both twins say the other is aging more slowly. Once the moving twin (moving relative to the universe of stars) slows down and stops then you see it is he (or she) that has aged more slowly ie is younger. Acceleration and deceleration were covered by Gen Rel not SR.

Last edited: Dec 19, 2011
3. Dec 19, 2011

### ghwellsjr

Are you sure the textbook says, "the clock at rest in a reference frame appears to tick slower to an observer moving with respect to that frame of reference"? Although this is true, it's also true the other way around and usually the equation you stated is for the other way around. In other words, t is for time in a reference frame and t' is for time in a second reference frame moving with respect to the first one.

Here's how I would explain it: pick a reference frame. Any clock at rest in that reference frame ticks normally. Any clock moving in the reference frame ticks slower according to the equation you quoted.

You have to be careful when you talk about time intervals between two clocks moving with respect to each other because they are at different locations between the beginning and ending of the interval and relativity of simultaneity comes into play. The proper way to deal with such things is to use the Lorentz transform but that's probably beyond your interest right now.

4. Dec 19, 2011

### Markus Hanke

The time dilation happens due to the relative speed of the two observers in relation to each other, and can only be detected as such. What this means is that none of the two observers will notice anything special if and when they move; it is only when, after having moved at some speed close to the speed of light relative to each other, you put them both to rest and compare their clocks that they will notice a difference in the time readings. Remember - there is no absolut frame of reference, the relative speed between the two clocks is all that counts.

5. Dec 19, 2011

### ghwellsjr

Your last two sentences are not correct.

First, you can't talk about "moving relative to the universe of stars" because they are all moving with respect to each other and there is no such frame we can identify as a preferred one based on the universe of stars.

Second, even when one of the twins slows down and stops with respect to the other twin, SR is still ambiguous about which twin ages less. It is true that in a frame in which they are now both at rest, the one that "slowed down and stopped" is the only one that experienced time dilation and so is younger, but other frames will not agree. For example, in a frame in which the "traveling" twin was at rest, it's the other twin that is now younger because he was always moving but the "traveling" twin was at rest for a period of time making him the one that aged more. This is the reason the Twin Paradox always has the twins reunite because it is only then that all frames will agree on their age difference.

Third, acceleration and deceleration of observers/objects/clocks are covered in SR, it's just that you can't use the Lorentz Transform to transform an inertial Frame of Reference to an accelerating Frame of Reference, it only works between non-accelerating (inertial) Frames of Reference.

6. Dec 19, 2011

### PrincePhoenix

If the first observer's clock seems slow to the second, shouldn't the second's clock seem faster to the first?

When you bring both of them to rest, in the same reference frame and compare their clocks, who's clock had ticked slower?

Just like above, if the pion thinks my clock slowed down and I think the pion's clock slowed down, then why is it only the pion's decay time increases from its value at rest. Why inly its time slows down in the end?

7. Dec 19, 2011

### PrincePhoenix

That made it simpler. :)

What happen when you bring both clocks to rest in one frame of reference? I'm confused about how can my clock be slow for the pion and the pion's clock slow for me at the same time? As the pion's life increases in my view does my life also extend bya few nano-seconds according to the pion? (Asking w.r.t the pion example in my original post).
Thank you.

8. Dec 19, 2011

### Staff: Mentor

Thats the thing about relativity, no observer is special. All observers agree that light travels at the same speed (deemed true via the Michelson/Morley experiment).

Observers using identical clocks such as the light clock. I see my clock bounce up and down and I see your clock take a longer time to bounce up and down because of your motion and conclude that you are living more slowly.

Conversely, you see your clock bounce up and down but see my clock take longer and conclude the same thing.

Who is right? Both of you are, until one of you decelerates and you both are now stationary to one another comparing your clocks. The one who decelerated with respect to the universe has the slower clock. SR doesn't describe acceleration or deceleration for that you need to study Gen Relativity where it says clocks in an accelerated frame of reference tick more slowly than ones in an inertial frame of reference.

On the Earth we do not normally live in an inertial frame of reference, unless we are falling from a mountain, building or sky... When you fall you are weightless and so are in an inertial frame of reference.

Here's a reference to the twin paradox on wikipedia:

http://en.wikipedia.org/wiki/Twin_paradox

9. Dec 19, 2011

### nitsuj

The pion "experienced" the same amount of proper time as they always do. If it's lifetime is 1 second when at rest, it's 1 second when in motion; according to it's own clock. According to your clock the pions lifetime while in motion is longer then 1 second.
I'd say this happens because the observation is semetric, to the point of your text "...textbook mentions that [inertial] motion and rest are relative.".

As far as observations go, who is moving the pion scenario? It is relative (motion), so both conclude the other is moving, simular to the "slower ticking", both see the others clock as ticking slower.

So it's really no more a leap then the concept motion is relative, and you haven't asked how both observers can be at rest at the same time if one is moving . This issue is the same for all observations (measurements+calculations) the observers make of each other. i.e. the other length contracted, has more kenitc energy ect.

If you would like to see more clearly (visually) how upon return to the "rest frame" the at near c traveller has definitively aged less check out doppler effects. For me that resolved the twin paradox absolutely. And I found it really clarifies the effect direction of motion has on this kinda "paradox".

I tried to think of a better answer to "how can my clock be slow for the pion and the pion's clock slow for me at the same time?" incorporating your use of the term "at the same time". I couldn't think of one, but maybe someone here could elaborate on this; It seems to me a point of SR could be that this type of question is nonsensical because there is no "same time" that the two frames share (i.e. a proper time). Said differently it is only an observation (measurement+calculation) the others clock is ticking slower. Get them at rest with each other and it becomes diffinitive who's clock was "actually" was ticking slower.

Ah I think I got it, when you say "at the same time" it implies a preffered frame, while the question explicitly says there isn't one. In that sense it's nonsensical. Your question is understandable for sure, but I think an answer should include this explination/distinction, it maybe the term "at the same time" is why it doesn't make sense to you that dilated time is relative.

Last edited: Dec 19, 2011
10. Dec 19, 2011

### harrylin

Hi yes, and ghwellsjr already mentioned relativity of simultaneity. It may be useful to elaborate how both clocks can appear to be slow relative to the other without any contradiction.

Imagine first a system at rest S with two clocks:

C1 ---------- C2

An observer who is using that system, will synchronise the clocks such that according to these clocks, radio signals take the same time both ways.

Now imagine a second, moving system S', also with two clocks:

C1' ---------- C2' -->v

Those two clocks have also been synchronised such that according to those clocks, radio signals take the same time both ways.
Consequently, according to the rest observer, C1' will be advanced on C2'.

Now, suppose that the systems pass each other at close distance as illustrated below. To determine the rate of a clock in relative motion, its "time" must be measured at two different positions:

C1' ---------- C2' -->v
----------------C1 ---------- C2

The consequence for a moving observer in S' (from the perspective of the observer in S):

First the time of C2' will be compared with for example C1, and next the time of C1' with C2. The difference of the moving clock readings will be exaggerated due to the "false" synchronisation, and be more than the advance of clock C1. Thus the moving clocks will appear to have a higher rate (according to S') thanks to the different clock synchronisation.

As a matter of fact, I now see that this has been explained with a detailed calculation example here:

https://www.physicsforums.com/showthread.php?t=250340

Harald

11. Dec 19, 2011

### ghwellsjr

Although in a rest frame for the pion, your clock will be running slower and its clock will be running normally, that is not why it can survive the trip, rather, it's because of length contraction--the distance it has to travel is much shorter.

12. Dec 19, 2011

### nitsuj

So the interval (lifetime of pion) is the same. From my rest frame it is because the pions time appears dilated that it was able to travel such a far distance. Isn't that right? (of course the pions proper time is the same, so it must be distance travelled that is reduced from it's FoR, as you mentioned)

Could you let me know if I got that right ghwellsjr?

13. Dec 19, 2011

### ghwellsjr

Yes, that is correct.

14. Dec 20, 2011

### PrincePhoenix

So the answer to my questions is, any clock at rest in a reference frame ticks normally. Any clock moving in that reference frame ticks slower. The moving clock's observer in its reference frame will see the same for the other clock. Both observers are correct in their own reference frames. And we can only compare whether one ticked slower than the other when both are brought to rest in the same reference frame. And that is studied in General Relativity.
Did I get it right?

15. Dec 20, 2011

### harrylin

Except for the last sentence, that is roughly* correct: you don't need general relativity for this, special relativity suffices.

*roughly correct: such expressions as "ticks slower" are absolute statements; however, the contrary is claimed by the other observer and a clock cannot tick both slower and faster than another clock. To avoid contradictions, it is common practice to write instead "appears to tick slower". However, when comparing two clocks at the same location then everyone will agree which clock is ahead.

16. Dec 20, 2011

### PrincePhoenix

When both clocks are brought to rest in the same frame of reference, which will have ticked more?

17. Dec 20, 2011

### nitsuj

It'd be which ever clock had to accelerate the least to be at "rest".

18. Dec 20, 2011

### PrincePhoenix

Thank you. That clears it up for me. Thanks to everyone who posted here and helped me out.

19. Dec 20, 2011

### PAllen

Actually, it is tricky to specify a non-math based criterion. For example, if one twin (A) is inertial except for a brief, extreme acceleration and deceleration, while the other twin (B) had a more moderate turnaround acceleration after a long separation, then a long path back to the first twin, B could still age less than A. Or, you could set it up so B and A had identical acceleration profiles, but still different elapsed times on being brought together.

An accurate statement that could be made is that whichever history (world line) deviated most from straight (geodesic) over the whole path is the one that aged more. [Edit: less not more].

I like an analogy with lines on a piece of paper. Between two points, you draw two arbitrary paths (one longer than the other). Where is (or what causes) the extra length in one? You can see this is meaningless. The only thing you can say is whichever deviates more from a straight line is longer - but this is really no different from saying the longer one is longer.

With relativity, the situation is the same except that greater deviation from straight produces less time between the two start and stop events, while for paths on paper, it is more length.

The difference in behavior between length and proper time is due to what analog of pythagorean theorem applies (called the metric). On a plane, you have:

distance = sqrt ( (delta x) ^2 + (delta y)^2)

For special relativity you have:

(proper) time = sqrt ((delta t) ^2 - (delta x)^2)

Last edited: Dec 20, 2011
20. Dec 20, 2011

### nitsuj

my "it's too general for PF!" alarm was going off when deciding to post it lol.

Being unsure of which scenario the poster was refering too, I came up with that response. It seemed like it covered most scenarios well.

But your statement "...whichever history (world line) deviated most from straight (geodesic) over the whole path is the one that aged more."

That's much much better. It makes it more clear for me too! Thanks

21. Dec 20, 2011

### PAllen

Of course, see my self correction to #19. Typed too fast ...

22. Dec 20, 2011

### nitsuj

Thanks for that post PAllen.

Looks like it will take some time for me to absorb but will clarify the difference.

23. Dec 21, 2011

### dgwsoft

Good question. It depends how you bring them into the same frame of reference. If you accelerate clock A until it is stationary relative to clock B, then clock A will have ticked fewer times (since the time at which they passed each other, and were in close proximity).

However if you If you accelerate (or decelerate) clock B until it is stationary relative to clock A, then clock B will have ticked fewer times.

That is the essence of the twins paradox.

Everything is symmetrical between the two frames of reference in SR. Any observer (provided they are not accelerating) is entitled to think of themselves as at rest. The rule that "moving clocks go slow" applies in every frame of reference. If the clocks don't accelerate you can compare them directly only once (as they pass each other). If one clock accelerates, so they can meet up again, then that is the "moving" one.

It is hard, at first, to see that this is consistent. Here are a couple of ways to see it.

1) Mathematically. The Lorentz transformation gives (x', y', z', t') in terms of (x, y, z, t; v). Invert that to get
(x, y, z, t) in terms of (x', y', z', t'; v). You will find the expressions are identical except v changes to -v (if I am moving at velocity v relative to you, then you are moving at velocity -v relative to me). That proves
the relationship between observers is symmetrical.

2) Draw some space-time diagrams. Different reference frames have different axes, and so slice up space and time differently. Under Galilean physics, only the space coordinates change for a moving observer, but under SR both axes move. "At the same time" means something different for different observers, as well as "at the same place".

With regard to the twins paradox, I find it useful to think about the path between two points on a plane.
The straight line is the shortest path (A->B) If you take a dog-leg path (A->C->B) that is a longer path. The plane is symmetrical. Distances increase "at the same rate" in whatever direction you chose to go.
But the dog-leg path is objectively the longer route between A and B. That is the path of the travelling twin. You can consider any one direction you like to be "forward" but you can't consider both legs of a dog-leg to be "forward".

The difference between the plane and SR is that in SR the direct path is the *longest* path (as measured by a clock) That is because the invariant on the plane is x^2 + y^2. In SR it is x^2 + y^2 + z^2 - c^2 t^2. The minus sign makes the difference.

PS That picture also clarifies for me why it is wrong to think that the "loss of time" for the travelling twin occurs during the short period as he turns around. That is no more true than saying that the "extra distance" on the dog-leg occurs at the turning point. The time (distance) is a function of the path as a whole.

Last edited: Dec 21, 2011
24. Jan 2, 2012

### PrincePhoenix

Another small confusion/question related to time dilation which I think doesn't need a thread of its own.

When we say a moving clock ticks slower, does it mean it will count less number of seconds than a stationary clock? e.g 5s in my stationary clock , I'll observe less than 5s in a moving clock?

25. Jan 2, 2012

### bobc2

I'm not sure if this will help. But, sometimes it is useful to view the situation from the standpoint of the geometry in a 4-dimensional universe. We'll just focus on the clocks that exist in two different frames. The clocks are thought of as 4-dimensional objects; in the sketch below they are extended along world lines defining the 4th dimension for each frame. Here, we have red and blue observers, each carrying a clock, moving away from each other with the same speed with respect to the black rest system.

The thing that happens with special relativity is that the X1 and X4 coordinates in the moving frames rotate such that the 45-degree line (photon) always bisects the angle between X1 and X4 for any moving observer, no matter what the speed (that's why every observer sees light as traveling at speed c). So, in our sketch below, red and blue literally live in two different 3-D worlds. Each 3-D world is a cross-section of the 4-D universe, but the cross-sections cut across the 4-dimensional universe at different angles for observers having different velocities with respect to some reference frame.

Thus, you can see in the sketch why red sees blue's clock ticking at a slower rate, and blue sees red's clock clicking at a slower rate. We have selected an instant of time labeled as t2 in red's and blue's frames. The t2 identifies a simultaneous world for red and for blue (two different worlds--and, again, they are not the same worlds). The clock times, t2 in red and t2 in blue read exactly the same time. Also, the clock times, t1 in red and t1 in blue read exactly the same time.

Last edited: Jan 2, 2012
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