# Time Dilation

1. Apr 26, 2007

### spaghed87

Hello, I'm Eddie and this is my first post here. I'm almost a Junior in college going for a B.S. In chemistry but I must say that time dilation is one of the most interesting things in science. I was wondering if anyone has actual figures with regards to the difference in time between a stationary and moving atomic clock at certain speeds? Please don't yell at me if this is already posted because I'm new here and since I'm in a rigorous field I don't have time to study much other than chemistry. If so just redirect me to that post please. Thanks for reading/responding to my question.

2. Apr 26, 2007

### pmb_phy

Welcome to the forum spaghed87.

I'm sorry but off hand I don't have that data. If you're interested in the theory behind it then you can take a look at a web page I created for that puyrpose. The physics is simple and it only require simple algebra to understand. Meanwhile let's hope someone can get you that hard data.

See - http://www.geocities.com/physics_world/sr/light_clock.htm

Best wishes

Pete

3. Apr 26, 2007

### Jonathan Scott

According to Special Relativity, the rate at which time elapses for a clock moving at speed v compared with one at rest is given by sqrt(1-v^2/c^2) where c is the speed of light in vacuum. When v much smaller than c, the fractional difference is about 1/2 (v^2/c^2). That is, for every 1 second of elapsed time, the moving clock loses 1/2 (v^2/c^2) seconds. For example, if the moving clock is going at about 300 m/s (as for a fast subsonic aircraft) and c is assumed to be 300,000,000 m/s, then the speed is about a millionth of the speed of light and the moving clock is slower by 1 part in 2,000,000,000,000 (around 43 nanoseconds per day).

4. Apr 26, 2007

### Jonathan Scott

Sorry, I may have misunderstood your question. If you want the actual experimental figures, look up the "Hafele-Keating" experiment. There is for example an entry in Wikipedia which includes the results. This also needed to take into account General Relativity for differences in gravitational potential.

5. Apr 26, 2007

### Staff: Mentor

6. Apr 27, 2007

### longshinewoole

Difficult language

[I'm sorry but off hand I don't have that data. If you're interested in the theory behind it then you can take a look at a web page I created for that puyrpose.
See - http://www.geocities.com/physics_world/sr/light_clock.htm

Best wishes

Pete[/QUOTE]

Pete, it was difficult to understand your language. In the web page you created, you used the words "clock" and "measure". Such language must mean that the time intervals will be read from the clocks, not calculated by means of the classic equation t = d/v. For instance, using a clock to measure a time interval, it means a departure time and an arrival time will be read from the clock, and the arrival time minus the departure time will give us the time interval. Since you did not use the tool, the clock, how can you tell us the tool worked slower or faster.

By the word "measure", you meant you would take such a measuring action, reading a departure time and an arrival time. But you did not.

Would you explain why?

7. Apr 27, 2007

### pmb_phy

That is correct. The measurement of the time as read in the rest frame of the clock is what you'd intuitively think of doing. Record the reading on the clock as it passes through the first event (i.e. flash of light is emitted). Then record the reading as it passes through the second event (i.e. flash of light is detected). All this is accomplished in the rest frame, S, of the clock. Then a similar procedure is done in the rest frame of the observer, a frame in which the clock is moving. Call this frame S'. In S' (as in all frames) there are many clocks which are all synchronized (Note: While the clocks are synchronized in one frame they need not be synchronized in another). Observers in S record the reading on the S' clock which is ideally co-located with the first event. Then the recording on another clock in frame S' is taken when the clock at rest in S passes through the second event. The difference of these time readings is called the time difference.

It is self explanitory. In fact you're the first person who has ever made such a comment amoung many people who have read that page over the last 5 years or so.

Pete

8. Apr 27, 2007

### Garth

Hi Eddie, and also welcome from me to these Forums!

In General Relativity gravitational fields also cause time dilation between two clocks at different altitudes (potentials) in the field.

See Gravitation Research Using Atomic Clocks in Space and Space Probe to Test Einstein's "Space-Time Warp" Theory from the Gravity Probe B website. (Just click on the underlined links)

At an altitude of 10,000 km Gravity Probe A measured a time dilation of 7 parts per 1010.

Garth

9. Apr 27, 2007

### spaghed87

Thanks for the replies. I now have a better understanding. Since General Relativity gravitational fields also cause time dilation are they considered for mathmatical formulas that also include the effect Special Relativity has on time dilation?

10. Apr 27, 2007

### Garth

Yes the field equations of GR are fully covariant which means they naturally subsume SR effects, the GR metric tends to the Minkowski SR metric when the gravitational field tends to zero.

Garth

11. Apr 27, 2007

### longshinewoole

difficult in language

Pete,
If you were truely reading time intervals from the clocks, then you were unable to write those 4 equations of yours. You were only able to write: departure time - arrival time = time interval. For instance, the train departed at 1 o'clock and arrived at 2 o'clock. 2 - 1 =1 hour.

In your web page, you used shorter and longer distances to obtain shorter and longer time intervals by means of t = d/v. You did not read from the clocks. If you would use distances to obtain time intervals, you should tell us so from the beginning.

If you told us so from the beginning, I believe you would not be able to tell us time was running slower or faster in such and such frames; you were only able to tell us this: a shorter distance resulted in a shorter time interval, while a longer distance resulted in a longer time interval.

12. Apr 27, 2007

### bernhard.rothenstein

Time dilation is a fundamental relativistic effect proved by experiment.
t=(t'+Vx'/cc)/sqrt(1-VV/cc) (1)
and to mention that in (1) t represents the reading of a clock C of the I frame t' representing the reading of a clock C' of the I' frame when they are instantly located at the same point in space. Clocks C and C' are synchronized with the other clocks of theirs reference frames and all the clocks read a zero time when the origins of the two frames are located at the same point in space.
Presenting (1) as
dt=(dt'+Vdx'/cc)/sqrt(1-VV/cc) (2)
dt and dt' represent non-proper (coordinate time intervals) measured as a difference between two clocks of the involved reference frames respectively located at the points where distant events take place.
For dx'=0 (2) becomes
dt=dt(0)/sqrt(1-VV/cc) (3)
dt representing a coordinate time interval whereas dt(0) represents a proper time interval measured as a difference between the readings of a single clock of I' present at the point where the involved events take place. I think that without stating all that the discussion will never end.
soft words and hard arguments

13. Apr 28, 2007

### MeJennifer

A comment about "clocks slowing down"

I like to make a comment here, I think that this notion of "clocks slowing down" that is frequently presented here on PF is an incorrect description which only leads to more confusion about relativity.

Suppose there are two beacons pulsating a green light each second. Each beacon can also measure the pulse frequency and the light frequency of the other beacon. What is called "time dilation" is when each beacon measures, after discounting the Doppler effect, a reduced pulse frequency and a lower light frequency from the other beacon.

But clearly the light that each beacon sends out is green, and the clock triggers one pulse each second. Obviously something does not slow down or changes frequency just because it is in relative motion with something else.

I think it is correct to state that the rate and frequency of the light signals is somehow observed differently at the destination than at the source, in other words the light signals appear to be "time dilated". But I think it is incorrect to state that the rate and frequency of the light signals at the source are changed.

The other case where "clocks slowing down" is incorrect is when two observers travel between two events on different spacetime paths. If one observer records less time it does not mean that his clock runs slower it simply means that he took a shorter path in spacetime.

A clock that really slows down is a clock that is obviously defect!

Last edited: Apr 28, 2007
14. Apr 28, 2007

### pmb_phy

longshinewoole

Each web page I create was created to help someone with a specific problem that either came up, kept comming up, or I thought was interesting enough to discuss. Each page does not describe relativity in its entirety as this page doesn't either. The reader is assumed to know what it means to take a time measurement and what a time interval is etc. Everything in that page is as it should be. The "t" in that expression is an actual time interval and it is that time interval that is calculated from readings on clocks. In fact that is how I defined it, i.e. Let "t" equal the time time between these two events as measured in S'.. Two readings are taken from clocks. One reading is what the clocks in S' read at event #1 and what the clocks in S' read at event #2. The difference of those time measurements will then be the time interval as measured in S'. The calculations are then expressed in terms of recorded measurements such as speed, that time interval, the proper time interval etc. The "t" is not deduced from a calculation t = d/v. That is just a relationship between physical quantities which are required in the calculations. I could have written "delta t" but I really didn't like that delta in this page. It was merely a matter of aestetics for me.

If you'd like I could create another web page to describe all of this to you and others who are having a difficult time with it. The light clock web page would then have a link to that page in an appropriate place and will be required reading. As of now I don't see a need to change anything.

If the people reading this are unable to understand the calculations and how the physical quantities are obtained then I'll consider creating another page on the meaning of recording events etc. But neither of you seem to have a problem following the derivation, or have you?

Pete

Last edited: Apr 28, 2007
15. Apr 28, 2007

### longshinewoole

Pete,

In your light clock setup, you showed that light departed from the clock at its second position, light emitted, event 1, while the clock was moving to the right at v. Your setup also showed light arrived back at the clock at its 4th position, light detected, event 2. It did not arrive back at the first postion. Such a departure time and arrival time must give a time interval equal to the time interval that the light took to cover the two hypotenuses, that should be t, not r.

Hence I think your r was flawed because the light did not arrive back at the first position. Instead, the light arrived back at the clock when it has reached the 4th position.

16. Apr 28, 2007

### pmb_phy

You do realize that the light clock is of several a standard ways to derive the time-dilation relation, right? It can be found in many text books.

That is its first position. You're looking at two diagrams right next to each other.
Actually its the third position.
It wasn't supposed to. If you measure the time interval between two events and the two events occur at the same location in one frame then they won't occur at the same location at the other. This fact is very important in understanding time dilation.

I highly recommend that you surf the internet for a "Light Clock" since I, of course, did not originate this idea. This is a standard derivation for the time-dilation formula as I mentioned above. If you come across a web site which does a much better job at it than I do then please post the URL so that I can see a better way to explain it. Okay? Thanks.

Pete

Last edited: Apr 28, 2007
17. Apr 28, 2007

### bernhard.rothenstein

time dilation, light clocks and wrist watches

Pete, it was difficult to understand your language. In the web page you created, you used the words "clock" and "measure". Such language must mean that the time intervals will be read from the clocks, not calculated by means of the classic equation t = d/v. For instance, using a clock to measure a time interval, it means a departure time and an arrival time will be read from the clock, and the arrival time minus the departure time will give us the time interval. Since you did not use the tool, the clock, how can you tell us the tool worked slower or faster.

By the word "measure", you meant you would take such a measuring action, reading a departure time and an arrival time. But you did not.

Would you explain why?[/QUOTE]
The light clock is involved with some "wristwatches" located located on each mirror in each of the involved inertial reference frames I and I' respectively. Let C'(0) and C(0) be the clocks on the mirrors located on the overlapped OX(O'X') axes in I and I' respectively. When they read a zero time they are located at the same point in space and the light ray is emitted. When the light rays (emitted and reflected) return to the OX axis clock C'(0) arrives in front of a clock C of the I frame syncronized with C(0) using Einstein's clock synchronization procedure. Clock C'(0) reads now t' whereas clock C reads t. Clock C'(0) measures a proper time interval t'-0 whereas in I observers measure a nonproper time interval t-0 related by
(t-0)=(t'-0)/sqrt(1-VV/cc)
a relationship which can be derived without using the light clock as shown in Asher Peres, "Relativistic telemetry," Am.J.Phys. 55 943-946 (1987).
soft words and hard arguments

18. Apr 29, 2007

### robphy

I disagree.
Given an inertial worldline, a light clock [constructed with a light source and receiver and an inertial distant mirror travelling alongside] functions as a wristwatch for that worldline. There is no need to consider a "wristwatch" on the distant mirror for this light clock.

19. Apr 29, 2007

### bernhard.rothenstein

light clock and wrist watch

I agree with you. But for a more sophisticated reader, the clock on the distant mirror in the moving light clock is synchronized with the clock on the ground mirror by the outgoing light signal in the moving inertial reference frame I'. Detecting that synchronization from the stationary reference frame I, we arrive at a tringle to which we can apply Pithagoras' theorem in the reference frame of the stationary light clock

cctt=vvtt+dd (1)
d invariant distance between the mirrors, ct distance travelled by the light signal that performs the synchronization in the stationary reference frame, vt distance travelled by the wrist watch on the distant mirror of the moving light clock detected from I.
From (1)
ct=d/sqrt(1-VV/cc)=ct'/sqrt(1-VV/cc). (2)
Somebody on the Forum told me that when it is about special relativity we are all childish. Am I?
Thanks and regards

20. Apr 29, 2007

### bernhard.rothenstein

light clock and wrist watches again

Thanks. As you can see the "wristwatch" on the distant mirror is not taken into account in my message.

21. Apr 29, 2007

### robphy

Right... that's why we don't need to add unnecessary complications.

This reminds me of a short conversation with a math professor of mine.
I said "A rectangle is a parallelogram with four right angles."
He stopped me and said "A rectangle is a parallelogram with one right angle."
While what I said is correct, it has unnecessary complication... and follows from his statement [which is easier to prove].

22. Apr 29, 2007

### longshinewoole

language problem

Pete, it is not necessary to look at any other light clocks because all are the same: language problem. Saying one thing and doing quite the opposite. Namely saying to use clocks to measure time intervals, but instead using t=d/v to produce them.

Let us look at yours again. Light departed from the clock. There must be a departure time, say det. Light arrived back at the clock and there must be an arrival time, say art. To derive the time interval is to deduct det from art, namely t = art - det. Since there is only one departure, only one arrival, there can be only one time interval.

But yours or anybody else showed us two different time intervals. These were produced from a shorter distance and a longer distance by means of t=d/v. You turned around and said these were read from the clocks. I am sorry I think there must be something wrong with your doing that.

23. Apr 29, 2007

### JesseM

But this is the whole point, in relativity there is no single way to measure the time between any two events, if you use clocks in different states of motion you'll get different answers, and you have no basis for saying one clock is measuring the "real" time while the others are not.

In the frame where the light-clock is in motion, clocks at rest in that frame will measure t = art - det to be greater than the time between these same two events as measured in the frame where the light-clock is at rest.

24. Apr 29, 2007

### bernhard.rothenstein

wrist watch kight clock

A single question. We use clocks in order to assingn a time coordinate to an event. Please define the point where takes place the event to which we assign the time displayed by a light clock?
Regards

25. Apr 29, 2007

### robphy

We assume a suitable distant mirror of the light clock [as described earlier].
Call the emission event on the inertial worldline of the source "tick 0".
Call the first reception event (after reflection) on the inertial worldline of the source "tick 1".
Call the second reception event (after reflection) on the inertial worldline of the source "tick 2".
etc...

Note: the units of my clock are in "ticks" [of this light clock].

Last edited: Apr 29, 2007