# How does the time dilation is being measured using atomic/quartz clocks?

1. May 7, 2012

### geordano

I understand time dilation exists according to relativity theory.
But what I'm not able to understand is this statement, "clock is ticking faster/slower on a moving object compared to observer".
How is that possible to measure it? or how does the clock (atomic/quartz) it experience the time difference? Because, lets say for atomic clock, the International System of Units (SI) has defined the second as the duration of 9192631770 cycles of radiation corresponding to the transition between two energy levels of the caesium-133 atom. Does number of cycles change when the atomic clock moves in order to observe the difference in time? Or more simply (lets say quartz clock) I'd imagine, frequency of quartz crystal (the physical characteristics) would be same no matter how fast it moves? Then how can we measure the time difference accurately. I'm sure I'm missing something and I'd really appreciate any insight to that.

2. May 7, 2012

### omega_minus

Well the number of cycles of the atomic vibrations does not change. What changes is the rate of time itself. If you watched someone traveling at 0.866c you would see the time between their "ticks" as twice as long as yours. It is not a mechanical effect on the clocks but a 'stretching' of the time dimension (accompanied by a compression of the spatial dimension along the direction of travel) So if your friend on the ship can blink his eyes 5 times a second while he is at rest with respect to you, then when you watch him travel in the ship he will blink them 5 times in the same time it takes his atomic clock to 'tick' 9192631770 times. But it would be 2 seconds in your time. Another thing to contemplate is that he sees you as moving, not him. Therefore your clocks are moving slowly in his frame of reference. Moreover, neither of you would be wrong!

3. May 7, 2012

### geordano

Thanks for the reply, one question though. Realistically, how are we measuring these differences?

4. May 7, 2012

### Mentz114

If an observer measures the frequency shift in an oscillatory signal coming from a transmitter which is in motion wrt to it, the shift is given by γ(1-v/c). This is different from the non-relativistic value (1-v/c) by the factor γ. I don't think a direct measurement of the time dilation factor is possible, but I could be wrong. Someone will soon tell us if this is the case.

5. May 8, 2012

### yuiop

In 1971 they carried out an experiment flying clocks around the world to test relativity. At the Equator the rotational velocity of the Earth is about 1700 kph in the Eastward direction. By putting clocks on aircraft and flying them around the world they expected clocks going Eastward to lose time relative to a clock that remained on the ground and clock going Westward to gain time. They also expected an additional effect due to General Relativity and altitude that would speed up the clocks equally in both directions. The actual results were close to the the theoretical predictions of relativity. The elapsed time of the clocks flown Eastward was less than the elapsed time on the reference clock that remained on the ground. See http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html#c1

6. May 8, 2012

### ghwellsjr

The factor γ(1-v/c) is identically equal to √((1-v/c)/(1+v/c)), the expression for Relativistic Doppler effect for two observers moving directly away from each other at v/c. So for the case of v/c = 0.866, that factor is 0.268. This means that as each observer compares the other ones clock rate to their own, they see it ticking at 0.268 of their own or for each of their own ticks, the other one ticks 3.732 times. This is the same period that omega-minus described when you take into account the light transit time. The period is the reciprocal of the rate. Measuring the period or the rate of the moving clock allows you to calculate the time dilation factor which is 1/γ or √(1-v/c).

To see how this is done, I'm going to use the common term β for v/c and D as the Doppler factor. So for the case of the observers moving away from each other, we start with the Relativistic Doppler Factor:

D = √((1-β)/(1+β))

But we need to know β from the measured Relativistic Doppler Factor D, so we do a little algebra:

D2 = (1-β)/(1+β)
(1+β)D2 = (1-β)
D2+βD2 = 1-β
β+βD2 = 1-D2
β(1+D2) = (1-D2)
β = (1-D2)/(1+D2)

So this tells us that if we measured a Relativistic Doppler Factor of 0.268, we can calculate the speed to be:

β = (1-D2)/(1+D2) = (1-0.2682)/(1+0.2682) = (1-0.0718)/(1+0.0718) = 0.9282/1.0718 = 0.866

Now we want to plug that value of β into the Time Dilation Factor:

√(1-β2) = √(1-0.8662) = √(1-0.75) = √(0.25) = 0.5

That's how the measurement of time dilation would be done (along with some calculation) for the specific case of two observers traveling inertially away from each other.

If the observers are traveling toward each other, then the Relativistic Doppler effect is inverted:

D = √((1+β)/(1-β))

which means they each see the other ones clock ticking faster than their own but when they factor out the light transit time, the time dilation factor is the same.

So let's do a little more algebra:

D2 = (1+β)/(1-β)
(1-β)D2 = (1+β)
D2-βD2 = 1+β
-β-βD2 = 1-D2
-β(1+D2) = (1-D2)
β = -(1-D2)/(1+D2)

Now this is almost the same thing we got for the observers moving away from each other, the only difference being the sign change but when we calculate the Time Dilation Factor, it comes out the same due to the value of β being squared. So even though in the case of the two observers approaching each other and they see each others clock ticking faster than their own, when they factor out the light transit time, they get the same value for time dilation.

7. May 11, 2012

### the_emi_guy

8. May 12, 2012

### ghwellsjr

As this thread has come to the fore, I see that I made a mistake in my previous post which is fixed here:

9. May 12, 2012

### Bob S

Physicists at Brokkhaven National Laboratory stored muons (lifetime 2.2 microseconds) at a gamma of about 29.4 in a storage ring, and stretched out (dilated) the lifetime to about 64 microseconds and measured it. See http://www.g-2.bnl.gov/hepex0401008.pdf

10. May 12, 2012

### Naty1

It is a phenomena that SEEMS strange when you first meet!! In some sense it will always
remain strange, but you'll accept it...a little like like a 'strange' friend you get used to....and you to them!! Two views of things.

It is measured by COMPARING the elapsed times of clocks from different frames.
A good clock carried by YOU always ticks at the same rate. A good clock carried by someone else always ticks at the same rate for that observer, locally. But when one clock either accelerates and returns, or one clock is placed within a different gravitatonal potential, and then returned for comparsion at the same place and time, different times will have elapsed! Those coincident clocks, back together again will then remain ticking at the same rate.

You can visualize different tick rates as seen by observers in different frames here:

http://en.wikipedia.org/wiki/Time_d...nce_of_time_dilation_due_to_relative_velocity

Some math and other examples are also provided in the article.

Regarding GPS clock issues, When I used to boat in Maine where there is loads of frequent fog, I elected two GPS receivers and separate displays to insure reliability when navigating in fog, one was WAAS and the other was DGPS...They use different satellites and different clock correction schemes, so I figured I'd have better reliability against failure. About once every other year or two one or the other would fail to plot my location and direction for maybe 15 minutes to even and hour or more...never figured out why and even restarting [like a home computer] did not help. I even talked with a US Coast Guard officer at Station Rockland...they had no record of a system failure...go figure!

Last edited: May 12, 2012