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It is said that we can verify time dilation by flying a very accurate clock on a fast jet or spaceship and prove that it registers less time than the clocks on earth. However, the clocks on earth would be moving relative to the clock on the spaceship, and since time always dilates and never goes faster regardless of the direction of relative motion, the clocks on earth should register less time than the clock on the spaceship.

Is this true? Whenever there is a fast-moving object such as a rocket do all clocks on earth really become slow?

If the rocket with the clock landed after moving at relativistic speed, would its clock and the earth's clock again show the same time since during its travel both appeared slow to each other?

Or is all this just an illusion, ie. the clocks just appear to be slow to each other but in actually run at normal speed, and neither is behind when the rocket actually lands?

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I will add up to what mentioned below, that if you are using GPS in your mobile phone, then be informed that it uses time corrections due to this deletion ! so you need no super jets to experience that ;) – TMS Feb 4 at 17:54
@TMS Nice example.. – Sachin Shekhar Feb 4 at 19:57

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Time dilation is real and is measured every day. For example the lifetime of a muon produced in the lab at low velocity is 2.2$\mu$s. However the lifetime of muons generated by high energy cosmic rays in the atmosphere is around 11$\mu$s. Their lifetime is extended by their high speed.

Calculating the time dilation of a plane flying around the Earth is complex because you have to take into account the gravitational time dilation as well as the plane's speed. However it has been done, and indeed experiment shows that if you put an atomic clock in a plane you do indeed measure the predicted time dilation.

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And, gravitational time dilation would beat relativistic time dilation because spaceships move very slowly. – Sachin Shekhar Feb 4 at 20:26

You're referring to what is commonly known as the twin paradox. The Wikipedia page provides several different ways of analysing the situation, but one way to look at it is this -

When the clock on the spaceship leaves earth, it'll experience an acceleration (even if it's a really small acceleration for a long time, or a huge acceleration for a small time) to reach relativistic speeds, and will experience an acceleration again when it has to turn around to return to earth.

Special relativity only claims that inertial frames of reference are equivalent. Since one clock experiences acceleration, and hence is in a non-inertial frame momentarily, the two situations aren't equivalent.

If you view the time dilation due to acceleration as a gravitational acceleration (principle of equivalence) and then do the calculations, the results obtained for both the clock on the earth as well as the clock on the spaceship agree. Again, the wikipedia link contains in detail this argument, as well as other arguments.

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If the rocket with the clock landed after moving at relativistic speed, would its clock and the earth's clock again show the same time since during its travel both appeared slow to each other?

For the rocket to depart from Earth and later return is a scenario in which a loop is closed.

A loop-closing scenario comes out symmetrical only when both participants have traveled the same spatial distance from departure to rejoining.

In an Earth-and-a-spacecraft-on-a-relativistic-journey scenario the spacecraft travels a much longer spatial distance than the Earth.

According to special relativity for the clock that has traveled a longer spatial distance less proper time will have elapsed, as seen when comparing clock times on rejoining.

The comparison on rejoining is a direct comparison, so it's clear and unambiguous. It's actually unhelpful to try an visualise what will be observed during the journey; because of transmission delays those raw observations are not a good perspective on what is happening.


It may be a surprise for you that difference in spatial distance traveled matters in special relativity. You may figure 'in space any individual spacecraft cannot count how many miles it has traveled'. And that is the case: any individual spacecraft cannot count its own mileage. More forcefully, no such individual mileage exists. However, special relativity does imply that you can always evaluate difference in spatial distance traveled. That difference in spatial distance traveled must be thought of as something that is relative between the two participants.

The relativity of special relativity is not a sweeping any motion is relative to something else. There is room for structure and buildup, giving rise to non-symmetrical scenario's, such as the one you ask about.

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