# SR time dilation real or perceived

1. Feb 4, 2010

### cbd1

I have a simple question. If we put a clock in a satellite in orbit for some amount of time, and then brought it back to earth, will the clock's reading of time passed be less than a clock on earth's measured passed time. *(This is strictly ignoring any general relativistic time dilation!)

In other words, I know that it appears that the clock on the moving satellite is ticking less frequently due to special relativistic effects, while viewing from earth, but is that clock actually ticking more slowly? That is, will less time have passed for the clock that was in orbit (after correcting for GR time dilation) once it is brought back to earth?

[What I am really asking, in regards to theory, from the viewpoint of the satellite, it would appear that the clock on earth is the one in motion. In that respect, an observer on the satellite would expect less time to have passed on the Earth clock when returning. Both cannot be true! So, it must be merely a percieved effect, with neither clock actually ticking more slowly than the other (due to relative velocity).]

2. Feb 5, 2010

### hamster143

If you take a rocket ship and use engines to make it travel in a circular path in gravity-free environment, the clock on that ship would run slower than a clock that's at rest.

If you replace the rocket ship with a satellite in a circular free-falling orbit, the clock on the satellite is now in an inertial frame, and the clock on the ground undergoes accelerated movement. Therefore, the clock on the satellite would run faster.

On top of that, there will be a GR effect due to the fact that the two clocks are at different elevations. The GR effect will likely dominate.

3. Feb 5, 2010

### cbd1

So, what you are saying is that time is passing more quickly on the satellite? Remember, this is ignoring general relativistic time dilation.

Edit: It cannot be both that if you look from the satellite time is passing more slowly on earth and also that if you look from earth time is passing more slowly on the satellite.

Last edited: Feb 5, 2010
4. Feb 5, 2010

### atyy

Yes. The tick rate of a clock is frame dependent. Tick rate means you compare the clock's ticks with a multitude of other clocks it encounters on its journey, and those other clocks must somehow be synchronized, which can only be done in a frame dependent way.

5. Feb 5, 2010

### yuiop

This is correct. If the instantaneous velocity of the satellite relative to the "stationary stars" is greater than the surface of the Earth relative to the stationary stars, then time will pass more slowly on the satellite than on the Earth surface and this will be confirmed when the clock is brought back down to Earth. (This is strictly ignoring any general relativistic time dilation!)
No, it's not just percieved.
Sorry, this is not correct if you are saying that the clock on the ground will run slower due to accelerated movement, you are already getting into the realms of GR and you seem to be assuming that the satellite will not also experience gravitational time dilation, which is not true. The time dilation of the satellite is a combination of velocity time dilation and gravitational time dilation due to its height and the same is true for the clock on the ground.

Have a look at this article on the Hafele and Keating experiment http://hyperphysics.phy-astr.gsu.edu/HBASE/Relativ/airtim.html#c3 which breaks down the time dilation of clocks flown around the world into velocity and gravity components. This is not exactly the same situation as a satellite clock, as the aircraft are not in freefall, but the principles are basically the same.

Last edited: Feb 5, 2010
6. Feb 5, 2010

### cbd1

How can any stars be "the stationary stars"? All stars are in motion. You cannot pick any one reference frame to be the master reference frame. Aren't they all relative to one another, with none being able to logically be argued more stationary than the other.

7. Feb 5, 2010

### yuiop

You are basically correct as far as linear motion is concerned, but rotation has an absolute nature. You can be sure whether you are rotating or not, where as you can not be sure if you are stationary or moving in a straight line in the linear sense. You probably can find a master reference frame for rotation.

Consider for example a geostationary orbit. The satellite stays above the same spot on the Earth and so it appears to have no velocity relative to the Earth's surface and yet it stays in orbit because it has rotational motion relative to the non rotating stars. On the other hand an object could be stationary with respect to the non rotating background and have a very high velocity relative to the Earth's surface, but it can not stay in orbit. What is important as far as orbits are concerned is the real rotational velocity relative to the background and not the velocity relative to the Earth's surface.

How does an observer on the surface of the Earth explain the the bulge of the Earth at the equator if he decides to imagine the Earth is stationary?

Last edited: Feb 5, 2010
8. Feb 5, 2010

### cbd1

This would mean that if a plane is flying around Earth in the opposite direction of it's rotation, that the plane is moving with slower velocity than the ground-based clock (according to the reference of the sun)!

If it is true, then the direction that the plane flies will either cause the rate of time on the plane to slow or speed up..

9. Feb 5, 2010

### yuiop

Exactly!

In the link http://hyperphysics.phy-astr.gsu.edu/HBASE/Relativ/airtim.html#c3 I gave earlier the results of the actual experiment were:

Predicted: Time difference in ns
Eastward
Gravitational 144 +/- 14
Kinematic -184 +/- 18
Net effect -40 +/- 23
Observed: -59 +/- 10

Predicted: Time difference in ns
Westward
Gravitational 179 +/- 18
Kinematic 96 +/- 10
Net effect 275 +/- 21
Observed: 273 +/- 21

Looking at only the kinimatic (velocity) effect and ignoring the gravitational effect, the East going plane lost 184 ms and the West going plane gained 96 ms relative to the ground clock.

Last edited: Feb 5, 2010
10. Feb 6, 2010

### cbd1

Okay, thank you for the extra info.

Now I'm wondering why the west going clock went faster than the ground clock. By special relativety, wouldn't both planes be considered to move the same speed (roughly) when compared only to the clock on the ground. That is, from the clock on the ground's perspective, it thinks it is stationary with both planes moving away from it with the same speed, only in different directions. So, it would expect to see that the time on those clocks to be ticking more slowly.

Where am I confused here? I know there is reference to a hypothetical clock at the center of the Earth, but that was never an actual comparison by my understanding..

Last edited: Feb 7, 2010
11. Feb 6, 2010

the planes are different, they may share the same Hz but the distance traveled increases as the radius does, so to match Hz for Hz the clock on a larger radius has to increase it's velocity proportionally.

12. Feb 6, 2010

this is also an inertial frame of reference as opposed to a non-inertial where at large enough radius the frame of reference would exceed the limit of c.

13. Feb 6, 2010

### cbd1

To clarify, my understanding of SR would say that the planes have relative velocity, so they should go slower than the ground based clock no matter the direction they fly.

Last edited: Feb 7, 2010
14. Feb 6, 2010

### Staff: Mentor

Remember that the surface of the earth is not at rest in an inertial frame. So you have to account for the rotation of the earth. That means that in the earth centered inertial frame the eastward clock is slower than the rest clock, but the westward clock is faster.

15. Feb 10, 2010

### jason12345

The clocks tick at the same rate, but they take different paths through space-time. The length of the paths connecting the same events will be generally different for each.

16. Feb 10, 2010

### JesseM

As kev said, this is actually incorrect, although you're right that the orbiting clock is moving inertially...as you probably know, all freefall paths are geodesics which are locally inertial (so an observer traveling on that path feels weightless just like an inertial observer in special relativity) due to the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html [Broken] for an explanation from physicist Sean Carroll on the "Cosmic Variance" blog.

Last edited by a moderator: May 4, 2017