# Time dilation on Earth

1. Apr 14, 2015

### Philjhinson

Does time go slower on the hemisphere of earth that is currently rotating counter to earths orbit than the side the side that is rotating with earths orbit

2. Apr 14, 2015

### A.T.

Depends on the reference frame.

3. Apr 14, 2015

### Ibix

It depends where you are watching from. Time dilation isn't an absolute - it's something that is measured by an observer in motion relative to the thing being observed. So, an observer hovering above the Earth (not orbiting, just hovering) would see Earth's clocks ticking at the same rate at all times of day because they are always travelling at the same speed relative to him. An observer hovering above the Sun (i.e. not orbiting the Sun) would see clock rates varying depending on the local time, yes.

4. Apr 14, 2015

### rootone

For anyone actually standing on the Earth, they are not moving in relation to themselves, (obviously).
So their experience of time and the speed of their clocks does not change as far as they are concerned.

5. Apr 14, 2015

### Philjhinson

I get that but if you we're stood on the equator and could look through the earth to clocks on the other side, presumably their relative speed and clocks would be different to your own, sorry if I'm asking a dum question just starting to get my head round this

6. Apr 14, 2015

### Staff: Mentor

The speed between them is zero. By the way, you don't have to say "relative" because there is no other kind.

7. Apr 15, 2015

### Ibix

It's not a dumb question. Rotating frames are complicated in a Newtonian world, and doubly so in a relativistic one.

The clock on the other side of the world must tick at the same rate as one on this side of the world, or a time difference would accumulate such that eventually it was day on both sides of the planet (among other paradoxes). Relativity is strange, but not completely crazy!

The reason for your intuition that the two should tick at different rates is that (leaving gravity out of this) you could fire clocks in opposite directions, so that one touches the equator next to you and momentarily at rest with respect to the ground, and one touches the equator on the opposite side of the world, also momentarily at rest with respect to the ground. These two clocks would have a relative velocity of around 3400kph, and each would report the other ticking slowly. So shouldn't the clocks stuck to the Earth, which are instantaneously moving the same as the straight line clocks, also show time dilation?

In relativity there is no globally applicable definition of "now" at any place except "right here". Time dilation is inseparable from this fact. The earth-bound accelerating clock next to you and the low-flying inertial clock also next to you have different notions of "now, on the other side of the world" and those notions evolve differently. The end result of those differences is that you don't get time dilation between clocks in the rotating case while you do in the straight-line case.

...or so I understand. I might not have it completely right. Here is a thread where some of the more experienced physicists here discuss rotating frames.

8. Apr 15, 2015

### A.T.

What is "speed between them"? In the instantaneous inertial frame of each clock the other clock has non-zero speed.

9. Apr 15, 2015

### BruceW

The problem here is that earth rotates quickly, relative to the time it takes to orbit the sun, so although one clock is placed on the day-time side of the earth, it will soon swing into the night-time side, before we could possibly detect any time dilation.

Maybe a more interesting question, is if we placed one clock on the moon, on the side nearest the earth, and one clock on the moon, on the side furthest from the earth. The moon orbits and rotates in such a way that the same side of the moon is always facing us. So the clock which we initially place on the side of the moon facing us will stay facing us. Therefore, we could leave that clock there, and potentially detect some difference in time dilation, due to the slightly different orbits taken by the two clocks.

edit: actually, it is not obvious to me which clock would tick faster. The clock on the far side of the moon or the near side of the moon... Neither of them are moving along geodesics. Also, they would both be equally close to their nearest Lagrangian points (L1 and L2).

Last edited: Apr 15, 2015
10. Apr 15, 2015

### Staff: Mentor

Yes. Because the guy on the other side of the earth is moving relative to you, you will find that his clocks are running a bit slow (time dilation) and his meter sticks will be a bit short (length contraction) compared with yours.

But please do remember that as far as he's concerned, he's the one who is at rest and you're the one who is moving; he finds that you're the one whose clocks are running slow and and whose lengths are contracted. You're both right.

11. Apr 15, 2015

### Staff: Mentor

I can't parse that. Are you creating an earth-centered, non-rotating frame for one and having the other move with respect to it? Why one and not both? Why would you do that? When I say the speed is zero, I mean nothing more or less than that the distance between them is not changing with time.

12. Apr 15, 2015

### Staff: Mentor

What am I missing here: are you saying that if I watch his clock over a very long period of time, they will accumulate different elapsed times? How is that possible?

13. Apr 15, 2015

### jbriggs444

The velocity of the second clock is zero in the earth-centered rotating frame in which the first clock is constantly at rest. It is non-zero in the inertial frame in which the first clock is instantaneously at rest.

In the instantaneous co-moving inertial frame in which the first clock is at rest, the second clock is running slow. And vice versa. In the earth-centered rotating frame (and using an earth-centered inertial synchronization convention) it is clear that both clocks tick at the same rate.

14. Apr 15, 2015

### Staff: Mentor

OK...so if I drill a hole through the earth and watch the other clock with a telescope, what do I see? Are our clocks deviating in elapsed time/tick rate or not? I don't think they are.

It sounds to me like the "instantaneous co-moving inertial frame in which the first clock is at rest"* changes from one instant to the next, so I'm not seeing a lot of value to using it to describe a situation where I can watch the other clock and see that its tick rate never deviates from my clock's tick rate.

*Just to be sure I'm clear on what this looks like, this frame is drawn with me at its center, rotating in place, right? So at any instant the other observer has a particular velocity in this frame. The catch is that he also has an acceleration that we are ignoring for the instantaneous picture, right?

Last edited: Apr 15, 2015
15. Apr 15, 2015

### A.T.

Constant distance over time doesn't imply equal clock rates, and is also frame dependent.

16. Apr 15, 2015

### 1977ub

Points on or in the Earth are not generally inertial frames in their movement. The center of the Earth is a better approximation of an IF than points on the surface. The Sun is better than the Earth, the barycenter is better than the Sun, and the Galactic core is better than the solar system barycenter.

The question looks like we are attempting to take the POV of someone in the Sun (or solar system barycenter, or some other frame which more approximates an inertial frame than earth-surface points). The earth has points where the rotational velocity and the sun-orbital velocity are additive (relative to the Sun's frame) and some where they are subtractive, leading to differences in the momentary velocities of those points as determined by an inertial/Solar observer. Then those earth surface points where the velocities combine would be a faster moving momentary trajectories with slower clocks than the opposite earth surface points.

This is an effect that would not be relevant to an observer in the center of the Earth, especially if the earth were spinning in space by itself.

17. Apr 15, 2015

### 1977ub

If we are hovering over the Earth, then all of the equator clocks tick at the same rate, but not toward the poles, eh? These would have less and less dilation approaching the poles.

18. Apr 15, 2015

### 1977ub

This is an interesting question to me - I realize that in the simplified version where we treat both my location and the opposite point as inertial frames we can look through across at each other and see in the moment that the other guy's clock is slow. But over time this would not make sense, we have to see each other's clocks as moving at the same rate - and this must have something to do with the fact that in the long run we are *not* moving in inertial frames, and so regular SR questions & answers don't apply...

19. Apr 15, 2015

### Philjhinson

I guess we all spend the same amount of time going faster and slower that it never makes a difference. Certainly can't use this as an explanation of why the nights are getting shorter to my 4 year old

20. Apr 15, 2015

### Staff: Mentor

It's not that regular SR doesn't apply - SR works just fine in accelerated and non-inertial frames (although the math gets more complicated without bringing in any fundamental new insights so you seldom see these cases covered in introductory texts).

What's going on is the we've agreed use to one standard time across the entire surface of the earth because there is no practical way of running a planet-wide civilization without it. As result, the proper time measured by local timepieces anywhere on or around the earth drifts a bit from the coordinated universal time. For example, clocks at the equator and the poles run at slightly different rates relative to one another so they cannot both agree with the standard time, and in practice neither one does. Instead, they have to be reset occasionally to keep them close enough to one another and the standard time.

We seldom notice, both because the discrepancies are small (no one cares if an airliner takes off a few nanoseconds early or late) and because many modern clocks are smart enough to adjust themselves using broadcast time signals, internet time servers, or the 50 and 60 Hz line frequencies in the power grid. However, the GPS system does notice (lose a microsecond and a ship is aground on a reef instead of safely in the channel a few hundred meters away) and much work has gone into making the GPS system smart enough to understand that clocks in different locations on and around the planet tick at slightly different rates.

Indeed, the effectiveness of the GPS system is one of the most convincing arguments for the basic correctness of special and relativity. If those theories didn't work to as many decimal places as we can measure, neither would the GPS system.

21. Apr 15, 2015

### 1977ub

I meant only in the specific sense of comparing tick rates with the other side of the rotating Earth. SR might suggest that these opposite-moving frames will find the other guy's clock to tick more slowly, but it would take extended time spent on these IF trajectories for these measurements to occur and make sense.

22. Apr 15, 2015

### Staff: Mentor

Guys, I think this is being over-complicated and as a result, the OP is walking away with a wrong answer. The way I read the OP's question is that he is asking if the rotation speed and orbit speed with respect to the fixed stars add/subtract to generate time dilation between observers on opposite sides of the earth. In other words, if you are on the equator at sea level on Earth and you drill a hole through the Earth and watch someone else's clock who is also on the equator at sea level and compare it to yours, over the course of days, would you see the other person's clock speed-up and slow down compared to yours as you each move faster and slower with respect to the fixed stars?

If that is indeed the OP's question, it doesn't "depend". The answer is just no. Right?

23. Apr 15, 2015

### Staff: Mentor

OK, that's fair... Although as the GPS example shows... with sensitive enough equipment the "extended time" required may be a matter of microseconds.

24. Apr 15, 2015

### 1977ub

What does "see" mean here? If we accept that neither location is moving in an inertial frame, then any calculation of faster or slower clocks using SR principles would be incorrect, right?

25. Apr 15, 2015

### Staff: Mentor

See means observe. You of course could not use your eyes, you'd have to use electronic means such, like GPS satellites do. One device sends a continuous stream of time signals, the other device compares them to a local clock.
For now I'd like to set aside the every increasingly complexity reference frame choices and just answer the question of what the result of this experiment would be.