Time Dilation on Earth: Why Hemisphere Matters

[Philjhinson]

Does time go slower on the hemisphere of Earth that is currently rotating counter to Earth's orbit than the side the side that is rotating with Earth's orbit

 

[A.T.]

Does time go slower on the hemisphere of Earth that is currently rotating counter to Earth's orbit than the side the side that is rotating with Earth's orbit

Depends on the reference frame.

 

[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 traveling 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.

 

[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.

 

[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

 

[russ_watters]

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...

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

 

[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.

 

[A.T.]

The speed between them is zero.

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

 

[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).

 

[Nugatory]

if you 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

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.

 

[russ_watters]

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

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.

 

[russ_watters]

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.

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?

 

[jbriggs444]

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?

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.

 

[russ_watters]

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.

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?

 

[A.T.]

When I say the speed is zero, I mean nothing more or less than that the distance between them is not changing with time.

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

 

[1977ub]

Does time go slower on the hemisphere of Earth that is currently rotating counter to Earth's orbit than the side the side that is rotating with Earth's orbit

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.

 

[1977ub]

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 traveling 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.

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.

 

[1977ub]

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

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...

 

[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

 

[Nugatory]

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...

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.

 

[1977ub]

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).

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.

 

[russ_watters]

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?

 

[Nugatory]

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.

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

 

[1977ub]

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?

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?

 

[russ_watters]

What does "see" mean here?

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.

If we accept that neither location is moving in an inertial frame, then any calculation of faster or slower clock susing SR principles would be incorrect, right?

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.

 

[1977ub]

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.

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.

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.

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.

So you don't mean "determine" then way SR normally compares the signals between frames, using Einstein synch etc. You are asking if they sit there with their eye to their telescope and record the ticks, are they slowed down. If that is what you mean, then I think that the ticks will not be regular, but on average they will come at the same rate. I mean, if the observer and source are both speeding up and slowing down wrt an IF, and the light from the source to the observer only travels reliably at C in an IF, then there are small complications to the overall regular rate (accepting that the average speed wrt IF of both ends is the same).

 

[jbriggs444]

So you don't mean "determine" then way SR normally compares the signals between frames, using Einstein synch etc. You are asking if they sit there with their eye to their telescope and record the ticks, are they slowed down. If that is what you mean, then I think that the ticks will not be regular, but on average they will come at the same rate. I mean, if the observer and source are both speeding up and slowing down wrt an IF, and the light from the source to the observer only travels reliably at C in an IF, then there are small complications to the overall regular rate (accepting that the average speed wrt IF of both ends is the same).

It's simpler than that. The tick rate that is "seen" will be exactly regular, always. All inertial reference frames will agree that the "seen" tick rate will match the proper time of the [non-inertial] observer.

This is clear from symmetry. If you picture the Earth rotating in the middle of nowhere, no one time is more special than the next. There is no particular time when the seen clock rate will be slower than average and no particular time when it will be faster. If the laws of physics predict a "seen" tick rate at all, that rate must be constant.

 

[1977ub]

It's simpler than that. The tick rate that is "seen" will be exactly regular, always. All inertial reference frames will agree that the "seen" tick rate will match the proper time of the [non-inertial] observer.

This is clear from symmetry. If you picture the Earth rotating in the middle of nowhere, no one time is more special than the next. There is no particular time when the seen clock rate will be slower than average and no particular time when it will be faster. If the laws of physics predict a "seen" tick rate at all, that rate must be constant.

If someone is in an IF writing down the time of light pulses as they are seen from a regularly emitting beacon, and that beacon is *not* in an IF, the ticks will be received at irregular intervals. Right? Then doubly so if observer is not in an IF.

If you read above, nobody on the surface of the Earth is really in an IF - and moreover a particular earther's the motions relative to an IF are complex, factoring the spinning of the earth, and the orbit of the Earth about the Sun, etc.

But in idealized Earth spinning by itself, with photons coming through an optical fiber through the middle, then no irregularity. But OP had asked about impact of the complex motions of points of the earth.

 

[jbriggs444]

If someone is in an IF writing down the time of light pulses as they are seen from a regularly emitting beacon, and that beacon is *not* in an IF, the ticks will be received at irregular intervals. Right? Then doubly so if observer is not in an IF.

First, there is no such thing as being "in an inertial frame". There is such a thing as being "at rest in an inertial frame".

Second, it is not true that an inertial observer will necessarily see irregular ticks when observing a non-inertial beacon. Take, for instance, the case of an observer sitting inertially in a hole in the center of a transparent earth. The tick rate that he "sees" from an object in orbit may be slowed down. But it will be regular.

Third, it is not true that a non-inertial observer will necessarily see irregular ticks when observing an inertial beacon. For instance, put the beacon at the center of a transparent Earth and put the observer in orbit.

Fourth, it is not true that two irregularities must add to produce an increased irregularity. They can cancel.

But in idealized Earth spinning by itself, with photons coming through an optical fiber through the middle, then no irregularity. But OP had asked about impact of the complex motions of points of the earth.

As long as the Earth is moving in free fall, those complex motions are, at least to first order, irrelevant. And way way above the level of OP's concern.

 

[1977ub]

Second, it is not true that an inertial observer will necessarily see irregular ticks when observing a non-inertial beacon. Take, for instance, the case of an observer sitting inertially in a hole in the center of a transparent earth. The tick rate that he "sees" from an object in orbit may be slowed down. But it will be regular.

Since

First, there is no such thing as being "in an inertial frame". There is such a thing as being "at rest in an inertial frame".

Second, it is not true that an inertial observer will necessarily see irregular ticks when observing a non-inertial beacon. Take, for instance, the case of an observer sitting inertially in a hole in the center of a transparent earth. The tick rate that he "sees" from an object in orbit may be slowed down. But it will be regular.

Third, it is not true that a non-inertial observer will necessarily see irregular ticks when observing an inertial beacon. For instance, put the beacon at the center of a transparent Earth and put the observer in orbit.

Fourth, it is not true that two irregularities must add to produce an increased irregularity. They can cancel.As long as the Earth is moving in free fall, those complex motions are, at least to first order, irrelevant. And way way above the level of OP's concern.

As for your center-of-earth examples, I could have added "generally" to my statements and been correct. I accept the very special case you have presented. Of course my mind had been framed by the OP and his sources & observers on the surface of the Earth.

Your canceling irregularities will be extremely rare. I would find it interesting if you could produce such a case.

The OP went out of his way to include these motions as his concern - explicitly asked about them. Anyone who had said "there are some variations, but they won't be very large" would have been helpful but if someone said such a thing, I didn't see it.

 

[A.T.]

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?

Right. But it doesn't follow from them being at constant distance apart, but rather from symmetry (same distance from an inertial center of rotation and of a spherically symmetrical mass).

 

[Philjhinson]

Thanks again for the replies,so my (uneducated) understanding now is that:-
from an observers point of view on earth, as we are all rotating at the same rate around the centre there would be no difference in time /tick rates. (the orbit around the sun can be ignored, as from the Earth bound POV the sun may as well be orbiting us??sorry if this is complete rubbish!)
However from a point of view outside of the earth, the orbits and velocities observed are complex and there would be constant accelerating and decelerating, with the associated time dilation effects. These would obviously average out over a 24hr period. Am i on the right track?

 

[rootone]

Sounds to me like you are on the right track, although the actual amount of time dilation noticed by your observer outside Earth would be miniscule.
For most purposes time dilation is of no practical consequence at the speed of Earth's rotation, though as Nugatory pointed out, the GPS system does have to be synchronised by a few nanoseconds per day to take account of it, and they are not traveling anywhere near light speed relative to Earth. (They travel at around 14,000 km/hour relative to Earth;s surface

 

[russ_watters]

Right. But it doesn't follow from them being at constant distance apart, but rather from symmetry (same distance from an inertial center of rotation and of a spherically symmetrical mass).

Constant distance apart was a poor wording: it was better when I said they aren't moving with respect to each other (clearly, that's in the Earth centered, rotating frame). If you prefer to say that both are revolving symmetrically around an inertial frame of reference, that's an equally correct, arbitrary choice. I prefer mine because I expect it is easier for the OP to understand. The entire issue here is that invoking a frame where they are both moving implied to the OP an oscillating time dilation. Pointing out that with respect to our common everyday frame they aren't moving makes it clear to the OP that there should be no time dilation. It was a lack of recognition of the principle of relativity (and implication of absolute motion!) that led to the error.

 

[A.T.]

it was better when I said they aren't moving with respect to each other (clearly, that's in the Earth centered, rotating frame).

No, this isn't any better, as a reason for equal clock rates. A clock at the Earth's center would also be at rest in the above frame, but it would run at a different rate than the surface clocks.