# A thought experiment general relativity

1. Dec 14, 2014

### Shahar

Let's assume the earth was perfect sphere with radios of 6,750,000 meters.
There are 3 clocks , one will not move one will move west in velocity V and the other will move east in the same velocity.
The earth is still rotating. So one clock is going with the earth rotation and the outher is going counter earth rotation.
So when we compare the clocks what will be the result , which clock has ran the slowest?
I think about this problem for quite a time and i just can't find a solution.

2. Dec 14, 2014

Staff Emeritus
3. Dec 14, 2014

### Shahar

Thats raise another question, what every clock sees.
Because i still can't understand how the clock on ground sees the two clocks moving at the same speed for the same time but the results doesn't fit for that view.

4. Dec 14, 2014

Staff Emeritus
Every clock sees itself ticking one second per second. As far as what the other clock sees, two can only be compared when they are next to each other.

5. Dec 14, 2014

### A.T.

In a rotating frame of reference equal speeds don't imply equal proper times.

6. Dec 14, 2014

### Shahar

Can you explain a bit more or send a link for a website that explain this ?

7. Dec 14, 2014

### A.T.

http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]

Page 4, last paragraph starts with the metric for a rotating frame. Note the last term which contains a non-squared $\omega$ (rotation of the frame relative to an inertial frame). The sign of that term depends which way around you move (sign of $d\phi$ : west vs. east), compared to the frame's rotation $\omega$.

Last edited by a moderator: May 7, 2017
8. Dec 15, 2014

### ghwellsjr

Your question, if it has an answer, must be determined according to an Inertial Reference Frame (IRF) in which each of the clocks has a constant speed and that would be the rest frame of the center of the earth. (I'm assuming that you are ignoring all other heavenly bodies.) Then the answer is: whichever clock has the highest speed according to that IRF.

Since you said that one clock will not move but the earth is still rotating, I presume you mean that the clock is rotating along with the earth and has some speed which you can calculate. As long as the other two clocks are at the same latitude as the first clock, then the speeds of the other two clocks are either added or subtracted from that speed accordingly. The clock that is moving easterly would have its speed added to that of the fixed clock and so it would have the highest speed and therefore the lowest tick rate. The clock that is moving westerly would have its speed subtracted from that of the fixed clock and so it would have the lowest tick rate.

Why do you say that "the clock on ground sees the two clocks moving at the same speed for the same time"? Remember, it's the speed of each clock according to the earth-centered IRF which is not rotating with the earth that determines the tick rate of each clock. Therefore, the tick rate of the ground clock will be between the tick rates of the other two clocks. Remember too, that the trip takes longer for the easterly clock because it has to "chase" the ground clock and will have to encircle the earth (according to the (IRF) more than once, probably about one and a half times. The trip for the westerly clock will be shorter, maybe only half way around the earth (according to the IRF). This is consistent with the kinematic prediction in the referenced wikipedia article. So the clock on the ground won't see the other two clocks moving at the same speed for the same time. Does that help you understand what's going on?

9. Dec 15, 2014

### Shahar

Yeah , thanks

10. Dec 15, 2014

### ImperialThinker

The only thing that matters here is if v <<< c then no effect will be observed. The rotation of the earth is meaningless here , because the earth's speed of rotation doesn't add up in any of these cases. As all 3 are moving relative to the motion of the earth.

11. Dec 15, 2014

### A.T.

The OP is obviously interested in speeds where an effect can be observed.

Wrong. See post #7 and:
http://en.wikipedia.org/wiki/Hafele–Keating_experiment#Results

12. Dec 15, 2014

### ImperialThinker

I read that honey, there is a difference, Those clocks were on airliners so they are not really in earth's frame of reference. But here he didn't mention them in airplanes, so normally they would be considered on earth unless stated otherwise.

13. Dec 15, 2014

### ghwellsjr

That's why I told him to look at just the kinematic prediction.

Also, everything is in every frame of reference so your statement that the airliners are not really in the earth's frame of reference is nonsense. Although problems like this can be analyzed in any frame of reference, some are easier than others. I'm not sure if you mean a rotating frame of reference for the earth but that would be very difficult to analyze the problem in. I suggested the IRF in which the center of the earth is at rest as the easiest.

14. Dec 15, 2014

### ImperialThinker

Yes a rotating frame of reference for earth, and as those airliners are flying high above the earth's surface. I am sure you mean that due to earth's pull, the atmosphere also recieves a 'drag' which includes it in earth's frame of reference. But at higher altitudes, that would not be the case. So the earth's motion would be apt for the airliners, because they are not in its frame. but objects on earth are.

The problem with choosing the centre of the earth as reference is that when the quartz atom was being jumped it wasn't done in outer space , it was done on earth. So naturally, the earth's rotation has affected the tradional ''one second'' that we know. So easier would be to choose a rotating frame so that we get the exact numbers while actually calculating

15. Dec 15, 2014

### ghwellsjr

As I said before, everything is in every frame. Usually, when people talk about the frame of something, they mean the frame in which that something is at rest. Sometimes people use the phrase "in a frame" to mean the same thing. But since these clocks are not at rest with respect to each other, there is no frame that we can call "their" frame or the frame that they are in, if you mean by that, a frame in which they are at rest.

Yes, as I said, the clock at rest on the earth's rotating surface would not be at rest in the IRF centered on the earth but rather would be traveling at some speed. The other two clocks would be traveling at different speeds, one higher and one lower. It's a very simple problem to analyze in the earth-centered IRF.

Are you saying that the result I got is wrong or are you just saying that you think there is a better way to solve the problem?

16. Dec 15, 2014

### A.T.

You should also read up on reference frames, if you think they are somehow spatially constrained, and don't contain the airliners. The airliners might not be at rest in the frame of the Earth, but they certainly can be analyzed in that frame. And the rotation of that frame does matter for the analysis of kinetic time dilation.

To OPs misconception is to think that equal speeds as measured in the Earth frame should result in equal time dilation. But that is not true because of the rotation of that frame.

17. Dec 15, 2014

### ImperialThinker

@ghwellsjr : Earth's centre as RF , you have to account for rotatiom of the earth.. if you consider a rotating RF you dont have to account for it. But in case of airliners, you'll have to account for earth's rotation no matter what. Yes i feel this is a better way.

@A.T. : Yes you can consider airlines in rotating reference wrt earth, but earth's motion would not affect its motion so it would ruin the point of taking it as reference because the whole point of it is to make our observation simpler. Same as the sitting in the car, flying over the car, trivial example.

18. Dec 15, 2014

### A.T.

Which is what the OP asks about: same speeds wrt Earth, but different time dilation. And that's because the Earth rotates.

19. Dec 15, 2014

### ghwellsjr

I did account for the rotation of the earth. In the earth-centered IRF, it requires a trivial calculation to determine the speed of the surface of the earth at whatever latitude the OP desires. I went on to say that the easterly clock would have the highest speed and the slowest tick rate.

You are claiming that in the rotating RF you don't have to account for the rotation of the earth. It would seem to me that since both clocks are traveling in opposite directions at the same speed relative to the surface of the earth, then you must be claiming that they will slow down by the same amount and therefore my answer is wrong. Is that what you are claiming?

20. Dec 15, 2014

### ImperialThinker

If 2 people are standing on a plank and moving on opposite direction with the same speed, and the plank itself is moving in one directing with some speed, an observer not on the plank will see one moving faster and one moving slower. While and observer on the plank will see both moving with the same speed in opposite directions.

This is the dilemma we are facing. I am saying, your answer is WRONG, your concept isn't. Because the quartz clock reads 1 second in rotating earth's frame. The time dilation would be different at the centre of the earth (earth's gravity also causes dilation) So, its best do use a frame which we know results about.