A thought experiment general relativity

[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.

 

[Vanadium 50]

You can find the answer here: http://en.wikipedia.org/wiki/Hafele–Keating_experiment

 

[Shahar]

You can find the answer here: http://en.wikipedia.org/wiki/Hafele–Keating_experiment

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.

 

[Vanadium 50]

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.

 

[A.T.]

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.

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

 

[Shahar]

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

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

 

[A.T.]

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

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

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 $\omega$.

 

[ghwellsjr]

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.

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.

You can find the answer here: http://en.wikipedia.org/wiki/Hafele–Keating_experiment

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.

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?

 

[Shahar]

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?

Yeah , thanks

 

[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.

 

[A.T.]

The only thing that matters here is if v <<< c then no effect will be observed.

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

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

 

[ImperialThinker]

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

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.

 

[ghwellsjr]

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.

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.

 

[ImperialThinker]

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.

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 receives 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

 

[ghwellsjr]

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 receives 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.

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.

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

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?

 

[A.T.]

I read that honey, there is a difference, Those clocks were on airliners so they are not really in Earth's frame of reference.

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.

 

[ImperialThinker]

@ghwellsjr : Earth's centre as RF , you have to account for rotatiom of the earth.. if you consider a rotating RF you don't 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.

 

[A.T.]

Yes you can consider airlines in rotating reference wrt earth,

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

 

[ghwellsjr]

@ghwellsjr : Earth's centre as RF , you have to account for rotatiom of the earth.. if you consider a rotating RF you don't 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.

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?

 

[ImperialThinker]

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?

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.

 

[ghwellsjr]

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.

In this case, all observers/objects/clocks are inertial and there are different answers to the question: which clock runs the slowest depending on the IRF we use. There is no unique answer. I think this is what you are saying and I think we agree.

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.

Nobody's at the center of the Earth and we don't care about any effect of gravity since all clocks are on the surface of the earth. Since you claim my answer is WRONG, what is the correct answer?

 

[ImperialThinker]

We care what's at the center of the earth, if you say you take center as RF then your observer is at the centre, and every calculation you do from there on is wrt that observer !

 

[A.T.]

if you say you take center as RF then your observer is at the centre

No, it just means that center is at rest in the chosen RF. It says nothing about any observers.

 

[ImperialThinker]

Since we are talking about relativity shouldn't we only be considering observational frame of reference ...

 

[ghwellsjr]

Since we are talking about relativity shouldn't we only be considering observational frame of reference ...

No, so-called observers can't observe what's going on throughout the frame of reference anyway until light from remote events reaches them. If you insist on an actual observer being located at the spatial origin of an IRF then what happens when you perform the Lorentz Transformation process on the coordinates of the events in one IRF to determine the coordinates of another IRF moving with respect to the original one? There will be no observer at the spatial origin of the new IRF. Does that bother you?

There is no requirement for an actual or virtual observer to be located at the spatial origin of an IRF.

 

[PeterDonis]

Since we are talking about relativity shouldn't we only be considering observational frame of reference ...

You can use any coordinates you like in SR, as long as spacetime is flat or you can ignore any curvature of spacetime that is present. In the case under discussion, since all the objects are on the Earth's surface and we are idealizing the Earth as spherical (or putting all the objects on the equator, which amounts to the same thing), we can ignore the fact that spacetime around the Earth is actually not flat and analyze the scenario using an "IRF" in which the Earth's center is at the spatial origin. (Note that this frame is not rotating, so an object sitting on the Earth's equator and at rest relative to the Earth will be moving in this frame.) But, as others have pointed out, that does not require that there are any observers at the spatial origin. A coordinate chart is just a way of assigning coordinates to events; it doesn't in itself require any observers anywhere.

If you want to use a rotating frame to analyze this scenario, you can, but you will have to make sure to use the correct SR equations for a rotating frame. You will end up getting the same answer as in the IRF, just with more work.