Exploring Time Dilation Due to Velocity in Special Relativity

[sisoev]

In a frame of reference at rest with respect to the center of the earth, the clock aboard the plane moving eastward, in the direction of the Earth's rotation, is moving faster than a clock that remains on the ground, while the clock aboard the plane moving westward, against the Earth's rotation, is moving slower.

Why "at rest with respect to the center of the earth"?
Both clocks are in relative velocity.
Does the mass of the frame of reference has something to do with the determination of the frame at rest?

 

[sylas]

Why "at rest with respect to the center of the earth"?
Both clocks are in relative velocity.
Does the mass of the frame of reference has something to do with the determination of the frame at rest?

Because everything works much more easily when analyzed in an inertial frame. A clock on the surface, or on an aircraft flying in either direction, is an accelerating frame, not an inertial frame.

Cheers -- sylas

 

[sisoev]

Because everything works much more easily when analyzed in an inertial frame. A clock on the surface, or on an aircraft flying in either direction, is an accelerating frame, not an inertial frame.

Cheers -- sylas

But if the speed of the aircraft is constant don't we treat it as inertial frame?
On the other hand if we treat it as accelerating frame, isn't it correct to say that the results of the clocks are valid only for accelerating frames?
In this case aren't we suppose to take in consideration the different velocity escape in both directions which I think will give different gravity on both air-crafts, because they kind of have different speed at the same latitude and altitude.

 

[sylas]

But if the speed of the aircraft is constant don't we treat it as inertial frame?

No; only if the velocity is constant. Because the aircraft is moving in a circle, it is an accelerated motion and hence the aircraft is not at rest in any inertial frame.

On the other hand if we treat it as accelerating frame, isn't it correct to say that the results of the clocks are valid only for accelerating frames?

I don't know what you mean by this. A clock measures proper time along its world line, regardless of what motions it undergoes. That is, a clock is ALWAYS valid. The easiest way to get proper time, which is what the clock measures, is to have the motions given with respect to some inertial frame.

In this case aren't we suppose to take in consideration the different velocity escape in both directions which I think will give different gravity on both air-crafts, because they kind of have different speed at the same latitude and altitude.

As long as you have the speeds with respect to an inertial frame, you can do the calculation. The methods used for the aircraft are described at the hyperphysics website. See Hafele and Keating Experiment. This page takes you through the correct way to do the calculations.

We describe all the motions with reference to an inertial frame in which the center of the Earth is at rest. Technically, even this is not inertial, but the circular motions around the Sun are not significant over the few days of the experiment.

Cheers -- sylas

 

[sisoev]

On the other hand if we treat it as accelerating frame, isn't it correct to say that the results of the clocks are valid only for accelerating frames?

I don't know what you mean by this. A clock measures proper time along its world line, regardless of what motions it undergoes. That is, a clock is ALWAYS valid. The easiest way to get proper time, which is what the clock measures, is to have the motions given with respect to some inertial frame.

An inertial frame of reference is a non-accelerating frame of reference.
In accelerating frame of reference we have different g-force which will affect the clocks.
Then the different time of the clocks will be related to the gravitational force in the accelerating frame. Obviously both air-crafts are in accelerating frames with different speed.

As long as you have the speeds with respect to an inertial frame, you can do the calculation. The methods used for the aircraft are described at the hyperphysics website. See Hafele and Keating Experiment. This page takes you through the correct way to do the calculations.

Thanks, I did read the results, and that is why I posted my question
I don't see in the results the change of the gravitational force due to the speed of the accelerating frames.

We describe all the motions with reference to an inertial frame in which the center of the Earth is at rest. Technically, even this is not inertial, but the circular motions around the Sun are not significant over the few days of the experiment.

How would we do that with two REAL inertial frames?
They both can be considered at rest and although the observers in them could see different running time in the other frame, both clocks will show the same value.
Am I correct?

Thank You

 

[sylas]

An inertial frame of reference is a non-accelerating frame of reference.
In accelerating frame of reference we have different g-force which will affect the clocks.

That's not a particularly good way to think about it. G-forces don't affect clocks. For example, consider a clock inside a massive spherical shell; or else one at the center of the Earth. The G-force is zero. Consider another clock infinitely far from the Earth. The G-force is again zero. There is, however, a time dilation difference between the two clocks related to their differing positions within the gravitational fields.

Then the different time of the clocks will be related to the gravitational force in the accelerating frame. Obviously both air-crafts are in accelerating frames with different speed.

That's not the way to calculate it The pseudo-gravitational forces experienced in an accelerating frame are not what relates to the different times. It is related to the differences in velocity... entirely independent of accelerations and g-forces.

Thanks, I did read the results, and that is why I posted my question
I don't see in the results the change of the gravitational force due to the speed of the accelerating frames.

That's because you don't use that. You use the differing gravitational potential from the Earth's gravitational speed, for a gravitational time dilation; and the different speeds, for the kinetic time dilation. The page illustrates the correct way to do this calculation.

How would we do that with two REAL inertial frames?
They both can be considered at rest and although the observers in them could see different running time in the other frame, both clocks will show the same value.
Am I correct?

Thank You

When you have two clocks which are a rest in different inertial frames, you are not going to be able to bring them back to a single point to compare them directly. Hence you will not be able to speak about them showing the same value with indicating one specific inertial frame within which you define the time at which the value are taken for comparison.

The aircraft clocks can be compared, because they use accelerated motions to travel in circles and come back to the same point again.

Cheers -- sylas

 

[yuiop]

But if the speed of the aircraft is constant don't we treat it as inertial frame?

No; only if the velocity is constant. Because the aircraft is moving in a circle, it is an accelerated motion and hence the aircraft is not at rest in any inertial frame.

This is not correct. A satellite moves in a circle and yet it is an inertial frame. A slow flying aircraft even when flying in a straight line is not an inertial reference frame. As long as you can feel a force acting on your butt when you sit in the aircraft then it is not an inertial frame. An aircraft that follows a special elliptical trajectory can aproximate an inertial frame for a short while and astronauts are trained to experience weightlessness using this method. See http://en.wikipedia.org/wiki/Vomit_Comet

 

[Ich]

This is not correct. A satellite moves in a circle and yet it is an inertial frame.

Sylas is talking about SR here. SR knows nothing about gravity or the equivalence principle, so gravity is a force like any other.
You can then additionally correct for gravitational effects, which are described by Newtonian gravitational potential.

This is definitely not a GR treatment; it's rather a patchwork of weak field approximations. But it's enough.

 

[sylas]

Sylas is talking about SR here. SR knows nothing about gravity or the equivalence principle, so gravity is a force like any other.
You can then additionally correct for gravitational effects, which are described by Newtonian gravitational potential.

This is definitely not a GR treatment; it's rather a patchwork of weak field approximations. But it's enough.

Exactly. (And the term I would use for what kev is describing is a timelike geodesic; not inertial frame.)

 

[yuiop]

Sylas is talking about SR here. SR knows nothing about gravity or the equivalence principle, so gravity is a force like any other.
You can then additionally correct for gravitational effects, which are described by Newtonian gravitational potential.

This is definitely not a GR treatment; it's rather a patchwork of weak field approximations. But it's enough.

The OP is obviously talking about the Hafele and Keating experiment http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html#c1 and in that link it can be seen that in the aircraft flying in the Westward direction, the time dilation due to gravity is greater than the time dilation due to velocity and the Westward clocks actually speed up, rather than slow down as you would expect from SR considerations alone. This experiment definitely involves GR and treating the gravitational effects as minor corrections is misleading when they are the greatest effect.

 

[Ich]

it can be seen that in the aircraft flying in the Westward direction, the time dilation due to gravity is greater than the time dilation due to velocity and the Westward clocks actually speed up, rather than slow down as you would expect from SR considerations alone.

But that's actually what you expect from SR considerations, as sylas tried to explain. It has nothing to do with gravity effects being greater, as both have the same sign. Look at the next box in your link.

From a pedagogical point of view, gravitation is rather a distraction in this experiment, not an essential ingredient.

 

[yuiop]

Why "at rest with respect to the center of the earth"?
Both clocks are in relative velocity.
Does the mass of the frame of reference has something to do with the determination of the frame at rest?

What they mean is that if you compare the velocities of the aircraft relative to the clock on the surface of the Earth you have to allow for the velocity of the clock on the surface of the Earth due to the Earth's rotation. By considering velocities relative to a frame "At rest with respect to the centre of Earth" they are in effect considering velocities relative to a non rotating frame which coincides roughly with a frame at rest respect to the distant stars. In such a frame a clock on the surface of the Earth has a velocity and this has to be allowed for.

 

[yuiop]

But that's actually what you expect from SR considerations, as sylas tried to explain. It has nothing to do with gravity effects being greater, as both have the same sign. Look at the next box in your link.

From a pedagogical point of view, gravitation is rather a distraction in this experiment, not an essential ingredient.

OK, agreed, we can take gravity out of the equation by considering an airport on top of a very high mountain. One aircraft flies East and the other flies West around the world. Both maintain an altitude equivalent to the height of the mountain and both return to the mountain airport. A clock at the airport is the reference. The clock of the West going aircraft shows more elapsed time than the time elapsed on the airport clock when it lands.The East going aircraft lands later, but its clock shows less elapsed time than the time elapsed on the airport clock when it lands. The reason the West going clock shows more elapsed time than the airport clock is because the airport clock has velocity due the rotation of the Earth. Basically it is the Sagnac effect.

 

[sisoev]

An inertial frame of reference is a non-accelerating frame of reference.
In accelerating frame of reference we have different g-force which will affect the clocks.

That's not a particularly good way to think about it. G-forces don't affect clocks. For example, consider a clock inside a massive spherical shell; or else one at the center of the Earth. The G-force is zero. Consider another clock infinitely far from the Earth. The G-force is again zero. There is, however, a time dilation difference between the two clocks related to their differing positions within the gravitational fields.

Well, I cannot imagine that and know the result but even if I could imagine it, it wouldn't help me, since there is no acceleration in the given example.
I can imagine though that the gravitational force in a certain altitude will be different for objects having different speed (in our case, acceleration). Clocks are physical objects and they should be affected by that.

Then the different time of the clocks will be related to the gravitational force in the accelerating frame. Obviously both air-crafts are in accelerating frames with different speed.

That's not the way to calculate it The pseudo-gravitational forces experienced in an accelerating frame are not what relates to the different times. It is related to the differences in velocity... entirely independent of accelerations and g-forces.

That is part of my question - why that is not the way to calculate it?
Why the pseudo-gravitational forces experienced in an accelerating frame are not what relates to the different times? Aren't this forces affecting the clocks the way Earth gravitation affects them?
If I can feel that forces on my butt (as kev said ) why the clocks wouldn't feel it?

When you have two clocks which are a rest in different inertial frames, you are not going to be able to bring them back to a single point to compare them directly. Hence you will not be able to speak about them showing the same value with indicating one specific inertial frame within which you define the time at which the value are taken for comparison.

Then we don't know nothing about this issue in inertial frames, do we?
Is it only assumption which is "proven" with accelerating frame experiment?
Can we actually use this experiment and apply it for inertial frames?
Are we allowed to do this assumption if by definition inertial frame is a non-accelerating frame.
Aren't we braking some rules here?

 

[sylas]

What they mean is that if you compare the velocities of the aircraft relative to the clock on the surface of the Earth you have to allow for the velocity of the clock on the surface of the Earth due to the Earth's rotation. By considering velocities relative to a frame "At rest with respect to the centre of Earth" they are in effect considering velocities relative to a non rotating frame which coincides roughly with a frame at rest respect to the distant stars. In such a frame a clock on the surface of the Earth has a velocity and this has to be allowed for.

Yes, exactly. Then there is also a small additional dilation from the different altitudes of the clocks, which is where they use what is (I think) the weak field approximation to take that into account.

Cheers -- sylas

 

[sisoev]

What they mean is that if you compare the velocities of the aircraft relative to the clock on the surface of the Earth you have to allow for the velocity of the clock on the surface of the Earth due to the Earth's rotation.

I understand that
That is the reason they used two air-crafts.

By considering velocities relative to a frame "At rest with respect to the centre of Earth" they are in effect considering velocities relative to a non rotating frame which coincides roughly with a frame at rest respect to the distant stars. In such a frame a clock on the surface of the Earth has a velocity and this has to be allowed for.

Thank you, I understand it now
So basically they set with that words a non rotating frame, for the purpose of the experiment.
Is that right?

But the experiment still don't use two inertial frames, because we cannot consider the aircraft at rest, yes?

 

[starthaus]

But the experiment still don't use two inertial frames, because we cannot consider the aircraft at rest, yes?

This is WHY the experiment is judged from the perspective of the INERTIAL frame centered in the center of the Earth. That was your original question, remember?

 

[yuiop]

This is WHY the experiment is judged from the perspective of the INERTIAL frame centered in the center of the Earth. That was your original question, remember?

This is a good point and the word INERTIAL is key here. It means the Earth centered frame has to be non-rotating and gyroscopes or Sagnac devices spread out in that frame should show no rotation.

 

[starthaus]

This is a good point and the word INERTIAL is key here. It means the Earth centered frame has to be non-rotating and gyroscopes or Sagnac devices spread out in that frame should show no rotation.

Interestingly, if we executed a Sagnac-type experiment (or a Michelson-Gale type experiment) very close to the center of the Earth (think "Journey to the center of the Earth" !), we wouldn't get a null result. We would get the same exact result as on the surface since the Earth is spinning. So, you need to assume that the z axis is aligned with the Earth rotational axis while the x-axis is tangent to the revolution trajectory. For short revolution distances, this approximates a non-rotating system of coordinates since the Earth trajectory approximates a straight line.

 

[sisoev]

This is WHY the experiment is judged from the perspective of the INERTIAL frame centered in the center of the Earth. That was your original question, remember?

Yes but my original question was with the wrong assumption that they treat all frames as inertial.
Now I understand this part of the problem.

I still don't have convincing answer about the different gravitation on the airplanes due to the different velocity in relation to Earth.
I connected this question in the OP with the different escape velocity in both directions just as a point that my assumption is not based on fictitious (pseudo) force.

Some help, please

 

[starthaus]

I still don't have convincing answer about the different gravitation on the airplanes due to the different velocity in relation to Earth.

Your question makes no sense. There is no connection between "different gravitation" and "different velocity". So, the "different gravitation" is NOT "due to different velocity".
Did you even try to read the page on the Haefele-Keating experiment that sylas linked in for you?

 

[sisoev]

Correct, if we executed a Sagnac-type experiment (or a Michelson-Gale type experiment) very close to the center of the Earth (think "Journey to the center of the Earth" !), we would get a null result.

I think that "null result" can be measured only on the fictitious axis of the Earth

 

[yuiop]

I still don't have convincing answer about the different gravitation on the airplanes due to the different velocity in relation to Earth.
I connected this question in the OP with the different escape velocity in both directions just as a point that my assumption is not based on fictitious (pseudo) force.

Some help, please

As I tried to explain in #13, if you keep all clocks in the experiment at the same height you can pretty much ignore the gravitational effects as it will be equal for the clocks involved.

I think what you are talking about is something that is sometimes called "effective potential" which combines the effects of angular velocity with gravitational potential into one combined effect, but that is probably over complicating things at this stage and you should try and keep gravitational and time dilation effects segragated.

Here is a link to page that discusses GR effects in terms of effective potetial http://www.fourmilab.ch/gravitation/orbits/

As you can see the maths is a bit more involved than that used in the link about the Hafele Keating experiment given to you by Sylas.

 

[sisoev]

As I tried to explain in #13, if you keep all clocks in the experiment at the same height you can pretty much ignore the gravitational effects as it will be equal for the clocks involved.

I have no problem with the gravity of the Earth taken in consideration in the experiment results.
That is not my point.

I think what you are talking about is something that is sometimes called "effective potential" which combines the effects of angular velocity with gravitational potential into one combined effect, but that is probably over complicating things at this stage and you should try and keep gravitational and time dilation effects segragated.

OK, let me explain how do I see the "velocity-gravity" problem which bugs me
Every orbiting around the Earth object tends to "free fall" on the surface.
If two objects are free falling and pass through a point at certain altitude with different speed they'll measure different gravitational force at that point.
The two airplanes have to maintain same altitude with different speed(velocity).
That should result (at least I see it that way) in different gravitational force on both airplanes.

Here is a link to page that discusses GR effects in terms of effective potetial http://www.fourmilab.ch/gravitation/orbits/

As you can see the maths is a bit more involved than that used in the link about the Hafele Keating experiment given to you by Sylas.

I rate you as the best helper in Physics Forums by now
Thank You!

 

[starthaus]

The two airplanes have to maintain same altitude with different speed(velocity).
That should result (at least I see it that way) in different gravitational force on both airplanes.

No, on both accounts:
- Gravitational force is not a function of the speed.
- The elapsed time on the clocks is not a function of gravitational force, it is a function (among other things) of gravitational potential

Why don't you read the wiki page sylas provided for the Haefele-Keating experiment? It explains everything in simple terms.

 

[sisoev]

No, on both accounts:
- Gravitational force is not a function of the speed.
- The elapsed time on the clocks is not a function of gravitational force, it is a function (among other things) of gravitational potential

Why don't you read the wiki page sylas provided for the Haefele-Keating experiment? It explains everything in simple terms.

If I didn't know about the experiment and the results in that link, I wouldn't have my questions, starthaus
I think that you don't understand my point or what bugs me for that matter

 

[starthaus]

If I didn't know about the experiment and the results in that link, I wouldn't have my questions, starthaus
I think that you don't understand my point or what bugs me for that matter

What bugs you is your lack of basic knowledge. In order to fix that, you need to learn.

 

[yuiop]

I rate you as the best helper ...

Thanks! Recognition at last!

More seriously let's take a look at the gravity velocity relationship in more detail.

The two airplanes have to maintain same altitude with different speed(velocity).
That should result (at least I see it that way) in different gravitational force on both airplanes.

Consider this thought experiment:
A smooth level road has been built around the equator. Two rocket cars race off in opposite directions. The one going West finds that when it gets to about 1000 mph relative to the road, it is effectively stationary in the Earth Centered Inertial Frame (ICIF). At this velocity it feels the maximum force of gravity pressing down on it as measured by stress sensors in the suspension system. The rocket car going East finds that at about 35,000 mph the force pressing down on it is zero and it is effectively orbiting. Now in some sense it seems like the cars are experiencing gravity differently, but that is not true as far as time dilation is concerned.

The time dilation due to velocity is given by 1/sqrt(1-v^2) and the time dilation due to gravitational potential is 1/sqrt(1-2GM/r) in units where c=1. The West going car is effectively stationary in the ECIF so the time dilation in the ECIF can be calculated by using 1/sqrt(1-2GM/r) only. Now it might be tempting to think that the East going car is effectively weightless and not subject to gravity and conclude that its time dilation in the ECIF can be calculated from 1/sqrt(1-v^2) only, but that would be wrong. To calculate the time dilation of the East going car you need to use 1/sqrt(1-v^2)*1/sqrt(1-2GM/r). The time dilation due to gravitational potential has to be taken into account, whatever the velocity of the object. Its the gravitational potential and not the gravitational force, that is important as far as time dilation is concerned.

 

[sisoev]

Thanks! Recognition at last!

More seriously let's take a look at the gravity velocity relationship in more detail.

Consider this thought experiment:
A smooth level road has been built around the equator. Two rocket cars race off in opposite directions. The one going West finds that when it gets to about 1000 mph relative to the road, it is effectively stationary in the Earth Centered Inertial Frame (ICIF). At this velocity it feels the maximum force of gravity pressing down on it as measured by stress sensors in the suspension system. The rocket car going East finds that at about 35,000 mph the force pressing down on it is zero and it is effectively orbiting. Now in some sense it seems like the cars are experiencing gravity differently, but that is not true as far as time dilation is concerned.

The time dilation due to velocity is given by 1/sqrt(1-v^2) and the time dilation due to gravitational potential is 1/sqrt(1-2GM/r) in units where c=1. The West going car is effectively stationary in the ECIF so the time dilation in the ECIF can be calculated by using 1/sqrt(1-2GM/r) only. Now it might be tempting to think that the East going car is effectively weightless and not subject to gravity and conclude that its time dilation in the ECIF can be calculated from 1/sqrt(1-v^2) only, but that would be wrong. To calculate the time dilation of the East going car you need to use 1/sqrt(1-v^2)*1/sqrt(1-2GM/r). The time dilation due to gravitational potential has to be taken into account, whatever the velocity of the object. Its the gravitational potential and not the gravitational force, that is important as far as time dilation is concerned.

Haha
Very entertaining thought experiment
I have to think about it, but on a first look, why the car should reach 1000 mph in order to become stationary in the Earth Centered Inertial Frame?
That would be true if orbiting the earth.
On the Earth it would be achieved by not moving, isn't it?

 

[yuiop]

I have to think about it, but on a first look, why the car should reach 1000 mph in order to become stationary in the Earth Centered Inertial Frame?
That would be true if orbiting the earth.
On the Earth it would be achieved by not moving, isn't it?

Nope, when you are stationary wrt the surface of the Earth, you are moving at about 1000 mph relative to the ECIF. The Earth's surface is not stationary according to the ECIF! That is sort of the whole point of this thread.

 

[sisoev]

Nope, when you are stationary wrt the surface of the Earth, you are moving at about 1000 mph relative to the ECIF. The Earth's surface is not stationary according to the ECIF! That is sort of the whole point of this thread.

OK, I understand your point, but you confused me even more now.
If ECIF is the whole point in this discussion, and the Earth surface is not stationary to it, then the clock on the Earth surface was not stationary to the air-crafts in the Hafele and Keating Experiment. So, according to your explanation all clocks were in accelerating frames with different speed. No wonder then why they moved differently

 

[sisoev]

KEV, I'm back hungry for knowledge

Why in your experiment the gravitation in the rotating frame of reference will change when becomes stationary to the ECIF?
Shouldn't we look at the velocities in your experiment as in http://en.wikipedia.org/wiki/Rotating_reference_frame#Relation_between_velocities_in_the_two_frames", where the forces are fictitious and have nothing to do with Hafele-Keating experiment?

In the Hafele and Keating Experiment we don't have rotating frames and my suggestion about the different gravitation due to different velocity has nothing to do with your experiment. Is that correct?

 

[stevmg]

At an instant in time, whatever an "instant" is, assuming the ground speed of both aircraft are the same but in opposite directions, the eastbound plane for that instant is traveling faster than the westbound plane (the Earth moves at a thousand miles an hour near the equator) in an eastbound direction. Thus, by time dilation, the 1/gamma facor is greater going eastward rather than westward and thus a greater slowing.

On the Earth's surface, there is no difference in velocity and with reference to the Earth, the 1/gamma's would be the same and the time dilation would be the same.

 

[sisoev]

On the Earth's surface, there is no difference in velocity and with reference to the Earth, the 1/gamma's would be the same and the time dilation would be the same.

Good point, stevmg.
Thank you

 

[yuiop]

Food for thought.

KEV, I'm back hungry for knowledge

Why in your experiment the gravitation in the rotating frame of reference will change when becomes stationary to the ECIF?
Shouldn't we look at the velocities in your experiment as in http://en.wikipedia.org/wiki/Rotating_reference_frame#Relation_between_velocities_in_the_two_frames", where the forces are fictitious and have nothing to do with Hafele-Keating experiment?

In the Hafele and Keating Experiment we don't have rotating frames and my suggestion about the different gravitation due to different velocity has nothing to do with your experiment. Is that correct?

Unfortunately, we do have rotating frames in the Hafele and Keating Experiment. The reference clock on the ground is in a rotating reference frame. See comments below.

On the Earth's surface, there is no difference in velocity and with reference to the Earth, the 1/gamma's would be the same and the time dilation would be the same.

Good point, stevmg.
Thank you

This is only aproximately true locally, for example if the aircraft turn around after a few miles and return to base. When the airscraft fly around the world, the situation is different. Ignoring gravitational effects, the East going clock loses 184 nanoseconds (runs slower) relative to the ground reference clock and the West going clock gains 96 nanoseconds relative to the ground reference clock (runs faster) - ref http://en.wikipedia.org/wiki/Hafele–Keating_experiment
This is not what is predicted if you assume the gamma factor is the same just because the speed of the aircraft is the same relative to the ground clock. The reason for the discrepancy is that the ground clock is in a rotating reference frame.
Only when you consider all the velocities relative to a true inertial (i.e non-rotating) reference frame such as the ECIF can you calculate the gamma factors using a simple application of 1/sqrt(1-v^2). Calculations are always simpler in a true inertial frame. Locally, any coordinate system can be aproximated by Special Relativity in flat space (i.e Minkowski coordinates).