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Time dilation due to velocity

  1. Apr 6, 2010 #1
    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?
  2. jcsd
  3. Apr 6, 2010 #2


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    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
  4. Apr 7, 2010 #3
    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.
  5. Apr 7, 2010 #4


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    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.

    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.

    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
  6. Apr 7, 2010 #5
    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.

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

    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 :smile:
  7. Apr 7, 2010 #6


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    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.

    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'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.

    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
  8. Apr 7, 2010 #7
    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
    Last edited: Apr 7, 2010
  9. Apr 7, 2010 #8


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    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.
  10. Apr 7, 2010 #9


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    Exactly. (And the term I would use for what kev is describing is a timelike geodesic; not inertial frame.)
  11. Apr 7, 2010 #10
    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.
  12. Apr 7, 2010 #11


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    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.
  13. Apr 7, 2010 #12
    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.
  14. Apr 7, 2010 #13
    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.
    Last edited: Apr 7, 2010
  15. Apr 7, 2010 #14
    Well, I cannot imagine that and know the result :smile: 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.

    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 :biggrin: ) why the clocks wouldn't feel it?

    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?
  16. Apr 7, 2010 #15


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    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
  17. Apr 7, 2010 #16
    I understand that :smile:
    That is the reason they used two air-crafts.

    Thank you, I understand it now :smile:
    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?
  18. Apr 7, 2010 #17
    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?
  19. Apr 7, 2010 #18
    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.
  20. Apr 7, 2010 #19
    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.
    Last edited: Apr 7, 2010
  21. Apr 7, 2010 #20
    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 :smile:
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