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Rotating object time dilation

  1. Mar 15, 2014 #1
    As I've red, different points on the rotating disk, or rotating object have different velocities and so they should time dilate differently. Points on equator are for instance quicker than those in Scandinavia. When we take an inertial frame that has zero velocity with respect to the Earth as a whole (the Earth is rotating wrt to that frame but isn't moving inertially in any direction), we can observe that different points have different tangential velocities, so that points on Equator have different directions of motions, but same speed.

    So my question is, if we take a frame that is travelling with some velocity wrt to Earth, how will the velocities add. I mean all points are travelling with a different direction, so how can we add their velocities to the velocity of our moving frame, and how will the time dilate wrt to that moving frame that I previously mentioned? And how will simultaneity be defined on that moving inertial frame considering the Earth as a worldtube that we're focused to? How will simultaneity differ between the moving frame and the frame which has the rotating points at rest?

    I hope you understand my question.
    Regards, analyst
  2. jcsd
  3. Mar 15, 2014 #2


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    You'll get time dilation that depends on the current position of the object. Integrated over one rotation of earth, all points at the same latitude get the same time dilation.
    Simultaneity is seen by an inertial frame, without issues from the rotating earth. Yes it will differ from inertial frames where a point on the surface is at rest (for a certain moment in time - it cannot stay at rest in an inertial frame as it is rotating around earth).
    Summary: You don't want to consider earth in a frame where the center of earth is moving, it just makes everything messy.

    By the way: for objects on earth, time dilation due to the gravitational potential is more relevant than the small rotation speed.
  4. Mar 17, 2014 #3


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    Suppose you have observers i=1,2,3,... with velocities [itex]\mathbf{v}_i[/itex] w.r.t. an inertial frame S. Then you can calculate their proper times [itex]\tau_i[/itex] in terms of the coordinate time t. For a world [itex]C_i[/itex] of an observer i you'll get

    [tex]\tau_i = \int_{C_i}d\tau = \int_0^T dt\,\sqrt{1-\mathbf{v}^2_i(t)}[/tex]

    For constant linear velocity with speed [itex]v_i[/itex] this is simply

    [tex]\tau_i = \sqrt{1-v^2_i} \,\cdot\,T[/tex]

    As usual the coordinate time T can be associated with the proper time of the inertial observer in S.


    Now consider rotation w.r.t. a fixed axis. For the the speed of rotating observers i at radius [itex]r_i[/itex] and with i-independent angular velocity [itex]\omega[/itex] you get

    [tex]\mathbf{v}^2_i(t) = r_i^2\omega^2[/tex]

    and therefore

    [tex]\tau_i = \int_{C_i}d\tau = \int_0^T dt\,\sqrt{1-r_i^2\omega^2} = \sqrt{1-r_i^2\omega^2} \,\cdot\,T[/tex]

    As usual the coordinate time T can be associated with the proper time of the inertial observer in S.


    Now the next step would be to add linear motion and rotation. This is straightforward, but rather messy in detail. First you start with translation [itex]\mathbf{v}^\text{trans}[/itex] w.r.t. the inertial frame of the earth S, then you add the above mentioned rotation [itex]\mathbf{v}_i^\text{rot}(t)[/itex] w.r.t. the moving inertial frame S'. Note that both velocities are vectors.

    You have to use the relativistic addition formula

    [tex]\mathbf{v}_i(t) = \mathbf{v}^\text{trans} \oplus \mathbf{v}_i^\text{rot}(t^\prime)[/tex]

    which you can find here


    This can be rather complex depending on the orientation of the vectors. You should consider some examples like
    a) translation along the z-axis and rotation around z'-axis
    b) translation in the xy-plane and rotation around z'-axis

    Note that due to the rotation you'll get a time-dependent total velocity and you have to use the general formula

    [tex]\tau_i = \int_{C_i}d\tau = \int_0^T dt\,\sqrt{1-\mathbf{v}^2_i(t)}[/tex]

    Caveat: don't forget that rotation around the z'-axis will be expressed in terms of t' like

    [tex]\mathbf{v}_i^\text{rot}(t^\prime) = r_i\omega\left( \begin{array}{c} \sin (\omega t^\prime) \\ -\cos (\omega t^\prime) \\ 0 \end{array} \right) [/tex]

    Before you can perform the dt-integration you have to transform to coordinate time t.
  5. Mar 17, 2014 #4


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    You can easily determine the speed of any point on the surface of the Earth just from the geometry. That determines the Time Dilation of these points. Assuming that all the points were synchronized to time zero at the Coordinate Time zero of the inertial frame, you can easily determine the Proper Time for any point (event) as a function of the Coordinate Time by simply dividing by gamma which is a constant for each point. Once you determine the Proper Time for any event, it will be the same Proper Time in all other frames.

    You don't need to calculate the velocities to determine the Proper Time at each point as a function of the Coordinate Time in the new inertial frame, Just use the Lorentz Transformation process. As I said before, each event has the same Proper Time in all frames. So if you care about Time Dilation in this new frame, you just compare the delta Coordinate Time to the delta Proper Time for any pair of events.

    Simultaneity is defined as two events having the same Coordinate Time.

    Simultaneity has a well-defined meaning in inertial frames and we can easily transform to any other inertial frame and establish a new well-defined set of simultaneities. But if you want to go to a non-inertial rest frame for the rotating points, YOU have to define what you mean by simultaneity. You can't ask us to answer a question for which you have not provided adequate information. I hope you realize that there is not a single answer to your question.

    And I would like to ask you a question: why do you care about simultaneity for the rotating points at rest?
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