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Slicing up spacetime

Discussion in 'The Material World' started by Ostronomos, Jan 29, 2021.

  1. Ostronomos

    Ostronomos Active Member

    May 31, 2018
    Slicing up Spacetime

    To see how this works, here are three observers in relative motion in a spacetime.


    [Image: rel_sim_1.gif]

    First we have an observer whose worldline runs vertically up the page.
    The next observer moves to the right with respect to the first.


    [Image: rel_sim_2.gif]


    [Image: rel_sim_3.gif]

    The third observer moves to the left with respect to the first.
    Notice how differently they slice up the spacetime into spaces of simultaneous events. That difference simply is the relativity of simultaneity. It is expressed in the tilting of the hypersurfaces of simultaneity as we move the judgments of simultaneity of events from inertial observer to inertial observer.
    In looking at the three slicings as they are drawn above, it is easy to fall into the trap of imagining that the first slicing is somehow the "right" one and the second and the third are distortions due to the observers' motion. That would be a mistake. The principle of relativity assures that all three observers are equally good. The judgments of simultaneity of any one is just as good a those of the other two and each of the figures is an equally good way of dividing the spacetime into sets of simultaneous events.
    The fact that one observer's worldline is drawn as a vertical line and the others are oblique is just an accident of the way we chose to draw the diagram. Correspondingly, the fact that one observer's hypersurfaces are perfectly horizontal and the others are tilted is again an accident of the way we drew the figure. We could redraw the figures so that the third observer's worldline, say, is vertical. Then the third observer's hypersurfaces of simultaneity would be drawn as horizontal; the worldlines of the other two observers would be diagonals; and their hypersurfaces of simultaneity would be tilted.
    Two points to watch when you are drawing this tilting of hypersurfaces.
    First, setting an observer into motion to the right will tilt the observer's world line to the right; and the hypersurface of simultaneity will also tilt up on the right side to meet it.
    Second, if one follows the usual convention of drawing light lines at 45o, then the angle of the observer's worldline to the vertical will be the same as the angle of the hypersurface of simultaneity to the horizontal.


    [Image: rel_sim_4.gif]

    Propagating Times through Space
    The tilting of the hypersurfaces gives us a simple picture of how inertial observers in relative motion assign times to events.
    An inertial observer carries a clock that marks the time of events along the observer's worldline as "1," "2," "3," ... That settles the time of events only on the worldline for the observer. What time should be assigned to events not on the observer's worldline? The observer's hypersurfaces of simultaneity answer.

    Consider the hypersurface that passes through the event of the clock showing "1." All these events are simultaneous in the judgement of the observer. Therefore all these events are assigned time "1.

    The same applies for the remaining hypersurfaces that pass through the events of the clock ticking "2" and "3." All the events on those hypersurfaces are assigned times "2" and "3," respectively.


    [Image: prop_time_1.gif]


    [Image: prop_time_2.gif]

    The same analysis obtains for a new inertial observer who moves relative to the original observer. The new observer's clock assigns times to events on the observer's world line. The observer's hypersurfaces of simultaneity are then used to propagate the times throughout the spacetimes.

    Clearly the original and new observer will differ on the times each assigns to the same event in almost every case. Is there a sense in which one is assigning times correctly and the other not? There cannot be. The prinicple of relativity requires each observer's frame to be equivalent. If the procedure is good in one inertial frame, then it is equally good in all. This reminds us once again that there is no frame independent notion of simultaneity in a Minkowski spacetime.

    Why Tilt?
    Just why is it that hypersurfaces of simultaneity tilt when we change the state of motion of the observer or reference frame? A simple construction in spacetime geometry shows how it comes about.
    The construction represents a physical procedure for identifying simultaneous events using light signals in a familiar way. An inertial observer stands at the midpoint of a large platform. the observer sends out two light signals, in opposite directions. They are reflected back to the observer by mirrors at each end of the platform. The light signals arrive back at the observer at the same moment.


    [Image: Synchrony_procedure.svg]

    Since the observer is at the midpoint of the platform and the signals arrive back at the same moment, the observer infers that the two reflection events occurred simultaneously. Call these events A and A'. By placing the mirrors at other positions on the platform, the observer can use the procedure to determine other pairs of simultaneous events, B and B'; and C and C'; and so on. These judgments of simultaneity enable the observer to map out a full set of simultaneous events that form a hypersurface of simultaneity.
    The figure opposite depicts this procedure in a spacetime diagram. We represent the inertial trajectory of this inertial observer by a timlelike worldline, drawn vertically in the diagram.

    Light signals leave the observer at the same moment, are reflected at event A and A' and arrive back at the observer's worldline at the same moment.

    The observer can then infer that the reflection events A and A' are simultaneous and lie on the same hypersurface of simultaneity.

    This same reasoning applies to the remaining events shown in the figure: B', C', B and C. They are produced by reflections from mirrors suitably located at other positions on the platform. The observer judges B and B' to be simultaneous; and C and C' to be simultaneous.

    If the emission of the light signals for reflections B/B' and for C/C' are suitably delayed (as shown), then all six of these events will lie on the same hypersurface of simultaneity that passes through the event O on the observer's worldline. The are thus simultaneous with O. The totality of simultaneous events like these form a flat plane perpendicular to the observer's worldline.


    [Image: why_tilt_1.gif]

    The light signals reflected at A' and A arrive back at the observer at the same time. That they must do this is obvious from the symmetry of the procedure. The light signals left the observer at the same moment. They are reflected back after they have traveled the same distance. Therefore they must arrive back at the same time. If they did not arrive back at the same moment, something would clearly have gone wrong in the procedure. This remark will become important in a moment!
    Now consider a second case in which a new inertial observer moves relative to the original inertial observer. The new observer's worldline will be drawn below as a tilted line. Which events will that observer judge as simultaneous with an event O on that observer's worldline?
    Although the worldline is now tilted, the same procedure just described can also be used to pick out the events simultaneous with O. Indeed the principle of relativity requires this, for otherwise we would have some intrinsic difference between the first inertial frame and the second; only in the first could this procedure be used.


    [Image: why_tilt_2.gif]

    Spacetime and the Relativity of Simultaneity