Editing Spacetime (section)

=== Mutual time dilation and the twin paradox ===
{{Main|Twin paradox}}

==== Mutual time dilation ====
Mutual time dilation and length contraction tend to strike beginners as inherently self-contradictory concepts. If an observer in frame S measures a clock, at rest in frame S', as running slower than his', while S' is moving at speed ''v'' in S, then the principle of relativity requires that an observer in frame S' likewise measures a clock in frame S, moving at speed −''v'' in S', as running slower than hers. How two clocks can run ''both slower'' than the other, is an important question that "goes to the heart of understanding special relativity."<ref name="Schutz" />{{rp|198}}

This apparent contradiction stems from not correctly taking into account the different settings of the necessary, related measurements. These settings allow for a consistent explanation of the ''only apparent'' contradiction. It is not about the abstract ticking of two identical clocks, but about how to measure in one frame the temporal distance of two ticks of a moving clock. It turns out that in mutually observing the duration between ticks of clocks, each moving in the respective frame, different sets of clocks must be involved. In order to measure in frame S the tick duration of a moving clock W′ (at rest in S′), one uses ''two'' additional, synchronized clocks W<sub>1</sub> and W<sub>2</sub> at rest in two arbitrarily fixed points in S with the spatial distance ''d''.
: <small>Two events can be defined by the condition "two clocks are simultaneously at one place", i.e., when W′ passes each W<sub>1</sub> and W<sub>2</sub>. For both events the two readings of the collocated clocks are recorded. The difference of the two readings of W<sub>1</sub> and W<sub>2</sub> is the temporal distance of the two events in S, and their spatial distance is ''d''. The difference of the two readings of W′ is the temporal distance of the two events in S′. In S′ these events are only separated in time, they happen at the same place in S′. Because of the invariance of the spacetime interval spanned by these two events, and the nonzero spatial separation ''d'' in S, the temporal distance in S′ must be smaller than the one in S: the ''smaller'' temporal distance between the two events, resulting from the readings of the moving clock W′, belongs to the ''slower'' running clock W′.</small>

Conversely, for judging in frame S′ the temporal distance of two events on a moving clock W (at rest in S), one needs two clocks at rest in S′.
: <small>In this comparison the clock W is moving by with velocity −''v''. Recording again the four readings for the events, defined by "two clocks simultaneously at one place", results in the analogous temporal distances of the two events, now temporally and spatially separated in S′, and only temporally separated but collocated in S. To keep the spacetime interval invariant, the temporal distance in S must be smaller than in S′, because of the spatial separation of the events in S′: now clock W is observed to run slower.</small>

The necessary recordings for the two judgements, with "one moving clock" and "two clocks at rest" in respectively S or S′, involves two different sets, each with three clocks. Since there are different sets of clocks involved in the measurements, there is no inherent necessity that the measurements be reciprocally "consistent" such that, if one observer measures the moving clock to be slow, the other observer measures the one's clock to be fast.<ref name="Schutz" />{{rp|198–199}}

{{multiple image|perrow = 1|total_width=250
| image2 = Spacetime Diagrams of Mutual Time Dilation B.png  |width2=300|height2=300
| image4 = Spacetime Diagrams of Mutual Time Dilation D.png  |width4=300|height4=300
| footer = Figure 2-10. Mutual time dilation }}
Fig. 2-10 illustrates the previous discussion of mutual time dilation with [[Minkowski diagram]]s. The upper picture reflects the measurements as seen from frame S "at rest" with unprimed, rectangular axes, and frame S′ "moving with ''v''&nbsp;>&nbsp;0", coordinatized by primed, oblique axes, slanted to the right; the lower picture shows frame S′ "at rest" with primed, rectangular coordinates, and frame S "moving with −''v''&nbsp;<&nbsp;0", with unprimed, oblique axes, slanted to the left.

Each line drawn parallel to a spatial axis (''x'', ''x''′) represents a line of simultaneity. All events on such a line have the same time value (''ct'', ''ct''′). Likewise, each line drawn parallel to a temporal axis (''ct'', ''ct′'') represents a line of equal spatial coordinate values (''x'', ''x''′).
: <small>One may designate in both pictures the origin ''O'' (= {{′|''O''}}) as the event, where the respective "moving clock" is collocated with the "first clock at rest" in both comparisons. Obviously, for this event the readings on both clocks in both comparisons are zero. As a consequence, the worldlines of the moving clocks are the slanted to the right ''ct''′-axis (upper pictures, clock W′) and the slanted to the left ''ct''-axes (lower pictures, clock W). The worldlines of W<sub>1</sub> and W′<sub>1</sub> are the corresponding vertical time axes (''ct'' in the upper pictures, and ''ct''′ in the lower pictures).</small>
: <small>In the upper picture the place for W<sub>2</sub> is taken to be ''A<sub>x</sub>'' > 0, and thus the worldline (not shown in the pictures) of this clock intersects the worldline of the moving clock (the ''ct''′-axis) in the event labelled ''A'', where "two clocks are simultaneously at one place". In the lower picture the place for W′<sub>2</sub> is taken to be ''C''<sub>''x''′</sub>&nbsp;<&nbsp;0, and so in this measurement the moving clock W passes W′<sub>2</sub> in the event ''C''.</small>
: <small>In the upper picture the ''ct''-coordinate ''A<sub>t</sub>'' of the event ''A'' (the reading of W<sub>2</sub>) is labeled ''B'', thus giving the elapsed time between the two events, measured with W<sub>1</sub> and W<sub>2</sub>, as ''OB''. For a comparison, the length of the time interval ''OA'', measured with W′, must be transformed to the scale of the ''ct''-axis. This is done by the invariant hyperbola (see also Fig. 2-8) through ''A'', connecting all events with the same spacetime interval from the origin as ''A''. This yields the event ''C'' on the ''ct''-axis, and obviously: ''OC''&nbsp;<&nbsp;''OB'', the "moving" clock W′ runs slower.</small>

To show the mutual time dilation immediately in the upper picture, the event ''D'' may be constructed as the event at ''x''′&nbsp;=&nbsp;0 (the location of clock W′ in S′), that is simultaneous to ''C'' (''OC'' has equal spacetime interval as ''OA'') in S′. This shows that the time interval ''OD'' is longer than ''OA'', showing that the "moving" clock runs slower.<ref name="Collier" />{{rp|124}}

In the lower picture the frame S is moving with velocity −''v'' in the frame S′ at rest. The worldline of clock W is the ''ct''-axis (slanted to the left), the worldline of W′<sub>1</sub> is the vertical ''ct''′-axis, and the worldline of W′<sub>2</sub> is the vertical through event ''C'', with ''ct''′-coordinate ''D''. The invariant hyperbola through event ''C'' scales the time interval ''OC'' to ''OA'', which is shorter than ''OD''; also, ''B'' is constructed (similar to ''D'' in the upper pictures) as simultaneous to ''A'' in S, at ''x''&nbsp;=&nbsp;0. The result ''OB''&nbsp;>&nbsp;''OC'' corresponds again to above.

The word "measure" is important. In classical physics an observer cannot affect an observed object, but the object's state of motion ''can'' affect the observer's ''observations'' of the object.

==== Twin paradox ====
Many introductions to special relativity illustrate the differences between Galilean relativity and special relativity by posing a series of "paradoxes". These paradoxes are, in fact, ill-posed problems, resulting from our unfamiliarity with velocities comparable to the speed of light. The remedy is to solve many problems in special relativity and to become familiar with its so-called counter-intuitive predictions. The geometrical approach to studying spacetime is considered one of the best methods for developing a modern intuition.<ref name="Schutz1985">{{cite book| last1=Schutz |first1= Bernard F. |title=A first course in general relativity|date=1985|publisher=Cambridge University Press|location=Cambridge, UK|isbn=0-521-27703-5|page=26}}</ref>

The [[twin paradox]] is a [[thought experiment]] involving identical twins, one of whom makes a journey into space in a high-speed rocket, returning home to find that the twin who remained on Earth has aged more. This result appears puzzling because each twin observes the other twin as moving, and so at first glance, it would appear that each should find the other to have aged less. The twin paradox sidesteps the justification for mutual time dilation presented above by avoiding the requirement for a third clock.<ref name="Schutz" />{{rp|207}} Nevertheless, the ''twin paradox'' is not a true paradox because it is easily understood within the context of special relativity.

The impression that a paradox exists stems from a misunderstanding of what special relativity states. Special relativity does not declare all frames of reference to be equivalent, only inertial frames. The traveling twin's frame is not inertial during periods when she is accelerating. Furthermore, the difference between the twins is observationally detectable: the traveling twin needs to fire her rockets to be able to return home, while the stay-at-home twin does not.<ref name="Weiss" /><ref group=note>Even with no (de)acceleration i.e. using one inertial frame O for constant, high-velocity outward journey and another inertial frame I for constant, high-velocity inward journey – the sum of the elapsed time in those frames (O and I) is shorter than the elapsed time in the stationary inertial frame S. Thus acceleration and deceleration is not the cause of shorter elapsed time during the outward and inward journey. Instead the use of two different constant, high-velocity inertial frames for outward and inward journey is really the cause of shorter elapsed time total. Granted, if the same twin has to travel outward and inward leg of the journey and safely switch from outward to inward leg of the journey, the acceleration and deceleration is required. If the travelling twin could ride the high-velocity outward inertial frame and instantaneously switch to high-velocity inward inertial frame the example would still work. The point is that real reason should be stated clearly. The asymmetry is because of the comparison of sum of elapsed times in two different inertial frames (O and I) to the elapsed time in a single inertial frame S.</ref>

[[File:Introductory Physics fig 4.9.png|thumb|Figure 2–11. Spacetime explanation of the twin paradox]]
These distinctions should result in a difference in the twins' ages. The spacetime diagram of Fig.&nbsp;2-11 presents the simple case of a twin going straight out along the x axis and immediately turning back. From the standpoint of the stay-at-home twin, there is nothing puzzling about the twin paradox at all. The proper time measured along the traveling twin's world line from O to C, plus the proper time measured from C to B, is less than the stay-at-home twin's proper time measured from O to A to B. More complex trajectories require integrating the proper time between the respective events along the curve (i.e. the [[Line integral|path integral]]) to calculate the total amount of proper time experienced by the traveling twin.<ref name="Weiss">{{cite web |last1=Weiss |first1=Michael |title=The Twin Paradox |url=http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html |website=The Physics and Relativity FAQ |access-date=10 April 2017 |archive-date=27 April 2017 |archive-url=https://web.archive.org/web/20170427202915/http://www.math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html |url-status=live }}</ref>

Complications arise if the twin paradox is analyzed from the traveling twin's point of view.

Weiss's nomenclature, designating the stay-at-home twin as Terence and the traveling twin as Stella, is hereafter used.<ref name="Weiss" />

Stella is not in an inertial frame. Given this fact, it is sometimes incorrectly stated that full resolution of the twin paradox requires general relativity:<ref name="Weiss" />
{{blockquote|A pure SR analysis would be as follows: Analyzed in Stella's rest frame, she is motionless for the entire trip. When she fires her rockets for the turnaround, she experiences a pseudo force which resembles a gravitational force.<ref name="Weiss" /> [[#Figure 2-6|'''Figs.&nbsp;2-6''']] and 2-11 illustrate the concept of lines (planes) of simultaneity: Lines parallel to the observer's ''x''-axis (''xy''-plane) represent sets of events that are simultaneous in the observer frame. In Fig.&nbsp;2-11, the blue lines connect events on Terence's world line which, ''from Stella's point of view'', are simultaneous with events on her world line. (Terence, in turn, would observe a set of horizontal lines of simultaneity.) Throughout both the outbound and the inbound legs of Stella's journey, she measures Terence's clocks as running slower than her own. ''But during the turnaround'' (i.e. between the bold blue lines in the figure), a shift takes place in the angle of her lines of simultaneity, corresponding to a rapid skip-over of the events in Terence's world line that Stella considers to be simultaneous with her own. Therefore, at the end of her trip, Stella finds that Terence has aged more than she has.<ref name="Weiss" />}}

Although general relativity is not required to analyze the twin paradox, application of the [[Equivalence Principle]] of general relativity does provide some additional insight into the subject. Stella is not stationary in an inertial frame. Analyzed in Stella's rest frame, she is motionless for the entire trip. When she is coasting her rest frame is inertial, and Terence's clock will appear to run slow. But when she fires her rockets for the turnaround, her rest frame is an accelerated frame and she experiences a force which is pushing her as if she were in a gravitational field. Terence will appear to be high up in that field and because of [[gravitational time dilation]], his clock will appear to run fast, so much so that the net result will be that Terence has aged more than Stella when they are back together.<ref name="Weiss" /> The theoretical arguments predicting gravitational time dilation are not exclusive to general relativity. Any theory of gravity will predict gravitational time dilation if it respects the principle of equivalence, including Newton's theory.<ref name="Schutz" />{{rp|16}}
{{anchor|Gravitation}}