# Reciprocal Time Dilation in Special Relativity [duplicate]

I'm trying to understand theory of special relativity, but there is one thing that really makes me confused which is reciprocal time dilation in special relativity.

In special relativity, the time dilation effect is reciprocal: as observed from the point of view of either of two clocks which are in motion with respect to each other, it will be the other clock that is time dilated. (This presumes that the relative motion of both parties is uniform; that is, they do not accelerate with respect to one another during the course of the observations.) - Wikipedia http://en.wikipedia.org/wiki/Time_dilation

This paragraph tells us that: as your friend flying on a high-speed moving rocket passes around you (who is at the rest in space), you see her age more slowly than yourself. She in turn will see you age more slowly.

This conclusion seems rather contradicting to me. What does this conclusion gives as the result of age-relationship between you and your friend? Does it mean if the rocket were forever in the uniform motion going away from you, you will be always older than your friend, and friend will be always older than you depending on which frame of reference you take?

## marked as duplicate by WillO, Rory Alsop, user36790, John Rennie special-relativity StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 2 '16 at 6:43

• If we walk away from each other, I see you getting smaller, but you also see me getting smaller! That seems just as contradictory -- who's really getting smaller? – knzhou Jul 1 '16 at 17:53
• The solution for both paradoxes is the same. You can only compare size (or age) when you're at the same place. To do this, somebody has to turn around, and that acceleration breaks the symmetry. Search 'twin paradox' for more detailed resolutions. – knzhou Jul 1 '16 at 17:54
• – user108787 Jul 1 '16 at 17:54
• The key to the whole issue is the relativity of simultaneity. You can't understand the theory until you accept that. – dmckee Jul 1 '16 at 17:54
• Also see this excellent graphic. @Mashu, it might be confusing since it looks like we're all saying different things. Actually, none of us are contradicting each other -- special relativity is just a really rich and cohesive theory, and there are always lots of ways to 'see' the same thing. (I think the graphic is the easiest one, though.) – knzhou Jul 1 '16 at 18:33

In relativity time is no longer a universal concept, it is a quantity specific to a frame of reference. It isn't meaningful to compare the "age" of two objects in two different frames of reference using a single "frame time." "Frame time" denotes time as measured in a specific frame of reference. This frame of reference could be that of either object for example or perhaps even a third party observer. There is one way to compare ages of objects that's meaningful though but it requires a different concept of time to be measured. Proper time is the time measured in a frame of reference where the observer measures him or herself to be at rest. This definition is rather bizarre since it doesn't allow arbitrary points in space to have a proper time. Only observers with a defined velocity (or motion of any kind) can have a measured proper time. It is still a useful concept as it allows us to directly compare what different observers measure from their own frame of reference.

With your example we can use proper time to see how old the travelers are as measured in their own time. By the definition of proper time, traveler A and traveler B age in terms of proper time at the same rate, that's the whole point of proper time! Now what happens when they try to measure each other's age using their own time? If they do this they're not measuring the other traveler's age in terms of proper time, but rather their own frame time. They would need to calculate the proper time of the other traveler that they're observing. But when traveler A observes traveler B at a proper time of 1 minute, does traveler A see traveler B at a proper time of 1 minute or some other time? Obviously there will be some delay for the light of traveler B to reach A, but even accounting for this effect, do we expect A to still measure a difference in her observation of B's proper time? The answer is yes, we do, because A's notion of simultaneity is not the same as B's.

When A looks at B, A does not see B at the same proper time A sees herself. Because they are moving at different relative speeds, A's frozen snapshot of space at a given time doesn't look like B's. We could imagine depicting A's frozen snapshot of space as a line where at each point we give the value of A's clock for that point. It will look like this for A say at time = 0:

...-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-...

Every point exists at the same time. Now from B's perspective we could depict A's frozen time snapshot assigning those same points a time, t', based on B's clock instead. It will look like the following for example:

...-0-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20-...

A's frozen time is frozen at a different time for each point in space from B's perspective. Likewise from A's perspective, B's frozen snapshot looks the same as A's did from B's perspective. This means measuring each other's age at a frozen snapshot is not a fair depiction of how A or B would measure their own age. A doesn't see B at her own value of proper time and likewise B doesn't see A at his own value of proper time. They are just as delayed from each other's point of view. This is the same thing as depicted in the graphic knzhou posted in the comments. The diagonal line of the driver represents B's simultaneity as seen from A. All along this line, A has different values of her clock even though it is a "frozen snapshot."

So what happens if A and B decide to meet in the same frame of reference, i.e. have 0 relative speed? Then they will have the same frame time and when they measure each other's proper time, it will match their own proper time. So what will their relative ages be? It depends on how they get there. If they both agree to do equal work to match speeds, they will see each other age the same amount over the time it takes to match speeds. If A does more of the work, then she will come out younger for it, if B does more of the work, then he will be younger.

Yes, Einstein's postulates entail symmetrical (reciprocal) time dilation, but in 1905 Einstein deduced, invalidly (in the sense that this does not follow from the postulates), asymmetrical time dilation - when the moving clock passes the stationary one, the former lags behind the latter. Nowadays you can often hear the same incorrect conclusion: "Moving clocks run slower than stationary ones", "Time slows down for you if you start moving" etc.