# Time Dilation in two reference frames

This is slightly related to the twins paradox, but doesn't include the part where the twin on the journey makes a U-Turn and comes back to earth (i.e. I'm asking about where the twin is at a constant velocity speeding away from earth)

So one twin stays on earth and the other leaves on a spaceship at 0.8c, from the inertial reference frame on earth, the time on the spaceship is running slower relative to earth's time, therefore the twin on the journey is aging less than the twin on earth. However, from the spaceship's reference frame, the earth is speeding away from them at 0.8c, and therefore earth's time is running slower than the spaceship and therefore the twin on earth is aging less than the twin on the journey.

This seems to be contradictory, while the twin on the spaceship is speeding away from earth, which twin is aging slower? Because, when looking at both reference frames, both twins seems to age less than the other? How can this be, and what have I missed?

I am going to answer your questions using a geometric analogy featured in my visualization:

https://www.desmos.com/calculator/ti58l2sair Go there and set the E-slider to -1

This shows a unit circle and two radial directions, the black one horizontal and the red one sloped with a positive angle. Think of each radial direction as the straightline path of a surveyor (each making his own map of the plane) and their dashed meterstick that each carries perpendicular to his path. Perpendicularity is defined using tangency to a circle.

The grid is already aligned with the black surveyor.

The black surveyor would assign t-coordinate=1 to every point on his dashed meterstick, in particular the point (1.0, 0.6) where the red path intersects the black dashed meterstick. However, the red surveyor would have assigned that point a t-coordinate larger than 1 on his map since he has already passed the unit circle.

Similarly... The red surveyor would assign t-coordinate=1 to every point on his dashed meterstick, in particular the point where the black path intersects the red dashed meterstick. The black surveyor would have assigned that point a t-coordinate larger than 1 since he has already passed the unit circle. (By symmetry, the value of this larger-than-1 t-coordinate is the same as the value from the red surveyor. In fact, this larger-than-1 value is 1/cos(arctan(slope)). With slope=0.6, that value is 1.1662.)

Presumably, no problem here because we are used to Euclidean geometry.

Now, in special relativity, the situation is analogous.

In the visualization, Go there and set the E-slider to +1

Rather than surveyors, the radial lines are now worldlines of inertial observers (making spacetime diagrams [position-vs-time graphs]).

These hyperbolas are the analogues of the circles in Euclidean geometry. They mark off all of the events of "one-tick" for each wristwatch traveling inertially [each with different velocities] from a common starting event O.

When an observer's worldline meets that unit hyperbola [marking one tick of that observer's clock], the tangent at the intersection defines the events simultaneous with the intersection event according to that observer.

So, the black dashed line has t-coordinate=1 on the diagram drawn by the black observer, in particular the event (1.0, 0.6) where the red worldline intersects the black dashed meterstick. However, the red observer would assign that event a t-coordinate smaller than 1 on his diagram since he has not yet reached the unit hyperbola. That value is 1/cosh(arctanh(slope))=0.8 . Physically, this is 1/timeDilationFactor.

...and similarly for the red dashed line.

So, the special relativity case is an analogue of what happened in the Euclidean case.

The "everyday common sense notion of time" turns out to be the special case...

In the visualization, Go there and set the E-slider to 0 , the Galilean case.

This "circle" in the Galilean case is a vertical line-- the t=1 line. All tangent lines to this circle coincide. All observers agree on assigning t=1. This is absolute simultaneity.

You are right. What happens is that their time axes do not coincide anymore. If two events (A,B) are simultaneous in one's frame of reference, one of those events will happen before the second one in the other frame of reference. In a sense, time dilation is really a measure of how much the time axis of the moving frame is rotated relative to the stationary one.

• So which twin ages less... Oct 9, 2017 at 10:49
• @Anonymous: "So which twin ages less?" I assume you mean more slowly? You've already answered this. In Twin A's frame, Twin B ages more slowly. In Twin B's frame, Twin A ages more slowly. If you think this is contradictory, try writing down exactly why, accounting for Arthur's answer, and you'll find the "contradiction" dissolves in front of you. Oct 9, 2017 at 13:32