__Return to current section__**: ****Time Dilation and Desynchronization****.**

**Exercises:**

(1) Compare ages of the twins when the trip is over. Discuss implications of this result.

The following interactive diagram will help explain why the stay-at-home twin is older on the astronaut twin's return,

no matter whose viewpoint or perspective (inertial frame) is used. In this diagram, the astronaut's speed relative to

the earth is chosen to be 0.8*c* on the trip out and − 0.8*c* on the way back. The *γ* factor works out to be

1 / sqrt(1 - 0.8^{2}) = 5/3 > 1. The last equation from the previous page (use the back button or the link below)

tells us t = *γ*τ which mean the earth twin ages 5/3 as the much as the astronaut during the trip. This diagram also

assumes the astronaut's destination is 2 light years away. At the speed of 0.8*c*, it takes 2.5 years to get there and

2.5 years to return (by the earth clocks), so the earth twin ages 5 years. We can calculate the spacetime interval,

using the earth perspective, for either branch of the trip to figure out the astronaut twin's proper time:

(ct)^{2} - x^{2} = (2.5)^{2} - 2^{2} = 6.25 - 4 = 2.25 = (cτ)^{2} ⇒ cτ = 1.5

So, the astronaut ages 3 years on the trip, which satisfies our previous statements.

In this interactive diagram, we can drag the *X*1 event around to analyze the trip from different viewpoints, as seen

from different inertial frames. The default (RESET) viewpoint has the earth twin drawing the spacetime map. That

twin's history is then the vertical axis, that twin is at rest and the astronaut's map velocity is 0.8*c* as described above.

Or, we could draw the spacetime map for the astronaut on the first leg of the trip. Then, the astronaut would be at

rest and the earth twin would be falling away at − 0.8*c*. In between, we can envision drawing the events from the

perspective of an observer traveling parallel to the astronaut's getaway, but not as fast. Then, the earth would seem to

be moving backwards and the astronaut forwards. We change perspectives by dragging the *X*1 event. Each time we stop

dragging *X*1, we are seeing things from a different frame of reference.

But, how did we create the diagram? How are we calculating how the event vectors change? There are a couple of

constants we can rely on. The ** proper times** of the astronaut on each leg and the earth twin

**, because**

*are constants*no matter how another inertial observer sees things, a clock at rest in my frame is still a clock at rest in my frame

and should continue ticking at my same rate no matter who is zooming by reading my clocks. The lower hyperbola

in the diagram defines points of constant proper time (= 1.5) for the outgoing astronaut, since the equation above

used to calculate it is the equation of a hyperbola. The upper hyperbola corresponds only to earth proper times of 5.

Therefore, the

*X*1 and

*X*2 event vectors must ride along these hyperbolae. We're choosing our viewpoint by dragging

*X*1, so all that is left is to find the direction of the green

*X*2 vector. That's actually not so easy here. It requires

knowing how relative velocities seem to change with perspective (we'll get the answer to this a few sections later

when we learn the velocity composition formulas). For now, we ask for trust that we have used those formulas

correctly to get the green vector. Finally, the red vector connecting

*X*1 and

*X*2 simply completes the triangle.

However, all we really need in order to understand the current problem is two viewpoints, from the earth twin's

perspective or from the astronaut twin's perspective. Then, one of the twins is at rest, the relative speeds are

simply 0.8*c*, and the purple and green arrows are easy to place correctly. The coordinates of *X*1 and *X*2 are updated

on the screen, so any intervals, distances, and frame velocities can be calculated for practice and checking.

The parable of the twins is historically called ** the twin paradox**, since it offered one of relativity's first confusions

(and confusing oneself is often too easy). An enjoyable quote comes from David Rohrlich in his book

**Classical**

Charged Particles(p.278): "Since the author has great difficulty constructing paradoxes from clear mathematical

Charged Particles

facts, he will not attempt to do so". The apparent paradox came from asking "if we imagine the earth as making

the trip, wouldn't the earth twin end up younger?" The diagram, however, shows the twins' histories really can

not be exchanged simply by exchanging which twin we take as at rest and which is in relative motion.

**, while the astronaut's path, no matter who observes**

*From any*

perspective, the earth history is a simple straight lineperspective, the earth history is a simple straight line

it, cannot be straigtened out. The worldline kink where outgoing velocity changes to incoming never goes away.

The velocities outgoing and incoming may appear to have different values to different observers, but the kink

in the path can never be made to go away.

Now, in the interactive diagram, drag *X*1 to the central axis, so we see things from the perspective of the

astronaut during the first leg of the trip. The astronaut is at rest in this frame and earth appears to be receding

at a velocity of − 0.8*c*. We can check out a few things in the diagram.

You can simply read the astronaut's first leg trip proper time from the *ct*-axis (it is 1.5).

From *X*1, go horizontally left to the green vector to read how far away the earth seems to be to the

astronaut who just finished the first leg of the trip. It should be 1.2 (from the earth's perspective, it

was 2 light years). The reduction factor is (guess what?) the *γ* factor (5/3, the same factor as initially,

because the relative speed is again equal to its initial value). This is an example of *Lorentz Contraction*

(the next topic).

We can visually check the earth's receding velocity, since the slope of the green line is just 4/5 leftwards.

We can use the coordinates to make sure, as advertised, that all three proper times haven't changed.

For the first leg, the astronaut stayed at rest while the earth flew away. Now, for the second leg, the

astronaut has to really move to overtake and land back on earth. You can see from the diagram's

yellow text updates that this speed is 0.98*c*. That is indeed pretty fast, but not as fast as Galileo

might have predicted (1.6*c*). Once again, we'll be able to predict this value after presenting the

relativistic velocity composition section later.

(2) Prove that the longest path between two points separated by a timelike interval is a straight line.

** **Consider the second figure from the previous page (use the back button or the link below). The proper time

along a timelike curve is the path's arclength. The path can be approximated by straight line segments

(more segments give a better approximation). A first approximation for the path's arclength in that diagram

would be ∆τ_{0}, a better approximation would be ∆τ_{1} + ∆τ_{2}, etc. Now, the previous exercise above showed

that the earth twin's straight line proper time was greater than the sum of the proper times of the segments

of the kinked astronaut twin's path. It didn't matter which observer was calculating these proper times.

That is, it didn't matter how the single earth segment was tilted. Therefore,

∆τ_{0} > ∆τ_{1} + ∆τ_{2} > ... > Arclength of any curved timelike path.

__Return to current section__**: ****Time Dilation and Desynchronization****.**