Special Relativity

The Twin Paradox

The apparent asymmetry of relativistic motion and its resolution

This essay examines the twin paradox in detail, showing how two observers can each regard the other as moving while nevertheless measuring different elapsed times when they reunite. The apparent asymmetry is resolved directly from the Lorentz transformations and the different spacetime descriptions of the journey.

Prerequisites

Linear algebra and inner-product spaces
Multivariable calculus
Fundamental principles of special relativity
Lorentz transformations

Twin paradox
Time dilation
Length contraction
Relativity of simultaneity
Lorentz transformations
Proper time

I. Introduction
II. Description of the Experiment
1. Acceleration to $v$
2. Trip to $G$
3. Trip Back to Earth
4. Events and the Apparent Asymmetry of Time Dilation

I. Introduction

In this essay we will discuss the twin paradox, which arises in relativity when two objects, initially at the same position and each equipped with its own clock, move away from each other and later return to the same position. We assume that the two clocks are initially synchronized and therefore measure exactly the same time. When the two objects meet again, however, the clocks will in general have measured different elapsed times.

In the twin experiment, one of two identical twins embarks on a space journey, moving at a speed close to the speed of light, while the other remains at home and waits for the traveling twin to return. When the twins meet again, the twin who remained at home has aged more than the one who traveled. This result appears paradoxical in two different ways. First, it is very counterintuitive to accept that simply moving differently can change the amount of time experienced by the two twins. Second, and more importantly, there appears to be a breaking of symmetry. Why should one twin age differently from the other when there is no privileged system of reference? It seems as though the younger twin must somehow be different from the twin who remained at home. Yet, during their relative motion, each twin may regard the other as moving. We will examine in detail how this apparent asymmetry arises and why there is no contradiction with the principle of relativity.

Below we will make use of formulas derived in the essay Lorentz Transformations . In particular, we will need the transformation between two inertial reference frames

F

and

F^{'}

, where

F^{'}

moves with velocity

v = v e_{1}

relative to

F

. The Lorentz transformation for this boost is

Eq:Lorentz_transformation

Λ (v) = γ ∣ e_{1}, 0 ⟩ ⟨ e_{1}, 0 ∣ - γ \frac{v}{c} ∣ e_{1}, 0 ⟩ ⟨ 0, 1 ∣ + ∣ e_{2}, 0 ⟩ ⟨ e_{2}, 0 ∣ + ∣ e_{3}, 0 ⟩ ⟨ e_{3}, 0 ∣ - γ \frac{v}{c} ∣ 0, 1 ⟩ ⟨ e_{1}, 0 ∣ + γ ∣ 0, 1 ⟩ ⟨ 0, 1 ∣ .

Here,

Eq:Lorentz_factor

γ = \frac{1}{1 - \frac{v ^{2}}{c ^{2}}} .

We will restrict our analysis to the case in which the relative velocity is directed along the

x

-axis. In this case, the corresponding transformations of space and time are

Eq:Lorentz_transformation_along_x

x^{'} y^{'} z^{'} t^{'} = γ (x - v t), = y, = z, = γ (t - \frac{v}{c ^{2}} x) .

As shown in the essay Composition of Velocities , the two frames measure relative velocities of equal magnitude and opposite direction:

Eq:Reciprocal_relative_velocities

u ∣_{u^{'} = 0} = v, u^{'} ∣_{u = 0} = - v, ∣ u ∣ = ∣ u^{'} ∣ = v .

These equations provide the kinematic relations needed to compare the elapsed times measured along the twins' different journeys.

II. Description of the Experiment

We will describe the experiment in terms of physical events . An event is a definite physical occurrence, such as the departure of the rocket or its arrival at planet

G

. The event itself is independent of the reference frame, although different reference frames generally assign different space and time coordinates to it. Consequently, a distance, an elapsed time, or a statement of simultaneity is meaningful only after the events used in the comparison have been specified.

Let us consider the Earth as the origin of the spatial coordinates of a reference frame with coordinates

(x, y, z)

, and let us consider a planet

G

located at a distance

l

along the

x

-axis of this frame.

We assume that one of the twins remains at the origin of the Earth reference frame and is equipped with a clock that he uses to measure time

t

. This twin does not change his position with respect to this reference frame and measures spatial distances using the Earth coordinate system.

We now consider the other twin, who will travel toward the planet

G

. We assume that, at the beginning of the journey, the two twins occupy exactly the same position. At this instant, the twin on the rocket introduces a spatial coordinate system

(x^{'}, y^{'}, z^{'})

whose origin and axes coincide with those of the Earth frame. The time measured by the clock carried by this twin will be denoted by

t^{'}

Before departure, the twins synchronize their clocks and denote the departure event by

(P_{0})

, choosing it to correspond to

t = t^{'} = 0.

At this instant their relative velocity is still zero, and the two spatial coordinate systems coincide. They therefore both agree that the planet

G

is located at a distance

l

from Earth along the positive

x

-axis.

The twin on the rocket now begins his journey toward the planet

G

, following a straight trajectory along the direction of the common

x

-axis.

Departure event

1. Acceleration to $v$

We now consider the initial portion of the journey during which the rocket accelerates until it reaches the cruise velocity

v .

We use the following two events to delimit this phase:

•
$(P_{0})$ The rocket leaves Earth and begins its initial acceleration.
•
$(B)$ The rocket reaches the cruise velocity $v$ .

The twin on Earth assigns to event

(P_{0})

the coordinates

(x, y, z, t) = (0, 0, 0, 0),

and to event

(B)

the coordinates

(x, y, z, t) = (x_{1}, 0, 0, ϵ),

where

ϵ

is the time required, according to the Earth clock, for the rocket to reach its cruise velocity, and

x_{1}

is the corresponding position of the rocket relative to Earth.

The clock carried by the rocket is also set to zero at event

(P_{0})

. Let

ϵ^{'}

denote the elapsed time measured by this clock between events

(P_{0})

and

(B)

. Since the rocket is accelerating during this interval, there is no single Lorentz transformation that relates the Earth frame to the rocket over the entire acceleration phase. Instead, at every instant we consider the Lorentz transformation whose velocity parameter is equal to the instantaneous velocity of the rocket.

For a Lorentz transformation with velocity parameter

v

along the

x

-axis, the differential of the time coordinate is

Eq:Lorentz_transformation_differential_time

d t^{'} = γ (v) (d t - \frac{v}{c ^{2}} d x) .

Looking at the differential above, we may wonder why no term proportional to

d v

appears, since the velocity of the rocket is itself changing with time. The reason is that the velocity

v

appearing in the Lorentz transformation is a parameter of the transformation. For each Lorentz transformation it is therefore held fixed when the differential with respect to the spacetime coordinates is taken. Thus,

d t^{'} = γ (v) (d t - \frac{v}{c ^{2}} d x),

contains no contribution proportional to

d v

During the acceleration phase, the value of the parameter

v

changes along the trajectory of the rocket. We therefore use, at each event of the trajectory, the Lorentz transformation whose velocity parameter is equal to the velocity of the rocket at that event. In this sense,

d t^{'}

represents the infinitesimal time measured in the corresponding instantaneous rocket frame; it should not be interpreted as the differential of a single inertial time coordinate

t^{'}

extending over the entire acceleration phase. Rather, we consider a family of Lorentz transformations

Λ (v (t)),

and at each instant we use the member of this family whose fixed parameter is equal to the instantaneous velocity of the rocket. The parameter labeling the transformation changes along the trajectory, but it does not contribute to the differential within each individual transformation.

Along the trajectory of the rocket,

d x = v (t) d t,

and therefore Lorentz transformation differential time becomes

Eq:Rocket_clock_differential

d t^{'} = γ (v) (1 - \frac{v ^{2}}{c ^{2}}) d t = \frac{d t}{γ ( v )} .

The elapsed time measured by the clock carried by the rocket during the acceleration phase is consequently

Eq:Acceleration_elapsed_time

ϵ^{'} = \int_{0}^{ϵ} \frac{d t}{γ ( v ( t ))} = \int_{0}^{ϵ} 1 - \frac{v ^{2} ( t )}{c ^{2}} d t .

The precise relation between

ϵ

and

ϵ^{'}

therefore depends on the acceleration profile

v (t)

of the rocket. No assumption is made here that the velocity is small compared with the speed of light; on the contrary, the cruise velocity may be arbitrarily close to

c

We will assume that the acceleration phase is very short compared with the duration of the complete journey. More precisely, if

T

and

T^{'}

denote characteristic time scales of the journey as measured in the Earth frame and by the clock carried by the rocket, respectively, we assume that

Eq:Negligible_acceleration_times

ϵ ≪ T, ϵ^{'} ≪ T^{'} .

The times spent accelerating are therefore negligible compared with the times spent traveling at cruise velocity. In the remainder of the analysis we will neglect both

ϵ

and

ϵ^{'}

, while allowing the cruise velocity

v

to remain fully relativistic.

We further assume that the distance

x_{1}

traveled by the rocket during the initial acceleration phase is very small compared with the distance

l

between Earth and planet

G

x_{1} ≪ l .

We will therefore also neglect

x_{1}

in the remainder of the analysis.

2. Trip to $G$

We now consider the journey of the rocket to the planet

G

. Since the motion takes place entirely along the direction of the common

x

-axis, we will omit the coordinates

y

and

z

in what follows.

We consider the following two events:

•
$(P_{0})$ The rocket departs from Earth.
•
$(P_{1})$ The rocket reaches the planet $G$ with velocity $v$ and begins to decelerate.

In light of the approximations introduced for the initial acceleration phase, we neglect both the time and the distance between events

(P_{0})

and

(B)

. Thus, for the cruise-phase calculation, event

(B)

is treated as coincident with

(P_{0})

to the accuracy of our approximation. Accordingly, both twins assign to event

(P_{0})

the coordinates

(x, t) = (0, 0), (x^{'}, t^{'}) = (0, 0) .

The twin on Earth assigns to event

(P_{1})

the coordinates

(x, t) = (l, t_{1}),

where

t_{1}

is the time required for the rocket to travel from Earth to planet

G

. Since the rocket moves with constant speed

v = ∣ v ∣

during this portion of the journey,

Eq:Outbound_time_earth

t_{1} = \frac{l}{v} .

During the outbound cruise, let

F_{out}^{'}

denote the inertial frame comoving with the rocket. The rocket moves with velocity

+ v

relative to the Earth frame

F

, while Earth and planet

G

move with velocity

- v

relative to

F_{out}^{'}

. The twin on the rocket assigns to event

(P_{1})

the coordinates

(x^{'}, t^{'}) = (0, t_{1}^{'}),

where

t_{1}^{'}

is the elapsed time measured by his clock during the journey to planet

G

. There are two different ways in which he can determine

t_{1}^{'}

The first is to use the Lorentz transformation directly. Applying it to event

(P_{1})

, whose coordinates in the Earth frame are

(x, t) = (l, \frac{l}{v}),

we obtain

Eq:Outbound_time_rocket_lorentz

x^{'} t_{1}^{'} = γ (x - v t) = γ (l - v \frac{l}{v}) = 0, = γ (t - \frac{v}{c ^{2}} x) = γ (\frac{l}{v} - \frac{v l}{c ^{2}}) = \frac{1}{γ} \frac{l}{v} .

Therefore,

Eq:Outbound_time_relation

t_{1}^{'} = \frac{t _{1}}{γ} .

There is, however, another way for the twin on the rocket to understand this result. As soon as the rocket has reached its cruise velocity

v

, the distance between Earth and planet

G

, measured in the rocket frame, is smaller than the corresponding distance

l

measured in the Earth frame.

To understand why, let us first consider how the twin on the rocket measures a length belonging to an object at rest in the Earth frame. As an example, consider the diameter of the Earth along the direction of motion. Let its diameter measured in the Earth frame be

d = 2 r .

A measurement of this diameter in the rocket frame requires the positions of the two opposite endpoints of the diameter to be determined simultaneously in the rocket frame . We therefore introduce two events:

•
$(E_{+})$ The position of the endpoint of the Earth on the positive side of the $x$ -axis is measured.
•
$(E_{-})$ The position of the endpoint of the Earth on the negative side of the $x$ -axis is measured.

Since these two measurements are simultaneous in the rocket frame, their coordinates can be written as

Eq:Earth_diameter_events_rocket

E_{+} E_{-} : (x^{'}, t^{'}) = (x_{+}^{'}, t_{+}^{'}), : (x^{'}, t^{'}) = (x_{-}^{'}, t_{-}^{'}) .

The diameter measured by the twin on the rocket is therefore

Eq:Earth_diameter_rocket_definition

d^{'} = x_{+}^{'} - x_{-}^{'} .

It is important to emphasize that the measurement of a spatial distance in a given reference frame requires the positions of its endpoints to be measured at the same time in that frame. The events

E_{+}

and

E_{-}

are simultaneous in the rocket frame, but they are not simultaneous in the Earth frame.

Since the Earth is at rest in the Earth frame and centered at the origin, the same two events have coordinates

Eq:Earth_diameter_events_earth

E_{+} E_{-} : (x, t) = (r, t_{+}), : (x, t) = (- r, t_{-}) .

Because the two events are simultaneous in the rocket frame,

t_{+}^{'} - t_{-}^{'} = 0,

using the Lorentz transformation for the time coordinate, we obtain

0 = γ (t_{+} - \frac{v}{c ^{2}} r) - γ (t_{-} - \frac{v}{c ^{2}} (- r)) = γ [t_{+} - t_{-} - \frac{v}{c ^{2}} (2 r)] .

Since

d = 2 r

, it follows that

Eq:Earth_diameter_relativity_simultaneity

t_{+} - t_{-} = \frac{v}{c ^{2}} d .

Thus, although

E_{+}

and

E_{-}

occur simultaneously for the twin on the rocket, event

E_{+}

occurs after event

E_{-}

according to the twin on Earth. This is a direct manifestation of the relativity of simultaneity.

We can now use the spatial part of the Lorentz transformation to determine the diameter measured in the rocket frame. Taking the difference between the transformed positions of the two events gives

d^{'} = x_{+}^{'} - x_{-}^{'} = γ [(r - (- r)) - v (t_{+} - t_{-})] .

Using earth diameter relativity simultaneity , we obtain

Eq:Length_contraction_example

d^{'} = γ (d - \frac{v ^{2}}{c ^{2}} d) = \frac{d}{γ} .

The twin on the rocket therefore measures a smaller longitudinal diameter for the Earth than the twin at rest with respect to it. More generally, the separation between two points that are at rest in the Earth frame and separated along the direction of relative motion is measured in the rocket frame to be

Eq:Earth_planet_distance_rocket

l^{'} = \frac{l}{γ} .

In particular, Earth and planet

G

are both at rest in the Earth frame and are separated by the distance

l

. The twin on the rocket therefore measures their separation to be

l^{'}

The following figure summarizes the outbound journey in both reference frames. In the Earth frame, Earth and planet

G

are at rest and separated by the distance

l

, while the rocket moves between them with speed

v

. In the rocket frame, the rocket is at rest, while Earth and planet

G

move with the same speed

v

in the opposite direction, but their separation is reduced to

l / γ

. In the rocket-frame diagram, the rocket is shown midway between Earth and planet

G

only for visual convenience. This midpoint has no special physical significance. At the instant represented in the figure,

x_{E}^{'} = - \frac{l ^{'}}{2}, x_{G}^{'} = \frac{l ^{'}}{2},

with

l^{'} = \frac{l}{γ} .

At any other instant during the cruise, the individual coordinates of Earth and planet

G

would be different, while their simultaneous separation in the rocket frame would remain equal to

l^{'}

$Outbound trip: Earth frame Fversus the outbound rocket frame F'_{\mathrm{out}}.$

Outbound trip: Earth frame

F

versus the outbound rocket frame

F_{out}^{'}

As shown previously in reciprocal relative velocities , if the rocket moves with velocity

v

relative to Earth, then Earth and planet

G

move with velocity

- v

relative to the rocket. The magnitude of the relative velocity is therefore the same in both descriptions and is equal to

v

. Consequently, according to the rocket twin, planet

G

approaches him with speed

v

, and the time required for the planet to reach him is

Eq:Outbound_time_rocket_length

t_{1}^{'} = \frac{l ^{'}}{v} = \frac{1}{γ} \frac{l}{v} .

This is identical to the result obtained directly from the Lorentz transformation in outbound time rocket lorentz . Since

γ > 1

for

v > 0

, we also have

t_{1}^{'} < t_{1} .

Thus, the two twins agree on the events marking the beginning and the end of this portion of the journey, but they assign different spatial distances and different elapsed times to it. For the twin on the rocket, the distance between Earth and planet

G

is shorter, and correspondingly less time elapses before the rocket and the planet meet.

3. Trip Back to Earth

Once the twin on the rocket reaches planet

G

, he reverses his motion and begins the journey back to Earth. As in the initial acceleration phase, we assume that the time required to reverse the motion is very small compared with the duration of the journey, and we neglect both this time interval and the corresponding displacement.

We denote by

(P_{2})

the event at which the traveling twin reunites with the twin on Earth. For the return cruise, let

F_{in}^{'}

denote the inertial frame comoving with the rocket. The rocket moves with velocity

- v

relative to the Earth frame

F

, while Earth and planet

G

move with velocity

+ v

relative to

F_{in}^{'}

. The magnitude of the relative velocity is therefore again

v

. Moreover, for the same reason discussed in the previous section, the distance between Earth and planet

G

measured in the inbound rocket frame is

l^{'} = \frac{l}{γ} .

Consequently, the elapsed time measured by the twin on the rocket during the return journey is

Eq:Return_time_rocket

t_{2}^{'} = \frac{l ^{'}}{v} = \frac{1}{γ} \frac{l}{v} .

For the twin on Earth, on the other hand, the rocket travels the distance

l

from planet

G

back to Earth with speed

v

. The elapsed time for the return journey is therefore

Eq:Return_time_earth

t_{2} = \frac{l}{v} .

The outbound and return portions of the journey have the same duration in each corresponding description,

t_{2} = t_{1}, t_{2}^{'} = t_{1}^{'} .

Within the approximation that the launch and turnaround durations and displacements are negligible compared with the two cruise portions, the total elapsed time measured by the clock carried by the rocket is therefore

Eq:Total_journey_time_rocket

T^{'} = t_{1}^{'} + t_{2}^{'} = 2 \frac{1}{γ} \frac{l}{v},

whereas the total elapsed time measured by the clock on Earth is

Eq:Total_journey_time_earth

T = t_{1} + t_{2} = 2 \frac{l}{v} .

Since

γ > 1

for any nonzero relative velocity,

Eq:Twin_elapsed_time_difference

T^{'} < T .

Thus, at the reunion event

(P_{2})

on Earth, the clock carried by the traveling twin records a smaller elapsed time than the clock that remained on Earth. Since each twin's clock measures the time experienced by that twin, the traveling twin is younger when the two are reunited. The quantity

T^{'}

is the elapsed time recorded by the same continuous onboard clock throughout the entire journey; it is not a coordinate time belonging to one global inertial frame

F^{'}

. The complete journey is summarized in the following figure. The upper row represents the motion in the Earth frame, while the lower row represents the corresponding description during the constant-velocity portions of the rocket's motion. The short turnaround interval is shown explicitly but is assumed to be negligible compared with the duration of the two cruise portions.

$Complete journey in the Earth frame Fand in the corresponding outbound andinbound rocket frames F'_{\mathrm{out}}and F'_{\mathrm{in}}.$

Complete journey in the Earth frame

F

and in the corresponding outbound and inbound rocket frames

F_{out}^{'}

and

F_{in}^{'}

As in the previous figure, the precise positions shown during the cruise are schematic. What is physically relevant is that Earth and planet

G

are separated by the distance

l

in the Earth frame and by the contracted distance

l / γ

in the corresponding outbound or inbound rocket frame.

4. Events and the Apparent Asymmetry of Time Dilation

We will now return to the question raised at the beginning of the essay and address the apparent asymmetry in the motion of the two twins. One of the apparent contradictions comes from the familiar statement that a clock observed from a reference frame in which it is moving appears to run more slowly. It therefore seems reasonable that, when the rocket reaches planet

G

, the clock carried by the rocket has recorded less time than the clock on Earth. By symmetry, however, we might also expect the twin on the rocket to find that the clock on Earth has recorded less time than his own.

The resolution of this apparent contradiction requires us to be very precise about the events being compared. The familiar formulas for time dilation do not apply to arbitrary events: they apply to particular pairs of events selected by specific conditions.

Let us first consider the arrival of the rocket at planet

G

from the point of view of the twin on Earth. The arrival event is

P_{1} : (x, t) = (l, \frac{l}{v}) .

At the same time in the Earth frame, the twin on Earth is still at the origin of

F

. We therefore consider the event

Q_{1} : (x, t) = (0, \frac{l}{v}) .

Thus, in the Earth frame, the two events

P_{1} and Q_{1}

are simultaneous. The first occurs at planet

G

, while the second occurs at the position of the twin on Earth. In this description, the Earth twin is at rest and his position remains fixed at

x = 0

Let us now describe the same arrival event

P_{1}

from the outbound rocket frame

F_{out}^{'}

. Its coordinates are

P_{1} : (x^{'}, t^{'}) = (0, \frac{l}{γ v}) .

At this same time in

F_{out}^{'}

, the twin on Earth is not at a fixed spatial position. He is moving toward the negative

x^{'}

direction with velocity

- v

. Since the distance between Earth and planet

G

measured in the rocket frame is

l^{'} = \frac{l}{γ},

at the instant at which the rocket reaches planet

G

, the Earth twin is located at

x^{'} = - \frac{l}{γ} .

We therefore introduce the event

R_{1} : (x^{'}, t^{'}) = (- \frac{l}{γ}, \frac{l}{γ v}) .

Thus, in the rocket frame, the two events

P_{1} and R_{1}

are simultaneous. Again, one event is the arrival of the rocket at planet

G

, while the other occurs at the position of the twin on Earth at that same time according to the rocket frame. The difference is that, in this description, the Earth twin is moving rather than remaining at a fixed spatial coordinate.

We can now determine how the event

R_{1}

is recorded in the Earth frame by using the inverse Lorentz transformation,

x = γ (x^{'} + v t^{'}), t = γ (t^{'} + \frac{v}{c ^{2}} x^{'}) .

For the spatial coordinate we obtain

x (R_{1}) = γ (- \frac{l}{γ} + v \frac{l}{γ v}) = 0,

as expected, since

R_{1}

is the position of the twin on Earth. For the time coordinate,

t (R_{1}) = γ (\frac{l}{γ v} - \frac{v}{c ^{2}} \frac{l}{γ}) = \frac{l}{v} (1 - \frac{v ^{2}}{c ^{2}}) = \frac{l}{γ ^{2} v} .

Since the elapsed time measured by the rocket clock between departure and

P_{1}

t_{1}^{'} = \frac{l}{γ v},

we may equivalently write

t (R_{1}) = \frac{t _{1}^{'}}{γ} .

We therefore recover the familiar reciprocal time-dilation result: according to the outbound rocket frame, the clock on Earth has advanced by less than the clock carried by the rocket.

There is nevertheless no contradiction with the result obtained previously. The Earth event

Q_{1}

used in the Earth-frame description and the Earth event

R_{1}

used in the rocket-frame description are not the same event. Indeed,

t (Q_{1}) = \frac{l}{v}, t (R_{1}) = \frac{l}{γ ^{2} v} .

Thus,

Q_{1} \neq = R_{1} .

The two events are selected by different simultaneity conditions. In the Earth frame,

t (P_{1}) = t (Q_{1}),

while in the outbound rocket frame,

t^{'} (P_{1}) = t^{'} (R_{1}) .

The two reference frames therefore associate different events on the Earth twin's reference system with the same arrival event

P_{1}

This distinction is the key point. The arrival event

P_{1}

is not arbitrary: it is constrained by the physical setup of the experiment. Planet

G

remains at the fixed position

x = l

in the Earth frame, while the rocket travels with speed

v

. Therefore the rocket can meet planet

G

only when

l = v t,

which fixes the Earth-frame time of the arrival event to be

t (P_{1}) = \frac{l}{v} .

The event selected by the reciprocal time-dilation comparison in

F_{out}^{'}

is instead

R_{1}

, a different event on the Earth reference system. The two apparently contradictory statements therefore compare different pairs of events.

The role of events in the comparison of elapsed times.

This illustrates a principle that is fundamental throughout special relativity: before applying a formula for time dilation, length contraction, or simultaneity, we must first identify precisely the events to which that formula refers.

The Twin Paradox

The apparent asymmetry of relativistic motion and its resolution

Prerequisites

Contents

Contents

I. Introduction

II. Description of the Experiment

1. Acceleration to $v$

2. Trip to $G$

3. Trip Back to Earth

4. Events and the Apparent Asymmetry of Time Dilation

Contents

I. Introduction

II. Description of the Experiment

1. Acceleration to v

2. Trip to G

3. Trip Back to Earth

4. Events and the Apparent Asymmetry of Time Dilation

1. Acceleration to $v$

2. Trip to $G$