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rolling stone

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- TL;DR Summary
- A simple "thought experiment” based on a circular trajectory is investigated by elementary notions of general relativity. The result is surprising: three different observers draw three different conclusions about the same scenario.

Consider a mass ##M## that generates a curvature of the space-time and an observer ##O##, fixed and positioned at such a great distance from ##M## that the time ##t## of its clock is not affected by ##M##.

Suppose that the observer ##O##, in his polar coordinates reference system centred at ##M##, observes a spacecraft (of negligible mass with respect to ##M##) that rotates with constant linear speed ##v## in the equatorial plane at his feet. Suppose also that on this spacecraft there is spaceman ##S_1## equipped with a clock whose time is ## t_1##.

THE SCENARIO SEEN BY THE OBSERVER ##O##

According to the observer ##O##, the infinitesimal line element, in the Schwarzschild's metric and spherical polar coordinates, of the spacecraft moving in the circular path is:

(1)## ~~~~~~~~~~ds^2=-(1-r_s/R)c^2 dt^2 + R^2 dφ ^2##

where ##G## is the gravitational constant, ##r_s=2GM/c^2## is the Schwarzschild radius, ## φ## is the azimuth angle and ##R## is the radius of the circular path.

If the spacecraft rotates with constant angular speed ## dφ / dt = v /R## :

(2)## ~~~~~~~~~~ ds^2= - (1- r_s / R) c^2 dt^2 + v^2 dt ^2##

In particular if ##R=GM/v^2##, the spacecraft moves by inertia along a geodetic path, where the centrifugal acceleration is compensated by the centripetal one.

Since the same infinitesimal line element computed by spaceman ##S_1## is:

(3)## ~~~~~~~~~~ ds^2=- c^2 dτ_1^2 ##

where ##τ_1## is the proper time of ##S_1##, from eq.(2) and eq.(3) it follows that:

(4)## ~~~~~~~~~~ - c^2 dτ_1^2 =-(1-r_s/R) c^2 dt^2 + v^2 dt ^2 ##

(5)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##

Let us now consider a second spacecraft , with the spaceman ##S_2## on board and a clock whose time is ## t_2##, moving in the same circular path as ##S_1##, with linear velocity ##-v## in the opposite direction.

NOTE: in order to avoid any clash between the two spacecraft s we can imagine, for example, that the common trajectory is a single virtual train track on which the two spacecraft s may gently slide away, face up and face down, respectively.

By the same considerations previously made for ##S_1## , the observer ##O## may deduce that for the spaceman ##S_2## too:

(6)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##

where ##τ_2## is the proper time of ##S_2##.

Eq.(5) and eq.(6) show that, according to the observer ##O##, the time on the spacecraft s of ##S_1## and ##S_2## flows exactly at the same rate.

Therefore if we assume that ##S_1## and ##S_2## are twins who, at the initial reference time ##t_1= t_2=0##, are positioned with their spacecraft s at the same point in the circular trajectory, according to the fixed external observer ##O## they will continue to have the same age regardless of the number of laps they will run around the mass## M##.

THE SCENARIO SEEN BY THE SPACEMAN ##S_1##

Moving in a geodetic path, each of the two spacemen may believe that his spacecraft is stationary while the twin's one is in motion. No physical experiment, on either of the two spacecraft s, will ever demonstrate the opposite, since their motion is inertial with constant speed. Consequently, according to spaceman ##S_1## , ##S_2## is moving away with relative speed ##–2v##.

In order to express, according to ##S_1## , the infinitesimal line element traveled by ##S_1## , the eq.(1) must be rewritten in terms of the clock time ##t_1##, remembering that ##(1-r_s/R) dt^2= dt_1^2## (because the clock of ##S_1## is located at distance ##R## from the centre of ##M##) and that ## dφ / dt_1 = - 2 v /R ## .Thus:

(7)## ~~~~~~~~~~ds^2=-c^2 dt_1^2 + R^2 ( dφ/dt_1)^2~dt_1 ^2= - c^2 dt_1^2+4v^2 dt_1 ^2##

But, according to spaceman ##S_2##:

(8)## ~~~~~~~~~~ ds^2= - c^2 dτ_2^2 ##

Now combining eq.(7) with eq.(8) :

(9)## ~~~~~~~~~~- c^2 dτ_2^2 = - c^2 dt_1^2+4v^2 dt_1 ^2##

one obtains:

(10)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - 4~v^2/c^2 } ~ dt_1 ##

So, according to spaceman ##S_1## , the time of the clock on board of the spacecraft of ##S_2## flows more slowly than his own; therefore after one lap his twin ##S_2## will be younger than him.

THE SCENARIO SEEN BY THE SPACEMAN ##S_2##

By repeating, on behalf of the spaceman ##S_2##, the same previous reasoning one obtains:

(11)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - 4~v^2/c^2 } ~ dt_2 ##

So, according to spaceman ##S_2##, the time of the clock on board of the spacecraft of ##S_1## flows more slowly than his own; therefore after one lap his twin ##S_1## will be younger than him.

DISCUSSION

In conclusion: after the two spacecraft s have run one lap around the mass ##M##,

- according to the external observer ##O##: “the twins ##S_1## and ##S_2## will continue to have the same age" [cfr. eq.(5) and eq.(6) ]

- according to spaceman ##S_1##: “##S_2## will be younger than ##S_1##” [cfr. eq.(10)]

- according to spaceman ##S_2##: “##S_1## will be younger than ##S_2##” [cfr. eq.(11)].

Questions:

- What is wrong in the previous reasoning?

- If nothing is wrong, may we assume that in the theory of relativity the description of the same scenario depends on the reference frame system from which the scenario is observed?

- If so, isn't Schopenhauer right when he says that "the world is a representation"?

Suppose that the observer ##O##, in his polar coordinates reference system centred at ##M##, observes a spacecraft (of negligible mass with respect to ##M##) that rotates with constant linear speed ##v## in the equatorial plane at his feet. Suppose also that on this spacecraft there is spaceman ##S_1## equipped with a clock whose time is ## t_1##.

THE SCENARIO SEEN BY THE OBSERVER ##O##

According to the observer ##O##, the infinitesimal line element, in the Schwarzschild's metric and spherical polar coordinates, of the spacecraft moving in the circular path is:

(1)## ~~~~~~~~~~ds^2=-(1-r_s/R)c^2 dt^2 + R^2 dφ ^2##

where ##G## is the gravitational constant, ##r_s=2GM/c^2## is the Schwarzschild radius, ## φ## is the azimuth angle and ##R## is the radius of the circular path.

If the spacecraft rotates with constant angular speed ## dφ / dt = v /R## :

(2)## ~~~~~~~~~~ ds^2= - (1- r_s / R) c^2 dt^2 + v^2 dt ^2##

In particular if ##R=GM/v^2##, the spacecraft moves by inertia along a geodetic path, where the centrifugal acceleration is compensated by the centripetal one.

Since the same infinitesimal line element computed by spaceman ##S_1## is:

(3)## ~~~~~~~~~~ ds^2=- c^2 dτ_1^2 ##

where ##τ_1## is the proper time of ##S_1##, from eq.(2) and eq.(3) it follows that:

(4)## ~~~~~~~~~~ - c^2 dτ_1^2 =-(1-r_s/R) c^2 dt^2 + v^2 dt ^2 ##

(5)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##

Let us now consider a second spacecraft , with the spaceman ##S_2## on board and a clock whose time is ## t_2##, moving in the same circular path as ##S_1##, with linear velocity ##-v## in the opposite direction.

NOTE: in order to avoid any clash between the two spacecraft s we can imagine, for example, that the common trajectory is a single virtual train track on which the two spacecraft s may gently slide away, face up and face down, respectively.

By the same considerations previously made for ##S_1## , the observer ##O## may deduce that for the spaceman ##S_2## too:

(6)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - v^2/c^2 - r_s/R } ~ dt ##

where ##τ_2## is the proper time of ##S_2##.

Eq.(5) and eq.(6) show that, according to the observer ##O##, the time on the spacecraft s of ##S_1## and ##S_2## flows exactly at the same rate.

Therefore if we assume that ##S_1## and ##S_2## are twins who, at the initial reference time ##t_1= t_2=0##, are positioned with their spacecraft s at the same point in the circular trajectory, according to the fixed external observer ##O## they will continue to have the same age regardless of the number of laps they will run around the mass## M##.

THE SCENARIO SEEN BY THE SPACEMAN ##S_1##

Moving in a geodetic path, each of the two spacemen may believe that his spacecraft is stationary while the twin's one is in motion. No physical experiment, on either of the two spacecraft s, will ever demonstrate the opposite, since their motion is inertial with constant speed. Consequently, according to spaceman ##S_1## , ##S_2## is moving away with relative speed ##–2v##.

In order to express, according to ##S_1## , the infinitesimal line element traveled by ##S_1## , the eq.(1) must be rewritten in terms of the clock time ##t_1##, remembering that ##(1-r_s/R) dt^2= dt_1^2## (because the clock of ##S_1## is located at distance ##R## from the centre of ##M##) and that ## dφ / dt_1 = - 2 v /R ## .Thus:

(7)## ~~~~~~~~~~ds^2=-c^2 dt_1^2 + R^2 ( dφ/dt_1)^2~dt_1 ^2= - c^2 dt_1^2+4v^2 dt_1 ^2##

But, according to spaceman ##S_2##:

(8)## ~~~~~~~~~~ ds^2= - c^2 dτ_2^2 ##

Now combining eq.(7) with eq.(8) :

(9)## ~~~~~~~~~~- c^2 dτ_2^2 = - c^2 dt_1^2+4v^2 dt_1 ^2##

one obtains:

(10)## ~~~~~~~~~~ dτ_2 = \sqrt {1 - 4~v^2/c^2 } ~ dt_1 ##

So, according to spaceman ##S_1## , the time of the clock on board of the spacecraft of ##S_2## flows more slowly than his own; therefore after one lap his twin ##S_2## will be younger than him.

THE SCENARIO SEEN BY THE SPACEMAN ##S_2##

By repeating, on behalf of the spaceman ##S_2##, the same previous reasoning one obtains:

(11)## ~~~~~~~~~~ dτ_1 = \sqrt {1 - 4~v^2/c^2 } ~ dt_2 ##

So, according to spaceman ##S_2##, the time of the clock on board of the spacecraft of ##S_1## flows more slowly than his own; therefore after one lap his twin ##S_1## will be younger than him.

DISCUSSION

In conclusion: after the two spacecraft s have run one lap around the mass ##M##,

- according to the external observer ##O##: “the twins ##S_1## and ##S_2## will continue to have the same age" [cfr. eq.(5) and eq.(6) ]

- according to spaceman ##S_1##: “##S_2## will be younger than ##S_1##” [cfr. eq.(10)]

- according to spaceman ##S_2##: “##S_1## will be younger than ##S_2##” [cfr. eq.(11)].

Questions:

- What is wrong in the previous reasoning?

- If nothing is wrong, may we assume that in the theory of relativity the description of the same scenario depends on the reference frame system from which the scenario is observed?

- If so, isn't Schopenhauer right when he says that "the world is a representation"?