# TEMPORAL DILATION

## (IF THE SPEED IS CONSTANT)

I propose to readers another article, where the characters may very well be two friends: ** Alpha** and

**.**

*Beta**(and not necessarily twins!)*

**decides to leave with his spaceship, first runs a reconnaissance lap increasing his speed and, just as**

*Beta***recognizes him by raising his eyes to the sky, he is seen by the latter darting at a speed equal to**

*Alfa***.**

*v = 0.866*c**(about **90%** of the speed of light indicated by **c**)*

** Alpha** knows that

**is headed for a planet**

*Beta***away and that the spaceship will not stop in the vicinity of the planet itself, continuing to travel in a uniform straight line (the two friends thus agreed before the departure of**

*17.32 light-years***, thus avoiding referring in the discussion to phases of acceleration and deceleration); in the example proposed it is in fact required that the spaceship cover the entire**

*Beta**Earth-planet distance*at a constant speed.

First, what does it mean that the distance is ** 17.32 light-years**? It means that the light takes

**to cover it all. (**

*17.32 years***correspond to almost**

*0.32 years***, actually**

*four months***and just over**

*3 months***)**

*25 days***is not as fast as light and will therefore take longer to travel that distance; inverting the simple formula**

*Beta’s spaceship*

*v = s/t**(because the speed is equal to the ratio between the distance traveled and the time taken)*, we obtain that

*t = s/v***.**

*(t_Alfa = 17,32/0,866 = 20 years)*** t_Alfa = 20 years** is the travel time with respect to the Earth’s frame and

**, although not seeing his friend, knows that he will pass in the immediate vicinity of the planet when**

*Alpha***have elapsed for the clocks in the Earth’s frame.**

*20 years*We have so far analyzed the journey with respect to ** Alpha**, the hourly equation of uniform rectilinear motion dictates that to travel a distance of

**at a constant speed of**

*17.32 light-years***takes**

*0.866 * c***.**

*20 years*

*(and it can’t be otherwise!)*Now let’s consider the travel time measured with respect to the spaceship’s frame; if we hold the traveler friend still, it is not only the Earth that moves with ** Alpha**, but the whole Universe does it (and every distance that apparently moves towards the spaceship with opposite speed is contracted by the

**, that is minor);**

*Lorentz Transformations***is well aware of this also because she knows that the planets and stars are not ellipsoids (contracted spheres) so “accentuated”. At velocity**

*Beta***, when the reciprocal of**

*v=0.866*c***, the Earth-planet distance is**

*the Lorentz factor is 0.5***for**

*8.66 light-years***(half of**

*Beta***, where this last is the length to travel with respect to the frame of the Earth)**

*17.32 light-years*Still considering the hourly equation of uniform rectilinear motion, for** Beta** the travel time is

**.**

*t_Beta = 8.66/0.866 = 10 years*

*(half of t_Alfa, and even here it cannot be otherwise!)*If desired, one can imagine that a rigid rod with a length equal to the *Earth-planet distance* (and with a direction parallel to the Earth-planet distance itself) is attached to the tail of Beta’s spaceship. When the spaceship is in motion, the friend left on Earth sees the rod itself in motion (therefore contracted), and it is precisely when he sees the passage of the rod end in front of his eyes that he will know what the arrival time indicated by the ** Beta** stopwatch.

When the auction ends, ** 10 years** have passed for

**(in the Earth’s frame) and**

*Alpha***for**

*5 years***(in the spaceship’s frame)**

*Beta***has not yet reached the planet with respect to the Earth’s frame: the spaceship has covered half the effective Earth-Planet distance.**

*, Beta*

*(8.66 light-years)**Alpha and Beta are not the same age. (When ten years have passed for Alfa, only five have passed for Beta in the spaceship’s frame!)*

*Conclusions*

The frequent mistake is to think that, if the spaceship is believed to be stationary, the “apparent travel” time of ** Alfa** with respect to the spaceship’s frame must be slowed down and this is what creates

*the “false” paradox***and therefore, just as in Classical Physics, if the travel time with respect to**

*(as if there were a misleading symmetry that does not allow us to understand for which of the two observers the time elapsed is actually shorter)*;*we still remember that for the spaceship the whole Universe is in motion (not only the Earth)***is**

*Beta***, also the “apparent travel” time of the Earth-planet distance (until the planet “reaches” the spaceship) is**

*10 years***. Sometimes, erroneously, considering**

*10 years**Minkowski’s space-time graphs*, the astronaut is considered to be stationary and an attempt is made to ideally trace the universe line of the Earth’s apparent motion with respect to the spaceship’s frame, in this case it is wrong do it because there is no time dilation (the time duration of the two motions is the same).

When I refer to the universe lines I mean the lines drawn in the Minkowski diagram.

*However, it would be different if the astronaut remained stationary floating in Space and if only the Earth (for some strange reason) began to move. So yes, **Alpha’s** travel time would be less than travel time measured by **Beta**! If only the Earth moved (whatever the length of the distance), **Beta **would see only the Earth contract (while it is in motion, and not the whole Universe as seen above), and for this he would get older than his friend!*

**I conclude, stating that Lorentz Transformations are symmetrical and do not generate contradictions (you just need to pay attention), a body in motion in a reference frame will always be “younger” than the other observers who are stationary in the frame itself!**

*Massimiliano Dell’Aguzzo*