In the twin paradox or twins paradox what do the clocks of the twin and the distant star he visits show when he's at the star?

Question

In the twins paradox of relativity one twin stays on earth while the other travels to a star ten light years away, and then immediately flies back. Because his rocket travels at just under the speed of light, the entire (including the return) journey takes just under twenty years as measured by clocks on earth, but only a day, say, according to the clocks on the rocket. Furthermore, the twin on earth has physically aged by twenty years, while the traveling twin has aged only a day.

If the clocks at the distant star are synchronized with those on earth, (Einstein synchronized, if that helps, which just means that looking at across the ten light year gap, each would see ten year old light indicating a time that was ten years less than one's own clock shows, a symmetrical situation) what do they say the time is when the twin arrives there and what do the clocks on the rocket say?

Note, "paradox" might be a bit of a misnomer, as it's really just counterintuitive.

Answer 1

This is a straightforward calculation. We assume the spaceship and Earth synchronise their clocks so the spaceship leaves at time zero for the 10 light year trip to Tralfamador, and the clocks on Tralfamador have been (Einstein) synchronised with those on Earth so that the Earth and Tralfamador clocks always show the same time. The spaceship's speed measured in Earth coordinates is given by some function $v(t)$ and we are not assuming any particular form for the function $v(t)$.

In Earth (and Tralfamador) coordinates the arrival time $T$ at Tralfamador is then given by:

$$ \int_0^T v(t) dt = 10~\text{light years} \tag{1} $$

Now to find the elapsed time on the spaceship clock we have to compute the proper time $\tau$ for the journey using the Minkowski metric:

$$ c^2 d\tau = c^2 dt^2 - dx^2 \tag{2} $$

Since the proper time is the same for all coordinate systems we can do this in the Earth/Tralfamador coordinates. We do the calculation by noting that $dx = v(t)dt$ and substituting this into our equation to give (after some rearrangement):

$$ d\tau = \sqrt{1 - \frac{v^2(t)}{c^2}} dt \tag{3} $$

then we integrate to find the proper time:

$$ \tau = \int_0^T \sqrt{1 - \frac{v^2(t)}{c^2}} dt \tag{4} $$

The elapsed time on the spaceship's clock is equal to the proper time, so the spaceship's clock will show the time $\tau$ when it arrives at Tralfamador while the clocks on Tralfamador will show the time $T$.

The values of $T$ and $\tau$ will depend on the form of the function $v(t)$, but note that for any $v(t) \ne 0$ the square root in equation $(4)$ is less than one so integrating will give a value of $\tau < T$ i.e. the time has been dilated on the spaceship.

If we wanted to include the return journey we simply do the same calculation in reverse for the return leg.

Answer 2

This is an example of the relativity of simultaneity. Let's suppose the distances and speeds are such that the journey takes exactly 10 years in the Earth frame and 12 hours in the ship frame.

When the twin arrives at the star, the clock on the star will read 10 years, while the clock on the spaceship with show 12 hours.

The time dilation effect is entirely symmetrical, so in the traveller's frame, the clock on the distant star has advanced only 12 hours since the start of the traveller's journey. In the traveller's frame the time at the distant star when the traveller started their trip was minus ten years, so from the traveller's perspective the clock on the distant star was out of synch by ten years.

Answer 3

In the following "clock" means a clock that is just under 10 light years away from earth, and said clock reads zero at the start of the journey, and it shows the number of years passed. And I'm answering the question: "what do they say the time is when the twin arrives there". And finally, I changed the distance to just under 10 light years, to save some typing.

Earth twin says: The clock reads 10 years

Traveling twin says: The clock reads 10 years

Earth twin says: The travel took 10 years, during that time the clock proceeded 10 years.

Traveling twin says: "The travel took a half day, during that time the clock proceeded a few seconds. Except if we decide to say that the travel took a half day, during that time the clock proceeded 10 years, almost all of that proceeding happened during my acceleration".

As for reference, I refer to all the literature of the twin paradox. I mean all parts of twin paradox have been explained many times.

Answer 4

Some time ago I produced a series of animations of the "Twin Paradox" which show the view from the traveling twin, looking back at the starting point. There is a combined clock in the top left corner showing the traveler's proper time, coordinate time, and the time on a clock on the "home station" at all times during the journey. In addition, there is a line of clocks showing coordinate time along the entire distance. There are no mysteries, paradoxes or sleight of hand here; just look at the clocks!

Here is the explanatory text from YT (note, truncated links within are not working, go to the page to use them!):

The twin "paradox" (in quotes because it is NOT a real paradox!) is a basic teaching scenario for Special Relativity: https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_intro.html

This is a series of visualizations of the journey from the point of view of the traveling twin, who flies 20 light years away from his home station then returns. The journey consists of four parts, joined together. The first quarter is an acceleration away from the station. During the second and third quarters the ship accelerates towards the station (so that at the half way point the ship is stationary 20 light years away). In the fourth quarter the ship accelerates away from the station in order to come to rest there.

The total coordinate travel time (shown as a red dot in the top left HUD clock) is 43.711/58.918 years for acceleration at earth/moon gravity levels, whilst proper time (green dot) is 12.101/38.694 years. The yellow dot represents the time the traveler would see on the station clock face through a very powerful telescope! The octahedral stations (spaced one light year apart and one light year to the left of the flight path) are all synchronized to coordinate time and rotate once over the course of the whole journey. There is a 2x2 light year wall one light year beyond the far end of the journey, and large rectangular frames every 5 light years. Where a floor is shown it is 1 light year per stripe, and there are small 1 ly milestone spheres along the way, with a larger one every five light years.

The flights are rendered without relativistic effects and then for two values of acceleration (currently earth gravity and moon gravity). The distortion artifacts are due to aberration of light, the Doppler effect and the headlight effect. The earth gravity videos exhibit some nice Penrose-Terrel "rotation" effects. Magenta markers in a circle show where you really are in the scene (in the sense that things outside the circle are behind you, whilst those inside the circle are in front of you. Doppler shift = gamma for this circle), and grey markers show the circle where the Doppler shift is 1 (gamma is in theory directly observable on this circle).

These videos were made with POV-Ray (https://www.povray.org/) and avconv/ffmpeg (https://www.ffmpeg.org/) video encoding software.

POV and support files are available from https://github.com/m4r35n357/FirstPersonRelativity