Your Twin Sibling is More Youthful Than You..

[Jothi]

When one of your twin siblings has travelled to space and when he returns, he will be younger than the twin sibling.

 

Why is it so ?

 

Can anyone explain me on this ?

[Delta1212]

Although the classic twin scenario involves sending a twin out into space and having them return younger, it's not going into space that causes this. What causes it is the twin moving at very close to the speed of light, and then turning around and coming back. That's obviously a bit difficult to imagine happening on Earth for any appreciable length of time, so we imagine the twin flying away in space.

 

Now, the closer you get to the speed of light, the more time slows down and the shorter distances become. So if the twin on Earth watches the twin in space fly away at, let's say, about 86% of the speed of light. For every minute that passes for the Earth-bound twin, he will see only thirty seconds pass on his twin's spaceship.

 

Here comes one of the more counter-intuitive parts of relativity. Once the twin in the spaceship has achieved cruising speed and is no longer accelerating, for him there will be no difference between his ship rocketing away from a stationary Earth at 86% the speed of light and his own ship standing still while the Earth recedes at 86% of the speed of light. It's a bit like looking out the window of your train and seeing another train going by, and needing a minute to tell whether your train is moving or the other train is. On the train you can usually figure it out from the vibrations as the train rolls over bumps in the tracks, but on a spaceship there are no bumps. There's no way to tell who is really moving. And that includes observing the time dilation of Earth. Because if the twin in the spaceship looks back at Earth, he'll see Earth's time ticking away at half the rate, not his own.

 

So if both twins see each other aging half as fast, how do we end up with the twin in the spaceship being younger? Well, the situation is only symmetrical as long as no one is accelerating. Once the twin in the spaceship turns around and returns to Earth, he will have undergone an acceleration that returns him to the same rest frame as Earth (and his twin) and his clock will be synced up with what the twin on Earth observed. He'll have experienced half the amount of time that his twin did over the course of the trip.

 

On the other hand, if the twin on Earth decides he misses his space-bound twin and gets in his own rocket to go catch up with his twin, upon arriving at the original spaceship the twins will discover that the twin who initially stayed on Earth is the one who experienced less time than the twin who first took off in the spaceship.

[Janus]

Basically, it comes down to the fact that our twins, traveling at relative speed with respect to each other measure time and space differently.

 

So for our Earth twin, if his brother travels to a star 8.6 light years away at 86% of c, it will take 20 years for him to return. He will also note that during this time, his brother will have aged 1/2 as fast as he did and will only be 10 years old on return.

 

The spaceship brother measures the following: As referred to by Delta1212, distances also become shorter at relative speed. That ten light year distance between Earth and star is only 4.3 light years as measured by him when he is traveling at 86% of c with respect to the two. This means that by his reckoning, it only takes 10 yrs to complete the round trip to the star.

 

But as mentioned already, he should also measure the Earth clock as running slow compared to his as he travels to and fro to the star and thus should expect the Earth twin to only have aged 5 years rather than 20. So where does the extra 15 years come from? As alluded to, it happens when our Spaceship twin accelerates at the star to make his return trip.

 

There is another aspect to the difference in how the two brothers measure time in addition to its rate. This aspect deals with how each brother determines simultaneity. It turns out that not only will the brothers not agree as to their respective time rates, but they also disagree as to what event are simultaneous.

 

To illustrate, consider a string of clocks in a line between Earth and the star. They have been set up so that they are synchronized with each other and the Earth brother's clock. In other words, they all read the same time as each other. For the space ship brother, this will not be true. by his measurement, the clocks all read different times, with the clocks in the direction he is traveling reading further ahead of the ones in the opposite direction. the further apart the clocks are the greater the difference in their times. Thus if he is passing a clock when it reads 12:00, according to the Earth twin, all the clocks at that moment read 12:00, but according to the Spaceship twin all the clocks ahead of him read later than 12:00 with the difference increasing with distance, and all the clocks behind him read earlier than 12:00. This is known as the Relativity of Simultaneity.

 

So consider now what happens to these clocks as he travels to the star and back. Assume the clocks all reads Jan 2014 according to the Earth twin when he starts his trip (we will assume near instantaneous acceleration. The instant he is moving at 86% of c, the clock on Earth still reads Jan 2014, however, now the clock at the star( being in his direction of travel) already reads July 2021. As he travels to the star it take 5 years by his clock to travel the 4.3 light years he measures to the star. during which time. the Earth and star clock advance 2.5 years so that the Earth clock now reads July 2016 and the star clock reads Jan 2024.

 

He turns around to head back to Earth (again instantly). The star clock still reads Jan 2024, However whereas before turnaround the Earth clock was behind him and thus reading 7.5 years earlier than the star clock, it is now ahead of him and reading 7.5 yrs later than the star clock, so it is now July 2031 on Earth by the spaceship twin's reckoning. It takes another 5 years by the spaceship clock to return to Earth. during which time the Earth clock advances 2.5 years and ends up reading Jan 2034 upon the spaceship's arrival, 20 years later than it did when the spaceship left, while 10 yrs has passed for the spaceship.

[LaurieAG]

How do things work out if the spaceship travels in a circle at a constant velocity?

[swansont]

How do things work out if the spaceship travels in a circle at a constant velocity?

 

In kinematic effects only, the clock will run slow relative to an inertial frame for the orbiting craft. This is well-confirmed experimentally.

[xyzt]

How do things work out if the spaceship travels in a circle at a constant velocity?

If the clock in the inertial frame measures an elapsed time [math]T[/math], then, in the rocket frame, the measured proper time will be less:

[math]\tau=\int_0^T{\sqrt{1-v^2/c^2}dt}[/math].

[math]\tau<T[/math]

For uniform circular motion [math]v=\omega R[/math] so [math]\tau=T\sqrt{1-\omega^2R^2/c^2}[/math].

Edited December 7, 2014 by xyzt

[david345]

The correct answer probably involves general relativity but I will give a simple answer using special relativity. In 2d space the distance between two points is s^2=x^2+y^2. This is just Pythagorean theorem.In 3d it is s^2= x^2 +y^2+z^2. In spacetime it is s^2=t^2-(x^2). It should be noted that -3^2=9 and -(3^2)=-9. S is called the spacetime interval. The spacetime interval is invariant. Invariant=same for all observers. Relative=different for observers moving at different speeds. To simplify the equation I set c =1 (one unit of time equals 1 year and one unit of distance equals 1 light-year) and only consider movement in the x direction. We will now compare the two equations.

 

S^2=x^2+y^2

 

S^2=t^2-(x^2)

 

The twin who travels to space and returns will be called spaceman the twin who remains on earth will be called earthman. We will use the reference frame of earthman. Earthman see's his twin leave earth at half the speed of light and it takes 5 years to get to his destination. It takes another 5 years for him to return at half the speed of light. We will now calculate the spacetime interval for space man. (Note: time will be earthman time and distance will be earthman distance)

Away trip:

S^2=5^2-(2.5^2) S=4.33

Return trip:

S^2=5^2-(2.5^2) S=4.33

Spacetime interval for total trip:

4.33+4.33=8.66

We will now calculate the spacetime interval for earthman.

S^2=10^2-(0^2) s=10

Spacetime interval for spaceman=8.66

Spacetime interval for earthman=10

 

Spaceman: spacetime interval(invariant)=proper time(invariant)=8.66

 

Earthman: coordinate time(relative)=spacetime interval(invariant)=proper time(invariant)=10

 

If you use the time dilation equation for spaceman you will find that it gives the same 8.66 result as the spacetime interval.

Edited January 25, 2015 by david345