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Time Dilatation: The Twins Paradox

  1. Aug 5, 2014 #1

    We’ve all seen this little fellas up there. Two brothers, twins, in their 60’s. And they decide that one of them is going to take a trip into space, let’s say for a year, at 99% the speed of light. The other brother stays there, waiting for his twin to return, and looking at the light clock that his brother installed in the outside of the rocket, so he on earth could always take a look at it.

    Finally the day comes, and brother number 1 is looking at his pocket light clock as he took his morning coffee like every other day, before going onboard the rocket and leave his planet for one year. He sits in the little chair in the rocket, and while looking at his watch, gets launched into space at 0.99c

    On the other hand, brother number 2 is standing in the hill, looking at his brother going into space, and he quickly realizes that his brother’s light clock got slowed down, a lot, in relation to his owns, and he calculates that a year for his brother would be something like 30 for him. But he patiently sits there, and waits, and gets old.

    The day his brother arrives, they go running to see each other and for the surprise of brother number 1, his brother was now 90, while he is now 61.

    Brother number 2 in the other hand, get’s shocked at the image of his now young brother, but he already knew what was going to happen, as he calculated 30 years before.

    The problem for me here is the following:
    Brother number 1 traveled a year at 0.99c. When he departed, he was looking at his clock and glancing through the window of the rocket at his brother clock’s, and from his point of view, his brother’s clock got slower, as the earth was launched out of the rocket, which for him was stationary. For him, the clock on earth was moving away and getting slower, so he thought he was getting older faster than his brother.

    What happens to the end of this story from the point of view of the voyager, in this case brother number 1? This story for me seems to have two different ends. Am I alone in this?
  2. jcsd
  3. Aug 5, 2014 #2


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    The twin paradox is explicated ad nauseum on the internet. They are NOT symmetrical. The traveler has to turn around and that breaks the symmetry. Yes, on the way out, they "see" symmetrical time dilation but in the end they have traveled through different space-time paths, one of which has more distance and the other of which as more time.
  4. Aug 5, 2014 #3


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    When you think about what they actually see, you're talking about the Doppler Effect. That means slower if both are moving away from each other, but faster if they approach each other. This effect is indeed symmetric in special relativity.

    The difference between the twins is this: when B turns around, he sees A's clock immediately go faster than his own. On the other hand, A sees B turn around almost 15 years later, because the light from this event takes such a long time to travel to him.
    So while B sees A age more quickly for half of the time, A sees B age more quickly for just half a percent of the time. Their experiences explain perfectly what they see at arrival.

    Of course, this is not a rigid derivation of time dilation. It just shows how the story can have only one end. Here's a link to more detailed explanations, including a diagram of what I've just said.
  5. Aug 5, 2014 #4


    Staff: Mentor

    Here is a good FAQ on the twin paradox:

    Please go over it and let us know if there is something you still don't get afterwards. Pay special attention to the spacetime diagram analysis. Basically, the spacetime diagram analysis boils down to the fact that different paths have different lengths. That should not be too surprising, after all, the same is true in Euclidean geometry also.
  6. Sep 12, 2014 #5
    This is my understanding of time dilation and the twin paradox.

    Let's take a one way trip leaving Earth for Jupiter. We have synchronized clocks on Earth, the Spaceship and on Jupiter. And let's assume Jupiter is 100 light minutes away from Earth. At time equal zero the spaceship will accelerate to 0.99 C in one meter. This very rapid acceleration takes a very short amount of time for both an observer on Earth and for an observer on the Spaceship. After the acceleration completes the clocks on the Spaceship and on Earth both still read very very close to time equal zero. This is the initial condition for the folowing discussion.

    The Spaceship is traveling at 0.99 C, and Jupiter is 100 light-minutes from Earth. Using "Time = Distance / Rate" the Spaceship will pass Jupiter at Time 101.01 minutes on the Jupiter clock. Due to time dilation, the Spaceship clock will read 14 minutes when the Spaceship passes Jupiter. In this scenario there is no return trip and there is no acceleration, therefore the time dilation can not be attributed to acceleration or a return trip.

    There are two concepts here:
    • It is not possible to transfer information from point A to point B faster than the speed of light. We could give the Spaceship a faster speed, for example 0.999C or 0.9999C but the Spaceship could never have a speed greater than C.
    • The observer on the Spaceship, using the Spaceship's clock, can get to Jupiter in a shorter time by achieving a higher velocity when departing Earth. This is common sense, the faster you move the quicker you get to your destination.
    The Twin Paradox is that the 2 reference frames are symmetrical, the observer on the Spaceship sees Jupiter approaching at 0.99C and the observer on Jupiter sees the Spaceship approaching at 0.99C. So how could the Spaceship clock and the Jupiter clock be different when Jupiter and the Spaceship meet.

    A closer inspection shows an asymmetry. At time zero (when the Spaceship is departing Earth at 0.99C) the observer on the Spaceship will see Jupiter approaching. However, at time zero the observer on Jupiter will see the Spaceship on the lauch pad at the Spaceport on Earth. It is not until the time on Jupiter equals 100 minutes that the observer on Jupiter will see the Spaceship depart Earth.

    Let's now derive the time dilation equation ...

    The unknown that we want to solve for is the time on the Spaceship's clock when it passes Jupiter, this unknown will be referred to as "T". To solve for T, use the principal of Special Relativity that says there is not a preferred inertial reference frame. An observer on the Spaceship, viewing the clock on Jupiter through a powerful telescope, will see the Jupiter clock ticking very rapidly. And an observer on Jupiter, viewing the Spaceship clock will see the Spaceship clock ticking equally rapidly.

    From start to end of the trip:
    • The Spaceship clock advances from 0 to T. Duration = T minutes.
    • The observer on the Spaceship sees the Jupiter clock advancing from -100 to 101.01. Duration = 201.01 minutes
    • The observer on Jupiter sees the Jupiter clock advancing from 100 to 101.01. Duration = 1.01 minutes
    Using the equality "Spaceship Observation of Jupiter Clock Rate" equals "Jupiter Observation of Spaceship Clock Rate" we can solve for T, as follows:

    201.01 / T = T / 1.01

    T = sqrt(201.01 * 1.01) = 14.25 minutes

    We can replace the numbers with D, V, and C; and solve for T.

    (D / V + D / C) / T = T / (D / V - D / C)

    And using algebra to solve for T, we get:

    T = (D / V) * sqrt(1 - V^2/C^2)
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