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Time travel by going fast in a spaceship: questions

  1. Aug 18, 2011 #1

    First of all, I just want to say that I don't know much about general relativity. But my question concerns an example often found in textbooks about special relativity.

    On to my question. I have heard several times that one can travel forward in time (earth-time) by leaving earth, traveling close to c in a spaceship and then come back. But I don't understand this. I know the textbook reasoning behind it, but I'm not satisfied by the textbook explainations.

    The textbook explainations go like this:

    Say we have a mirror in the ceiling of the traveling spaceship, and send a light pulse straight up from the spaceship floor, let it reflect in the mirror, and come back. We measure the time of that, and call it t.

    From earth, the light appears to travels not straight up, but in a longer path since the spaceship moves relative to earth. But since c is the same for all reference frames, the time it takes will appear longer from earth. Let's call it t2. So t2 > t.

    And so time moves slower in the spaceship than on earth, so when the spaceship gets back, more time has elapsed on earth than in the spaceship.

    My question: What if the mirror experiment had been carried out on earth? Then the exact same reasoning could be made, and the conclusion would have been the opposite: that more time had elapsed in the spaceship than on earth. And without that particular explaination: Why would the time in one reference frame go faster than time in the other, given that the laws of physics are supposed to be the same in all inertial reference frames?
  2. jcsd
  3. Aug 18, 2011 #2


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    You can reverse the rolls - from the Earth's point of view, time is slow on the spaceship and normal on the Earth. From the spaceship point of view, time is slow on the Earth,and normal on the spaceship.

    This is possible because the definition of "at the same time" depends on the observer and his state of motion.

    If you draw a space-time diagram, it looks like the included figure below:


    The stationary observer uses the green lines of simultaneity, the moving observer uses the red lines. Each concludes that the other clock is slow when they compare clocks using their own notions of simultaneity (the green and red lines on the diagram).

    This is known as the twin "paradox". It's not a real paradox, of course, when the fact that simultaneity is relative is factored in. There's about two zillion posts on this, and just as much more written about it, if you care to read about it.

    One interesting case is when one twin accelerates, so that they eventually re-unite. In that case, in flat space-time, the twin that accelerates wil be the younger when they reunite.
  4. Aug 18, 2011 #3


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  5. Aug 18, 2011 #4
    I figured this question would have been asked a lot, but didn't know the name of the paradox.

    Anyway, I read up on wikipedia, and it makes more sense now. But the way they describe it in every textbook I ever read, they don't resolve the paradox. No wonder people are puzzled by it.

    Thank you!
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