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Time dilation

  1. Feb 1, 2007 #1


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    The twin paradox is very confusing, and even after reading the explanation, I still get questions. The only explanation of the paradox is when an object is first moving away from earth, and then moving towards the earth. They make it very complicated.

    How about just moving one direction, and then stop.

    Let's say that an object moves one lightyear from earth in the rest frame in a straight line for an average speed that makes the dialation for the object 0.5 in the rest-frame. The object will see the earth move away from it, and assume it is dialated by 0.5 as well. When both observe this, what will be true when the object stops? Will the object be "older" or will the earth be?
  2. jcsd
  3. Feb 1, 2007 #2
    But this is exactly why the situation ensues. Try to explain the following:

    1. If the twin doesn't turn around and come back, how will the two twins compare ages?

    2. If the twin does not turn around , how will the time differential occur? Have you looked at the wiki diagrams?

    The twin that "moved in one direction and then stopped" will be younger. The calculation is showed on the wiki page.

    see above. try reading the wiki explanation for accelerated motion, it will also clear (maybe) your confusion about why you need to use hyperbolic motion equations when describing accelerated motion in SR.
  4. Feb 1, 2007 #3


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    Ok. Nakurusil, I think you are incorrect at the other post about the accelerated motion and length contraction.

    What I wonder is, why would the earth see this object as "younger" if the object could say the same thing?

    What is the deal with acceleration, I don't understand the equations.
  5. Feb 1, 2007 #4
    Learn the math. How much do you know already? This stuff is 10x easier to get once you pick up the math. You can spend hours trying to work out who's older, or you can learn that geodesics minimize proper time & get some real information that's useable elsewhere.
  6. Feb 1, 2007 #5


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    Well, I know that we must use the lorentz transformation. And I am not dealing with acceleration here, so I guess this is what I can:

    The trip in the rest frame (earth) to a point in space is 1 lightyear.
    An object is sent out in a speed of 0.8c in the earths frame and is stopped when it has reached 1 lightyear.
    We are ignoring acceleration, gravitation and air resistance here.

    The time it will take for the earth will be: [tex]\frac{1}{0.8} = 1.25[/tex], this means that after 1.25 years the object will have stopped in the earth frame.

    The object is sent out at 0.8c which is 299 972 456*0.8 = 239977966.4 which is approximately 240 000 000 m/s.
    The object will see the earth move in the same speed, correct?

    (First time using tex :smile:)
    Ok, I put this in the relative [tex]\gamma[/tex] factor: [tex]\gamma = \frac{1}{\sqrt \frac{v^2}{c^2}}[/tex]

    So the [tex]\gamma[/tex] factor will be: 1.666


    Now, the time it will take for the object is [tex]\frac{1}{1.666} = 0.6[/tex] years.

    The space will also be contracted for the object, so it will only move
    [tex]\frac{1}{1.666} = 0.6[/tex] lightyears in it's frame. This is what I think is the connection between time dialation and length contraction. The object moves as fast as the earth percieve it, but it's clock is slowed down, showing it 1.666 times slower.

    Is it correct so far?

    Ok, moving on:

    So 1 lightyear away from the earth, after 0.6 years (in the objects frame) it has stopped. The object is 0.6 years old.

    For the earth it has gone 1.25 years, so the earth is 1.25 years old as it observe the object stop.

    Is this correct? If not, what has I been doing wrong?

    If it is correct:

    This shows that the object is younger than the earth.
    But what hinders me in turning this the other way around, making it the EARTH the object that is moving (because it is in the objects frame). In that case the earth would have aged 0.6 years and the object 1.25 years.
    THAT is the thing I don't understand.
    Last edited: Feb 1, 2007
  7. Feb 1, 2007 #6


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    Because the Earth was using a single inertial frame on both the outbound and the inbound leg, while you'd have to use two separate inertial frames if you always want to look at things from the perspective of the object's rest frame. These two frames have different definitions of simultaneity, so the reading on the Earth's clock the instant before the ship turned around, in the frame where the ship was at rest before turning around, will be totally different than the reading on the Earth's clock at the instant after the ship turned around, in the frame where the ship was at rest after turning around. If you forget to take into account this discontinuous jump in the Earth's clock-reading when you switch frames, you get the wrong answer.

    Alternately, you could analyze both the inbound leg and the outbound leg from just one of these frames--for example, you could continue to use the frame where the ship was at rest during the outbound leg after the ship had turned around and begun the inbound leg (at which point the ship's velocity will be even larger than the Earth's velocity in this frame), and in this case you will again find that the ship has aged less when it reaches the Earth (because although the Earth's clock was ticking slow during the first leg, the ship's clock was ticking even slower during the second).
    You have the number for the gamma-factor correct, but the equation should be[tex]\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
    Last edited: Feb 1, 2007
  8. Feb 1, 2007 #7


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    Let's set up an example and trace through it carefully. Let the Earth and Star Base Alpha be at rest with respect to each other in inertial reference frame S, 4 light-years (ly) apart. Their clocks are synchronized in frame S.

    The traveler starts out at rest in frame S, near the Earth. He looks out his window and synchronizes his clock with the clock on Earth. When the two clocks both read zero, he starts his trip to Alpha, accelerating very rapidly so he quickly reaches a cruising speed of 0.8c (0.8 ly/yr), before he's gotten very far. He is now at rest in a second inertial reference frame, S'.

    In frame S, he travels the distance of 4 ly at speed 0.8 ly/yr, arriving at Alpha at t = 5 yr, and decelerates quickly to a stop. During the trip, his clock runs slower in frame S by a factor of 3/5, so when he arrives at Alpha, his clock should read 3 yr.

    In frame S', of course, the traveler's clock runs at its normal rate, and the clocks on Earth and on Alpha run slower by a factor of 3/5. But those two clocks are also not synchronized in S', because of the relativity of simultaneity. In S', the Alpha clock runs "ahead" of the Earth clock by [itex]v \Delta x / c^2[/itex] = (0.8 ly/yr)(4 ly)(1 ly/yr)^2 = 3.2 yr. Therefore, just after the traveler begins his trip, in his new frame (S'), the Earth clock still reads (practically) zero, the Alpha clock reads 3.2 yr. The traveler might describe this by saying that during his acceleration, the distant clock on Alpha jumps ahead by 3.2 yr, whereas the nearby clock on Earth keeps the same reading.

    In S', the distance between Earth and Alpha is contracted to 2.4 ly, so his trip lasts (2.4 ly) / (0.8 ly/yr) = 3 yr according to him. During that time, the time-dilated Earth and Alpha clocks both advance by (3/5)(3 yr) = 1.8 yr. The Earth clock advances from 0 to 1.8 yr, and the Alpha clock advances from 3.2 yr to 5.0 yr.

    Finally, when the traveler decelerates to a stop at Alpha, he switches back to frame S. In this frame, the clock at Alpha (which is now right next to him) has the same reading as just before the deceleration, but the clock on Earth is now synchronized with it. The traveler might describe this by saying that during his deceleration, the distant clock on Earth jumps ahead by 3.2 yr, whereas the nearby clock on Alpha keeps the same reading.

    Therefore, when the traveler arrives at Alpha, his clock reads 3 yr and the Alpha clock reads 5 yr, according to both the traveler and the station manager who greets him at Star Base Alpha.
    Last edited: Feb 1, 2007
  9. Feb 1, 2007 #8
    Yes you are - see
    means that by definition the object has acceleration in some point of the journey. This breaks one of the assumptions of Special Relativity - that the laws of physics are the same in all inertial frames. If the object experiences any acceleration, it is no longer inertial.
  10. Feb 1, 2007 #9
    Doesn't matter what you think, you are obviously wrong. The others set you right.

    This is why is called a "paradox":blushing:

    This is why you need to take some classes, you can't learn from the posts if you don't have the basic knowledge. Because you can't understand the posts.
  11. Feb 2, 2007 #10
    I think most of the solutions are confusing - they give the right answers but they miss an easy analysis based upon a straight forward application of the invariance of the spacetime interval. Take your original premise (one way) - we assume no acceleration - the outbound twin is in Los angeles and heads outbound toward his brother in NY. He reaches his crusing speed before arriving at NY, and as he passes by NY he sets his clock to read the same as his brothers - then continues at this same velocity until he reaches VEGA. When the traveling twin reaches Vega he checks a local clock and finds his own clock reads less by the Gamma factor. Since the vega clock and the NY clock are in the same frame these two clocks will read the same - therefore at this point in the journey the traveler will have aged less than his stay at home brother. To find the total age difference if he returns by doing a quick turn around - you double the one way time loss.

    There is no need to consider acceleration. On the outbound trip the spacetime interval for the stay at home twin consists only of a temporal amount whereas the spacetime interval for the traveler comprises both a temporal factor and a space component (he moves a spatial distance vt and temporal excursion ct') whereas the stay at home twin only experiences a temporal excursion ct. Therefore ct > ct'
  12. Feb 2, 2007 #11
    He has a point I did it Irish, I got to grips with the paradox in my head through visualising the problem and then looked at the maths which pretty much represented how I was thinking, I have no doubt it's a lot less confusing the other way round, it takes a bit of mental juggling to get your head around length contraction and time dilation, but once you do you see the paradox isn't a paradox at all. Just evidence that you don't understand the theory involved.

    If you can say this is not a paradox because of x and understand it, your most of the way there to grasping Special relativity.
  13. Feb 2, 2007 #12


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    Thanks for your posts, they were very insightful.

    I think I understand the essence of it, and this is probably the post that made me:
    This was pretty detailed and complex, but as far as I understand it: In the frame S' the clock on the earth was 0 when the object had traveled a part of his journey already? You say it is zero when the object is close to point alpha, I don't make any sense of that :confused:

    But this is true, right?: In the frame S' the simoultanous points of time in S is not the same at all.
    Wouldn't this make the direction of the object matter? I mean that the object would not read the alpha to be near it's own if it had traveled excactly the same journey, only in the opposite direction. So in a way, you could say that the direction of the object is relevant to what it will read on other clocks.

    Yes, that's the equation I meant, thanks.

    But I don't get what you mean by inbound, and outbound leg :confused:
    But thanks for the post.
  14. Feb 2, 2007 #13


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    I was also a bit confused with the twin paradox, until I wrote my own explanation:
    I hope that someone else will find it usefull too.
    See, in particular, Eqs. (8) and (9) and their derivation and explanation.
    Last edited: Feb 2, 2007
  15. Feb 2, 2007 #14
    This is a very nice paper, I greatly enjoyed it.Very easy to read, complete and to the point. Congratulations.
  16. Feb 2, 2007 #15


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    Thanks! :smile:
  17. Feb 2, 2007 #16


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    Outbound leg = the phase of the trip when the ship was traveling away from the Earth at constant speed, inbound leg = the phase of the trip after the ship had turned around and was returning to Earth at constant speed, right up until the time it returns to Earth and the two clocks are compared.
  18. Feb 2, 2007 #17


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    Thanks for your link, I will read your article :smile:

    Ah, thanks.

    But JesseM, why do the object read the clock of the Alpha point different that the clock of the earth, moving in the frame S'.
  19. Feb 2, 2007 #18


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    Only a very very small part of the journey. The traveler accelerates so rapidly that he reaches his cruising speed after traveling an insignificant distance, and an insignificant amount of time has passed on his clock. For practical purposes of calculation, we can consider his speed to change instantly from 0 to 0.8c, so that he does not travel any distance at all in this process. Such instantaneous changes are not physically possible, of course, but they make the calculations much simpler, and we can think of them as the "limiting case" of more and more rapid acceleration.

    By "it" do you mean the clock on the earth? Where do you think I said that, in the part that you quoted? :confused:

    Yes. That's a good observation. If the traveler had been moving in the opposite direction, away from Alpha, then in that reference frame the clock on Alpha would run "behind" the clock on Earth by 3.2 yr. So for example if he starts his trip at the beginning of 2008, then after the "instantaneous" acceleration to his new reference frame, the Earth clock would still read 2008.0 but the Alpha clock would read 2004.8.

    (In the original situation, with the traveler moving towards Alpha, after his acceleration the Earth clock would still reads 2008.0 and the Alpha clock would read 2011.2.)
  20. Feb 2, 2007 #19


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    "the Alpha point" is where the ship turns around, right? If so, all frames will agree on what the ship's clock reads at the moment it reaches that point, and on what a clock sitting at the Alpha point reads at that moment--this is a local question about the readings of two clocks when they are right next to each other, so different frames can't disagree. But because frames define simultaneity differently, they disagree about clock readings for widely-separated events--for example, frames would disagree about which tick of the Earth-clock happened "at the same time" as the ship was reaching the Alpha point, and likewise they'd disagree about which tick of the clock at the Alpha point happened "at the same time" that the ship was leaving Earth, or returning to Earth.

    Does this answer your question? If not, can you elaborate on what you meant when you said the ship would "read the clock of the Alpha point different that the clock of the earth"?
  21. Feb 2, 2007 #20


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    I read through your original post, and I drew and calculated everything that was described in it. I think everything made sense, except a few things:

    - Why does the Alpha clock show 3.2 years in S'? If it is because it is moving towards Alpha, why doesn't the lorentz transformation take direction in consideration?

    - How can it be that if the objects clock in frame S' be nearly synchronized to Alpha in frame S, if the object are moving at 0.8c? This is not either taken in consideration in the lorentz transformation.

    - I have found out that both direction and position to the observer must be taken in consideration when measuring objects in relative speeds. How can this be when the lorentz transformation takes none of them in consideration?

    - And how can clocks jump several years during deceleration/acceleration (switching from S' to S). This was very confusing.

    But thanks again for a terrific post.

    I have misunderstood this. I have read your post carefully again.

    Is it maybe the direction of acceleration/deceleration that matters, not velocity? Or have I misunderstood the whole concept.
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