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Twin paradox and asymmetry

  1. May 29, 2015 #1
    Dear PF Forum,

    Can someone make it clear for me?
    Perhaps I should ask a very simple question. Concerning just one direction.

    Picutre02a.jpg
    The distance between earth and Star T is 100 lys.
    The clock is synchronized for A and B
    And as PeterDonis pointed out at my previous thread.
    Which I believe 87% is an easy number. Since ##87\%^2## ≈ 0.75
    B travels to star T at [itex]\sqrt{0.75}[/itex] c according to Lorentz transformation
    This is the Lorentz Factor.
    [itex]\frac{1}{\sqrt{1-\sqrt{0.75}^2}}[/itex] = [itex]\frac{1}{\sqrt{1-0.75}}[/itex] = [itex]\frac{1}{\sqrt{0.25}} [/itex] = [itex]\frac{1}{0.5}[/itex] = 2
    According to A: B needs [itex]\frac{1}{\sqrt{0.75}}[/itex] to reach T or 115 years.
    According to B: B needs [itex]\frac{1}{\sqrt{0.75}}[/itex] /2 to reach T or 57.75 years
    Okay...
    But motion is relative, right?
    A would have tought that it's A who moves, B stays.
    Picutre02c.jpg

    Or, please compare Pic 3
    Picutre02b.jpg
    Why B experience time dilation, while A not?
    Isn't motion relative?
     
  2. jcsd
  3. May 29, 2015 #2

    DEvens

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    Ah yes. Twins. Do not be alarmed if you don't get it. Lots of full profs with a decade or more experience teaching relativity do not "get" twins.

    The thing to notice is, you have a real problem comparing clocks when they are physically separated. To compare them they really should be located at the same physical location. If you try to compare them by sending signals you get into lots and lots of complexities. These can be dealt with, but there really is quite a lot involved.

    The usual situation in twins is for one twin to go on a journey and return. Only when the twin returns is there a good opportunity to compare clocks and note that one had a difference.

    And that is the key to understanding the symmetry here. And it goes like this. Imagine that twin B starts on his journey at 87% of c relative to Earth. But after 1 year of ship time he notices that he has left his favorite DVD at home. So he sends a laser-phone message to twin A. Twin A is really the nice twin, and sets out in an even faster ship to catch up to twin B, bringing the beloved DVD. Now he sets out at a speed such that the two twins see each other approaching at 87% of c.

    What will the two twins observe about each other's relative clock values when they get together?

    For bonus points: work out the actual values of the clocks, taking into account propagation of the light signal sent by the travelling twin, and the fact that he sent it after one year of ship time.

    And for even more bonus points: The two twins see each other leaving then approaching at 0.87c. What does an observer still on the Earth see with respect to the speeds of the two ships?

    Hint: Having a twin start out and the second twin go catch him is symmetric (in a particular way) with having one twin start out and then return. This is an important symmetry in relativity.
     
  4. May 29, 2015 #3
    There is no such symmetry because the movements are not symmetric. Remember than in special relativity there are some frames that are specials, the inertial frames. If you go outside those frames, then you are outside the special relativity.

    In this case, one of the twin stays in an inertial frame, while the other have an accelerated movement (at least at one point in his trajectory) .

    So, there is nothing wrong with computing different times for A and B since they movement where not qualitatively the same...that is what I think at least.
     
  5. May 29, 2015 #4

    Janus

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    You have to account for length contraction. In picture 3, where you have E and T moving while B is at rest, you have to apply length contraction to the distance between E and T and B will measure this distance to be 50 light years, so for him, T traveled 50 light years at 0.87c, which took 57.75 years. To complete the picture you need to take the relativity of simultaneity into a account also. If in picture 2 you have clocks on E and T that are synchronized to each other.( both read 0 when B leaves E and Both read 115 years when B reaches T), these clocks are not synchronized in picture 3, the frame in which B is at rest, In this frame E reads 0 when B leaves, but T already reads 86.125 years. As T travels towards B and E away from B, they both age 1/2 as fast as B for a total of 28.875 years. Thus when T reaches B its clock reads 96.125+28.875= 115 year and the Clock on E reads 28.875 years.
     
  6. May 29, 2015 #5

    Ibix

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    With your one-directional example, you still have asymmetry between your ship and planet frame: the origin and destination are defined in the planets' frame. This means that, in the ship frame, the distance between the planets is length contracted.

    So: in the planet frame, the ship has 100ly to travel at 0.866c, so it takes 100/0.866=115y to get there. However, the ship is in motion with a Lorentz gamma factor of 2. It is length contracted (which is irrelevant to this scenario) and time dilated - so its clocks will record 115/2=58y of travel time.

    In the ship frame, the planets are moving, so they and the distance between them are length contracted - so the destination planet only has 50ly to cross to reach the ship. At 0.866c, this takes 58y by the ship's clocks. The planets' clocks are time-dilated in this frame, so they accumulate 58/2=29y during the crossing.

    Note that both frames agree about the time accumulated by the ship's clock in the crossing (although their explanations for why it reads what it does differ). However, they disagree about the time on the planets' clocks. This is because relativity put "now" on the same footing as "here" - it means different things to different people. There's no ambiguity about the time taken according to the ship's clocks because you are using one clock. There's ambiguity in the planets' clocks because there are two clocks and two frames with different definitions of what "now" means. According to the planet frame, both planets started their stopwatches simultaneously and recorded 115 years while the rocket travelled. According to the ship frame, the destination planet started its stopwatch 86 years early and accumulated 29 years in the crossing, so it reads 115 years when the ship arrives. The origin planet started its stopwatch, accumulated 29 years during the crossing, but then left it ticking for another 86 years for some reason.

    You can't really analyse this completely using the time dilation and length contraction formulae - you need to use the full Lorentz transforms. And preferably draw a space-time diagram.
     
  7. May 29, 2015 #6

    pervect

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    You forgot a couple of important details, which are important to understanding the problem.

    You say the distance between the earth, which I'll call E, and the start T is 100 light years. You stated the distance as if it were an absolute fact, either forgetting or not realizing that in relativity, distances are frame dependent. Either way, you need to be a bit more specific if you're going to come to grips with the problem. You say that the distance is 100 light years. We will assume that what you mean is that the distance is 100 light years in the common frame that E and T share, which we can give a name, the ET frame.

    You also said that the clocks on the spaceship A and B were synchronized. Since Anad B are at the same place, there are no frame dependency issues.

    You didn't say that the clocks on E and T had to be synchronized, but in order to carry out the experiment, you need to make some such assumption. Because clock synchronization is also frame dependent in special relativity (as is distance), you need to again specify that E's clock and T's clock are synchronized in the ET frame.

    To analyze this problem properly, you need to answer the following questions:

    We have a new frame, B, associated with the moving observer B.

    1) What is the distance between E and T in B's frame? We know that the distance was 100 light years in the ET frame, but we need to figure out what the distance is in B's frame. It is possible to do this via the equations in the Wiki article you linked. I am not going to answer the question for you yet, but I'm going to ask you to make your best attempt. I will give one hint - that hint is "Lorentz conraction of distances".

    2) In a similar vein, we know that the clocks on E and T were synchronized in the ET frame. But we need to ask, are they still synchronized in the B frame? If they aren't synchronized, we need to ask "what is the reading on the T clock at the instant B and A meet in the B frame.

    I will again not answer this directly, but give one hint. That hint is that in special relativity, simultaneity is relative.

    3) You also need to account for time dilation on the part of B, which you've already mentioned.
     
  8. May 30, 2015 #7
    In this case I prefer to use 87% constant is [itex]\sqrt{0.75}[/itex]. And my calculations here are based on [itex]\sqrt{0.75}[/itex], not 87%.
    Vb = [itex]\sqrt{0.75}[/itex]
    Vba = [itex]\sqrt{0.75}[/itex]

    First we have to work out what is A speed so that B sees A approaching at 87%c, Vb-a
    Oh, that's easy.
    Newton: Va = Vb + Vba; Va = 87%c + 87%c = 176%c. Sorry, I was joking.

    GR: [itex]V_a = \frac{V_b+V_ba}{1+V_b * V_ba} = \frac{\sqrt{0.75}+\sqrt{0.75}}{1 + 0.75} = \frac{2\sqrt{0.75}}{1.75}[/itex] ≈ 0.9897 c (calculator here)
    [itex]V_a = \frac{2 \sqrt{0.75}}{1.75}[/itex]c

    LZaLorentz factor for A [itex] = \frac{1}{\sqrt{1-V_a^2}} = \frac{1}{\sqrt{1-\frac{2 \sqrt{0.75}}{1.75}^2}}
    = \frac{1}{\sqrt{1-\frac{4 * 0.75}{1.75^2}}} = \frac{1}{\sqrt{\frac{0.25^2}{1.75^2}}} = 7[/itex], so
    LAz = 7

    LZbLorentz factor for B [itex] = \frac{1}{\sqrt{1-V_b^2}} = \frac{1}{\sqrt{1-\sqrt{0.75}^2}} = \frac{1}{\sqrt{1-0.75}} = \frac{1}{\sqrt{0.25}} = \frac{1}{0.5} = 2 [/itex], so
    LZb = 2


    I short all calculations here

    What will the two twins observe about each other's relative clock values when they get together?
    So..., after 1 year of B time, B sends a laser-phone message to twin A
    B distance is LZB x Vb = 2 x 0.87 = 1.73 Ly from A
    B sends a message. The signal reaches A at 1.73 + 1.73 (the light still travels at light speed, not 1-87%, right?) = 3.46 years from Tzero or 3 years 5 months 17 days
    So, the distance between B and A now is 3.46 ly.
    A travels at 0.9897 c chasing B
    B travels at 0.87 c
    Where and when will they meet?
    So...
    Va * time = Vb * time + 3.46 year, sorry
    Va * time = Vb * time + 3.46 light year, right?
    [itex]\frac{2 \sqrt{0.75}}{1.75}[/itex] * time = [itex]\sqrt{0.75}[/itex] * time + 2(signal trip) * 2(LZB) * [itex]\frac{1}{\sqrt{0.75}}[/itex]
    Time = 112/3 years (earth time) ≈ 37 years, 4 month.


    Anwer:
    A travels at the speed [itex]\frac{2 \sqrt{0.75}}{1.75}[/itex]c
    A start travels at time: [itex]2 * \sqrt{3}[/itex] years or Third year, 17th of May.
    They'll get together after 112/3 years or 37 years, 4 month, earth time.
    They'll get together at [itex]\frac{56}{\sqrt{3}}[/itex] lys from earth
    LAz = 7
    LZb = 2

    A clock after they met:
    A travel for: [itex]\\frac{112}{3} - 2 *\sqrt{3}[/itex] years in relativistic speed
    = 33.8692 years
    Sorry, no latex here, too clumsy:
    Divide by Lorentz factor: 4.8385 years
    A clock at start traveling + A clock at travel time: 8.3026 years:
    B clock: after adjusting by Lorentz factor: 112/3 / 2 = 28/3 years
    After they met:
    A Clock: 8.3026 years:
    B Clock: 9.3333 years

    What will the two twins observe about each other's relative clock values when they get together?
    For bonus points: work out the actual values of the clocks, taking into account propagation of the light signal sent by the travelling twin, and the fact that he sent it after one year of ship time.


    A start travel at: [itex]2 * \sqrt{3}[/itex] years or Third year, 17th of May.
    They will meet at 112/3 years Earth Time:
    A Clock: 8.3026 years
    B Clock: 9.3333 years


    And for even more bonus points: The two twins see each other leaving then approaching at 0.87c. What does an observer still on the Earth see with respect to the speeds of the two ships?
    A travels at the speed [itex]\frac{2 \sqrt{0.75}}{1.75}[/itex]c

    Arghhh, you're trying to kill me with those calculations :nb)
    Actually I want to ask about the universe frame of reference. But I have to understand this twin paradox asymmetry first.
    Now, I DON'T want to ask about universe frame of reference. So that I don't get confused, the twin paradox, first.
    So, did I calculate correcty?

    Thanks DEvens
     
  9. May 30, 2015 #8
    Thanks pervect for your help.

    Yes, that's true.
    The distance between E, T at rest is 100 lys.

    Yes, all the clocks are synchronized.
    At initialization, all objects are at rest. Synchronizing clocks in E and T is done by Einstein clock synchronization.

    ET distance in B's frame is 50 lys.
    Lorentz contraction factor for 87% c is 2

    Now, this question I can't answer.

    "are they still synchronized in the B frame? "

    This, I don't understand. Care to explain more to me?
    Are you trying to say that:
    1. Does clock synchronization for 2 points at rest (or at the same frame) depends to another frame?
    For example: Does synchronizing E and T clock at ET frame differ from synchronizing E and T clock at B frame?


    "what is the reading on the T clock at the instant B and A meet in the B frame."

    This, I don't understand either.
    What do you mean by B and A meet at the B frame? Is it that B and A meet at Earth after B returns, and B stop, or B still traveling pass E from T?

    If, after B returns to E and stops, will the reading at T is minus 100 year that of E?
    If, after B returns to E and still going, will the reading at T is minus 100 year that of E?
     
  10. May 30, 2015 #9

    Janus

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    Yes.

    Consider two clocks at rest with respect to the A frame. They are not running and are set to zero. to synchronize them, you emit a light signal from a point halfway between them that will start each clock when it hits them like this:

    synch1.gif
    The light expands out as a circle, striking both clocks simultaneously and starting both.

    Now consider the same situation as viewed from the B frame which is in motion with respect to the A frame. The same light is emitted from the point half way between the two clocks. because the speed of light is invariant. it expands out in a circle from the point of emission. However, now one clock is moving towards the source point and the other away from it and one clock is struck by the light and starts running before the other like this.

    synch2.gif

    Once running both clocks run at the same rate but out of step with each other. These are the same clocks and the same light as in the first animation, they are just being considered from a different frame of reference.
     
  11. May 30, 2015 #10
    Good, very good!. Thanks Janus.
    Even both locations at same frame or reference will behave differently if observed by another moving observer.
    This I can understand, or at least barely grasp.

    Judging from this picture
     
    Last edited by a moderator: May 7, 2017
  12. May 30, 2015 #11

    Janus

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    Both clocks will be green.

    The easiest way to look at this is from the frame in which the clocks are not moving. In this case, both clocks receive the other clock's signal 25 min after it is sent, so it will get a clock reading of 25 min. ago. This does not change no matter which frame you are viewing from.
    One thing that should be pointed out is that my animations only took relativity of simultaneity into account. To fully analyze events as seen from a frame moving with respect to the clocks, you also have to take time dilation and length contraction into account. If the clocks were moving at 0.866c with respect to our frame, then they will run 1/2 as fast and the distance between them will be only 12.5 light min.

    The difference between the clocks according to this moving frame will be
    [tex]\frac{vx}{c^2}[/tex]

    Where v is the relative speed between frames and x is the distance between the clocks as measure in the rest frame of the clocks.
    This gives.
    [tex] \frac{0.866c(25 light min)}{c^2} = 21.65 min[/tex]

    Thus, according to B, the left clock will read 1:21:39 when the right clock reads 1:00 and emits its signal.

    it will take
    [tex]\frac{12.5 light min}{c+0.866c}= 6.7 min[/tex]

    or 6 min 42 sec for the light and the left clock to meet the signal. During which time the left clock runs 1/2 as fast and accumulates 3 min 21 sec.

    This, added to the 1:21:39 the clock started at when the signal left the right clock mean that the left clock gets the 1:00 signal from the right clock when it reads 1:25.

    This is important in Relativity. events that happen in any frame ( such as the left clock receiving a signal from the right clock with a 1:00 time stamp) have to occur in all frames. So if you want to work out the particuliars of any event, just choose the frame in which it is the easiest to do so. (for example, in the above scenerio, it is easiest to work from the frame in which the clocks are at rest.)
     
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