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Time paradox

  1. Jan 10, 2013 #1
    I'm studying special relativity and I can't understand the following.

    Let's imagine me and my twin brother making the following experiment: I stay in Earth and my brother travels in a spaceship with velocity 0.5c. For moving referentials the time passes more slowly, but for the first principle of special relativity, the Physics laws are the same for any inertial referential. So for me, my brother are in slow motion (I became older). But for my brother, he became older. When he arrives here, who will be older?
  2. jcsd
  3. Jan 10, 2013 #2


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    You will be older.

    As you say, the first postulate of relativity is that the laws of physics are the same in all INERTIAL frames. Your brothers frame is not inertial.
  4. Jan 10, 2013 #3


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    This has been discussed on this forum approximately 9,487 times. Do a forum search for the "twin paradox".
  5. Jan 10, 2013 #4


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    Huh. I am approximately 10 minutes older than my twin sister. Given 7 months and 1 week of gestation, I wonder what speed I would have needed to travel to get that difference in age over that time period. My mother must have had one hell of a pregnancy with all the relativity going on inside her.
  6. Jan 10, 2013 #5
    Roflmao maybe you should ask her
  7. Jan 11, 2013 #6

    Given phinds' point, I surprised your variation of the twin paradox hasn't been asked / discussed.
  8. Jan 11, 2013 #7


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    There is a very simple way to answer this question. It relies just on the first principle of special relativity that says that you and your brother will each see the other ones clock ticking slower than their own by exactly the same amount during the outbound portion of the trip, plus the fact that light from different sources travels at the same speed (without identifying what that speed is, which is Einstein's second postulate). On the way back, your brother will see your clock ticking faster than his as soon as he turns around. This leads to the conclusion that if your brother travels away from you at the same speed that he travels back to you, then his observation of the ratios at which your clock ticks compared to his during each half of his trip will be reciprocals of each other.

    Hermann Bondi's book, Relativity and Common Sense, pages 77 to 80, describes the process to show that this is true.

    So whatever ratio your brother sees of your clock compared to his on the way out plus its reciprocal on the way back added together and divided by two gives us the final average ratio of your clock compared to his when you reunite, since the times for his two halves of the trip are the same because he is traveling at the same speed over the same distance (although we aren't specifying what that speed, distance or time are). That average ratio is always greater than one.

    Let's say that that ratio is R for the return trip and 1/R for the outbound trip. Adding these together and dividing by two gives us (R + 1/R)/2 or (R2+R)/2R. For any value of R greater than 1 this evaluates to a number greater than 1. Try it and see.
  9. Jan 11, 2013 #8
    You are right. I've just read the twin paradox at wikipedia. They say like 20 times that the explanation beacause there is no contradition is that the event is not symmetrical, as only my brother has experienced acceleration.

    But isn't acceleration relative too? When I say my brother is accelerating at acceleration a, shouldn't he say the same of me?
  10. Jan 11, 2013 #9
    George, I know this is largely a matter of taste, but I think the relativistic Doppler effect is the least clear way of explaining the twin paradox. It takes a lot of work when first explaining relativity to get the point across that when we talk about events, we're not discussing when these events appear to happen in different reference frames but when they actually happen, taking into account the relativity of simultaneity. I think emphasizing what the twins actually see is unnecessarily confusing and risks giving the false impression that time dilation is just some kind of visual trick. Since the times when clock ticks are seen are governed by the relativistic Doppler effect, which is just the non-relativistic Doppler effect plus time dilation, you are ultimately just starting with time dilation, adding in the lag effect on finite propagation of light (i.e. the non-relativistic component of the Doppler effect), and then subtracting the lag effect out again (when showing how each twin would use their observations to calculate their sibling's age). I think that's a lot of needless clutter.

    Personally, I find the clearest explanation is to not bother with any talk of when various age milestones are seen in each twins' frame and just focus on when they happen in each frame—i.e. the spacetime diagram approach. It's easy as pie to show that, while time dilation is indeed symmetric on each leg of the trip (separately!), relativity of simultaneity means that when the traveling twins makes her about-face, her brother ages a large amount in her frame in a very small amount of time—instantaneously in the limit of an instantaneous turnaround—and this more than makes up the difference. Everyone has their pedagogical preference, but I really think not futzing around with super telescopes does a much better job of showing precisely how the asymmetric aspect of the twins' experience (one inertial reference frame vs. two) directly leads to the correct calculation in both frames.
  11. Jan 11, 2013 #10


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    No. Acceleration* is not relative. Your brother and you can each measure your acceleration using accelerometers without any reference to the other. If you do so then unambiguously your accelerometer reading will be 0 but his will not.

    *The type of acceleration which is not relative is called "proper acceleration". There is also a type of acceleration called "coordinate acceleration" which is relative to some specified coordinate system. Coordinate acceleration cannot be measured by an accelerometer and doesn't have any physical effects, only proper acceleration does. So usually when people just say "acceleration" they mean "proper acceleration" which is not relative.
  12. Jan 11, 2013 #11
    No, acceleration is not relative because you can feel it. Either you or your brother felt a force due to the acceleration and whoever did is simply the one who accelerated—no fooling about. All inertial observers will agree on whether or not something is accelerating and anyone who disagrees is, by definition, not an inertial reference frame.
  13. Jan 11, 2013 #12


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    No. It's somewhat surprising, but acceleration (change in velocity) is not relative although velocity is. You can measure your own acceleration without reference to anything else: If you are standing on a scale, the weight it measures will increase if you're accelerated upwards (think of high-G-force rocket launches); if you're holding a spinning gyroscope it will resist acceleration; and so forth.

    We can build black boxes called accelerometers which display measure the acceleration they're being subjected to, and you can't do the same thing with speed (think about how an automobile speedometer "knows" that the car is moving relative to the roadway).
  14. Jan 11, 2013 #13

    I think I've got what you mean. Pretend I'm sending messages to my brother year by year, and he does the same . At the moment of the about-face my brother will receive many messages of mine, and when he arrives in Earth, I will be older.

    But does't it mean that in the accelerated referential of the spaceship (at the moment of the about-face) see the light would aprroximating with a velocity bigger than c?

    I know that c is constant for intertial referentials, I don't know how it works for accelerated referentials. Is it possible?
  15. Jan 11, 2013 #14


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    It works the same way - at any particular moment the accelerated twin can choose a reference frame in which he is at rest; it's just that the math gets more complicated because in the next moment he won't be at rest in that frame. That's why we try to work with inertial frames when we can, and why most textbook examples are worked in inertial frames.
    (One unfortunate side effect of this tendency is that it's easy to get the impression that special relativity only works for inertial frames, and you need general relativity to handle acceleration . Although widely repeated, that is not true - you only need GR if the spacetime is not flat).
    Last edited: Jan 11, 2013
  16. Jan 11, 2013 #15


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    As Nugatory mentioned, for an object in any state of motion there exists, at each instant in time, a specific inertial reference frame where the object is (at least momentarily) at rest. This is called the momentarily co-moving inertial reference frame (MCIRF). The MCIRF is an inertial reference frame so light moves at c in it, but the object may only be at rest in it for one instant.

    There are also non-inertial reference frames, such as a rotating referece frame. In non-inertial frames the usual laws of physics take different forms unless you write them using tensors. Specifically, this means that unless you use tensors then you may get that the speed of light ≠ c in a non-inertial frame. Again, consider a rotating reference frame, in such a frame even nearby planets or the moon may be moving faster than c.
  17. Jan 11, 2013 #16


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    [This is an interesting point worth further discussion. It's also a digression from the original thread, so if it looks like it's going to take on a life of its own some harried, overworked, and underappreciated moderator (is there any other kind?) might want to split it out into a thread of its own]

    As L1S says, this is a matter of taste, and we all know that de gustibus non disputandum est... but there is much pleasure to be had in discussion, as opposed to dispute.

    My experience has been that there are two basic approaches to SR thought experiments: Start with the actual observable physical behavior of the light signal, as ghwellsjr does; and start with the spacetime picture and Lorentz transforms to construct consistent histories of events in each reference frame, as L1S does.

    I find that many people naturally gravitate towards one style or the other, and find the other one somehow sneakily unsatisfying and unconvincing.

    For example, I've never found the light behavior explanations to be gut-level satisfying; I feel as as if I could do something just a bit more clever with my moving mirrors and light sources I could somehow subvert the experiment. (This suspicion may be what's motivating the posters who show up asking whether relative effects are just an "optical illusion"). I prefer t work through the spacetime diagram and Lorentz transforms to satisfy myself that no matter how I manipulate the experiment, it all has to come out just as relativistic doppler and similar phenomena say it will.

    On the other hand, I also know from endless friendly discussions that there are people who find the coordinate-based description to be completely non-fundamental; it's all full of coordinate artifacts and abstract mathematical relationships meaningful only if they connected to some real physical observers.

    My personal opinion on the subject:
    1) You don't really understand until you're comfortable using either style of description. (It's worth noting that Einstein, and just about any serious stdent of relativity after him, are effortlessly fluent in both styles).
    2) When solving problems for yourself, use whichever style you're most comfortable with. When reading someone else's analysis in the style that you don't prefer, consider transforming it to the one that you do prefer. It's good practice for you and may help someone else understand.
    3) When explaining to someone else, start with the style that you're most comfortable with. But be alert for signs that it's not working, and be prepared to switch to the other style. This doubles your chances of getting the magical "Aha - now I get it!" moment that is the goal of all explanation.
  18. Jan 11, 2013 #17


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    It's not a matter of taste--the relativisitic Doppler effect is the problem that "other" explanations need to explain.
    No, you've got it backwards. What each observer sees is what is actually happening. These are all local events. The assignment of coordinates to remote events according to different reference frames are not "when they actually happen".
    Time Dilation is not visual so it cannot be a visual trick. It's a mathematical calculation and dependent on the chosen frame and changes with each arbitrarily selected frame. How can it be visual? Relativistic Doppler is visual, at least in the sense that it is the result of transmitted light, but it never tricks us.
    You could say that the relativistic Doppler effect is the time dilation plus the changing lag effect on the propagation of the light but since both of these are arbitrarily determined by the selected reference frame, neither one can be said to be governing.

    Also, the twins don't need to do any calculation, they just watch their siblings age (or their clocks) and when they return, they each agree on what actually happened. We need to do some calculation to determine what they will see, but that's a different matter and it's very easy because it doesn't involve any understanding of Special Relativity or any other theory. We don't have to learn about synchronizing clocks or defining an Inertial Reference Frame or what the Lorentz Transformation is all about.
    It's not like we have an either/or situation here, we can show both the relativistic Doppler and the time dilation on the same spacetime diagram which is what I do all the time. You can do a search on "diagram" with my name and find lots of them.
    Yes, if you keep the two legs separate in two different Inertial Reference Frames, one for the traveling twin's outbound leg and one for the inbound leg, but when you try to conflate them into one spacetime diagram, you're in for all kinds of complications that are totally unnecessary and do not provide a single bit of insight into what is happening.
    Then please show me how you actually draw the spacetime diagrams according to your pedagogical preference.
  19. Jan 11, 2013 #18
    We're doing SR, not GR. Spacetime coordinates are not arbitrary decisions observers make. Each inertial observer in SR has an intrinsic notion of simultaneity. If you are genuinely arguing (what seems to be the extremely bizarre position) that events occurring on the space-like hypersurface defining simultaneous events to some time for some observer don't actually, in a meaningful sense, happen at that time in that reference frame, then you will definitely be the first person I've ever met to take that position. I suppose it's something of a philosophical point since they're spacelike separated events, but I still find it odd.

    This one, or a similar diagram you'll find in just about every other overview of the twin paradox. The lines of simultaneity clearly show how when the traveling twin turns around, she very rapidly/instantaneously finds herself in a new inertial frame that's simultaneous with a much older twin back home.

    As I said in my original post, I was merely commenting on what I believe to be the educational value of one approach over the other. Frankly, I find it more than a bit insecure that you instantly decided my preference for a different teaching method reflects a "backwards" understanding of relativity. Do you assume that everyone who teaches things differently than you do does so because they don't understand it as well as you?
  20. Jan 11, 2013 #19


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    Even in SR coordinates are arbitrary and any coordinate system may be adopted for any reason or none at all. Specifically, there is no requirement that you use a coordinate system where you are at rest. Also, there is no requirement that you must use an inertial coordinate system. Furthermore, there is no requirement that you use Cartesian coordinates for your spatial coordinates.

    That said, I have no particular preference for the Doppler explanation. My favorite explanation is the spacetime interval one. I actively dislike the switching reference frames explanation.
  21. Jan 11, 2013 #20
    That's irrelevant. You are always at rest in your own reference frame. Whether or not you want to use another reference frame to do your calculations is up to you. However, your reference frame is the one in which you are at rest. I can't make my watch tick faster as I look at it by doing some fancy maths on a paper. My proper time is my time. That's what 'proper' means, for goodness sake.

    That is an egregious decontextualization of what I said. The full quote was, "We're doing SR, not GR. Spacetime coordinates are not arbitrary decisions observers make. Each inertial observer in SR has an intrinsic notion of simultaneity." I was very specifically talking about the natural choice for the time coordinate observers inherit by virtue of them being at rest in their own reference frame. Would replacing the second period with a colon have helped? Obviously spatial coordinates are arbitrary. I didn't say, "Each inertial observer in SR has an intrinsic coordinate system," I said, "Each inertial observer in SR has an intrinsic notion of simultaneity."
  22. Jan 11, 2013 #21


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    It isn't irrelevant. It directly refutes the point that coordinates are not arbitrary decisions in SR. They are arbitrary decisions.

    Sure, that is the definition of the term "your own reference frame". You are stating a tautology

    Exactly. So it is an arbitrary decision whose reference frame to use. You can use your frame, you can use my frame, you can use anybody's frame, or you can even use nobody's frame.
    Last edited: Jan 11, 2013
  23. Jan 11, 2013 #22
    The entire point of issues like the twin's paradox is that a misunderstanding of how inertial frames works leads one to think that if each twin works out what will happen in their reference frame they arrive at contradictory results. Obviously in general you are free to use whatever coordinate system you want when doing calculations in SR, however when they question is to reconcile what seems (due to a misunderstanding of the principles) to be happening according to two observers' reference frames, you have to—no kidding—show what really happens according to their reference frames.

    I'm utterly at a loss to understand how you believe choices of reference frame are arbitrary when the question being asked is: what happens in these two reference frames? It's like someone is asking you, "How would I get around the city if I visited New York?" and you said, "You know, you don't have to visit New York, where you travel is arbitrary. Why don't you visit Peru?"
  24. Jan 11, 2013 #23


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    No, your brother will not receive many messages from you at the moment of his about-face. His about-face will not cause him to receive any messages from you. What's going to happen is that for the first half of his trip, he will receive messages from you at a slower rate than he sends them (1/R as I said in my first post to you), then for the second half of his trip he will receive messages from you at a faster rate than he sends them (R).

    So for your example of your brother traveling at 0.5c, we can use the Relativistic Doppler formula to calculate what R is:

    √((1+β)/(1-β)) = √((1+0.5)/(1-0.5)) = √((1.5)/(0.5)) = √3 = 1.732

    And 1/R is the reciprocal, 0.57735.

    This means that he will see your yearly messages coming to him slower than his during the first half of the trip. In fact it will take 1.732 years before he sees your first message.

    And for the last half his trip, he will see your messages arriving more often than once per year according to his clock. It will only take 0.57735 years between each of your messages.

    Now without knowing how long the trip will take, we can average these two numbers:

    (1.732+0.57735)/2 = 2.30935/2 = 1.154675

    This is the final ratio of your two clocks when he returns. How ever many years it took him according to his clock, yours will be 1.154675 times that amount.

    So let's say your brother travels away at 0.5c for 13 years and then takes 13 years to get back at the same speed. Here is a spacetime diagram to show what is happening according to your rest frame. I show you as a thick blue stationary line with dots every year and your messages going out as thin blue lines traveling at c. I show your brother as a thick black line traveling at 0.5c with black dots every year and his messages coming back to you as thin black lines:


    Now let's see how the previous calculations based on Relativistic Doppler fit in with this diagram. First off, I said that the rate at which your brother receives your yearly messages take 1.732 of his years. Can you see that on the graph? For example, at about his year 12, just before he turns around, he is just receiving the message you sent at your year 7. Can you follow that? If we divide 12 by 7 we get 1.714 which is about right. (We don't expect it to be exact because he didn't receive your message exactly at his year 12.)

    Furthermore, if you look at your year 12, you can see that you are just receiving his message from year 7. It's symmetrical.

    Now you should be able to see that after he turns around, he starts receiving your messages faster than one per year. In fact, from about his year 19 (you'll have to count his dots) to when you meet at his year 26, he will have received your messages from year 18 to 30. That is a ratio of the differences of (26-19)/(30-18) = 7/12 = 0.583, close enough to 0.577.

    And in a similar manner, you can see that from his year 14 (just after he turns around) until you meet (12 years of messages from him), you will see them from your year 23 to your year 30 (7 years) and the reciprocal ratio applies.

    Now that we can look at a spacetime diagram, we can see that the reason why the two of you age differently is because your brother sees these two ratios for half of his total trip time each but you see the smaller ratio for three-quarters of the time and the higher ratio for just one-quarter of the time. This means you are seeing him age less for a longer time while he sees you age less for half the time.

    The last thing we want to notice is that ratio of your final age difference is 30/26 or 1.1538, very close to the actual 1.154675. (Again, these numbers are not exact because we're eyeballing them off the diagram.) This ratio is the famous value of gamma which is also the time dilation factor which shows in the diagram as the ratio of the coordinate time for your brother compared to his actual time on his clock. Can you see that?

    Now I want to show you what the exact same information presented in the first diagram looks like in two more diagrams based on the IRF's in which your brother is at rest, first during his outbound leg and then during his inbound leg. First the outbound leg:


    Notice how your brother's time is not dilated during the outbound leg (because he is at rest) but yours is. Note also that he has to travel at a higher speed than 0.5c (look up "veloctiy addition" in wikipedia to see that this higher speed is 0.8c) when he turns around and therefore now has more time dilation than you have. Nevertheless, all the signals between the two of you continue to travel at c and arrive at exactly the same times according to your own clocks as they did before. Does this all make sense to you?

    Finally the diagram for the IRF in which your brother is at rest during the inbound leg:


    This is very similar to the previous diagram so I won't go into any more explanation except that I want to point out that when your brother turns around, in no case does that have any bearing on what you see, until some time later and even then, each diagram shows accurately what you actually see and what your brother actually sees during the entire scenario.

    Any questions?

    Attached Files:

    Last edited: Jan 11, 2013
  25. Jan 11, 2013 #24


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    Excellent. That was my point.

    Sure, but that is due to the specific question asked. It is not a restriction imposed by SR, as you stated. SR is not restricted to specific coordinates.
  26. Jan 11, 2013 #25
    As much as I may prefer the simultaneity explanation, that's about as good as it gets for the Doppler approach. Nicely done, George, especially the diagrams.
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