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Scenario where time dilation seems to be paradoxal

  1. Jun 24, 2011 #1
    Imagine the following scenario:
    There is a train, that crosses the entire planet, and has the lenght of the earth diameter (has no start or no end, it's like a "ring").
    This train has in the floor a treadmill that runs in the opposite direction of the train. Also, the treadmill has the lenght of the train.
    The train travels at near speed of light.
    The treadmill runs at exact same speed of the train.
    Now imagine three observers, each one with a personal clock. The first (A) is outside the train, the second (B) is inside the train, but not over the treadmill, and the third (C) is standing over the treadmill.
    The observer A sees observer B running at near speed of light.
    The observer B sees observer C running at near speed of light.
    The observers A and C can see each other all the time, from their point of view they are not moving.
    If observer A looks at observer B running at near speed of light, then time runs more slow for observer B, when comparing to A. (B < A)
    If observer B looks at observer C running at near speed of light, then time runs more slow for observer C, when comparing to B. (C < B)
    If B < A and C < B, then we could assume that C < A, in other words, time for observer C will pass much more slow when comparing to observer A.
    But the thing is that observers A and C see each other not moving at all, so it seems the time for both should have no dilation.

    Somebody can explain what am I missing? Time dilation by speed doesn't care about direction, but in this example it seems that direction is relevant.
     
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  3. Jun 24, 2011 #2

    ghwellsjr

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    You have just complicated things with your treadmill since it is no different than the roadbed under the train. This makes A and C equivalent as you have pointed out so get rid of C. Your long train incircling the earth is also unnecessary. All you have is one observer, B, traveling in a circle and a second observer, A, stationary. Now you have the origin of the Twin Paradox that Einstein described in his 1905 paper at the end of section 4. It turns out that B's clock will run slower, overall than A's clock, that is everytime they meet, they both will agree that B's clock has accumulated less time. However while they are passing, they each observe the other one's clock to be running slower than their own.

    In order for each observer to continually see the other one's clock running slower than their own, they must continue in a straight line. When one of them accelerates, either by changing his speed or by changing his direction, then you cannot assume that the time dilation is reciprocal.
     
  4. Jun 24, 2011 #3
    Sorry, but I have "complicated things" for a reason. You can't just remove things from my scenario and transform it into a regular example.
    You said that A and C are equivalent. So, how do you explain that C's clock runs slower than B's clock, and B's clock runs slower than A's clock, wich takes us to belive that C's clock runs slower than A's clock? That's the whole point of my question.
     
  5. Jun 24, 2011 #4

    Janus

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    Which clock is running faster than another depends upon who is measuring the difference.

    C's clock runs slower than B's clock as measured by B
    B's clock runs slower than A's clock as measured by A.
    B's clock runs slower than C's clock as measured by C
    A's clock runs slower than B's clock as measured by B
    A and C's clocks run at the same speed as measured from A and C.

    Time measurement is relative and frame dependent.

    It's like the following analogy:

    You have two men A and C, standing one in front of the other and facing the same direction. To their left is another man B, who is facing the opposite direction. According to A, B is to his left, and according to B, C is to his left. But this does not mean that you could conclude that C is to the left of A. It is the same type of thing with time dilation.
     
    Last edited: Jun 24, 2011
  6. Jun 24, 2011 #5
    Thank you Janus, your analogy is interesting. I'm still trying to get the concept, maybe the full understanding could be unreachable for me now, with my current background knowledge. But let me try one more question. If A, B and C syncronize their clocks before each one take their positions, then the train (and treadmill) turns on for some period of time (one month, one year, whatever), and finally everything stops, the three observers meet together and compare their clocks, what will they see? Note that I'm not asking about measurement during the experiment, I'm talking about after, wich I think eliminates the issue of multiples frames, because the comparison is being made with 3 elements together at the same space and time, before and after the experiment.

    One possible result will be:
    A: 09:00
    B: 08:59
    C: 09:00
    Assuming that A and C had no relative change in time, but B had passed the experiment time with clock slower than A.

    In the other hand, we could get:
    A: 09:00
    B: 08:59
    C: 08:58
    Assuming that B had passed the experiment time with clock slower than A and C had passed the experiment with clock slower than B.
     
  7. Jun 24, 2011 #6

    Dale

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    B is not an inertial observer at any point so none of the usual symmetry between inertial observers ever exists in this scenario. All three observers agree that B's clock runs slower than A or C at all points.
     
  8. Jun 24, 2011 #7
    Alright, it seems correct to think that A and C clocks will mark the same time. But they should show less or more time than B?
    Maybe what is confusing me is some more basic concept. Let's see this basic example:
    I syncronize two clocks and send 1 clock in a spaceship to travel at near speed of light, then when it comes back I put the two clocks together again. As far as I now, the clock that was in the spaceship will show less time than the clock that was in earth all the time. But if speed is relative, what rule defines that the spaceship clock is the one that runs slower? Again, I'm trying to objectively compare the clocks always together in the same space and time, before and after the experiment. I'm trying to not think about twin brothers, or observers that has to agree or disagree, or frame dependency. I look to the 2 clocks, send 1 to space, wait it comes back and I look to the 2 clocks again. I'll see something objectively.
    I understand that time has dilation, but I also understand that speed is relative, in other words, doesn't matter if X is running away from Y, or Y is running away from X, what matter is the relative speed between X and Y. So, who defines what clock run slower?
     
  9. Jun 24, 2011 #8

    Dale

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    They show more time. B's clock runs slow.

    The spaceship clock is the one that underwent proper acceleration. Same with B in your above example. A clock that undergoes proper acceleration will always show less time than one that goes inertially between the same two events.

    Geometrically this is analogous to the fact that the shortest distance between two points is a straight line, and therefore any line between two points that is bent must necessarily be longer.
     
  10. Jun 25, 2011 #9
    Note that from B's perspective, when looking at C, B is standing and C is moving away. Analysing B and C only, imagining that A don't exist, even so is correct to say that B's clock run slower?

    But acceleration isn't also relative? Let's say my car is accelerating over a road, isn't the same to say that my car is stationary and the road is accelerating backwards? We would need a third reference to determine who is standing and who is accelerating. So, I still don't get how we can say what clock will run slower.

    I understand this geometric concept, but I don't get the analogy at all. I don't want to bother anyone (I'm happy that I found a place to have this discussion), so you don't need to waste much time explaning this kind of thing, if you don't want it.
     
  11. Jun 25, 2011 #10

    DrGreg

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    In general this is true, but DaleSpam referred to proper acceleration, which is what an accelerometer measures. There is a difference between an inertial frame (where the proper acceleration of an object at rest is zero) and a non-inertial frame.
    No, in your car you feel a "g-force" as you accelerate; someone standing on the street feels nothing when you acclerate.
    Draw a spacetime diagram of distance travelled in an inertial frame versus time. For an inertial object (with zero proper acceleration) the graph is a straight line, for a non-inertial object the line is bent. The length of the line represents the time experienced by the object itself.
     
  12. Jun 25, 2011 #11

    Dale

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    Yes, B is non-inertial so B agrees that B's clock runs slower than C's.

    No, an accelerometer attached to your car will measure the acceleration and an accelerometer attached to the road will not. Both measurements can be taken without reference to any external reference frame. Acceleration (specifically proper acceleration) is absolute, not relative.

    In relativity we consider the position of an object as a function of time to be geometrically represented by a line in spacetime called a worldline. An inertial object has a straight worldline, an accelerating object has a curved worldline, and the time measured by a clock is the length of the worldline.
     
  13. Jun 25, 2011 #12

    Janus

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    A big part of your problem is that you are trying to approach this by only considering time dilation when there are other effects that come into play, namely the relativity of simultaneity and length contraction.

    Let's say that you send the clock a distance of one light year away at some speed (w). According to you, the ship takes 1/w= x years to make the trip, during which the ship's clock runs slow by a factor of y. meaning that the ship clock will read x/y years when it returns.

    According to the ship, its clock runs at a normal rate, but due to length contraction, the distance that you measure as 1 light year is only 1/y light year, thus it only takes (1/y)/w= x/y years to make the trip. Here we see that you and the ship agree as to how much time has passed for the ship, but for different reasons.

    On the outbound leg and inbound legs, according to the ship, your clock will run slow by a factor of y. If this were all there was to it, the ship would expect your clock to read 1/y² upon his return. This is where the relativity of simultaneity comes in.

    The relativity of simultaneity basically states that if you have a string of clocks all synchronized in a given frame, according to an observer traveling along this string, the clocks will not be synchronized, the clocks in his "forward" direction will be ahead of the one he is next to, which will be ahead of the ones to his rear.

    When the spaceship reaches the turn around point of its trip, from its perspective, you go from being behind him to being in front of him. If you imagine a clock at rest with respect to you, synchronized to your clock(according to you) and sitting at that point, then according the ship, your clock goes from being behind this clock to being ahead of this clock when the ship turns around. In other words, your clock will have "jumped forward". If you add this forward jump in time to the time accumulated on your clock according to the ship on the outbound and return legs, you will find upon return, the ship will expect your clock to read x year.

    Again both you and the ship agree as to how much time passed on your clock, but for different reasons. According to you, your clock just ticked off time normally, while according to the ship, it ran slow, jumped forward, ran slow.
     
  14. Jun 25, 2011 #13

    Fredrik

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    That complication is unnecessary because the apparent paradox is there even if you only consider A and B: "B's clock is slow relative to A" appears to contradict "A's clock is slow relative to B". To understand the problem here, it's essential that you understand that these statements are actually defined to mean something different from what they appear to be saying. What they actually mean is this:

    "The coordinate system associated with A's motion assigns time coordinates to B's world line that increase faster along B's world line than the numbers displayed by B's clock"

    "The coordinate system associated with B's motion assigns time coordinates to A's world line that increase faster along A's world line than the numbers displayed by A's clock"​

    You might still find it surprising that the two longer statements can both be correct, but at least they're not slapping you in the face with a contradiction.

    The simplified statements at the start of this post are inaccurate, but you were considering a version of them that had been "simplified" to the point where they didn't even contain a hint about any coordinate assignments being involved at all. That's why they look like very obvious contradictions.
     
  15. Jun 25, 2011 #14
    Thank you all guys. Many things start to fit, and a lot things to think about. "Proper acceleration" may be one key I was missing.

    Fredrik, the proposed scenario may be unecessary complicated and may be simplified, but I was trying to know each step that allows it to happen. The hard thing about explain something is to guess what happens in the other peoples minds, and then what could be the best aproach to clarify the subject.

    In my example, if you imagine the train running but the treadmill off (so B and C will be equivalent), then you turn on the treadmill, C will receive an acceleration (proper acceleration) and will have a time dilation compared to B. I think this is what was confusing me. But now, with a better knowing of the elements that involve time dilation (it's not about speed only, acceleration is an important key), it seems easy to note that in order to C receive an acceleration by the treadmill, it must be on the train, thus it had to have recieved an acceleration in some moment before, in the opposite direction (when the train starts moving). So the two accelerations somehow nullify each other and we can say A and C are equivalents.

    Thanks again.
     
  16. Jun 25, 2011 #15

    Dale

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    Since B is undergoing uniform circular motion B is continuously experiencing proper acceleration (centripetal acceleration). When you turn on the treadmill C will stop receiving proper acceleration.
     
  17. Jun 26, 2011 #16

    ghwellsjr

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    In Special Relativity (where we ignore the effects of gravity), time dilation is only about speed, not acceleration. You had it correct back in post #7 where you said:
    But in Special Relativity, we pick any single inertial (non-accelerating) reference frame and we define the speeds of all object/observers relative to that reference frame (whether or not there are any objects/observers stationary in that reference frame). These object/observers can be accelerating in any manner we want. The rate of the passage of time for each object/observer is then determined by its instantaneous speed relative to the reference frame. We usually try to pick a reference frame that makes our explanation, analysis, and calculations the easiest for us, but we could also pick one that made our work torture.

    So, for your first example from post #1, we could pick the reference frame in which the earth is stationary. (We will ignore gravity and the rotation and other motions of the earth.) Then observers A and C will have zero speed and zero time dilation. Their clocks will be running at "normal" speed. But observer B is always traveling at a high speed in a circle that comes close to A and C on every lap. So because of B's high speed, he will have a lot of time dilation and his clock will run continually slow and all three observers will agree on this every time they "contact" each other. On each lap, B's clock will have less elapsed time on it than the times on A's and C's clocks (which always match each other). Don't confuse time dilation with this difference in times on the clocks. It's the time dilation (a clock running slower) accumulated over of period of time (one lap) that results in the clocks having different times on them when they meet again.

    Also note how you were correct in your first post when you said:
    It is only B's speed (relative to our defined reference frame) that determines his time dilation, even though he is constantly changing direction.

    Also, please note that there is no inertial frame of reference in which B is always stationary (because he is constantly changing direction which is an acceleration). There is no other frame of reference that we could pick for this example that would make it easier to understand what is going on.

    Now let's take a look at your second example from post #7:
    Let's assume that the clock in the spaceship is quickly accelerated at the start, middle and end of the trip so that we can ignore those short periods of changing speed. We have to define a reference frame. The easiest one to pick is the one in which both clocks start out and end up at rest and where they run at the same normal speed. During the outbound and inbound portions of the trip, the spaceship clock will be experiencing a high speed and therefore a lot of time dilation and will end up with a smaller time on it at the end.

    Now we could have picked the frame of reference in which the spaceship clock is stationary during the outbound portion of the trip. In this case, during this time, the spaceship clock will be running normally and the other clock will be experiencing a lot of time dilation because it will have a high speed. In fact it will have this high speed during the entire course of the example and therefore will be running slow all the time. But when the spaceship turns around to head for home, its speed relative to our defined frame of reference must be much greater than the other clock's speed or else it will never be able to get back to the other clock. And so during the inbound portion of the trip, the spaceship clock will be experiencing much more time dilation than the other clock so that when it finally reaches the other clock, it ends up with less time on it.

    So when you asked at the end of your example in post #7:
    Can you see that during the first portion of the trip, depending on your selected frame of reference, we can pick either clock to be running slower? But the clocks will be always getting farther apart and in order to bring them back together, at least one of them must change its speed and this causes it to experience more time dilation than the other one. So while the clocks are traveling apart, we cannot determine which of them is "really" running slower but when we finally bring them back together, it won't have mattered which frame of reference we used to defined the exercise in, they all yield the same result.

    So, I hope you can see that it is only the speed that determines time dilation, but in order to separate and bring two clocks back together, some acceleration must be involved and if it only occurs in one clock, that one will be the one with less accumulated time on it.
     
    Last edited: Jun 26, 2011
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