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B Time dilation problem question

  1. Nov 18, 2016 #1
    So there is a problem I don't get. I special relativity, things moving relative to me, I see them as experiencing more time/ slower time. Ok....so then I should also see it's velocity decrease. If I see it's clock running slower, so should I it's velocity. V=....m/s
    So if I see it's time slow down then V=.....m/2s. And if the answer is, distance also decreases, if I saw something run past me at 0.5c, I should see the path before it shrink? Distance decreases only from the point of view of that thing (lenght contraction), but not from mine.
    Thank you for help, I really don't get this.
     
  2. jcsd
  3. Nov 18, 2016 #2

    PeroK

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    The velocity you measure is simply the spatial displacement you measure divided by the time you measure. SR has nothing to do with that.
     
  4. Nov 18, 2016 #3
    Realize it is not your clock that slows. If something were moving relative to you than it is the clock of that thing that is slow relative your clock; your measure of their velocity or distance doesn't change. Their measure of distance does in fact disagree with yours and would be shorter and they also would observe your clock to be running faster than theirs.
     
  5. Nov 18, 2016 #4

    Ibix

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    You need to make measurements with your clocks and your rulers. They don't get contracted or dilated just because they're measuring someone who is moving. If you try to make measurements with someone else's clocks or rulers then you'll get strange answers - unless you are careful to account for simultaneity differences as well.
     
  6. Nov 18, 2016 #5

    PeroK

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    That's not correct. Something at rest relative to you will be length contracted to them. But, something at rest relative to them will be length contracted to you. The situation is symmetrical.
     
  7. Nov 18, 2016 #6

    Nugatory

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    And because the situation is symmetrical, they also find that your clock is running slower than yours, not faster. Yes, both observers find that the other clock is slower than their own, and they're both right.

    (PeroK already knows this, of course - I'm adding this for others following the thread)
     
  8. Nov 18, 2016 #7
    I was talking about the distance between them as the OP inferred he should see an apparent velocity change and that should imply the distance from his position would decrease also. I was not discussing that the object would appear lenght contracted in its direction of motion. (like a muon traveling at near light speed would measure the distance to Earth as being much less than an observer on Earth would measure to the muon.)
     
  9. Nov 18, 2016 #8

    PeroK

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    But, in the muon frame it's the Earth that is moving, so there is symmetry there too.

    For both they are ##vt## apart ##t## seconds before they meet, where each measures ##t## on their clock.
     
  10. Nov 18, 2016 #9
    Exactly and their clocks don't agree on the value of t.
     
  11. Nov 18, 2016 #10

    Mister T

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    You meant to say his rate of change of distance, not the distance itself. To say the distance decreases is to say there is motion towards to the observer.

    But anyway, to answer your original question ...

    Not if you use your own clock to measure their velocity. Which is really the only way to do it that makes sense.

    (If they use their own clocks to measure their own speed they get zero, since they're not moving relative to themselves. So in that sense you do see a decrease in velocity!)
     
  12. Nov 18, 2016 #11

    PeroK

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    So, which one thinks they are "further apart"? Muon or Earth? Which one's clock is running slower?
     
  13. Nov 21, 2016 #12
  14. Nov 24, 2016 #13

    stevendaryl

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  15. Nov 24, 2016 #14

    stevendaryl

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    Everybody has pointed out that this is incorrect, but I don't think you have acknowledged it.

    According to SR, if Alice and Bob are moving at constant velocity relative to one another, then
    • According to an inertial coordinate system in which Alice is at rest, Bob's clock runs slower than Alice's.
    • According to an inertial coordinate system in which Bob is at rest, Alice's clock runs slower than Bob.
     
  16. Nov 28, 2016 #15
    Forget "Alice and Bob" use the muon example in the Georgia State University link above. It represents a real observed phenomenon.
     
  17. Nov 28, 2016 #16

    stevendaryl

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    Geez. Okay:
    • According to a frame in which the muon is at rest, clocks on the Earth are running slow.
    • According to a frame in which the Earth is at rest, the muon is decaying slowly (that is, it has a longer lifetime than usual).
     
  18. Nov 28, 2016 #17

    stevendaryl

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    But I don't want to forget "Alice" and "Bob". I want to know what your answer is. I think you're still confused, but I can only find that out by asking you about the details of your understanding of SR. A link to a paper does not tell me what your understanding is, unless you are the author of that paper.
     
  19. Nov 28, 2016 #18
    It's not a paper, it's an example of the actual muon problem with a calculated solution as per SR. Just substitute Alice and Bob for muon and earth observers to ask your question or make your point (I don't want misunderstandings between us do to syntax i.e. "Alice and Bob").
     
  20. Nov 28, 2016 #19
    That appears to contradict this section of the example:

    "In the muon experiment, the relativistic approach yields agreement with experiment and is greatly different from the non-relativistic result. Note that the muon and ground frames do not agree on the distance and time, but they agree on the final result. One observer sees time dilation, the other sees length contraction, but neither sees both.

    mu5b.gif
    These calculated results are consistent with historical experiments." link: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

    Please explain?

    Agreed.
     
  21. Nov 28, 2016 #20

    stevendaryl

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    The underlined sentence is incorrect, but irrelevant. What fraction of muons reach the ground before decaying has nothing to do with how fast Earth-clocks are running in a frame in which the muons are at rest.

    Time dilation is seen from every frame, and length contraction is seen from every frame. But calculating how many muons make it to the surface of the Earth only involves time dilation in the Earth frame, and only involves length contraction in the muon frame.
     
  22. Nov 28, 2016 #21
    Not so, in the case of the twin paradox when the space traveling twin has turned around and accelerated back towards Earth they see a contracted distance to the Earth relative to the Earth twin (who does not); the space traveling twin also sees the Earth observers clock as running fast relative to their own while the Earth observers see the traveling twins clock running slow. Think of the traveling twin in this part as the Muon relative to the Earth.

    I disagree as it is also seen in the twin example above, and bear in mind, that site is hosted by the Department of Physic and Astronomy of Georgia State University while my only knowledge of your credentials is what I see as a forum poster here (and I don't know the criteria to get those acknowledgements).
     
  23. Nov 28, 2016 #22

    Ibix

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    No. Unless you are literally referring to what is seen and not correcting for the travel time of light. Both will determine that the other's clock ticks slowly if they do correct for that. Both will see the other's rulers as contracted. As @stevendaryl says, there is a minor and error in the paper that is irrelevant to its conclusions.
    Literally every text book on relativity will show you that Steven is right. It's a trivial application of the Lorentz transforms, and you are making a naive mistake in applying the time dilation formula to the travelling twin's perspective and drawing the conclusion you do - namely ignoring the impact of the change of simultaneity convention at turnover. Ultimately, you are implying the existence of an absolute frame of reference, because you have a global sense in which one clock is moving faster than the other and hence ticking slowly.
     
  24. Nov 28, 2016 #23

    Nugatory

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    Nonetheless, it is incorrect. It's clear that the author was not speaking precisely, because they used the word "see" - and we all understand (I hope) that neither time dilation nor length contraction are things that you see. I suspect that the author intended the word "see" to mean "use in our calculations of what's going on", in which case they're saying the same thing as Stevendaryl, except less precisely.
     
  25. Nov 28, 2016 #24

    stevendaryl

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    Okay, so I was right to suspect that you don't understand mutual time dilation and mutual length contraction. In the case of the twin paradox, there are three inertial frames that are relevant:
    • [itex]F_1[/itex]: The frame of the stay-at-home twin (on the Earth).
    • [itex]F_2[/itex]: The frame in which the traveling twin is at rest on his outbound journey.
    • [itex]F_3[/itex]: The frame in which the traveling twin is at rest on his return journey.
    According to the inertial coordinate system associated with [itex]F_1[/itex], the traveling twin's clock runs slower than Earth clocks during both legs of his journey. According to [itex]F_2[/itex], the traveling twin's clock runs faster than the Earth clocks during the first leg of his journey, but slower during the return leg. According [itex]F_3[/itex], the traveling twin's clock runs slower during the outward journey, but faster on the return journey.

    All three frames see time dilation and length contraction. They disagree about whose clocks are time-dilated when, but all three agree that the traveling twin is younger than the stay-at-home twin when the two get back together.

    I have never made any claims based on having particular credentials, but the particular line in question is either wrong or we're both misinterpreting it. Maybe what they mean is that there are two phenomena that can be used to explain the muon count reaching the Earth:
    1. Time dilation of the muons.
    2. Length contraction of the distance between the point of muon creation and the surface of the Earth.
    Only the Earth frame sees 1, and only the muon frame sees 2.
     
  26. Nov 28, 2016 #25
    Do you see the contradiction for a case of the Muon and Earth observers if "seeing" each others clock running slowly"; the actual decay rate observations from either observer necessitate only the muon clock to be running slow relative to the Earth clock as calculated and explained in the Hyperphysics. The same is true for the traveling twin on their inbound leg, but muon decay is an actual observed phenomenon (they decay too slowly relative to clocks on Earth).

    If we say both observers see the others clock running slow we break symmetry in the observations. The Muon decays at the expected rate according to its clock, but too quickly relative to how it sees the earth clock (seen as running relatively slower). Meanwhile the Earth observer sees muon decay as being to slower than expected according to his own clock, but as expected relative to how Earth sees the muon clock. On the other hand the Hyperphysics explanation is symmetrical, both observers see the proper decay rate relative to how they see the muon clock and a slow decay rate relative to how they see the Earth clock.
     
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