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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.
     
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  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.
     
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