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I Length contraction

  1. Oct 27, 2017 #1
    Hi

    Is this analysis right

    A space ship is travelling at a steady speed V in the direction shown. Inside the space ship is a simple clock of length L and consisting of a light source at “a” and a mirror at “b”. The light leaves “a”, bounces off the mirror at “b” and goes back to “a”, which counts as a single tick of the clock.

    As the space ship is moving the clock is shown at times t1, t2, and t3. At time t2 the clock which was originally at a and b will now be at a' and b'. Time t2 is the time the light reaches the mirror.

    The mathematical analysis for the time taken for one tick to occur is

    T = ((L – vt)/c) + ((L + vt)/c)

    (L – vt)/c = is the time taken for the light to leave the light source and travel to the mirror.

    (L + vt)/c = the time to do the return trip.

    vt = the distance travelled by the clock in the time it takes for the light to travel the length of the clock.

    do I have to allow for length contraction? Is the above correct?
     

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  3. Oct 27, 2017 #2

    kuruman

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    It looks right. Is this the time measured by an observer watching the spaceship go by or is it the time measured by an observer inside the spaceship. Does it matter?
     
  4. Oct 27, 2017 #3
    it is the time measured by both and the same would be observed by both. If that is the case it appears the observer inside the spaceship would observe the clock, and thus the spaceships, movement relative to the photons in the clock!!
     
  5. Oct 27, 2017 #4

    kuruman

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    Remember that the speed of light is c regardless of the frame of reference.

    You asked if you have to allow for the length contraction of the spaceship as seen by an observer who watches it go by. Obviously, from the viewpoint of that observer the spaceship is shorter by a factor of γ. If the spaceship is shorter and the speed of light is c, why does light take the same amount of time for the round trip in the two frames? I think the answer to that question is the lesson to be learned in this problem.
     
  6. Oct 27, 2017 #5
    ah yes, thanks for that

    will both observers observe the movement of the clock relative to the photons within the clock, it appears to me they will?
     
  7. Oct 27, 2017 #6

    kuruman

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    Please don't say things like "the movement of the clock relative to the photons". There is no such thing as a "rest frame" of the photons. So there is no movement "relative to them". The photons move at the speed of light, c, relative to the clock and relative to any other frame one might think of, including the observer and the spaceship. They also move at speed c relative to other photons.

    On Edit: The last sentence is withdrawn; it should not have been written. :oops:
     
    Last edited: Oct 28, 2017
  8. Oct 27, 2017 #7
    a rest frame is where one frame is at rest wrt another frame isnt it?

    But wont the observer external to the spaceship observe the clock is moving relative to the photons?

    How come photons are not affected by the change of time? Photons have time dependent properties that seem to be immune to the change in time. It appears other phenomena "know" when they are in a moving frame of reference but photons are completely oblivious!
     
    Last edited: Oct 27, 2017
  9. Oct 27, 2017 #8

    kuruman

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    The observer on the ground will see the clock move at speed v and the photons at speed c.
    The observer in the spaceship will see the clock at rest and the photons move at speed c.
    What kind of properties are you thinking of that are time-dependent?
    The laws of physics are invariant under a coordinate transformation. This means that two physicists moving relative to each other will deduce the same laws of physics after observing the same phenomena.
     
  10. Oct 27, 2017 #9
    time dependent velocity, frequency, I dont know that much about photons but there are probably many others, entropy, do photons age?
     
  11. Oct 27, 2017 #10

    kuruman

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    One more time, the photon velocity is not frame-dependent or time-dependent; it is fixed and equal to c. Frequency can change due to the Doppler shift. Photons do not "age" which or decay. Be patient and learn some more physics. By the time you get to Quantum Mechanics, you will be more comfortable with photons.
     
  12. Oct 28, 2017 #11

    Mister T

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    I think your last sentence contradicts what you say in your 2nd and 3rd sentences?

    .
     
  13. Oct 28, 2017 #12

    kuruman

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    I agree, I should have not written it.
     
  14. Oct 28, 2017 #13
    so just to clarify my analysis is right?. The person inside the space ship would use the same analysis?

    and length contraction is only observed by and observer moving relative to the space ship
     
  15. Oct 28, 2017 #14

    kuruman

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    No. The person inside the spaceship does not see the mirror move. Therefore, the time for the round trip is T = 2L/c however the "go" trip takes the same amount of time L/c as the "return" trip. That is is not true for the person on the ground. Although the time for the round trip is the same, the "go" trip takes less time than the "return" trip as you have calculated.
    Yes, although it is better to say that length contraction is observed in the frame with respect to which the spaceship is moving.
     
  16. Nov 25, 2017 #15

    jbriggs444

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    Cosmic ray muons which are detected at the Earth's surface in greater than expected numbers will have observed a length-contracted atmosphere.

    http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html
    It is part and parcel of Special Relativity. Length contraction, time dilation and the relativity of simultaneity are all part of the Lorentz transformation.
     
  17. Nov 25, 2017 #16

    kuruman

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