How to test length contraction by experiment?

Discussion in 'Physics & Math' started by PengKuan, Jun 17, 2019.

  1. Halc Registered Senior Member

    Messages:
    350
    Only if the ship isn't accelerating, yes, you can measure its proper length this way.

    No, that doesn't work. Take a ship that is 1 light second long moving at .866c relative to Earth, so gamma is 2.
    Measured on board, light takes 1 second each way, or 2 seconds to go round trip. From Earth's perspective, that light-clock is running at half speed so it takes 4 seconds to do the same round trip. More precisely, in Earth frame, it takes 0.268 seconds for the light pulse to go from front to the rear mirror, and 3.732 seconds it to get back to the front.
    If the ship was a full light second long, it would take 8 seconds to do that, but it is contracted to half its length, so it takes 4. So yes, you've managed to demonstrate its contracted length by this experiment.

    If you want to measure its length in a less complicated way, use a camera and take a picture of it when the middle of the ship appears to directly across from the camera. Use a nice grid with km markings on it as a background. Poof, you've measured it.

    Edit: Oops: that didn't get the answer right unless the camera is wicked far away.
     
    Last edited: Mar 12, 2020
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  3. Confused2 Registered Senior Member

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    Although I intuited my result it is a consequence of flogging through an analysis of a horizontal light clock done by David Waite some years ago on a now extinct forum. Unfortunately I can't get the result I want without cribbing and even if I could it would have no credibility.

    The fact that you can present numbers suggests you actually know what you are doing - any chance you'd be willing to show how you got those numbers?
     
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  5. Halc Registered Senior Member

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    350
    Pretty easy actually. Length in light seconds, so our ship of length 1 is actually 300,000 km. Time in seconds, speed in fraction of c.

    Speed is 0.866, or sqrt(.75) which gives a gamma of 2. That's the hardest math, and the only place that Lortentz transformation is used.
    So clocks of things moving that fast run at half speed. Ship length is contracted to 0.5. It's mass is doubled, but we don't care about that for this.

    For light moving from the front, the rear is coming back to meet it, so the combined speed is 1 for the light and 0.866 for the speed of the rear, so 1.866 of a light is covered each second.
    So time in that direction is Len/net-speed which is 0.5/1.866 = 0.268.
    In the forward direction, the front is running away at 0.866 and the light is moving at 1, so the difference is only 0.134c, so 0.5/0.134 = 3.732 seconds. Total is 4, which is exactly twice what the ship measured with its clock moving at half speed.

    If Newton tried to do the same calculation, all clocks would run at full speed and the ship would still be of length 1, so time expected would be 8 seconds (measured anywhere), not 2, so they knew something was wrong when the measurements got accurate enough that they still noticed no difference despite a 60 km/sec change in velocity, which was about the limit of the speeds achievable at the time.
     
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  7. Neddy Bate Valued Senior Member

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    2,548
    If you already know the length contraction of the spaceship should be by a factor of 1/gamma, then you already know the 1.000 light second long (proper length) spaceship should be 0.500 light seconds long as measured by the earth frame. Then you can calculate t1=0.500/(c+v)=0.500/(1.000+0.866)=0.268 seconds for front-to-back, and t2=0.500/(c-v)=0.500/(1.000-0.866)=3.732 seconds for back-to-front, and 0.268+3.732=4.000 seconds total. But you probably already knew that, and wanted to derive length contraction from that set up.

    If you already know that the entire measurement takes 2.000 seconds of time on the ship, then from the time dilation formula you also know that the entire measurement should take 4.000 seconds of earth time. Then you can assume the proper length of the ship, calculate t1=1.000/(c+v)=1.000/(1.000+0.866)=0.536 seconds for front-to-back, and t2=1.000/(c-v)=1.000/(1.000-0.866)=7.464 seconds for back-to-front, and 0.536+7.464=8.000 seconds total, from which you can conclude that the length you assumed must be too long by a factor of 8.000/4.000=2.000 which is gamma, and there you have it.

    EDIT: Halc answered at the same time I did, but our methods appear to be slightly different.
     
    Last edited: Mar 12, 2020
  8. PengKuan Registered Senior Member

    Messages:
    136
    I think the length of the ship in the frame of the earth is the same problem than the length of the arm in the Michelson 's experiment, the computation of which is flawed because light and the motion of the ship do not obei the same law. See my post here #158
    http://www.sciforums.com/threads/ho...tion-by-experiment.162037/page-8#post-3623343
     
  9. PengKuan Registered Senior Member

    Messages:
    136
    Sorry, your “-c²t'²=v²/t²-c²t²” is wrong

    I'm talking about the time in the frame of the muon where the earth moves at 0.98c. Below is just a copy of your computation, but in the frame of the muon .

    In the frame of the muon, the ground is at xg

    Muon
    s²=xg2-c²t'²
    Earth
    s²=0-c²t²
    s²=s²
    so
    xg2-c²t'²=0-c²t²
    v=-xg/t (see Newton)
    v2t’2-c²t'²=0-c²t²
    t² = (1-v2/c²) t’2
    t’=6.8 micro second , muon's halflife
    So, Time of the ground is t=1.36 micro second.
     
  10. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Yes, but it doesn't have to be the only place the Lorentz transformations (LT) are used. Let the spaceship consider its rear point to be x'=0.000, so that when a light signal is sent from the rear, the event could have simple coordinates (x', t') = (0.000, 0.000) which easily transform to earth coordinates as (x, t) = (0.000, 0.000).

    Then, when the light signal is received at the front, the event has coordinates (x', t') = (1.000, 1.000). So you can plug those values into the LT to find the coordinates in the earth frame, as follows:

    v = 0.866c
    gamma = γ = 1 / √(1 - (v²/c²)) = 2.000

    t = γ(t' + (vx' / c²))
    t = 2.000*(1.000 + (0.866*1.000 / 1.000²)) = 3.732

    x = γ(x' + (vt'))
    x = 2.000*(1.000 + (0.866*1.000)) = 3.732

    -------------------------------

    Or let the spaceship consider its front point to be x'=0.000, so that when a light signal is sent from the front, the event could have simple coordinates (x', t') = (0.000, 0.000) which easily transform to earth coordinates as (x, t) = (0.000, 0.000).

    Then, when the light signal is received at the rear, the event has coordinates (x', t') = (-1.000, 1.000). So you can plug those values into the LT to find the coordinates in the earth frame, as follows:

    v = 0.866c
    gamma = γ = 1 / √(1 - (v²/c²)) = 2.000

    t = γ(t' + (vx' / c²))
    t = 2.000*(1.000 + (0.866*(-1.000) / 1.000²)) = 0.268

    x = γ(x' + (vt'))
    x = 2.000*(-1.000 + (0.866*1.000)) = -0.268

    -------------------------------

    Of course if someone insists that the ship must choose either the front or the rear as its x'=0.000 location, and if they insist that the actual reflection point must be used, then the actual coordinates would be different, as follows:

    Light signal sent from rear:
    (x', t') = (0.000, 0.000)
    (x, t) = (0.000, 0.000)

    Light signal arrives at the front and is reflected:
    (x', t') = (1.000, 1.000)
    (x, t) = (3.732, 3.732)

    Light signal arrives back at the rear:
    (x', t') = (0.000, 2.000)
    (x, t) = (3.464, 4.000)

    So the earth says that after the light signal was reflected, it traveled from x=3.732 to x=3.464 light seconds, which is a distance of 3.732-3.464=0.268 light seconds. And that was during the time the earth clocks went from t=3.732 to t=4.000 seconds, which is a time of 4.000-3.732=0.268 seconds.
     
    Last edited: Mar 12, 2020
  11. Confused2 Registered Senior Member

    Messages:
    609
    This the point where you need to look at events not what you (seem to) want to look at. Passing the top count station (muon frame event 1) the distance to the count station is zero. Some time later when passing the lower count station (muon frame event 2 ) the distance to the lower count station is also zero. The actual velocity of the count stations (or Earth) doesn't come into it - as far as the muon is concerned these events take place in the same place and are only separated by time.
     
  12. Confused2 Registered Senior Member

    Messages:
    609
    Imagine you ask google to perform a search. You get the result back in (say) 1 second. How far have you moved? In our local galaxy frame maybe 25,000 km. In the local galaxy clump maybe ? . In your frame how far have you moved?
     
  13. Halc Registered Senior Member

    Messages:
    350
    Trick question? Typically, 'X's frame' means the frame in which X isn't moving, so are you fishing for a different answer than zero?

    Maybe I accelerated after punching the search, in which case my frame is an accelerated one and distance moved is not particularly defined.
    The most I've ever accelerated in a second is to about 150 km/hr (over 4 sustained G's), but that's nowhere near my max unsustained acceleration of maybe 10 times that.
     
  14. Confused2 Registered Senior Member

    Messages:
    609
    I think PenKuan (as in the OP) and myself have sufficient difficulty with inertial frames - maybe look at accelerated frames after we have the inertial frames tied down.

    There are enough pitfalls in SR without the need to add them for no reason.
     
    Last edited: Mar 12, 2020
  15. Halc Registered Senior Member

    Messages:
    350
    I might still consider 'my frame' to be one in which I am moving. It is still an inertial frame.
    For instance, I consider the car I'm in to be moving, and the frame against which that motion is relative can reasonably be assumed to be inertial.
    Point is, saying 'your frame' doesn't necessarily imply that I'm stationary.
    'My frame' is typically the local ground around me, and I'm often in motion relative to that.
    This is why I asked if it's a trick question in my prior reply. The question as framed is arguably ambiguous.

    Assuming the frame in which I am stationary, the answer is zero.
     
  16. PengKuan Registered Senior Member

    Messages:
    136
    What is the distance to the ground when muon passes the count station and in the frame of the muon? 10km or 2km?
     
  17. Confused2 Registered Senior Member

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    609
    I am not convinced that your claim to be able to move relative to yourself is helpful or even physical.
     
  18. Halc Registered Senior Member

    Messages:
    350
    I didn't claim to move relative to myself (although waving my hands does this just fine).
    I claimed to be moving relative to my frame of choice, which is typically the frame of the ground under me.
     
  19. Confused2 Registered Senior Member

    Messages:
    609
    As far as I know there is no better way to define distance than distance=velocity x time
    so in the context of the muon experiment refenced earlier [ http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html ]
    In the muon frame the distance between count stations (10km in the Earth frame) is 2 km.
     
  20. Confused2 Registered Senior Member

    Messages:
    609
    I doubt that a drag racer would ever be given the position of observer on an (inertial) muon. Dress code. Do I really need to say more? No dresses.
    I found your earlier post re length contraction to be very helpful - I'm not sure how we got to drag racing.
    With hindsight I feel the need to apologise to the LGTBPRSNQ+ community also feminists for my implicit assumption that the muon command structure would not uphold the principles of equality in their choice of observers.
     
    Last edited: Mar 13, 2020
  21. PengKuan Registered Senior Member

    Messages:
    136
    So, in the frame of the muon the ground moves at 0.98c in 6.8 micro second to the muon. Then, the muon see the time in the frame of the ground slows 5 times.
     
  22. Confused2 Registered Senior Member

    Messages:
    609
    I believe that agrees with everything I think I know and think I understand about SR. So yes.
     
  23. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Yes, but the earth-clock located at the count station at 10km height will not be synchronised with the earth-clock located at the count station at 0km height, as measured by the reference frame in which the muon is stationary.

    So, if PengKuan tries to say that you have revealed a contradiction, because the earth frame is supposed to measure more time, not less time, compared with the muon's frame, then that is not a valid argument. Relativity of simultaneity allows the earth-clock located at the count station at 0km height to be well ahead of the earth-clock located at the count station at 10km height, as measured by the reference frame in which the muon is stationary. This interval reveals a longer total time as measured by the earth frame, even though both earth-clocks' TICK RATES are slower in the muon's frame. This is something anti-relativists get wrong over and over again.
     

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