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Light Clock query

  1. Jun 9, 2012 #1
    So my uncle recently asked me about a phenomenon that I have been unable to explain, but there that there must surely be a good answer for.

    A light clock is being observed on an airplane, with the light bouncing back and fourth between the plates of clock. The plates of the clock have the same velocity as the airplane, and the light is moving back and fourth at light speed. Why then, does the light not slowly move horizontally and escape the light clock?

    I have a few thoughts on the matter, but I would appreciate input so that I could come to a more concise answer.
     
  2. jcsd
  3. Jun 9, 2012 #2
    The light and the plates in the moving frame have 0 x-component of velocity, so it doesn't escape simply because its velocity doesn't point in a direction in which it CAN escape.

    In the earth's frame of references, the light pulse has a nonzero x-component of velocity, but it is exactly equal to the plate's x-velocity, so again the light simply isn't moving in a way that it is able to escape.
     
  4. Jun 9, 2012 #3
    The trajectory of a photon is determined by the orthogonal velocity of its source. The speed of the photon has nothing to do with the source, but the vector does.

    When the light is originally shot toward the other plate, because the source is moving with the plates, the light doesn't travel perpendicular to the "non-moving" frame of reference, but instead angles exactly enough to meet the other plate in the exact right location so as to reflect again at that same angle (all with respect to that "non-moving" frame).

    In other words, from the perspective of the non-moving frame, the light was never aimed perpendicular to the plates. If it had appeared to be aimed truly perpendicular, the plate would move out of its path.

    From the perspective of the airplane, it is the Earth that is moving under it. The light source is aimed directly at the plate, so the Earths movement has nothing to do with the light or plate.

    If you had shot the light originally while the plane wasn't moving, but could somehow accelerate the plane very quickly, the light would actually escape.
     
  5. Jun 9, 2012 #4

    Granted,but from the earth's frame of reference, when the plane is accelerating, does the light also appear to accelerate? That seems bizarre, because I thought the speed of light is constant.

    I understand there is some fundamental truth I am overlooking here, is it that the "speed limits" and "laws" for light only act when you are in the same reference point as the light?
     
    Last edited: Jun 9, 2012
  6. Jun 9, 2012 #5
    I like this explanation, but in my head I am picturing a nozzle where the light came from pointing downwards. I feel that if I looked at the same nozzle from outside the airplane passing by that it would still be point downwards, which begs the question: how did the light come out at an angle?
     
  7. Jun 9, 2012 #6
    I was picturing it shooting upward, but let's go with your picture!

    The nozzle does shoot it straight down- relative to the nozzle. But the nozzle is also moving to the side, and you can't remove that sideways velocity from the light. So we need to think of velocity as a vector, which separate components, and the total length of the vector velocity (which I'm calling speed) is always c... but because of the way velocities transform at speeds near c, the speed remains c even if it is acceleration or has components being added together.
     
  8. Jun 9, 2012 #7
    In essence, you're saying that the light trades some of its downward velocity for horizontal velocity? Also, it's horizontal velocity is such that it is equivalent to the speed of the craft, and that is because the craft was moving at a certain speed when it was fired? (This is all from the outside reference frame as I understand.)
     
  9. Jun 9, 2012 #8
    Well okay, if you need to get more into the detail;

    A photon is produced by a specific mechanism and its geometry, for example; when an electron falls from a higher orbit and releases a photon. The production of a photon is not an instantaneous event. Now if that geometry is in motion while producing that photon, each tiny bit of the production of that photon is shifting in time, thereby shifting the "front" of the photon from is "back".

    The amount of shifting of the position of the front and back is exactly equal to the amount of shifting of the atom as it was traveling. Thus the photon is actually manufactured already angled with consideration of the velocity of the atom that produced it. If the photon was released in the forward direction, the wavelength of the light would be shorter (blue-shifted). Frankly, I suspect that a "non-moving" frame would detect a polarization of the light coming out from the side.

    ..probably not the best explanation... pictures would help.
     
    Last edited: Jun 9, 2012
  10. Jun 9, 2012 #9
    That is actually brilliantly explained, thank you!
    But then... what happens when the plane accelerates @.@
     
  11. Jun 9, 2012 #10
    I like James' explanation, but to simplify a bit (in case it helps) imagine throwing a tennis ball in the air while sitting in a car. It travels with the same sideways velocity as the car! This is how kinematics works, and it's all due to the way vectors add.

    The interesting thing with light is that, no matter what the components are, the total vector sum in magnitude of the velocity comes out to be c.
     
  12. Jun 9, 2012 #11
    Acceleration changes the game. The light would escape if the plane accelerates. Merely the travel time between the plates could be enough to allow the plate to move out of the line of fire.
     
  13. Jun 9, 2012 #12
    Thank you both, I'm glad to finally have an answer! haha
     
  14. Jun 11, 2012 #13
    The situation is same as I am struggling with. So, I have a question here.

    If source's velocity vector is pointing to east. Light vector pointing to north-west locally to the source. Then for stationary observer, at some source speed, light vector pointing to north?
     
  15. Jun 11, 2012 #14
    Exactly!
     
  16. Jun 11, 2012 #15
    Ok, now we are with this. Please, look at below image.

    light_shpere_01.JPG

    If we think about spherical light source, then is above situation true?
    Left circle shows source light vectors locally. And right circle shows source light vectors while source is moving right relative to an observer.
     
  17. Jun 11, 2012 #16
    Yes, it's called the headlight or relativistic beaming effect. :tongue2:
    - https://en.wikipedia.org/wiki/Relativistic_beaming
     
  18. Jun 11, 2012 #17
  19. Jun 11, 2012 #18
    Right.
     
  20. Jun 12, 2012 #19
    Ok, If there are two intensity detector at front of light source and end of light source in light source's frame. And same two detector in observer's frame.

    light_shpere_03.JPG

    Will the source's detectors measure equal intensity? And observer's detectors measure unequal intensity?
     
  21. Jun 12, 2012 #20
    Yes.
     
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