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Detecting absolute motion?

  1. Feb 17, 2008 #1
    Set-up #1: A person is enclosed in a container traveling with a constant velocity in the x direction. Wouldn't a laser inside the container attached to the floor and aimed perpendicular to the direction of travel (i.e., y direction) illuminate a spot on the ceiling of the container that is BEHIND a point marked on the ceiling directly above the laser, thus allowing the occupant to detect uniform motion in the x direction without external reference? As a laser photon traverses from floor to ceiling the container is moving and since light is not effected by the speed or direction of the source it will travel straight up.

    Set-up #2: A person is enclosed is in a container traveling with a constant velocity in the x direction. In the middle of the container is a lit light bulb. One spectrum analyzer is attached on the inside front wall and another analyzer attached on the inside back wall of the container. Wouldn't occupants be able to see different phasing due to the doppler shift and thereby detect their motion without external reference?
     
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  3. Feb 17, 2008 #2

    Doc Al

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    Nope, that won't work. It will hit a spot behind the source, but only when viewed from another frame. You're in a moving frame right now: try it and see!

    Nope, that won't work either.
     
  4. Feb 17, 2008 #3
    Detecting absolute motion?

    No chance.
     
  5. Feb 17, 2008 #4
    It is the speed of light that is unaffected by the motion of the source but the direction can change. In the Michelson Morley experiment the photon appears to be going straight up and down the y arm to an observer at rest with the interferometer. To an observer with relative motion with respect to the interferometer the photon appears to following a zig zag path. In both cases the observers measure the speed of the photon to be c. In your example both observers will see the laser hit the spot on the ceiling directly above the source. To the external observer the laser photon is emmited at a forward diagonal angle rather than straight up.


    Doppler shift comes about as a result of difference in the velocities of the source and the receiver. In your example the bulb and the walls all have exactly the same velocity so no doppler shift will be observed.
     
  6. Feb 17, 2008 #5
    Are you saying that a photon emitted by a laser in the y direction has an x component? How can that be, it has no mass? Once emitted the photon continues on its vertical path.
    Einstein used a similar thought experiment where a light beam would appear to bend while traversing across an accelerating elevator. It would seem that in a non-accelerating elevator (i.e., one moving at a constant velocity) the beam would not bend but traverse the elevator diagonally.

    In set-up #2 it is true that the bulb and analyzers are traveling at the same velocity however, once the light leaves the bulb it travels at c in both directions and the light waves impacting the back wall will be more compressed than those impacting the front. I appreciate your patience.
     
  7. Feb 17, 2008 #6

    Mentz114

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    No, the walls are stationary wrt to the emitter.
     
  8. Feb 17, 2008 #7
    In the case of the elevator with constant motion the direction of the emitted photon is altered so that it hits the exact same spot on the opposite wall as it would when the elevator is stationary.

    When the elevator is accelerating the photon moves in a straight line just as in the first example but once emitted the photon is "committed" and can not alter its trajectory due the additional change in velocity of the elevator in the period between being emmited and received. In this scenario the photon misses the target and to the observer in the elevator the photon appears to be following a curved path.

    P.S. The full and complete general answer to your original post was given on post #3 ;)

    P.P.S. I was talking about an elevator being artificially accelerated by a rocket in flat space. If the elevator is free falling in a gravitational field then the photon and the elevator are acceleated equally by gravity and the photon hits its target. The observer inside an enclosed free fallin elevator will think he is stationary. An observer that is outside the elevator and stationary in the gravitational field will see the photon following a curved path while the observer inside the free falling elevator will see the photon following a straight path.
     
    Last edited: Feb 17, 2008
  9. Feb 17, 2008 #8

    russ_watters

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    Relativity tells us that the laws of physics are the same in all inertial frames of reference. That means that since the emitter is stationary with respect to the box, it is stationary with respect to the box. Period. That's all it cares about. So the beam does not have an x-component of motion with respect to the box.

    The beam can have an x-component of motion with respect to someone else traveling past the box.

    BTW, this part of relativity predates Einstein. The fact that you can throw and catch a baseball on an earth rotating with a speed of 1000mph is the same principle. Einstein extends the principle beyond that, though...
     
  10. Feb 18, 2008 #9
    Russ, Kev, et al - let's just stay inside the box for now. the emitter IS stationary with respect to he box however once a photon is emitted it is independent of both the emitter and the box and will travel in the y direction while the box continues to move in the x direction. unlike a baseball tossed up in a moving vehicle, the emitted photon does not have momentum in the x direction. likewise, in set-up#2, since the emitted flash of light is independent of the bulb and container the light will impact the front and back walls at different times.

    Kev, what alters the direction of an emitted photon in the elevator?
     
  11. Feb 18, 2008 #10

    Doc Al

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    aberration

    Bill, when viewed from within the box, the photon trajectory is along the y-direction. But that's not the case when viewing the photon from outside the box. From outside the box, the photon trajectory is at an angle (similar to, but not exactly like, a baseball). This effect is called the "headlight" effect or relativistic aberration. See: Relativistic aberration

    And yes, viewed from a frame in which the box is moving in the x-direction, the photon does have an x-component of momentum.
     
  12. Feb 18, 2008 #11
    Doc, if momentum (p) = mass x velocity and if a photon is massless, how is momentum in the x direction imparted by the container moving in the x direction the y direction?
     
  13. Feb 18, 2008 #12

    Doc Al

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    Momentum only equals mass x velocity for slow moving particles. For photons, which are massless, p = E/c (photon energy divided by the speed of light).
     
  14. Feb 18, 2008 #13

    Janus

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    The momentum of a photon is related to its energy and not its mass(or lack there of).
     
  15. Feb 19, 2008 #14

    russ_watters

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    Yes.
    No. The principle of relativity tells us that what is happening inside the box is not affected by the motion of the box. The light moves at C with respect to the box regardless of the box's speed.

    I realize this is just a thought experiment, but please be aware that real experiments have been done on this (such as the M&M exp cited above). So the answer is not in doubt. Yes, if light behaved more like sound, for example, the experiment would work as you describe. But it doesn't.
     
    Last edited: Feb 19, 2008
  16. Feb 19, 2008 #15
    Of course if you used a sound source in the box (filled with air) you still would not detect absolute motion (or a doppler shift in the sound) inside the box.
     
  17. Feb 19, 2008 #16

    russ_watters

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    The box would have to be open for it to be analagous to the ether.
     
  18. Feb 19, 2008 #17
    Bill

    Setup #1: I'm not convinced the laser will hit the spot. I would like some experimental evidence to show it doesn't lag behind.

    Setup #2: The spectrum analysers will both read the same frequency. Going forward - an increase in frequency would be countered by a decrease in speed, and vice versa.

    Good questions.
     
  19. Feb 19, 2008 #18
    The laser beam is vertical even in the system where laser moves. This is not exclusively relativistic phenomenon: imagine a juggler siting in a train and throwing balls straight up in his system. The balls travel verticaly in his system, but not in a system where train is moving.
     
  20. Feb 19, 2008 #19
    If we have sound source mounted on the centre of an open railway carriage and receivers mounted on the rear and front of the same carriage then we will not detect a change in frequency even when the train is moving and the air is moving relative to the carriage.

    There are any number of ways we can detect our motion relative to the air but unfortunately detecting doppler shift using emitters and receivers that are only mounted on the carriage is not one of them.

    The classical non relativistic doppler shift equation is often stated as

    [tex] f ' = f {(v \pm v_o) \over (v \mp v_S)} [/tex]

    where v is speed of sound in the medium and [itex] \vec v_o[/itex] and [itex]\vec v_S[/itex] are the velocities of the observer and source respectively, relative to the medium.


    Getting the signs correct when using that formula is a little tricky as they are determined by whether the source is moving away from or towards the observer. When the source and the observer are moving at the same velocity it is hard to decide which signs to use as there is no obvious "away" or "towards" in this case. (The secret is to see which way the vector arrows are pointing.) The classical doppler equation can be expressed in terms of vectors like this:

    [tex] f ' = f {( |\vec v| \pm (\vec v_o) )\over (|\vec v| \mp (\vec v_S))} [/tex]

    The difficulty of choosing the correct signs remain and information is lost by only using the unsigned value of the wave velocity [itex] |\vec v| [/itex] . If we take the medium as a coordinate system and express all velocities (including the wave velocity) as vectors relative to the medium then the doppler equation can be expressed like this:

    [tex] f ' = f {( \vec v - \vec v_o)\over (\vec v - \vec v_S)} [/tex]

    Using this form there is no difficulty getting the signs right and it immediately clear when [itex] \vec v_o= \vec v_S[/itex] that [itex] f ' = f [/itex]

    Doppler shift only detects relative velocity and is completely useless for detecting absolute velocity, whether the medium has relative motion or not.
     
    Last edited: Feb 19, 2008
  21. Feb 20, 2008 #20
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