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Absolute movement

  1. Mar 20, 2009 #1
    1. I understand that it is impossible to say if something is absolutely stationary, for example wrt the universe, and that there is no such thing as absolute movement

    now

    2. if you place a point source of light at the origin, “O”, of a Cartesian coordinate system. At t1 the point source of light simultaneously releases a single photon along the x,y and z axis.

    3. After a short time say t2 you simultaneously measure the distance from the origin to the photons on the x,y and z axis.

    4. As the speed of light is the same in all frames of reference if the distance to all three photons is the same then you can conclude that the point source of light is absolutely stationary. Is that right?

    5. Is this explanation the definition of absolutely stationary?

    6. If you conclude that the point source of light is absolutely stationary then anything moving relative to the point source of light is not relative movement but absolute movement is that right?
     
  2. jcsd
  3. Mar 20, 2009 #2
    7. Sorry I forgot to put if the distance is not the same then you can conclude that the point source of light is moving
     
  4. Mar 20, 2009 #3

    CompuChip

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    Statement 1 is correct.
    From 2 - 4 you cannot conclude that the point source of light is "absolutely stationary", you can just conclude that it is stationary with respect to you. However, if I come running along your x-axis at 90% of the speed of light, I can claim that you are moving together with the light source and do a similar experiment with my own lightsource (which will be moving for you and stationary for me).

    [In fact, if I am moving along your x-axis, I can simultaneously measure the single photons emitted in the y and z-direction by your light source, but the one on the x-axis has already passed me because I was moving toward it]
     
  5. Mar 20, 2009 #4

    JesseM

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    The speed of a photon is independent of the velocity of the source, so if you have a source resting at the origin that releases a photon in the x direction at time t1, and you have another source moving at high velocity along the x axis which crosses the origin at time t1 and also releases a photon at that moment, both photons will move side-by-side along the x axis at the same speed.
     
  6. Mar 20, 2009 #5
    Hello Dreads.

    If a number of observers moving relative to each other are present at the same point at the same time as the emission of a "light flash" , then, as a consequence of the speed of light being the same for all of them, each observer will see a sphere of light expanding away from them with the centre based at them. So they can all come to the same conclusion as to their state of motion or rest. In other words they can infer nothing about their state of motion from this situation.

    This quote is from W.Pauli, The Theory of Relativity:-
    -----let us take a light source L which moves relative to an obsever A with velocity v, and consider a second-observer B at rest with respect to L. Both observers must then see as wave fronts spheres whose centres are at rest relative to A and B, respetively. In other words, they see differtent spheres. This contradiction disappears, however, if one admits tlat space points which are reached by the light sinmultaneously for A, are not reached simultaneously for B. This brings us directly to the relativity of simultaneity------

    Matheinste.
     
  7. Mar 20, 2009 #6
    For now I am not interested in what an external observer would see. For a second just forget what an external observer would see.

    What I am saying is if the light source is moving a person at rest wrt the light source would get a different measurement to the photons
     
  8. Mar 20, 2009 #7
    is that right ?
     
  9. Mar 20, 2009 #8

    jtbell

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    Regardless of how the source is moving, the three photons will all travel the same distance in the same period of time.

    Therefore, the observer cannot come to any conclusion about how the source is moving, simply from measuring the distances traveled by the three photons.
     
  10. Mar 20, 2009 #9
    Hello Dreads.

    I am not sure what you mean by an external observer. However, for a point source of light the the wave fronts in ALL frames, (for ALL observers) form spheres with their centres at rest. If ANY observer, is present at this emission event that observer remains at the centre of this sphere. So ALL observers are at rest with respect to the centre of the expanding sphere of light no matter what their relative motion. Completely unintuitive but that's relativity.

    Matheinste
     
  11. Mar 20, 2009 #10
    I disagree with this quote if the frame of ref is moving in order to conserve momentum the wave fronts will not form spheres they will form skewed spheres ie some shape that is not a perfect sphere
     
  12. Mar 20, 2009 #11
    Incorrect. An observer at rest wrt the light source observes an expanding sphere of photons centered on the source, all traveling equal distances in equal times, because photons all travel at c.


    You are incorrect. All observers observe an expanding sphere of photons, traveling equal distances in equal times, because photons all travel at c.
     
  13. Mar 20, 2009 #12
    Sorry no I got that wrong it will form a sphere but the person at rest wrt to the light source will not be at the centre of the sphere
     
  14. Mar 20, 2009 #13
    in order to consrve momentum
     
  15. Mar 20, 2009 #14
    If you are in a moving frame of ref along the x axis and you fire a laser along the Y axis the photons will not move long the Y axis. To conserve momentum the photons will travell at an angle less than 90 degrees wrt the X axis its called Maxwells Symmetry theory or something like that
     
  16. Mar 20, 2009 #15
    The effect this creates is that the centre of the sphere will be slightly infornt of you.The faster the moving frame of ref is moving at the greater the effect. I imagine if you are at rest wrt a light source traveling at C the entire shpere will be infront of you
     
  17. Mar 20, 2009 #16
    or am I wrong ?
     
  18. Mar 20, 2009 #17

    JesseM

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    Call this person "Bob"--then in Bob's own rest frame, if he was next to the source when it emitted light in all directions, he will remain at the center of the sphere according to the measurements of his frame. If we consider a different observer "Alice" who is moving relative to Bob, then in Alice's frame the light will still form a sphere but Bob will move away from the center of it.
     
  19. Mar 20, 2009 #18

    DaveC426913

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    "moving frame of reference" wrt what??

    First, prove you're moving. You can't.
     
  20. Mar 20, 2009 #19
    That is incorrect and assumes absolute X and Y axes. You really need to get a textbook on relativity and learn about frames of reference and coordinate systems referred to them. All textbooks will, in the first few pages, deal with this and the meaninglessness of absolute rest or moton.

    Also I think you will find that momentum is frame dependent and is only required to be conserved within any SINGLE single frame.

    Matheinste.
     
  21. Mar 20, 2009 #20
    Ok so I am in a space ship and I want attempt to prove it is moving. The space ship will move along the x axis

    While it is on the launch pad I set up a laser along the Y axis so the laser is pointing at a spot on the opposite wall of the space ship. Th space ship now blasts off and obatins a constant velocity. Will the laser still be fixed on the spot? if so why?
     
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