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Harmless Lorentz transformation question

  1. Feb 6, 2005 #1
    I have a harmless question which involves using the Lorentz transformations. Suppose that I am an observer in reference frame S, which is stipulated to be an inertial reference frame, and i am located at the origin of the frame (0,0,0).

    A laser also located at (0,0,0) suddenly fires a photon along the x axis of frame S, in the direction of increasing x coordinates.

    Let t denote the time coordinate of inertial reference frame S.

    At the moment the photon is emitted, let t=0.

    Now, after some amount of time [tex] \Delta t [/tex] has elapsed, the photon will have travelled some distance L, as measured by the X-axis ruler.

    Now, let reference frame S` be attached to the photon, so that the photon is always at rest in reference frame S`. Furthermore, let the positive x` axis of frame S` coincide with the positive x axis of frame S, let the positive y` axis of frame S` coincide with the positive y axis of frame S, and let the positive z` axis of frame S` coincide with the positive z axis of frame S.

    My question is simple.

    If, after time amount of time [tex] \Delta t [/tex] has elapsed according to some clock which is permanently at rest in S, the photon has coordinates (L,0,0) in frame S, what is my x` coordinate in frame S`?


    For the sake of definiteness, suppose that exactly one second has ticked according to a clock at rest in frame S. Therefore, the location of the photon in frame S is given by (299792458 meters, 0,0).

    Let (M,0,0)` denote the location of the origin of inertial reference frame S, in reference frame S`, at the instant that the clock at rest in frame S strikes one. Solve for M.

    Thank You

    Guru
     
    Last edited: Feb 6, 2005
  2. jcsd
  3. Feb 6, 2005 #2

    jtbell

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    There is no such reference frame, at least not one that can be related to other reference frames via the Lorentz transformation. The Lorentz transformation is not defined for v = c.
     
  4. Feb 6, 2005 #3
    Obviously there are reference frames in which photons are at rest. Since you say that the Lorentz transformations wont allow me to transform coordinates correctly, can you tell me what coordinate transformations will?

    In other words, just tell me what M is please.

    Thank you

    Guru
     
  5. Feb 6, 2005 #4

    JesseM

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    You are free of course to use any coordinate transformation you like--the Galilei transform of Newtonian mechanics can give you a frame where the photon is at rest, for example. But all the most accurate known laws of physics are Lorentz-invariant, which means they will have the same form if you express them in terms of different Lorentzian coordinate systems, but they will not have the same form in different Galilean coordinate systems. This also insures that if you want your coordinate systems to have any physical meaning in terms of the readings on a network of rulers and clocks, you must use the Lorentz transform. This was actually how Einstein came up with the Lorentz transform, by imagining that each observer has a network of rulers and clocks at rest relative to himself throughout space, with each clock synchronized using the assumption that light travels at the same speed in all directions relative to himself. Then if each observer assigns space and time coordinates to different events by looking at the marking on the ruler right next to the event and the reading of the clock at that marker at the moment the event happened, the coordinates assigned to the same event by different observers will be related by the Lorentz transform. Check out my thread An illustration of relativity with rulers and clocks for some illustrations of how this works out.
     
  6. Feb 6, 2005 #5

    Hurkyl

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    A coordinate chart in which a photon is at rest will not be an inertial reference frame.

    I don't think it can even be a reference frame. As I recall, that requires four orthogonal, nondegenerate axes, and the time axis would be degenerate.

    They can't coincide. I presume you merely meant that they're parallel?

    Could be anything.
     
  7. Feb 6, 2005 #6

    Tom Mattson

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    You seriously think that the Lorentz transformation can be used to boost to the rest frame of the photon, when the Lorentz transformation was derived from the assumption that no such frame exists?

    No one said that.
     
  8. Feb 6, 2005 #7
    Can you prove the statement above?

    Regards,

    Guru
     
  9. Feb 6, 2005 #8
    Would the time axis be degenerate if the Galilean transformations are correct in this instance?

    Regards,

    Guru
     
  10. Feb 6, 2005 #9
    I meant coincide at the moment the laser fires the photon t=t`=0. Afterwards, the y, y` axis will no longer coincide of course, nor will the z, z` axes coincide, but the x,x` axes will still coincide.

    And I don't want S` spinning relative to S, so stipulate that at all moments in time the y axis is parallel to the y` axis, and the z axis is parallel to the z` axis.

    Sorry if I didn't make that clear.

    Regards,

    Guru
     
    Last edited: Feb 6, 2005
  11. Feb 6, 2005 #10
    Actually no I don't think so, I really just want to know what M is.

    Regards,

    Guru
     
  12. Feb 6, 2005 #11
    Where did he come up with that assumption... electrodynamics??

    Is special relativity all a consequence of:

    [tex] c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} [/tex]

    ?

    Regards,

    Guru
     
  13. Feb 6, 2005 #12

    JesseM

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    It depends how you define "inertial reference frame". Usually it's part of the definition that the laws of physics should work the same way in every inertial reference frame--if this is the case, then if all the most accurate laws are Lorentz-invariant, it shouldn't be hard to show that using some other transformation will always result in the laws of physics looking different in different frame's coordinate systems. On the other hand, if you just define "inertial reference frame" to mean that a point with a fixed set of space coordinates in one system (the origin, for example) should correspond to a point moving at constant velocity in other systems, then a coordinate system created by applying the Galilei transformation to some Lorentzian coordinate system could also be an "inertial frame" in this weaker sense.
     
  14. Feb 6, 2005 #13

    JesseM

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    Yes, basically. Since Maxwell's laws are not Galilei-invariant, then if they hold in one frame, they can't hold in a different frame related to the first by a Galilei transformation. So, physicists imagined that Maxwell's laws only hold exactly in one preferred frame, the rest frame of the "luminiferous ether", and that in other frames light would travel faster in one direction than the other. But Einstein showed that if each observer sees moving rulers shrink by [tex]\sqrt{1 - v^2/c^2}[/tex], and sees the length of time for a moving clock to tick extended by [tex]1/\sqrt{1 - v^2/c^2}[/tex], then if each observer synchronizes his clocks by assuming light moves at the same speed in all directions relative to himself, then Maxwell's laws can hold exactly in every observer's reference frame.
     
  15. Feb 6, 2005 #14
    Well what is an inertial reference frame?

    Regards,

    Guru
     
  16. Feb 6, 2005 #15
    I would appreciate if if you go a little slower, I'm not that smart.

    Lets starte here: What is the quickest proof that Maxwell's laws aren't Galilei-invariant.

    Regards,

    Guru
     
  17. Feb 6, 2005 #16

    Hurkyl

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    Yes. In inertial reference frames, light travels at c, thus cannot be at rest.
     
  18. Feb 6, 2005 #17

    krab

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    Your phrase "correct in this instance" betrays a lack of understanding of physical theory. Galilean tranformations are not correct, but a sufficiently good approximation when v<<c. Lorentz transformations are correct in all measured cases.
     
  19. Feb 6, 2005 #18

    JesseM

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    Maxwell's laws say that all electromagnetic waves travel at the same velocity c in all directions. But using the galilei transformation, it's easy to see that if something has velocity [tex]v_1[/tex] along the x-axis in one frame, then in another frame which is moving at [tex]v_2[/tex] along the x-axis of the first frame, that same object must have velocity [tex]v_1 - v_2[/tex]. So, if you use the Galilei transformation, an electromagnetic wave moving at velocity c along the x-axis in one frame could not be moving at c in other frames.
     
  20. Feb 6, 2005 #19

    dextercioby

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    Perform a Galilei Transf.on the wavefunction and then try to see if the new wave function verifies the Galilei transformed D'Alembert equation.
    Chose for simplicity
    [tex] \psi=\psi(x,t) [/tex]

    Daniel.
     
    Last edited: Feb 6, 2005
  21. Feb 6, 2005 #20

    jcsd

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    A reference frame in which the two postulates of special relativty hold true. Any frame travelling at constant velcoity rleative to an inertial frmae is also an inertial frame.
     
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