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Minkowski diagram questions

  1. Oct 16, 2015 #1
    I had some questions about how to understand the Minkowski diagram better:

    588px-Minkowski_lightcone_lorentztransform.svg.png

    1) One thing that I hear characterizes a light ray or a photon is that it "experiences no time." That is, it's time dilation is so extreme that at the "null lines" or light cone lines at the 45 degree mark, the photons, again, do not age or experience the progression of time (at least relative to the "lab frame"). However, if you look at the diagram, the photon (or light ray) is clearly moving through time and "aging." It just seems to be moving through space at the same rate it is moving through time. If the photon truly weren't aging, wouldn't it just be confined to the x-axis?

    This same argument, of course, applies to length contraction and in the exact same way. At the same 45 degree null lines, the length contraction is so extreme that the photon ostensibly has no size and "experiences no conception of space." However, is seems clear from the diagram that it is, indeed, moving through space as the null line has an increasingly (increasing) x-component value.

    So it seems I'm missing something here...Please do help explain what.
     
  2. jcsd
  3. Oct 16, 2015 #2

    Nugatory

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    You've heard that, but it is not correct. That argument is basically trying to calculate the length contraction and time dilation in a frame in which the flash of light is at rest - and there is no such frame, so the length contraction and time dilation formulas cannot be applied. We have a FAQ on this, about the "rest frame of a photon".
    Your interpretation of the diagram is correct (although I added a bit in bold). The flash of light is moving at speed c in all reference frames.
     
  4. Oct 16, 2015 #3
    I think I found the FAQ you are referring to:https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

    But I'm talking about my (lab) frame being at rest and watching the moving frame of the photon which is not at rest. In which case the Lorentz transformations would apply. That is, unless you are stipulating a condition where, even though I'm calculating what the Lorentz transformations mean for the photon according to my frame of reference, in order for that to be valid for the moving body (photon in this case), that moving body must be able to claim that it possesses it's own stationary rest frame. And since we cannot claim that it can do so (because it's a photon) then we cannot comment on what the photon experiences as the passage of time nor what it experiences as a conception of space. Is that what you are saying?
     
  5. Oct 17, 2015 #4
    "Frame of the photon" means a frame in which the photon is at rest. There can be no such frame, whether moving relative to you or not. The Lorentz transformation applies between two frames in relative motion at speed v < c, with emphasis on the strict inequality. You can choose v as close to c as you want but not equal to c.
    ("Photon" here to be taken as a classical point-particle of light not the QM object.)
     
  6. Oct 17, 2015 #5
    Ahhh, the old "You can push your transform as far asymptotically as you want but you can't actually put it on the asymptote" trick, huh? That's pretty sneaky..o_O

    The Lord is not subtle, he is malicious..Although the asymptote trick is a little subtley malicious.

    So it is basically as I characterized it in my post number 3, then?

    By the way, somebody better inform Neil deGrasse Tyson of this technicality, I don't think he got the memo:



    This guy didn't get the memo either:



    But he's apparently somewhere out in the deep forest, I don't think anyone's going to find him anyway to let him know. In fact, he's been out in the woods so long he's still talking about "relative mass" increase :biggrin:
     
  7. Oct 17, 2015 #6

    PeterDonis

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    These are just illustrations of the fact that pop science, even if it is being peddled by people with scientific credentials, is not science.
     
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