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I Time experience of light in Minkowski Diagram

  1. Apr 28, 2017 #1
    Hello,

    I noticed something peculiar when looking at the Minkowski Diagram and I'm not sure how to interpret this.
    Let observer A be the reference frame for the diagram and B someone who travels with respect to A.

    Diagram.jpg

    The red line is the displacement of light over time. The blue, yellow and green lines indicate fixed times of t = 2, 2.5 and 3 respectively according to someone who is traveling, such that if B intersects one of those lines at any ##x##, a time t=2, 2.5 or 3 has passed for B.

    This means that, if B has a sufficient velocity to travel exactly on one of those lines, he won't experience any time (such a velocity is ofcourse not possible). You can see that those lines eventually have the same slope as ##c## so that light also doesn't experience time.

    Now, let light leave A at t = 2 instead of t = 0. I'd then get the following diagram:

    Diagram 2.jpg

    The purple line is light but now transposed and leaving A at t=0. You can see that this line intersects the blue, yellow and green lines. This implies that if B is leaving A at t=2 at the speed of light, he'd experience time lapse of 2.5 and 3 seconds over different distances of x when traveling, thus he wil experience time.

    How is this possible while traveling at the speed of light shouldn't let someone experience time?
    I have a feeling the blue, yellow and green lines are only with respect to ##c## if it starts from t=0 and starting from t=2 at the speed of light would need its own lines. Is this correct?
    If so, then if a traveler C is leaving A at t=0 at ##c## (red line) and B is leaving A at t=2 also at ##c## (purple line), will C see B's clocking moving while B says that his own clock is not moving at all?
     
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  3. Apr 28, 2017 #2

    vanhees71

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    You can't travel with the speed of light, because you aren't massless!
     
  4. Apr 28, 2017 #3

    Orodruin

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    Also, the proper time of someone starting at x=0 at t=2 have very little to do with the hyperbolae of constant proper time separation from the event x=0, t=0. You are just confusing yourself and mixing up the nomenclature.
     
  5. Apr 28, 2017 #4
    So my previous question/statement...

    ..was the correct reason for this confusion. Right?

    If this is the case, then I'm really curious about my second question:

    I understand it's not possible to travel at ##c## but this is just for the sake of argument and not the point of my question. If it really concerns, then let's substitute observer B and C with two light beams and assign my question to that.
     
    Last edited: Apr 28, 2017
  6. Apr 28, 2017 #5

    Ibix

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    It does not make sense to talk about the perspective of anything travelling at c in Special Relativity. To do so, it would be necessary to describe an inertial frame in which light was stationary, which contradicts Einstein's postulates. Travelling faster than light is even more problematic because it allows causality violations.

    Thus, the blue, yellow and green lines do not represent paths that anything can follow. An object following the blue curve does not exist prior to t=2, then starts to move at infinite speed in both the positive and negative x directions and decelerates towards c in both directions. Your questions don't make sense because the situation you are describing doesn't make sense.

    The blue line is the set of all events where a clock that left the origin at constant velocity would read 2. That's all. For objects that did not pass through the spatial origin at t=0 it is nothing special at all.
     
  7. Apr 28, 2017 #6
    Even if we're talking about 2 light beams? I'm not asking about if an observer is following the blue curve.

    Let me rephrase it. Two lightbeams, one starting from t=0 (red line) and one starting from t=2 (purple line) are traveling. Can one say anything about the clock that the purple light beam is experiencing according to the red one theoretically?
     
  8. Apr 28, 2017 #7

    Ibix

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    How are you going to build a clock that moves at light speed?
     
  9. Apr 28, 2017 #8

    PeterDonis

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    There is no such thing as "the clock that a light beam experiences". Your question is meaningless.
     
  10. Apr 28, 2017 #9

    PeterDonis

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    But those intersections are physically meaningless, because the blue, yellow, and green lines are hyperbolas centered on the origin, not on the point t = 2, x = 0, where the purple line intersects the t axis. When you transpose the light beam, you need to also transpose the hyperbolas, so they are now centered on the point t = 2, x = 0. If you draw those hyperbolas, you will see that the purple line never intersects any of them.

    To put this another way: in order to make the comparison you are trying to make, to verify that a light ray is a null worldline (which is the correct way to say what you have been saying as "does not experience time"--the latter phrasing is incorrect and misleading, as your own questions about it indicate), you have to look at hyperbolas that the light ray is an asymptote for. The purple line in your diagram is not an asymptote for the hyperbolas in your diagram. Only the light ray through the origin of the diagram is.
     
  11. Apr 28, 2017 #10
    Yes exactly, that's what I indeed realised and stated in my OP as a second question. Got confused there. Thanks!

    As for my question in my post #6, I thought that this question could merely be answered theoretically based on the implications of the Minkowski Diagram, since the Minkowski Diagram is sometimes also drawn with other physical impossible aspects. Like this:

    Minkowski.png

    I understand that the lines below ##c## are physically impossible. So I was merely trying to understand what the answer to my question in post #6 would be according to the Minkowski diagram.
     
  12. Apr 28, 2017 #11

    PeterDonis

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    Your question in post #6 is meaningless. It's not "physically impossible". It's meaningless. It doesn't have an answer "according" to anything, because it's not even a well-defined question in the first place.

    This thread is closed.
     
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