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Constant speed of light

  1. Jul 17, 2012 #1
    Relativity says that the speed of light is constant for all observers regardless of their state of motion.

    So does that mean even if you are heading towards a light beam at a constant velocity that the light beam would take the same amount of time to reach you as when you are at rest relative to the light beam?
  2. jcsd
  3. Jul 17, 2012 #2
    Yes, everyone measures that same exact value for the speed of light, regardless of your frame of reference. Lorentz transformations (time dilation and length contraction) assure this.
  4. Jul 17, 2012 #3


    Staff: Mentor

    By "heading towards a light beam at a constant velocity", do you mean "heading towards the source of a light beam at a constant velocity"? Otherwise I would have a hard time making sense of your question.

    This is true, but if the answer to my question above is yes, then the answer to the OP's question doesn't involve just the speed of light. It also involves the relative motion of the observer and the light source. Consider two scenarios:

    (1) Observer and light source at mutual rest, separated by distance d. Then the light will take time t = d (in units where the speed of light is 1) to get from the source to the observer.

    (2) Observer moving towards light source with speed v. Then the light will take time t = d / (1 + v) to get from the source to the observer, i.e., *less* time than case #1 above.

    (Note: in both cases, "time" is the time in the rest frame of the light source. In case #2, the *proper* time experienced by the observer between light emission and reception will be different from the time in the rest frame of the source; it will be smaller.)
  5. Jul 17, 2012 #4
    OK I will put it another way. The source of the light and the travellers starting point are both at rest relative to each other and they have synchronized their clocks. At a specific time the source of light emits a constant light beam and the traveller starts moving towards the source of the light with a constant velocity. Would the amount of time taken to reach the beam of light be the same as if the traveller was still at the starting point?
  6. Jul 17, 2012 #5


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    Staff Emeritus
    Science Advisor

    ]"At a specific time" is ambiguious in relativity. Are you familiar with space-time diagrams? Those are unambiguous.

    There's an example of a space-time diagram for your problem (or rather, how I interpreted your problem) appended below.


    In this space-time diagram, as in most, the time coordinate goes up the page. There is one spatial coordinate, which goes across the page.

    At event A, a light beam is emitted to the right. The worldline of the lightbeam is represented on the space-time diagram by the world line L.

    At event B, the worldlines of two observers, observers S and M, cross. observer S, who is stationary, has his worldline goes straight up the page, and observer M, who is moving towards the source of the light beam , moves towards event A as time increases.

    The observer in the moving wordline, M, measures less time to reach the light beam L than the observer on the stationary worldline, S. However, the observer on worldline M also thinks that the distance to event A is shorter than observer S does, due to Lorentz contraction. So, observers S and M both compute the same value for the speed of light, but they'll disagree on things like distances and times.

    Attached Files:

  7. Jul 17, 2012 #6


    Staff: Mentor

    pervect gave a good answer to this, the only thing I would add is that the motion of the traveller causes two effects: (1) relativistic length contraction/time dilation (and also relativity of simultaneity, but I suspect that by "at a specfic time" you meant "time in the rest frame of the light source", since the traveller doesn't start moving until that event anyway); (2) change of distance to the source as seen in the source's rest frame, because of the traveller's motion relative to the source. The latter would change the time taken to reach the light beam even without relativistic effects.
  8. Jul 18, 2012 #7
    It gets messy if the state of motion of the clock is changed: according to SR the clock cannot be used as a reference clock of the earlier synchronized system except if you make corrections.

    But perhaps you meant that you set up a reference system in which a traveler moves straight towards a light source. Then of course the time for the traveler to meet up with the light source gets shorter the faster the traveler moves:
    Δt=Δx/(c+v) with Δt=time to meet up, Δx=original distance, c=speed of light, v=speed of traveler.

    PS. The problem may be due to ambiguous, abbreviated jargon. I think that you refer to special relativity, according to which the speed of light in vacuum is measured as the same constant c with all standard inertial reference systems, regardless of their velocity.
    Last edited: Jul 18, 2012
  9. Jul 21, 2012 #8

    Hi Codeman,

    PeterDonis found Pervect's explanation a good answer but I feel a bit uncomfortable with it.
    I'm a layman. I want to visualize what your observer M is talking about. A diagram helps a lot. But..... where in the diagram is the shorter distance to A, due to Lorenz contraction?

    Can we visualize what M thinks? I'll give it a shot with a Loedeldiagram. A loedel diagram looks a bit like your diagram, but the axes are into other directions to be able to measure the time and space distances directly from the diagram (which is not so obvious on a Minkowski diagram). On a Loedel diagram the lightbeam covers the same length on time and space axis (of same color). Pink is your M observer, blue is your S observer. Green is the lightbeam. At event E the lightbeam hits pink observer M.


    Possible answers for the <shorter distance of A>:
    1/ Blue length L? It cannot be, because here we mix time and space of two observers (not same color). Blue observer S might think that blue length is what pink obxerver M will measure, but that's wrong. Pink measures on pink lines (his worlds of simultaneous events).

    Furthermore this option 1 is wrong because the lengths are not equal, therefore not speed of light.

    2/ Pink L? No, it has not the same length as pink 0,57 time, therefore not a candidate to calculate speed of light at event E.

    3/ Black L? No, that's not even a contraction (shorter length) compared to the length between event A en event B! And black length L is not equal to pink 0,57 timelength at event E. No speed of light.

    Where is the Lorentz contraction/shorter distance to A to keep speed of light same for observers M and S?

    Because of Einstein's relativity of simultaneity, observer M and S live in different worlds (world = his collection of simultaneous events). In his pink world at event B the lightbeam is already at 'only' 0,57 distance from him (the distance between event B and crossing of pink L and green). That's why at event E the light will have travelled (in his pink world) pink distance 0,57 in 0,57 pink time.

    My point is now: is <M also thinks that the distance to event A is shorter than observer S does, due to Lorentz contraction> a correct and sufficient explanation to understand why in your little exercise (that I found not that easy to answer!) the speed of light is constant for both observers?
    I don't think so (sorry, PeterDonis).
    (And I wonder whether calculations only can make all that clear. Or lots of text only. Not to me. That's the power of a Loedeldiagram. I'm not sure a Minkoswki diagram could visualize how time and space lengths work out for your exercise. I didn't try.)

    Let me now get back to your question:
    Let's first consider the two observers were always moving relative to each other and crossing each other at event B. Your question involves accelleration ('starts moving') which makes the diagram a bit more complicated. I leave that for later. First the diagram without accelleration.
    Due to relativity of simultaneity the pink world of observer M is not the same as blue world of observer S. In pink world at event B (clocks synchronised at 0) the light is already on it's way.

    If you (or he) want(s) to know what pink measured from the moment event A happened in his pink world up to event E, you have to take the black L and black T (indeed het starts timing before he crosses observer S !). These 'black' lenghts are equal and therefore give him speed of light result at event E.

    Your question involves accelleration.
    Here I have incorporated accelleration in the Loedel diagram. (For the Loedel experts: I'm not 100% sure I am allowed to draw it that way, please correct me where I'm wrong...).


    Pink time and space are identical up to event B (when M starts walking).
    From the moment he starts walking the angle between the worlds of observer M and S increases, and stabalizes when he continues walking at a steady speed.
    In the drawing he accellerates very very fast, in a split second.

    In this context the previous option 'black L and black T' falls away.
    Your question (with accelleration): <Would the amount of time taken to reach the beam of light be the same as if the traveller was still at the starting point?>
    I would say 'NO'.
    For relative speeds as shown in loedel diagram:
    Lightbeam from event A to event B for Observer M: pink watch at 0,57sec.
    Lightbeam from event A to event B for Observer S: blue watch at 1 sec (top right of diagram).

    However: Will they measure the same speed of light with accelleration involved?
    How S (in his blue worlds) measures is obvious.
    But M (in his in pink worlds)...

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