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Question concerning speed of light in reference frames

  1. Oct 9, 2014 #1
    My friend posed a question to me that I was unable to succinctly answer (or answer at all for that matter). So I thought I'd make an account and ask people who know more about this than I do.

    Ok. So let's assume that Person A starts at Position 1. Person B is at position 2, which is 1 light year away. There also exists a giant reflective surface between them such that these three points form an equilateral triangle (where Position 1 is the bottom left corner, Position 2 is the bottom right, and the reflective surface is at the top).

    So Person A leaves Position 1 going 90% of the speed of light directly towards Position 2. At the same time, a beam of light is shot from Position 1 directly towards the reflective surface, such that when reflected, it will eventually reach Position 2 (in what would be exactly double the time at slower speeds).

    Person A, though he is traveling 90% of the speed of light, will still see that beam of light as going the speed of light relative to his speed, correct? If that is the case, from his frame of reference, the beam of light will reach Position 2 much earlier than he does. However, Person B standing at Position 2 would see Person A arrive much earlier than the beam of light. Therefore, Person A would see the beam of light arrive before him, but then be able to witness the light arriving as he stood next to Person B?

    This is the question he posed.

    Any help?
  2. jcsd
  3. Oct 9, 2014 #2


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    I think the question confuses itself by saying that person A can "see" the beam of light that is sent from 1 to the mirror to 2. This is not true. How would he "see" light that is nowhere near him?. When A arrives at 2, A and B will both see the light arrive, traveling at c, some time later.
  4. Oct 9, 2014 #3

    Simon Bridge

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    The description also asserts that the triangle is an equilateral one - which is only the case for one reference frame.
    Shall we guess that the equilateral triangle is in the reference frame in which person B is at rest?
    The initial speed of person A is not given - shall we guess that they are always moving at 0.9c?

    The trick with relativity situations is to be very careful about the setup.
    If I'm getting you correctly, this is what you are thinking of:

    In the reference frame where B is at rest:
    Then at ##[t_0]_B=0##, A is at (x,y)=(-d,0) and B is at (0,0) so that |AB|=d.
    There is a mirror at (x,y)=(x,h): ##h=d\sqrt{3}/2##.
    Person A is moving at 0.9c to the right and passes through position A at ##[t_0]_B=0##
    Also at position A, and stationary, is a box that emmits a pulse of light in all directions at ##[t_0]_B=0##.
    B receives the pulse of light first at ##[t_1]_B=d/c## directly from the source, and then the reflected pulse at ##[t_2]_B=2d/c##.
    From that information, B deduces that the direct and reflected pulses traveled along the sides of an equilateral triangle ARB
    Thus the point of refection is ##R=(d/2,h).##
    Further, A passes through point B at time ##[t_A]B = d/v = d/(0.9c) = (10/9)(d/c)##
    So ##[t_1]_B < [t_A]_B < [t_2]_B##
    i.e. B meets A between the two pulses.

    Now work all that out for the reference frame in which A is at rest.
    The times of interest being:
    ##[t_1]_A## ... the time on A's clock that the first pulse gets to position B (which will be different in A's frame.)
    ##[t_2]_A## ... the time when the reflected pulse reaches B.
    You'll have to figure out how A knows about these events though, that could make a difference ... perhaps A could deduce the times.
    i.e. B could hold up a sign that A can read at a glance when A and B are passing each other which just says how many light pulses B has seen before they met... since that's the issue right?

    ##[t_A]_A## ... the time when B reaches the position of A.
    (Remember, in this reference frame A is stationary and B is moving.)
  5. Oct 9, 2014 #4


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    I don't see how you concluded the beam of light would reach position 2 earlier than A does.
  6. Oct 10, 2014 #5
    It wasn't my conclusion. But I have what I need now. I figured it would be obvious. Thanks!
  7. Oct 10, 2014 #6

    Simon Bridge

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    What did you discover?
    (For the benefit of others wrestling with the same sort of thing?)
  8. Oct 10, 2014 #7
    Of course. It was simple simple oversight. I had trouble explaining the speed of light in a frame of reference to my friend. The light would obviously arrive later than Person A. The simplified example I've thought of the explain it to him is this:

    Think of a 2 dimensional grid. Let's say that light moves 10 units per second and Person A and his frame of reference is 2 units long. As Person A approaches the speed of light, his and his frame of reference's length is lowered significantly. So while the photon is still moving c as it slowly passes him to a third-party observer, it moves c past him in his frame of reference as his frame of reference is very compressed. So for example, let's say that the compression has brought his 2-unit length down to only 1 to an observer. His length is still 2 to him, so when the photon goes past, he measures it going 10 units per second. That does not mean it is going almost 20 as Person A could conclude from measuring the speed, as an observer sees him as only one unit (therefor it is going its normal 10 units per second in a neutral frame of reference). Photons aren't compressed and that's why this is possible.

    I've only taken three physics courses so I'm sorry. This was probably glaringly obvious to you all. Please correct me if any of my thinking is wrong.
  9. Oct 10, 2014 #8

    Simon Bridge

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