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The consistency of the speed of light

  1. Jul 5, 2013 #1
    Spaceship Alpha is travelling at 0.9c. It shines a light beam.
    An observer that is stationary observes that the light beam is travelling at c, not 1.9c.
    Explain.

    How can I answer this question? I don't want to just say because Einstein said c is the same regardless of frame of reference.
    Imagine that the spaceship was travelling at c (which is impossible) and shines a light, does this mean the person will not see the light?
    What led Einstein initially to come up with the idea that light speed is constant?
     
    Last edited: Jul 5, 2013
  2. jcsd
  3. Jul 5, 2013 #2

    phinds

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    There is no such thing as a stationary observer in what you apparently mean as an absolute sense. ALL observers see the light as traveling at c and all observers have motion only relative to something.

    Since, as you correctly point out, that is impossible, your question is meaningless.
     
  4. Jul 6, 2013 #3
    Well, you just said it. The stationary observer (SO) sees Spaceship Alpha traveling at 0.9c, say moving across his field of view from left to right. Alpha shines a beam of light into the direction it is traveling, and the SO sees this beam of light emanate from Alpha at a speed of 0.1c. Thus, the SO is witnessing Alpha moving at 0.9c and light moving at c. Incidentally, the passengers on Alpha will also see the beam of light moving at c, even though the SO only sees the light moving relative to Alpha at 0.1c. The reason this can happen is that the perception of time is slowed way down on Alpha relative to SO's perspective, slowed down to a degree where c is invariant for all observers. This is a governing principal of special relativity.

    As you and Phinds pointed out, this is impossible. However, if it were possible, than you are right, the SO probably would not see a beam of light as it would hug Spaceship Alpha too closely.

    This was one of the keys to Einstein's brilliance. He just made a stipulation that c was a universal constant and not time, and proceeded to develop SR from there. It was a bold iconoclastic move that paid off handsomely. Everyone else was fumbling in the dark trying to hold on to the Newtonian conception of time being an absolute and looking for anomalies in, for example, the results of the Michelson and Morley experiment in order to hold on to that tradition.
     
    Last edited: Jul 6, 2013
  5. Jul 6, 2013 #4

    jtbell

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    Relativistic "velocity addition":

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

    Let v = 0.9c, and u' = c. What do you get for u?
     
  6. Jul 7, 2013 #5

    ghwellsjr

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    OK. First off, you need to be aware that no one can observe a light beam. Once you emit it, it travels away from you and it is gone unless it hits something and reflects part of the beam back to you. This is what happens when you shine a laser beam and dust particles in the air reflect portions of the beam so that you can see its path.

    Secondly, it would be very difficult to measure the speed of a continuous light beam so let's agree that what you really want to do is have Spaceship Alpha emit a very short burst of light.

    Now in order to measure the speed of this burst each observer must place a reflector some measured distance away and measure how long it takes for the burst to leave his vicinity, bounce off the reflector and return back to him. Then he can calculate the average or roundtrip speed of the burst by taking twice the measured distance and dividing it by the measured time interval.

    Now to make things simple, we will have the spaceship emit its burst of light at the exact moment that it passes the stationary observer, OK?

    One other important point: each observer must have his own reflector that is stationary with respect to himself, so the one burst of outgoing light will be traveling differently for the return trip back to each observer.

    We are using units of feet and nanoseconds and assuming that the speed of light is 1 foot per nanosecond.

    Hopefully, this is all very clear or clear enough that you can follow what happens using some spacetime diagrams representing your scenario in various Inertial Reference Frames (IRF's), starting with the one in which the stationary observer (in blue) is stationary and Spaceship Alpha (in black) is traveling at 0.9c. The stationary observer's mirror is depicted in red, three feet away from him. The Spaceship's mirror is depicted in green, three feet away from it. The Proper Times for each observer/object clocks is shown in its corresponding color. The observer's clocks are both synchronized to read zero at the Coordinate Time of zero in this IRF and their corresponding reflector's clocks are synchronized to read zero in their respective rest frames:

    attachment.php?attachmentid=60114&stc=1&d=1373176315.png

    Spaceship Alpha emits the burst of light at the moment that it passes the stationary observer when both their Proper Times are 0 (and the Coordinate Time is 0). This burst is depicted as a thin black line going upward and to the right at a 45-degree angle. It first hits the stationary observer's red reflector at its Proper Time of 3 (and the Coordinate Time of 3) and travels back to the stationary observer (depicted as a thin red line going upward to the left at a 45-degree angle) at his Proper Time of 6 (and the Coordinate Time of 6). Later it reflects off Spaceship Alpha's green reflector at its Proper Time of 3 and travels back to the spaceship (depicted as a thin green line going upward to the left at a 45-degree angle) at its Proper Time of 6.

    Now as far as either observer can tell, the measurement of the roundtrip speed of light is identical. It takes from their own Proper Time of 0 nanoseconds to 6 nanoseconds to travel 3 feet away to a reflector and back again which comes out to 1 foot per nanosecond for the speed of light.

    Now you may be wondering why the black and green lines seem distorted. If we go to the IRF in which Spaceship Alpha is stationary, we get the diagram:

    attachment.php?attachmentid=60115&stc=1&d=1373178144.png

    You should note that this diagram was created by using a speed of 0.6c on the Lorentz Transformation for all the events in the original diagram. Now you can see that Spaceship Alpha looks undistorted in its own rest frame and the original "stationary observer" is moving at 0.9c in the negative direction.

    I have not bothered to draw in the Proper Times for each observer/object but I think you can easily see that the same description applies to this IRF as the original one. Especially notice that the Proper Time of the green reflector (which I did not indicate in this drawing) is zero at the same time as Spaceship Alpha's Proper Time is zero (at the origin of the IRF).

    Now one last IRF in which none of the observers or objects are at rest. This IRF is moving at 0.627c with respect to the first one and results in both objects moving at 0.627c in opposite directions:

    attachment.php?attachmentid=60116&stc=1&d=1373178144.png

    There are two different issues going on here. First Einstein affirmed that the measured roundtrip speed of light was equal to the universal constant c and second, he postulated that the unmeasurable one-way speed of light in any Inertial Reference Frame is also equal to c. If you look at all three of the IRF diagrams that I drew, the light bursts travel along 45-degree angles.

    Since it's impossible, there is no meaningful answer.

    As I said before, he didn't come up with the idea that light speed was a constant. That was something that was already measured to be constant. Einstein postulated that the propagation of light in an IRF was at a speed of c which we can see by making drawings resulting in a simple and consistent way to depict a scenario.
     

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    Last edited: Jul 7, 2013
  7. Jul 10, 2013 #6
    Thank you a lot for the replies everyone! Very helpful!
     
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