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Will we measure the velocity c?

  1. Aug 8, 2012 #1
    I had an experiment in my mind and I wonder if anyone could help me out with this, I'm just familiar with the basic concepts of relativity so all help is appreciated.

    Imagine the following case below in the image I have posted (it's possible this image will be gone as time moves forward).

    physicsforums1.jpg

    Imagine we are this stationary observer, we have 2 mirrors moving in one direction facing the reflective surface towards each other. They are both moving at the same velocity v=c/10 (the speed is just made up, made a simple number to use for calculation, in case it's needed).
    Light (or imagine a photon if you wish) is emitted from mirror1, this pulse of light is traveling towards mirror2 then bounces back to mirror1 and this process is repeated over and over again.

    I know that the speed of light is the same for all observers, but I'm just having problems accepting this in some cases and could anyone explain how this will sort out in this case I've painted above?
    1. Will we (the stationary observer) measure the speed of light of the photon?
    2. Will we (if we where mirror1 in one case, and mirror2 in another case) also measure the speed of light c, of the photon?
    If this is true, is it true for the photon going in both directions? Left and right on this image.

    I would say yes we would measure the speed of light of the photon in all cases no matter where we were placed in this situation. But I'm having problems with my basic knowledge of relativity to understand how - except accepting the fundamental statement that the speed of light is the same for all observers. (I know different speeds can't just be added on to each other like in the newtonian physics, the universe manage to solve this by changing time, right?)

    I hope you understand my explanation.

    Thanks in advance
    Robin Andersson
     
  2. jcsd
  3. Aug 8, 2012 #2
    Yes, you are correct in your presumption that both observers will measure the photon's velocity as c. In fact all observers will measure the photon's velocity as c.

    I can't really tell you why, and frankly no one really can. It is just the case that light is always moving at the same speed. What I can tell you, is that because of this curious phenomenon, you get things like length contraction and time dilation.

    I know this isn't a satisfactory answer, but the way I look at it is that we detect all this weird phenomenons (time dilation, strange velocity addition, length contraction....), and you can explain all these weird things in terms of one weird thing: the consistency of the speed of light.

    There is no good reason for why the universe does this, however.
     
  4. Aug 8, 2012 #3

    Nugatory

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    Staff: Mentor

    OK, let's try working through the math here. That won't explain WHY this happens, but it will explain HOW it happens and why it's all consistent and even makes a fair amount of sense.

    I'm going to make two changes in your scenario. First, I'm going to set v=.6c instead of v=.1c, just because it makes all the calculations come out more easily.

    Second, I'm going to say that the two mirrors are exactly one light-second (aka 186,000 miles, aka 300,000,000 meters, aka the distance that light travels in one second) apart, as measured by an observer riding along with the mirrors.

    Now, an observer sitting on the left hand mirror will see himself sitting on stationary mirror, with the second mirror at rest 186,000 in front of him. He also sees the guy in the green shirt with the funny hat (you've labeled this guy "stationary observer") zipping past him at .6c.

    The observer sitting on the mirror sees a photon leave at time zero. It reaches the remote mirror one second later (it would be convenient to have an observer with a synchronized clock on that mirror to record that event), is reflected, and gets back to the first mirror one second after that. So as far as our mirror observers are concerned, we have three events:
    1) Photon emitted at X=0 and T=0
    2) Photon reflected from distant mirror at X=1 and T=1
    3) Reflected photon gets back to first mirror at X=0 and T=2
    (When you're interpreting these X numbers, remember that we and the mirrors are at rest as far as we're concerned, so the left hand mirror is the left-hand mirror is stationary at X=0 and the right-hand mirror is stationary at X=1).

    Obviously and trivially, this observer on the mirror sees the photon traveling at speed c, one light-second per second. It traveled one light-second in the first second to get to the distant mirror, and it traveled one light-second in the second second to get back to the first mirror. speed=distance/time doesn't get any easier than this.

    Now let's look at things as viewed by the guy that you've labeled "stationary observer". Of course he's not stationary for us; he's moving to the left at speed v=.6c relative to the observer sitting on the mirror. The crucial equations here are the Lorentz Transformations, which relate our measurements of time and space to his:
    [tex]x' = \gamma(x-vt)[/tex][tex]t' = \gamma(t-vx)[/tex][tex]\gamma=\frac{1}{\sqrt{1-v^2}}[/tex](There should be some factors of c bouncing around in there as well, but I've left them out because when we're measuring time in seconds and distances in light-second, c is equal to 1, so these factors just clutter up the equations).

    So let's apply these formulas to see what the guy in the hat measures. v=-.6, and [itex]\gamma[/itex]=5/4=1.25 (and the wisdom of using .6c instead of .1c in the example should now be apparent).
    1) Photon emitted at X=0 and T=0; guy in hat sees x'=0 and t'=0
    2) Photon reflected from distant mirror at X=1 and T=1; guy in hat sees x'=2 and t'=2. That is, the guy in the hat saw the photon travel a distance of two light-seconds in two seconds to get to the distant mirror. This is a combination of two things: First, the mirror was moving away from him, and second he's measuring time and space differently. But, and this is the point of the whole exercise, when he calculates the speed of the photon (speed=distance/time) he gets the exact same answer as the guy on the mirror - one light-second per second, c.
    3) Reflected photon gets back to first mirror at X=0 and T=2; guy in hat sees this happening 2.5 seconds after the photon was first emitted, and he sees the first mirror at the point x'=1.5. So on the return trip the photon traveled a distance 2-1.5=.5, and it did in .5 seconds - again, the speed of the photon comes out right.

    It's worth taking the effort to draw a spacetime diagram, a graph with X on the horizontal axis and T on the vertical axis, and plotting the path of the photon between the two mirrors, just to see what it looks like to both observers.
     
  5. Aug 9, 2012 #4
    Thanks to both of you, you cleared up much fuzziness in my head!
     
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