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Special relativity: flashes of light

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data
    In a frame of reference A lights are on the x axis at x = D and x = -D, where D = 0.6 x109. They flash simultaneously at t = 0.

    There's also a frame of reference A' moving at v = 0.8c.

    i) Where and when do the flashes happen in A'?

    ii) Therefore when would observers at the origins of A and A' see the light?

    2. Relevant equations


    3. The attempt at a solution
    i) Well, I have to add D to the location in x to find the location in x'. Then I have to add something else. But I'm pretty confused about what to add. As far as an observer in A' is concerned, I think the light has to travel x' = D + vt'? But I can't use that because I don't know t' or x', so that's probably wrong...
    In A, both flashes happen at t = 0.6 x109 / 3x108 = 2 seconds. So I have t' as well:
    t' = γt where γ is the Lorentz factor, so maybe I do have t' actually, I think it's 10/3.

    Essentially, my question for the first part is whether x' = D + vt' is right.

    And then the problem is that haven't I already worked out when the observer at A sees the light? It's at t =2. And for the observer at the origin of A', at t = 10/3?
     
  2. jcsd
  3. Nov 16, 2015 #2
    What does the Lorentz Transformation tell you about where and when the flashes occur in A'?

    Chet
     
  4. Nov 16, 2015 #3
    Using x' = γ(x - vt) , t=0, x' = -1 x 109m for the bulb at -D, and 1 x 109 for the bulb at D.

    And as to when they happen - that should be at two different times. Using t' = γ( t - ##\frac{vx}{c^2}##), and subbing in t = 0,
    t' = -2.67 s for the bulb at +D, and 2.67 s for the bulb at -D. I think.

    So if I'm in frame F', it looks like F is moving, doesn't it? And that means anything measured in F will appear shorter to me, and seem like it happened either later/ earlier, depending on the direction of movement?
     
  5. Nov 16, 2015 #4
    And about the second part: I'm pretty sure the time at the origin of A is easy to calculate. It's just t = D/c and is the same for both bulbs.

    As for at the origin of A', for the bulb at D: That person thinks the flash happened at t = 2.67, and that the light travelled for t = ##\frac{1 \times 10^9}{c}##- 2.67 s, but I'm really not 100% sure about subtracting 2.67 seconds! Should I be adding instead? Or just doing nothing? Does that take account of the fact that F' is moving towards the bulb?

    And for the bulb at -D: it travels for ##\frac{1 \times 10^9}{c}## + ##\frac{x'}{v}## + 2.67? Maybe? Struggling to get my head round this.
     
  6. Nov 16, 2015 #5
    Forget about the F frame of reference for now. In the F' frame of reference, the two flashes occur at -1 x 109m, + 2.67 s and 109, -2.67 s. Since they are traveling at the speed of light in F', at what value of t' do each of these flashes reach x' = 0?
     
  7. Nov 16, 2015 #6
    t' = ##\frac{1 \times 10^9}{c}- 2.67 ## for x' = D?

    ##\frac{1 \times 10^9}{c} + 2.67 + \frac{x'}{v}## for x = -D. Possibly? Not sure I've got the adding/ subtracting of 2.67 the right way round. Or that I'm accounting for the movement of the frame correctly!
     
  8. Nov 16, 2015 #7
    Your answer for -2.67 is correct. Since frame F' doesn't know that it is moving, your answer for +2.67 is not correct. Lost the x'/v.
     
  9. Nov 16, 2015 #8
    F' thinks the light bulb at -D is moving away from it though, doesn't it? At the v = 0.8c assigned to F' relative to F? Although by that logic I should also have subtracted x'/v for x' = D as well. That's seriously confusing.
     
  10. Nov 16, 2015 #9
    Or is the -2.67 term accounting for the movement?
     
  11. Nov 16, 2015 #10
    Once the flash occurs at the indicated location and time, it doesn't matter what the light bulb does next. It is the light flash that is traveling toward x' =0 at the speed of light; the light bulb has nothing to do with this.
     
  12. Nov 16, 2015 #11
    Oh, I get it!

    Thanks for wading through my messy first post and replying. I really appreciate it! :)
     
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