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Light move away from the emitter at 1.00 c ?

  1. Jul 8, 2009 #1
    if an object moving at .99 c emits a light, does that light move away from the emitter at 1.00 c ?
     
  2. jcsd
  3. Jul 8, 2009 #2
    Re: .99c

    Yes, because the light does not know that the source is in motion.
     
  4. Jul 8, 2009 #3
    Re: .99c

    if the emitter suddenly stops, does the light know it? If the light doesn`t know it, the light will be moving at c plus the previous speed of the emitter.
     
  5. Jul 8, 2009 #4

    Hootenanny

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    Re: .99c

    No it won't. A stationary observer would still measure the speed of light as c. The invariance of the speed of light is a fundamental concept in special relativity.
     
  6. Jul 8, 2009 #5
    Re: .99c

    The light will not ever be moving at c plus the speed of the emitter. Light rays travel at c. Every light ray in vacuum will be measured to be traveling at c by every inertial observer. The emitter measures the light ray to recede from him at c. The receiver measures the light ray to approach him at c. Any and all passers-by to this experiment measure the speed of the light ray as c. It makes no difference whatsoever what the various observers' velocities are relative to each other; they all, each and every one of them, measure the speed of that light ray as c. Not c + the velocity of the emitter, not c + the velocity of the receiver, not c + anything. C.

    That is what it means for the speed of light to be constant.
     
  7. Jul 9, 2009 #6

    Fredrik

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    Re: .99c

    In a way it will, but with a different definition of "plus". In SR you have to add up velocities like this:

    [tex]u\oplus v=\frac{u+v}{1+\frac{uv}{c^2}}[/tex]

    If u=0.99c and v=c, the result is

    [tex]u\oplus v=\frac{0.99c+c}{1+\frac{0.99c^2}{c^2}}=\frac{1.99c}{1.99}=c[/tex]

    What if u isn't 0.99c? Let's try it again, with u arbitrary and v=c:

    [tex]u\oplus v=\frac{u+c}{1+\frac{uc}{c^2}}=\frac{c\left(\frac u c+1\right)}{1+\frac{u}{c}}=c[/tex]
     
  8. Jul 9, 2009 #7
    Re: .99c

    Fredrik, what equation is that? Does it have a name I can google? :)
     
  9. Jul 9, 2009 #8

    malawi_glenn

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  10. Jul 9, 2009 #9

    Doc Al

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    Re: .99c

    That's the relativistic addition of velocity formula.
     
  11. Jul 9, 2009 #10
    Re: .99c

    Thank you. That's handy.
     
  12. Jul 13, 2009 #11
    Re: .99c

    Does the [tex]u \oplus v[/tex] means the velocity of the emitter relative to the observer plus the velocity of the photon relative to the emitter? If I could try your patience for one more stupid question, what's a plus sign with a circle around it? Just a sign for adding vectors or what?

    Thanks again.
     
  13. Jul 13, 2009 #12

    malawi_glenn

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    Re: .99c

    v is the velocity relative the observer for objectA and u is the velocity of the objectB relative objectA, always. And [itex]
    u \oplus v
    [/itex] means the velocity of object B as measured by the observer. [itex]
    \oplus
    [/itex] means "composition law for velocities under boosts"

    In the wiki article this [itex]
    u \oplus v
    [/itex] is just called "s".
     
    Last edited: Jul 13, 2009
  14. Jul 13, 2009 #13
    Re: .99c

    Woah! Back to the old drawing board. Thank you for the explanation. I though [tex]u \oplus v[/tex] or "s" was the velocity of objectB relative to the observer!
     
    Last edited: Jul 13, 2009
  15. Jul 13, 2009 #14

    malawi_glenn

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    Re: .99c

    here is really good explanation what is included http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

    forget about my last post, I was not focused on this thread, Sorry for that.

    I have changed it now so it should be correct now, I had many irons in the oven :-) (and maybe I was confused by the non-standard notation \oplus ) ;-)

    The point of the calculation given by Fredrik is that light always move with speed c for all observers.
     
    Last edited: Jul 13, 2009
  16. Jul 13, 2009 #15

    DrGreg

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    Re: .99c

    Sorry, that's confused.

    If A, B and C are 3 objects all moving along the same straight line
    • u is the velocity of B measured by A
    • v is the velocity of C measured by B
    • [itex]u \oplus v [/itex] is the velocity of C measured by A
    The symbol [itex]\oplus[/itex] is not standard notation. Some may use it but others don't. It's just a way of denoting a different way of "adding" velocities. I'd prefer to call it "composition" rather than "addition".
     
  17. Jul 13, 2009 #16

    malawi_glenn

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    Re: .99c

    yup I know, it was an error, I fixed it.

    regarding "composition" vs. "addition" I actually mentioned that [itex]
    \oplus
    [/itex] stands for "composition" right? why adding that?

    Cheers
     
  18. Jul 13, 2009 #17

    DrGreg

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    Re: .99c

    Sorry, my intentions weren't clear. That comment was really aimed at Pesto in answer to post #11.

    In case anyone reading this thread is confused, I was correcting an error in post #12, but then malawi_glenn corrected his own error at the same time, making my comment out of date by the time I posted it. But it's too late to undo that now without causing even further confusion..... :frown:
     
  19. Jul 13, 2009 #18

    malawi_glenn

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    Re: .99c

    it's ok, it do happens sometimes that a person is editing his posts meanwhile someone is pointing out the misstake they did - no big deal :biggrin:
     
  20. Jul 13, 2009 #19
    Re: .99c

    Thanks again. Fixed my post too :).
     
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