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FTL Question

  1. May 5, 2004 #1
    If this equation is true:
    V=(V1+V2)/(1+(V1*V2)/c^2)

    Then why is it when you plug in c^2 for both V1 and V2 you get the total velocity as 2m/s
     
  2. jcsd
  3. May 5, 2004 #2

    chroot

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    You don't.. you get c.

    [tex]V = \frac{2 c}{1 + c^2/c^2} = \frac{2 c}{2} = c[/tex]

    - Warren
     
  4. May 6, 2004 #3

    turin

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    This will give a nonsensical result. You must plug in speed, not squared speed.
     
  5. May 11, 2004 #4
    What I meant to say was (3*10^8)^2
     
  6. May 11, 2004 #5

    arildno

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    It's still meaningless, besides wrong arithmetic:
    Putting your value into slots for V1 and V2 yields:
    [tex]
    \frac{2*(3*10^{8})^{2}}{1+(3*10^{8})^{2}}
    [/tex]

    To clarify, and make a more "accurate" argument:
    Let the "velocities" be some big, ugly number on the form: V=kc, k>>1
    Then you have by plugging in:
    [tex]\frac{2V}{1+(\frac{V}{c})^{2}}\approx\frac{2c}{k}[/tex]
    In your case, k=c; hence the approximate value of 2.
     
    Last edited: May 11, 2004
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