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How to prove that relative speed between a pair of Lorentz transformations is < c ?

  1. Jun 11, 2012 #1
    [ T'' ] = [ T ] [ T' ]

    [ T'' ] =

    [ ( γ'' ) , ( - γ'' β'' ) ]

    [ ( - γ'' β'' ) , ( γ'' ) ]

    γ'' = γ γ' ( 1 + β β' ) → no problem, proven to be b/w [ 1 , ∞ )

    β'' = ( β + β' ) / ( 1 + β β' ) → can't figure out how to prove it's b/w [ 0 , 1 )
     
  2. jcsd
  3. Jun 11, 2012 #2
    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    Each of the individual betas ought to be less than 1, right?
     
  4. Jun 11, 2012 #3
    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    Yes, for each β:

    0 ≤ β < 1
     
  5. Jun 11, 2012 #4
    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    So in the limiting case of each beta going to 1, what do you get for the final result?
     
  6. Jun 12, 2012 #5

    mfb

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    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    Hint: Assuming both β and β' are in [0,1), what happens to β'' if β (or β') increases?
    What about the limit β=β'=1?

    Can you combine both to a formal proof?
     
  7. Jun 12, 2012 #6
    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    Perhaps yet another way to go from here is to note that that looks awfully like the tangent addition formula ... and then making a substitution.
     
  8. Jun 12, 2012 #7
    Re: How to prove that relative speed between a pair of Lorentz transformations is < c

    I've figured it out.

    ( β + β' ) / ( 1 + β β' ) - 1 + 1

    [ ( β + β' ) - ( 1 + β β' ) ] / ( 1 + β β' ) + 1

    [ - ( 1 - β ) ( 1 - β' ) ] / ( 1 + β β' ) + 1

    1 ≤ ( 1 + β β' ) < 2

    0 < ( 1 - β ) ≤ 1

    0 < ( 1 - β ) / ( 1 + β β' ) ≤ 1

    0 < ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) ≤ 1

    -1 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) < 0

    0 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) + 1 < 1
     
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