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How do I get the velocity out?

  1. Jun 18, 2009 #1
    Alright, im not the best at algebra. So i was wondering if anyone could show me how to get the V out on its own in the lorentz factor equation

    l= 1/[tex]\sqrt{1-(v^2/c^2)}[/tex]


    I tried to first times both sides by bottom then manipulate it but i just cant get it right.

    Any help would be much appreciated!!!!
     
  2. jcsd
  3. Jun 18, 2009 #2

    Doc Al

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    Staff: Mentor

    Not exactly sure what you're trying to accomplish, but you can take:
    γ = 1/√(1 - v²/c²)

    and solve for v in terms of γ.

    Cross-multiply and then square both sides.
     
  4. Jun 18, 2009 #3
    What, you mean like this...

    [tex]\gamma[/tex]([tex]\sqrt{1-v^2/c^2}[/tex])=1

    then: finally finishing with;

    [tex]\sqrt{(1-1^2/\gamma^2)c^2}[/tex]=v

    ??
     
    Last edited: Jun 18, 2009
  5. Jun 18, 2009 #4
    your question was 1/[tex]\sqrt{1-v^2/c^2}[/tex]=1 ?
     
  6. Jun 18, 2009 #5
    My question is how do i solve for "v", that l should be an L lower case. as in lorentz factor "L". not a 1. sorry

    Like what should the final equation be?
     
  7. Jun 18, 2009 #6

    this is wrong.
    return to the previous step square both sides, you dont have to multiply gamma into the square root.
     
  8. Jun 19, 2009 #7
    ah, right i get you now. Thanks!
     
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