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Simplifying square root

  1. Feb 22, 2016 #1
    relativistic 4.jpg
    How is it equal to v in the end?
    I'm sorry for asking such questions. But I'm just trying to understand
     
  2. jcsd
  3. Feb 22, 2016 #2

    fresh_42

    Staff: Mentor

    Because addition is associative and commutative and ##c^2≠0##. For short: just calculate the left hand side.
     
  4. Feb 22, 2016 #3

    DrClaude

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    Staff: Mentor

    Can you simplify the denominator?
     
  5. Feb 22, 2016 #4
    Dear god I must be blind
    v^2-v^2 = 0
    c^2 and c^2 cancels
    And all thats left is a rooted v^2 which cancels itself

    But look at this
    relativistic 5.jpg
    This is how the same solution is illustrated in another book.
    (1-u^2/c^2) cancels each other
    The u^2 subtracts each other in the denominator like in the previous sollution
    Which leaves a rooted u^2 / c^2
    How does that simplify?
     
  6. Feb 22, 2016 #5

    DrClaude

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    Staff: Mentor

    Look to the left of the square root sign...
     
  7. Feb 22, 2016 #6
    I can see the c root What does it mean?
     
  8. Feb 22, 2016 #7

    DrClaude

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    Staff: Mentor

    No, it's c times the root.
     
  9. Feb 22, 2016 #8
    So are you saying that the large root will cancel the squares of u^2 / c^2 making it like u / c * c/1
    and c and c cancels???
     
  10. Feb 22, 2016 #9

    DrClaude

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    Staff: Mentor

    Yes. Note that the author there takes only the positive root, while in the OP the two roots are kept.
     
  11. Feb 22, 2016 #10
    Thank you very much
     
  12. Feb 22, 2016 #11

    Mark44

    Staff: Mentor

    This is actually incorrect. On the right side it should be |v|, not ##\pm v##. In other words, the square root evaluates to a single nonnegative number, not two numbers.
     
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