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When to square a sum and when not to

  1. Feb 14, 2010 #1
    Hey, I've got a question which might be really simple, i'm not too sure yet!

    Basically, I'm going over the compton scattering calculations, and there's a part where:

    v - v' + (mec2)/h = [tex]\sqrt{something else}[/tex]

    Basically, the next step is to square both sides of the equation.

    To do this, my lecturer squares the LHS as a sum (i.e. in the form (a+b)2 instead of doing a2 + b2... where a = (v - v') and b = (mec2)/h).

    My question is... why do you in some instances take the square of the sum, and in other instances take the square of the individual components. I'm assuming there is a non-arbitary reason for this.. but I don't know about it!

    Any help or pointers in the right direction would be much appreciated.

    Cheers,
    Andrew
     
  2. jcsd
  3. Feb 14, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hey Andrew! Welcome to PF! :smile:

    (have a square-root: √ :wink:)
    You always take the square of the sum. :smile:

    (Though there are a few cases where that is the same as taking the sum of the squares, for example if they are perpendicular components of vectors :wink:)
     
  4. Feb 14, 2010 #3
    Cheers for the welcome, and for the answer!
     
  5. Feb 14, 2010 #4
    If you got
    [tex]x = \sqrt{y}[/tex]
    and you square it to
    [tex]x^2 = y[/tex]

    If x happens to be a sum
    [tex]x = a + b[/tex]
    then you get
    [tex](a + b)^2 = y[/tex]
    and not
    [tex]a^2 + b^2 = y[/tex]

    The reason is simple. In general,
    [tex]a^2 + b^2 \neq (a + b)^2[/tex]
    because there's also the crossterm:
    [tex](a + b)^2 = a^2 + b^2 + 2ab[/tex]
     
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