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Can someone give me a hand with large root here

  1. Feb 22, 2016 #1
    relativistic.jpg
    i get the differentiation
    the halves cancel the 2 the h bar cancels the h bar square
    and to get rid of the root in the denominator the entire thing is squared
    But I cannot understand where the huge square root came from at the end where I circled.
    Can someone help me here
     
  2. jcsd
  3. Feb 22, 2016 #2

    fresh_42

    Staff: Mentor

    ##a = \sqrt{a^2}## for positive ##a## and ##\frac{a}{\sqrt{b}} = \frac{\sqrt{a^2}}{\sqrt{b}} = \sqrt{\frac{a^2}{b}}##
     
  4. Feb 22, 2016 #3
    Ok i understand what you are saying.
    But what if we square the entire thing
    relativistic 2.jpg
    And the root and the square gets cancelled like this
    relativistic 3.jpg
     
  5. Feb 22, 2016 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Exactly. If
    $$
    \left( \frac{pc^2}{\sqrt{p^2c^2 + m^2 c^4}} \right)^2 = \frac{p^2c^4}{p^2c^2 + m^2 c^4}
    $$
    then
    $$
    \frac{pc^2}{\sqrt{p^2c^2 + m^2 c^4}} = \sqrt{\frac{p^2c^4}{p^2c^2 + m^2 c^4}}
    $$
    which is what you have in original text.
     
  6. Feb 22, 2016 #5

    fresh_42

    Staff: Mentor

    You don't square the entire quotient. Set ##a=pc^2## and ##b = p^2c^2+p^2m^4##.
     
  7. Feb 22, 2016 #6
    Ok if one can get rid of the huge square root why keep it.
    I mean as you said in the first part, why is the solution choosing to keep the huge root if we can choose to remove it
     
  8. Feb 22, 2016 #7

    DrClaude

    User Avatar

    Staff: Mentor

    It is a question of preference. Some things can be easier to see if all the terms are on the same footing (e.g., all under the square root), but where to stop simplifying can be a matter of taste.
     
  9. Feb 22, 2016 #8
    Thank you very much
    I'll try it both ways once I get a hang of things
    Thank you very very much
     
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