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Why are these 2 derivatives not treated the same?

  1. Nov 18, 2011 #1
    In equation (5.35) the constant drops out when the derivative with respect to r is taken. However, in equation (5.36) the constant does not drop out. Does anybody know why?

    Equation (5.35)
    [tex]
    F_L=-\frac {d}{dr}\left(\frac {n^2h^2}{2\mu r^2}\right)=\frac{n^2h^2}{\mu r^3}
    [/tex]


    Equation (5.36)
    [tex]
    F_C=-\frac{d}{dr}\left(-\frac {e^2 }{4\pi \epsilon_0 r}\right)=-\frac{e^2}{4\pi \epsilon_0 r^2}
    [/tex]



    The equation can be found in context at the following link, which should take you to page 173, equation (5.35) & (5.36) is found on page 172.

    http://books.google.com/books?id=FnQ...page&q&f=false [Broken]

    Stochastic Simulations of Clusters: Quantum Methods in Flat and Curved Spaces By Emanuele Curotto
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 18, 2011 #2
    Do you see why

    [tex] \frac {d}{dr} ( \frac {1} {r^2} ) = - \frac {2} {r^3} [/tex]

    and

    [tex] \frac {d}{dr} ( \frac {1} {r} ) = - \frac {1} {r^2} [/tex]
     
  4. Nov 18, 2011 #3
    Wow, it has been a long time since I have performed a derivative.

    Yep, I think I understand.
    Does any constant goes to 1. Then what ever power r is raised to becomes a constant and then the r is raised to the next higher (lower) power.

    Is this example correct?
    [tex]
    \frac {d}{dr} ( \frac {1} {5r^\left(-4\right)} ) = - \frac {1} {\left(1\right)4r^\left(-5\right)}
    [/tex]


    Athough in your example the r^2 in the denoninator put a 2 constant in the numberator. This I don't understand.
    [tex]
    \frac {d}{dr} ( \frac {1} {r^2} ) = - \frac {2} {r^3}
    [/tex]


    Although in the book this equation
    [tex]
    F_C=-\frac{d}{dr}\left(-\frac {e^2 }{4\pi \epsilon_0 r}\right)=-\frac{e^2}{4\pi \epsilon_0 r^2}
    [/tex]

    should be:


    [tex]
    F_C=-\frac{d}{dr}\left(-\frac {e^2 }{4\pi \epsilon_0 r}\right)=-\frac{e^2}{\left(1\right)\pi \epsilon_0 r^2}
    [/tex]
     
  5. Nov 18, 2011 #4
    i'm going to use quick reply instead of taking the time to tex it.

    this may help: look at the 1/r as being r^-1 instead. when you take the derivative of something simple like that, you take the exponent (-1) and multiply that to the front then subtract one on the exponent so (-1) becomes (-2).

    so d/dr of r^(-1) is -(1)r^(-2) aka -1/r^2

    now for the stuff above:

    just take all that negative e squared over 4 pi epsilon-0 out since they aren't being acted upon with calculus ... and you have all that times the derivative of 1/r ... and we know from above that the derivative of 1/r is -1/r^2 ... so when you take the negative of that (where Fc wrote -d/dr) you'll have positive 1/r^2 still multiplied by all the junk in front ... yielding what the book has.

    hope that helps ... go E&M ... and calculus!
     
  6. Nov 21, 2011 #5
    Thanks bpatrick, your tip to factor out the unaffected variables will be quite useful.
    I checked some other references also and everything is starting to come back and clear up; however, I have one persistent question about equation (5.36)
    Equation (5.36)

    [tex]
    F_C=-\frac{d}{dr}\left(-\frac {e^2 }{4\pi \epsilon_0 r}\right)=-\frac{e^2}{4\pi \epsilon_0 r^2}
    [/tex]

    I understand when evaluating the derivative all constants go to 1 thus the 4 in the denominator went to 1. So, what I don’t understand is where the new 4 came from in the denominator as shown in the book. Anyway, here is what I get.

    [tex]
    F_c = (-1) (-\frac{(e^2)}{(\pi \epsilon_0)})
    \frac{d}{dr} (\frac{1}{4r})
    = (-1) (-\frac {e^2 }{\pi \epsilon_0 }) (\frac{ 1}{ (1) (r^2)})
    = (\frac {e^2 }{\pi \epsilon_0 }) (\frac{1}{r^2})
    [/tex]

    So where did the 4 come from AND why did the sign change?
     
  7. Nov 21, 2011 #6
    You can factor out all constants too (done in original tex). In your case, 1/4.

    Fc = (-1)[-e2/(4πε0)] d/dr (1/r)

    The sign should have changed because you can factor a negative one from (-1)[-e2/(4πε0)] = (-1)(-1)[e2/(4πε0)] = [e2/(4πε0)]
     
  8. Nov 21, 2011 #7
    Did not know about factoring out the constants; however, it presents a problem.

    It doesn’t seem right that the derivative can have 2 solutions.
    This:
    [tex]
    F_C=-\left(-\frac {e^2 }{4\pi \epsilon_0 }\right) \frac{d}{dr} \frac{-1}{r}
    =-\frac{e^2}{4\pi \epsilon_0 r^2}
    [/tex]

    Or this:
    [tex]
    F_C=- \left(-\frac{e^2}{\pi \epsilon_0 } \right) \frac{d}{dr} \frac{-1}{4r} =-\frac{e^2}{\pi \epsilon_0 r^2}
    [/tex]
     
  9. Nov 22, 2011 #8
    What?! [tex]\frac{d}{dr}[-r^{-1}]=r^{-2}[/tex]
    And [tex]\frac{d}{dr}[-\frac{1}{4}r^{-1}]=\frac{1}{4}r^{-2}[/tex]
    You leave the coefficients alone when you differentiate or integrate.
     
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