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Fraction Algebra Rusty Need Help on Equation

  1. Nov 16, 2011 #1
    I was reading and came across a formula that didn’t look right so I tried to simplify the left side to match the right side; however, my fraction algebra is a little rusty and could use some guidance. I did get the 2 fractions to a common denominator; however, the best I could do was factor out an r. Here is the formula:

    [tex]
    E_n=\frac{n^2h^2}{2\mu r_n^2}- \frac{e^2}{4\pi\epsilon_0 r_n}=-\frac{e^4\mu}{8\epsilon^2_0 n^2h^2}
    [/tex]

    The equation can be found in context at the following link, which should take you to page 173, equation (5.40) is on page 172.

    http://books.google.com/books?id=Fn...esnum=21&ved=0CJQBEOgBMBQ#v=onepage&q&f=false

    Stochastic Simulations of Clusters: Quantum Methods in Flat and Curved Spaces By Emanuele Curotto
     
  2. jcsd
  3. Nov 16, 2011 #2

    eumyang

    User Avatar
    Homework Helper

    It looks like you're supposed to substitute rn from Eq. 5.39 into Eq. 5.40. However, I've tried it two different ways and I am not getting the RHS. I end up getting this:
    [tex]\frac{n^2h^2}{2\mu r_n^2}- \frac{e^2}{4\pi\epsilon_0 r_n}=-\frac{e^4\mu}{32\pi^2\epsilon^2_0 n^2h^2}[/tex]
    Note that I'm not familiar with the material in the book you referenced, so it is quite possible that there is something that I am missing.
     
  4. Nov 17, 2011 #3
    Inserting the values for r was helpful because it gave us more variables to work with. Here is how far I was able to take it. Close, but still off a little, (might as well be a kilometer)


    [tex]
    E_n=\frac{n^2h^2}{2\mu r_n^2}- \frac{e^2}{4\pi\epsilon_0 r_n}=-\frac{e^4\mu}{8\epsilon^2_0 n^2h^2}
    [/tex]

    [tex]
    r_n=\frac{4\pi \epsilon_0 n^2h^2}{\mu e^2}
    [/tex]



    Now I inserted the value for r


    [tex]
    E_n=\frac {\left(n^2h^2\right)\left(\mu e^2\right)^2}{\left(2\mu \right)\left(4\pi \epsilon_0 n^2h^2 \right)^2} - \frac{\left(e^2\right)\left(\mu e^2\right)}{\left(4\pi \epsilon_0\right)\left(4\pi \epsilon_0 n^2h^2\right)} =-\frac{e^4\mu }{8\epsilon^2_0 n^2h^2}
    [/tex]


    Now expanding the squared brackets



    [tex]
    E_n=\frac {n^2h^2 \mu^2 e^4}{2\mu 16 \pi^2 \epsilon_0^2 n^4h^4 } - \frac{e^4\mu }{4\pi \epsilon_0 4\pi \epsilon_0 n^2h^2} =-\frac{e^4\mu}{8\epsilon^2_0 n^2h^2}
    [/tex]



    Now combining variables


    [tex]
    \frac {\mu e^4}{32\pi^2 \epsilon_0^2 n^4h^4} - \frac{\mu e^4 }{16 \pi^2 \epsilon_0^2 n^2h^2}= -\frac {e^4\mu}{8\epsilon_0^2 n^2h^2}
    [/tex]
     
  5. Nov 17, 2011 #4

    Deveno

    User Avatar
    Science Advisor

    this looks correct. in the book you quoted, the formulae for En and rn are in terms of hbar, not h, which turns the 32π2hbar2 into 8h2.
     
  6. Nov 17, 2011 #5
    There was a typo in the denominator in the first term of the last equation in Bluestar's last post. That equation should look like the following:

    [tex]
    E_n=\frac {\mu e^4}{32\pi^2 \epsilon_0^2 n^2h^2} - \frac{\mu e^4 }{16 \pi^2 \epsilon_0^2 n^2h^2}= -\frac {e^4\mu}{8\epsilon_0^2 n^2h^2}
    [/tex]


    Next factor out all variables in the left side that are found in the right of the equal sign
    [tex]
    E_n=\frac {\mu e^4}{8 \epsilon_0^2 n^2h^2} \left( \frac{1 }{24 \pi^2 } -\frac {1}{8\pi^2} \right) = -\frac {e^4\mu}{8\epsilon_0^2 n^2h^2}
    [/tex]

    Inverse the factored out variables and move to right side.

    [tex]
    E_n=\frac{1 }{24 \pi^2 } -\frac {1}{8\pi^2} = -\frac {e^4\mu}{8\epsilon_0^2 n^2h^2}\frac {8 \epsilon_0^2 n^2h^2}{e^4\mu }
    [/tex]


    Reduce right side to 1
    [tex]
    E_n=\frac{1}{24\pi^2}=\frac{1}{8\pi^2}=1
    [/tex]


    Simplify equation
    [tex]
    8 = 24 ...or...
    1 = 3
    [/tex]

    It would appear the equation in the book, referenced earlier does not balance out.
     
  7. Nov 17, 2011 #6
    Hey Deveno, I went back and looked at both formulas with my laptop and still didn’t see the hbar. Could be the resolution of the laptop is not so great. When first working this problems I wondered if those were Planck’s constants or Dirac’s constants. Subsequent to your note I posted some follow-up work on the formula and found it does not balance. Even though Dirac’s constant should have been used it still does not change the imbalance in the numeric constants. But, check it out if you want and lets see.
     
  8. Nov 19, 2011 #7

    Deveno

    User Avatar
    Science Advisor

    wait a minute...you factor 32 as 8*24 and 16 as 8*8?

    you should have:

    [tex]\frac{1}{4\pi^2} - \frac{1}{2\pi^2}[/tex]

    on the left, at this point


    no, it reduces to negative 1
    what i get that you ought to have, is:

    [tex]\frac{-1}{4\pi^2} = -\frac{\hbar^2}{h^2}[/tex]

    which makes perfect sense.
     
  9. Nov 19, 2011 #8
    What a silly mistake. All I can say is that it was late, I was in a rush and didn’t double check my work which I always try to do. Anyway, what you got does make sense and does balance out.
    From you answer I see you used h and hbar. Thus, I am still working with a handicap because I cannot determine which terms are h and which terms are hbar in the original equation in the book. Deveno, would you kindly post the original formulas (5.40) and (5.39) so I can see which terms contain h and hbar. Please... Thanks
     
  10. Nov 21, 2011 #9

    Deveno

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    Science Advisor

    here is a magnified version of the equations in question (click on the thumbnail to see the larger version).
     

    Attached Files:

  11. Nov 21, 2011 #10
    Deveno, THANKS!
    Appreciate the help.
     
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