Fraction Algebra Rusty Need Help on Equation

In summary, the conversation discusses a formula found in a book on page 173, which involves the substitution of a value for rn in equation 5.40 on page 172. The formula is used to calculate E_n, which is given by the equation 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}. However, after trying two different approaches, the result obtained does not match the right hand side of the equation. The conversation also discusses the possible use of Planck's constant or
  • #1
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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
 
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  • #2
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.
 
  • #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]
 
  • #4
bluestar said:
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]

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.
 
  • #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.
 
  • #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 let's see.
 
  • #7
bluestar said:
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]

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

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]

you should have:

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

on the left, at this point


Reduce right side to 1

no, it reduces to negative 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.

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.
 
  • #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
 
  • #9
here is a magnified version of the equations in question (click on the thumbnail to see the larger version).
 

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  • #10
Deveno, THANKS!
Appreciate the help.
 

1. What is the purpose of using fraction algebra in equations?

Fraction algebra is used to solve equations that involve fractions. It allows you to manipulate and simplify fractions in order to find a solution to the equation.

2. How do I determine the common denominator in a fraction algebra equation?

The common denominator is the lowest number that is a multiple of all the denominators in the equation. To find it, you can list out the multiples of each denominator and find the smallest number that appears in all lists.

3. Can I cancel out fractions in a fraction algebra equation?

Yes, you can cancel out common factors in both the numerator and denominator of a fraction. This will simplify the equation and make it easier to solve.

4. How do I solve for a variable in a fraction algebra equation?

To solve for a variable, you need to isolate it on one side of the equation. This can be done by performing the same operation on both sides of the equation, or by using inverse operations to undo operations that are already present.

5. What is the order of operations in fraction algebra equations?

The order of operations in fraction algebra equations is the same as in regular algebra equations. First, solve any operations inside parentheses, then simplify any exponents, followed by multiplication and division from left to right, and finally addition and subtraction from left to right.

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