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Very simple question on negative powers

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is stated in the attempt as a solution

    2. Relevant equations

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    3. The attempt at a solution

    Suppose you have

    [tex]n^{-1}[/tex] am I right in thinking this is just [tex]\frac{n}{1}[/tex] so for something like [tex]n^{-2}[/tex] would it be simply [tex]\frac{n}{2}[/tex]?
     
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  3. Sep 29, 2012 #2

    Dick

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    No. Why would you think that? n^(-1)=1/n, n^(-2)=1/n^2.
     
  4. Sep 29, 2012 #3
    Because in Dimensional analysis, we tend to use the notation ie. [tex]MTl^{-2}[/tex] which would mean that it is mass times time divided by length squared...


    mmm... has it got something to do with things like

    [tex]\epsilon \mu = c^{-2}[/tex]

    such that

    [tex]\sqrt{\epsilon \mu} = \frac{1}{c}[/tex]

    I could be totally off, mind my ignorance.
     
  5. Sep 29, 2012 #4

    Dick

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    I have no idea what you are confused about. The definition of n^(-k) where k>=0 is pretty simple. It's 1/n^k. Could you try and explain again?
     
  6. Sep 29, 2012 #5
    Let's try this again. Let us say, E is energy.

    What would be E^{-2} be?
     
  7. Sep 29, 2012 #6

    Dick

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    It would be 1/E^2. Since energy has units M*L^2/T^2, E^2 has units M^2*L^4/T^4. So 1/E^2 would have units T^4/(M^2*L^4). Wouldn't it?
     
  8. Sep 29, 2012 #7
    In fact let's do this another way. I was reading about the dirac operator

    [tex]D(\psi) = \hbar^2 R^{-2} \psi(x)[/tex]

    How does R^{-2} read to you? Just 1/R^2 like you said before?
     
  9. Sep 29, 2012 #8

    Ray Vickson

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    If R is a number, then YES, it is just 1/R2. If R is an operator, then NO, it means something different.

    RGV
     
  10. Sep 29, 2012 #9
    R is taken to be a radius or a length.
     
  11. Sep 29, 2012 #10

    HallsofIvy

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    Yes, "divided by length squared" which contradicts what you wrote! MTl2= MT/l2, Not "MTl/2".


    You are making this much too difficult- [itex]\epsilon\mu= \frac{1}{c^2}[/itex] which again contradicts what you said originally.
     
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