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Potentital integral

  1. Mar 4, 2006 #1
    I have a force and need to find the potential, but I'm stuch at a point. Any ideas how to integrate this: [tex] \frac{(r')^2 - 2 (r'') r}{r^2}[/tex]
     
  2. jcsd
  3. Mar 4, 2006 #2
    what are r' and r''?
     
  4. Mar 4, 2006 #3

    benorin

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    Likely unknown functions of an independent variable, say t,

    something close is

    [tex]\left( -\frac{r^{\prime}}{r}\right) ^{\prime} = \frac{(r^{\prime})^2-r^{\prime\prime}r}{r^2} [/tex]

    but this leaves the term [tex]-\frac{r^{\prime\prime}}{r}[/tex] unaccounted for

    also close is

    [tex]\left( -\frac{(r^{\prime})^2}{r}\right) ^{\prime} = r^{\prime}\frac{(r^{\prime})^2-2r^{\prime\prime}r}{r^2} [/tex]

    but has an extra factor of [tex]r^{\prime}[/tex] in it...

    Are you sure that is the term to be integrated?
     
  5. Mar 4, 2006 #4
    Mathematicians...

    Unfortunately, I'm sure this is the force to be integrated. It's a part of an exercise from Classical Mechanics, Goldstein, p32.
     
  6. Mar 8, 2006 #5

    samalkhaiat

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    [
    [/QUOTE]

    This is a scalar quantity! Force needs to be represented by a vector field.
    You are missing something.

    regards

    sam
     
  7. Mar 14, 2006 #6
    You can put a unit vector [tex]e_r[/tex] if you like. But it's so obvious that if we're talking about force, we're talking about a vector. And if you want a more precise definition, r is the distance of the particle to the center of force.

    Instead of being so pedantic and saying "hey question wrong, question is incomplete, etc", will someone give a correct answer? Such posts have no use.
     
    Last edited: Mar 14, 2006
  8. Mar 14, 2006 #7

    George Jones

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    I think it works out better if you split the given in the question as

    [tex] \frac{1}{r^2} \left( 1 + \frac{\left( r' \right)^{2}}{c^2} \right) - \frac{2}{c^2} \left( \frac{\left( r' \right)^{2}}{r^2} - \frac{r''}{r} \right)[/tex]

    However, I am quite tired right now, so I may have made mistakes while doing the problem

    Regards,
    George
     
  9. Mar 15, 2006 #8

    samalkhaiat

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    OK, here is a useful post. The potential you are after is;
    [tex]V= \frac{\dot{r}^2}{2r}[/tex]
    and you donot have to carry out any integration**. Just try to write your force in the form;

    [tex]F=-\frac{\partial{V}}{\partial{r}} + \frac{d}{dt}(\frac{\partial{V}}{\partial{\dot{r}}})[/tex]

    **Two months ago, I created a thread here called "Integrate This". The purpose of the thread was to share with people some of the tricks that I use in solving complicated integrals without going through a messy process of integration. Unfortunately, the thread was locked (I believe) for psychological reasons:biggrin: .Have a look at that thread, you may find some useful mathematical tricks in there.

    regards

    sam
     
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