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I want a second opinion

  1. Oct 7, 2007 #1
    [SOLVED] I want a second opinion

    We know the equation

    [tex] F=-\frac{dV}{dr}[/tex]

    we want to find the integral from [tex]r_{0}[/tex] to [tex]r[/tex].

    I have seen someone doing this

    [tex]\int^{r}_{r_{0}}Fdr'=-\int^{r}_{r_{0}}\frac{dV}{dr}dr'[/tex]

    I am a mathematician and the way I was doing at the university was

    [tex] F=-\frac{dV}{dr}\RightarrowFdr=dV[/tex]

    and then I integrate

    [tex]\int^{r}_{r_{0}}Fdr=-\int^{}_{r_{0}}dV[/tex]

    Since the potential depends on r we can integrate. So I would like someone who knows the subject to tell me if the first way is correct since I know the second is correct. Its difficult for me to accept the introduction of another variable r' while we have the r itself.
     
  2. jcsd
  3. Oct 7, 2007 #2

    arildno

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    Well, the first one is incorrect; it should be:
    [tex]\int_{r_{0}}^{r}Fdr'=-\int_{r_{0}}^{r}\frac{dV}{dr'}dr'[/tex]

    Now, the second integral simply equals [tex]V(r)-V(r_{0})[/tex], thus, we have identity between the expressions:
    [tex]\int_{r_{0}}^{r}\frac{dV}{dr'}dr'=\int_{V(r_{0})}^{V(r)}dV[/tex]
     
  4. Oct 7, 2007 #3
    ok the first way is what a I saw, exactly the way I wrote it. With your correction makes sense. Thank you very much for your answer.
     
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