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Derivation in bohm's Quantum Theory

  1. Aug 17, 2012 #1

    A_B

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    halfway page 41 Bohm obtains for the action variable
    [tex]
    J = 2\int_{a(E)}^{b(E)} dq \sqrt{2m[E-V(q)]}
    [/tex]
    Then he obtians the partial derivative to E "by a well-known theorem of the calculus":

    [tex]
    \frac{\partial J}{\partial E} = 2\left\{ \sqrt{2m\left[E-V(q)\right]} \right\}_{q=b} \frac{\partial b}{\partial E} - 2\left\{ \sqrt{2m[E-V(q)]} \right\}_{q=a} \frac{\partial a}{\partial E} + 2 \int_a^b \sqrt{\frac{m}{2[E-V(q)]}} dq.
    [/tex]

    What is this "well-known theorem"?

    Thanks,
    A_B
     
  2. jcsd
  3. Aug 17, 2012 #2
  4. Aug 17, 2012 #3

    A_B

    User Avatar

    thanks! ...Never came across that one before...

    A_B
     
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