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Prove The Following Obeys Hamilton's Equation...

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    05M0ndw.png

    2. Relevant equations
    Given in the above picture

    3. The attempt at a solution
    I have tried to rearrange the relationship between P and R to gain an expression for R, in terms of P. I subbed that into the expression for E and attempted to differentiate. I ended up with this expression.
    dnOEpuQ.png
    I can't see how this is going to lead to the relationship for V.. any help would be appreciated here because I'm not sure where I am going with this question
     
  2. jcsd
  3. Apr 18, 2017 #2

    stevendaryl

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    I think the expression for [itex]E[/itex] is wrong. I get the right answer for [itex]V[/itex] if

    [itex]E = \frac{\rho \kappa^2 R}{2} (ln(\frac{8R}{a_0}) - \frac{3}{2})[/itex]

    The E given doesn't work out, unit-wise. (It has the same units as momentum, while it should have the units of momentum times velocity)
     
  4. Apr 18, 2017 #3
    Oh dear..... this is really concerning because this is a problem that my lecturer gave my class from a textbook that he edited.
     
  5. Apr 18, 2017 #4

    George Jones

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    How are ##dE/dP## and ##dE/dR## related?

    I don't know about the typo, as I haven't done the calculation.
     
  6. Apr 18, 2017 #5
    Is it a chain rule? dE/dR * dR/dP=dE/dP?
     
  7. Apr 18, 2017 #6

    George Jones

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    Yes.

    How is dR/dP related to dP/dR?

    Then, put everything together.
     
  8. Apr 18, 2017 #7
    Okay so I have attempted to apply the chain rule, I have the following equations, it seems close but not quite there, am I along the right lines?
    8zfeuiY.png
     
  9. Apr 18, 2017 #8

    George Jones

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    What is

    $$\frac{d}{dx} \ln \left( ax \right)?$$
     
  10. Apr 18, 2017 #9

    stevendaryl

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    You're making a little mathematical mistake:

    [itex]\frac{d}{dR} ln(AR) = 1/R[/itex], not [itex]\frac{A}{R}[/itex]
     
  11. Apr 18, 2017 #10
    Ahhh yes!, so instead of -3/2 + a_0/8 it would be -3/2 + 1, hence the -1/2. Thank you, this is very helpful!
     
  12. Apr 22, 2017 #11
    Hiiiiiii :woot:
     
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