Prove The Following Obeys Hamilton's Equation....

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The discussion revolves around proving a relationship that adheres to Hamilton's equation, focusing on the expressions for E, P, and R. Participants express confusion over the correctness of the given expression for E, noting it does not match the expected units. There is a significant emphasis on applying the chain rule to relate derivatives dE/dP and dE/dR. A mathematical error in the differentiation of a logarithmic function is identified and corrected, which aids in clarifying the approach. Overall, the conversation highlights the challenges faced in deriving the necessary relationships and the collaborative effort to resolve them.
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Homework Statement


05M0ndw.png


Homework Equations


Given in the above picture

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
 
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CMJ96 said:

Homework Statement


View attachment 195547

Homework Equations


Given in the above picture

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.
View attachment 195548
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

I think the expression for E is wrong. I get the right answer for V if

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

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)
 
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stevendaryl said:
I think the expression for E is wrong. I get the right answer for V if

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

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)

Oh dear... this is really concerning because this is a problem that my lecturer gave my class from a textbook that he edited.
 
How are ##dE/dP## and ##dE/dR## related?

I don't know about the typo, as I haven't done the calculation.
 
George Jones said:
How are ##dE/dP## and ##dE/dR## related?

I don't know about the typo, as I haven't done the calculation.
Is it a chain rule? dE/dR * dR/dP=dE/dP?
 
Yes.

How is dR/dP related to dP/dR?

Then, put everything together.
 
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
 
What is

$$\frac{d}{dx} \ln \left( ax \right)?$$
 
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CMJ96 said:
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?
View attachment 195655

You're making a little mathematical mistake:

\frac{d}{dR} ln(AR) = 1/R, not \frac{A}{R}
 
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  • #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!
 
  • #11
Hiiiiiii :woot:
 
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