Homework Help: Prove The Following Obeys Hamilton's Equation...

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1. Apr 18, 2017

CMJ96

1. The problem statement, all variables and given/known data

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.

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. Apr 18, 2017

stevendaryl

Staff Emeritus
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)

3. Apr 18, 2017

CMJ96

Oh dear..... this is really concerning because this is a problem that my lecturer gave my class from a textbook that he edited.

4. Apr 18, 2017

George Jones

Staff Emeritus
How are $dE/dP$ and $dE/dR$ related?

I don't know about the typo, as I haven't done the calculation.

5. Apr 18, 2017

CMJ96

Is it a chain rule? dE/dR * dR/dP=dE/dP?

6. Apr 18, 2017

George Jones

Staff Emeritus
Yes.

How is dR/dP related to dP/dR?

Then, put everything together.

7. Apr 18, 2017

CMJ96

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?

8. Apr 18, 2017

George Jones

Staff Emeritus
What is

$$\frac{d}{dx} \ln \left( ax \right)?$$

9. Apr 18, 2017

stevendaryl

Staff Emeritus
You're making a little mathematical mistake:

$\frac{d}{dR} ln(AR) = 1/R$, not $\frac{A}{R}$

10. Apr 18, 2017

CMJ96

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. Apr 22, 2017

Hiiiiiii