Proving a Hamiltonian satisfies Hamilton's equations

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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 [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)
 
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stevendaryl said:
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)

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:

[itex]\frac{d}{dR} ln(AR) = 1/R[/itex], not [itex]\frac{A}{R}[/itex]
 
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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!