Units problem with my Hamilton's equations

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In summary, Homework Equations states that the new coordinates are Q(q,p) and P(q,p), and the new Hamiltonian is K=\frac{Q}{P}. The equations are solved for times t<P_0^2/2, and q(t) and p(t) are found for the initial conditions q(0)=0 and p(0)=1.
  • #1
fluidistic
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Homework Statement


Let the Hamiltonian with canonical variables be [itex]H(q,p)=\frac{\alpha ^3 e^{2\alpha q }}{p^3}[/itex] where alpha is a constant.
1)Given the generating function [itex]F(q,Q)=\frac{e^{2\alpha q }}{Q}[/itex], find the expression of the new coordinates in function of the old ones: [itex]Q(q,p)[/itex] and [itex]P(q,p)[/itex].
2)Find the expression of [itex]K(Q,P)[/itex] and the corresponding Hamiltonian equations.
3)With the initial conditions [itex]Q(t=0)=Q_0[/itex] and [itex]P(t=0)=P_0[/itex], solve these equations for times [itex]t<P_0^2/2[/itex].
4)Find [itex]q(t)[/itex] and [itex]p(t)[/itex] for the initial conditions [itex]q(0)=0[/itex] and [itex]p(0)=1[/itex].

Homework Equations


Lots of.

The Attempt at a Solution


1)I found out [itex]Q=\frac{\alpha e^{\alpha q}}{p}[/itex] and [itex]P=\frac{p^2}{\alpha ^2 e^{\alpha q}}[/itex].
2)The Hamiltonian in function of the new variables gave me [itex]K=\frac{Q}{P}[/itex]. This simple expression makes me feel I didn't make any mistake yet.
3)Hamilton equations gave me [itex]\dot P=-\frac{1}{P}[/itex] and [itex]\dot Q = -\frac{Q}{P^2}[/itex].
Solving the first equation gave me [itex]\frac{P^2}{2}=-t+\frac{P_0^2}{2}[/itex]. But... I am adding a time with a linear momentum squared ( kg times m /s )^2. How can this be right? Even in the problem statement, they write "[itex]t<P_0^2/2[/itex]", does this even make sense?
By the way I do not know how to answer to question 3. Can someone help me?
 
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  • #3
Liquidxlax said:
http://en.wikipedia.org/wiki/Generating_function_(physics)

you're are using the rules of F1 right?

it should fall out pretty easy from that

Yes I do, I actually solved part 1 and 2 (stuck on part 3).
Just a question... P and Q can in theory have any units? Because the problem statement compares time unit vs [itex]P^2[/itex] units. In other words, can P have units of [itex]\sqrt s[/itex]?
 
  • #4
fluidistic said:
Yes I do, I actually solved part 1 and 2 (stuck on part 3).
Just a question... P and Q can in theory have any units? Because the problem statement compares time unit vs [itex]P^2[/itex] units. In other words, can P have units of [itex]\sqrt s[/itex]?

I can't say for sure as I'm currently learning this as well, but i think you're right so long as the transformation is canonical

MJMT=J the units are negligible

I just had my midterm and 2 of the canonical transforms didn't have proper units yet i did get the questions right
 
  • #5
Ah ok thanks a lot.
Assuming what I found is right then for part 3) I find [itex]P(t)=\sqrt {-2t+P_0^2}[/itex] and [itex]Q(t)=e^{\frac{\ln (2t+P_0^2 )-\ln (P_0^2)}{2}+\ln Q_0}[/itex].
For 4), I assume they meant q(t) and p(t) as written and not Q(t) and P(t) that I just found. Otherwise the condition [itex]Q_0=0[/itex] would be a real problem.
 
  • #6
i figured since i can't explain it i'd actually do the problem and i did not get the same P as you

i got P = (p2e-2αq)/4α2

P = -dF/dQ = e2αq/Q2

maybe that is why you're not getting your desired units?

i'm not going to finish the problem because I've suffered enough with my homework and midterms lol
 
Last edited:
  • #7
My bad I made a typo when writing F here. It should be [itex]F(q,Q)=\frac{e^{\alpha q}}{Q}[/itex]. I do not see any other typo for now... sorry about that.
P.S.:No problem if you don't solve the problem. :) But now I'm convinced P and for that matter, p can have almost any possible units. Not necessarily the ones of linear momentum or so, as I previously thought.
 
  • #8
These are generalized coordinates, so p and P don't necessarily have to be linear momentum.
 
  • #9
ygolo said:
These are generalized coordinates, so p and P don't necessarily have to be linear momentum.

I see, thank you. Their name/letter mislead me.
 

1. What are Hamilton's equations?

Hamilton's equations are a set of equations used in classical mechanics to describe the evolution of a physical system over time. They are based on the principle of least action and are used to determine the trajectory of a system in phase space.

2. What is the units problem with Hamilton's equations?

The units problem with Hamilton's equations arises because the equations involve both position and momentum, which have different units. This can lead to confusion when trying to interpret the results of the equations in terms of physical quantities.

3. How can the units problem be solved?

The units problem can be solved by using a consistent system of units throughout the equations. This can be achieved by choosing a set of fundamental units and expressing all other units in terms of these fundamental units.

4. What are some common fundamental units used in Hamilton's equations?

Some common fundamental units used in Hamilton's equations include time, distance, mass, and energy. These units can be combined to derive units for other physical quantities, such as velocity, acceleration, and force.

5. Can the units problem affect the accuracy of the results obtained from Hamilton's equations?

Yes, the units problem can affect the accuracy of the results obtained from Hamilton's equations if the units are not consistent. This can lead to incorrect interpretations of the results and affect the overall validity of the calculations.

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