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Units problem with my Hamilton's equations

by fluidistic
Tags: equations, hamilton, units
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fluidistic
#1
Mar5-12, 07:10 PM
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P: 3,188
1. The problem statement, all variables and given/known data
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].
2. Relevant equations
Lots of.


3. 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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Liquidxlax
#2
Mar6-12, 01:14 PM
P: 321
http://en.wikipedia.org/wiki/Generat...tion_(physics)

you're are using the rules of F1 right?

it should fall out pretty easy from that
fluidistic
#3
Mar6-12, 01:26 PM
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Quote Quote by Liquidxlax View Post
http://en.wikipedia.org/wiki/Generat...tion_(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]?

Liquidxlax
#4
Mar6-12, 01:44 PM
P: 321
Units problem with my Hamilton's equations

Quote Quote by fluidistic View Post
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
fluidistic
#5
Mar6-12, 02:08 PM
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P: 3,188
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.
Liquidxlax
#6
Mar6-12, 02:17 PM
P: 321
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
fluidistic
#7
Mar6-12, 02:25 PM
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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.
ygolo
#8
Mar6-12, 07:04 PM
P: 30
These are generalized coordinates, so p and P don't necessarily have to be linear momentum.
fluidistic
#9
Mar6-12, 08:17 PM
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Quote Quote by ygolo View Post
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.


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