Units problem with my Hamilton's equations

[fluidistic]

Homework Statement

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

Homework Equations

Lots of.

The Attempt at a Solution

1)I found out $Q=\frac{\alpha e^{\alpha q}}{p}$ and $P=\frac{p^2}{\alpha ^2 e^{\alpha q}}$.
2)The Hamiltonian in function of the new variables gave me $K=\frac{Q}{P}$. This simple expression makes me feel I didn't make any mistake yet.
3)Hamilton equations gave me $\dot P=-\frac{1}{P}$ and $\dot Q = -\frac{Q}{P^2}$.
Solving the first equation gave me $\frac{P^2}{2}=-t+\frac{P_0^2}{2}$. 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 "$t<P_0^2/2$", does this even make sense?
By the way I do not know how to answer to question 3. Can someone help me?

 

[Liquidxlax]

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

 

[fluidistic]

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 $P^2$ units. In other words, can P have units of $\sqrt s$?

 

[Liquidxlax]

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 $P^2$ units. In other words, can P have units of $\sqrt s$?

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

MJM^T=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]

Ah ok thanks a lot.
Assuming what I found is right then for part 3) I find $P(t)=\sqrt {-2t+P_0^2}$ and $Q(t)=e^{\frac{\ln (2t+P_0^2 )-\ln (P_0^2)}{2}+\ln Q_0}$.
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 $Q_0=0$ would be a real problem.

 

[Liquidxlax]

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 = e^2αq/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]

My bad I made a typo when writing F here. It should be $F(q,Q)=\frac{e^{\alpha q}}{Q}$. 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]

These are generalized coordinates, so p and P don't necessarily have to be linear momentum.

 

[fluidistic]

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.