(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Units problem with my Hamilton's equations

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