# Canonical transformation such that Hamiltonian of a freely falling body becomes $H'(P,Q)=P$

Can someone please help me with this problem I am unable to find a suitable generating function?

The question says:

To find a canonical transformation such that Hamiltonian of a freely falling body with one degree of freedom becomes, $$H'(P,Q)=P$$.

The hint given with this was that, we need to use the fourth form of generating function $$F_4(P,p,t)$$ to find the transformed coordinates- $$Q_i(p,q,t)$$ and $$P_i(p,q,t)$$.

Here's how I proceeded- For freely falling body hamiltonian is given by

1. $$H=p^2/2m+mgq$$

Since we are finding $$F_4$$, I used the following equation to get $$\frac{dF_4}{dt}$$-

1. -$$\sum{\dot{p}_ iq_i} -H = -\sum{Q_i\dot{P_i}}- H' + \frac{dF_4}{dt}$$

Comparing the coefficients with the total time derivative of $$F_4(P,p,t)$$, I found

1. $$\frac{\partial F_4}{\partial t} = P-\frac{p^2}{2m}-mgq$$

After that I don't understand if I solve for $$F_4$$ integrating 3 and taking all other variables independent of time, the resulting generating function does not give the required transformed Hamiltonian. Can someone please help with what to do after step 3? Have I proceeded correctly?

• For homework-type questions it would be appropriate to show what you have tried and a specific issue that may be the root of your confusion. Dec 28, 2020 at 15:40
• Actually, I tried but I don't know whether I approached correctly or not. Dec 28, 2020 at 15:43
• @kaylimekay I reached up to finding the partial derivative of F_4 with respect to time, after that my thoughts froze because it wasn't giving any result. Dec 28, 2020 at 15:48
• please see here for guidelines on homework questions. Dec 28, 2020 at 15:56
• Then I will mention steps up to where I reached. Thanks. Dec 28, 2020 at 16:03

I haven't thought about canonical transformations in a while so I thought it would be fun to give some hopefully constructive comments on this question. Note that we can use simple physical reasoning to determine the relationship between the new and old coordinates $$(p,q)\leftrightarrow(P,Q)$$.

We are told that the new Hamiltonian $$H'$$ is only a function of $$P$$. This means that Hamilton's equation for $$P$$ is trivial: $$\dot P=0$$. In other words, $$P$$ is a conserved quantity. On the other hand, since $$H'$$ only depends linearly on $$P$$, Hamilton's equation for $$X$$ is just $$\dot X=1$$. In other words, $$X$$ is a quantity that is changing at a constant rate.

My questions for you are then:

• What conserved quantity does the falling body have?
• What quantity describing the falling body is changing at a constant rate?

You probably now know what the new coordinates are and how to express them in terms of the originals. From there, can you integrate to find $$F_4(p,P,t)$$? Feel free to edit this answer or post your own later if you have learned anything interesting you'd like to share.

• A falling body will have total energy conerved. And the change in momentum here $\dot{p}$ is describing the falling body's velocity is changing at a constant rate. Sure, I think I will clarify this question from my professor the make edits. Dec 30, 2020 at 11:26

The Hamiltonian of the vertically free-falling mass-point has one degree of freedom and it looks like this: $$H(q, p) \, = \, \frac{1}{2m}\, p^2 \, + \, mg\, q$$ Since it is a conservative, time-independent Hamiltonian, one should look for a generating function of a canonical transformation that is also time-independent. Since you want to change the canonical variables to new canonical variables in which the Hamiltonian is simply $$\tilde{H}(Q, P) = P$$, your generating function should have $$P$$ as one variable. Hence you could look for a function of the form $$S(q, P)$$ but since the variable $$q$$ is linear in the original Hamiltonian, it is going to be very easy to solve the Hamilton-Jacobi equation for a generating function that actually looks like $$S(p, P)$$ and canonical transformation $$q \, = \, \frac{\partial S}{\partial p}(p, P)$$ $$Q \, = \, \frac{\partial S}{\partial P}(p, P)$$ Then, the Hamilton-Jacobi equation for a function generating a desired canonical change of variables is $$H\left(\, \frac{\partial S}{\partial p}(q, P)\,, \,\, p \,\right) \, = \, P$$ or explicitly $$\frac{1}{2m}\, p^2 \, + \, mg\, \frac{\partial S}{\partial p} \, = \, P$$ Well, treating $$P$$ as a constant, rewrite the equation as $$\frac{\partial S}{\partial p} \, = \, \frac{P}{mg} \, - \, \frac{1}{2m^2g}\, p^2$$ and integrate with respect to $$p$$ $$S(p,\, P) \, = \, \int \, \left( \, \frac{P}{mg} \, - \, \frac{1}{2m^2g}\, p^2 \, \right) dp$$ Your job is now to integrate this expression, treating $$P$$ as a constant, and then write down explicitly the canonical change of variables from $$(Q, P)$$ to $$(q, p)$$ and check if the original Hamiltonian reduces to the target one under this change of variables.

• Yes, the hamiltonian transformed to $P$ from this method. It worked!! Dec 31, 2020 at 5:44