Can you clarify the question?What is the meaning of independent in this context?

[homology]

I'm working my way through Jose and Saletan's mechanics text and I'm at the end of chapter 5 which introduces Hamiltonian dynamics. I've just finished reading about 'types' of generating functions.

They work through an example (5.5) with the following transformation

$$Q=\frac{m\omega q +ip}{\sqrt{2m\omega}},\mbox{ } P=i\frac{m\omega q - ip}{\sqrt{2m\omega}}$$

for the Hamiltonian $$H=0.5m\omega^2 q^2+p^2/2m$$ (yeah this whole thing looks like quantum, but we're classical here).

They claim that its generating function is of Type 1, meaning that q and Q are independent of one another. However I'm bit perplexed on how the authors are using 'independent' in this context. Choosing a value for q seems to greatly restrict values for Q so they don't seem independent.

Can anyone clarify this matter? Thanks in advance.

 

[Bill_K]

I don't have the book, but I think what he means is, you *may* use a type 1 generating function, F₁(q, Q). If q and Q were uniquely related, Q = f(q), you couldn't do this. Instead you have Q = f(q, p). Given q, you don't know what Q is (without also specifying p) so in that sense they're independent.

 

[homology]

That was my first thought, but since Q is complex, choosing q amounts to restricting Q along a vertical line in the complex plane.

 

[vanhees71]

If you generating function of the canonical transformation is $$f(q,Q)$$ then

$$p=\frac{\partial f}{\partial q}, \quad P=-\frac{\partial f}{\partial Q}$$.

To find $$f$$ for the given transformation (why one should do this is, however, another question, since you have the transformation already in explicit form); to see whether it's really canonical you only need to prove the canonical Poisson-bracket relations for the new variables), you first need to solve for $$p$$ and $$P$$ in terms of $$q$$ and $$Q$$, and then integrate the partial differential equations to get finally $$f$$.

 

[homology]

that's true but not what I asked :)

 

[vanhees71]

Then I don't understand your question. Since the mapping $$(q,p) \mapsto (Q,P)$$ is one-to-one, any pair of variables can be chosen as to be "the independent variables".

 

[homology]

So, in particular, for the mapping I posted, how does this work? q and Q don't look independent. Could you be more explicit? Choosing q restricts Q, they are not independent, so it seems to me.

Thanks :)

 

[vanhees71]

You just take the two equations, defining the transformation and solve for $$p$$ and $$P$$ in terms of $$q$$ and $$Q$$. This is uniquely possible, and thus you can take $$q$$ and $$Q$$ as independent variables. The result reads

$$p=\mathrm{i} (m \omega q-\sqrt{2 m \omega} Q), \quad P=\mathrm{i}(\sqrt{2 m \omega} q-Q)$$.

Solving the set of partial-differential equations for $$f$$, given by the generator equations, leads to

$$f(q,Q)=\frac{\mathrm{i}}{2} (m \omega q^2-2 \sqrt{2 m \omega} q Q+Q^2)$$.

 

[fzero]

For a canonical transformation,

$$\frac{dF}{dt} = P \dot{Q} - \mathcal{H}(P,Q) - \left( p\dot{q} - H(p,q) \right). ~~(*)$$

If we compute the RHS of this transformation, it will be of the form

$$\frac{\partial F}{\partial t} + \frac{\partial F}{\partial S} \dot{S} + \frac{\partial F}{\partial s} \dot{s},$$

where $$S$$ is either of $$Q,P$$ and $$s$$ is either of $$q,p$$. The pair $$s,S$$ are what we call the independent variables. To determine the type of transformation we're dealing with, we need to actually compute the RHS of (*).

 

[homology]

Thanks for your responses so far, but how am I to tell by looking at it. The initial transformation doesn't look like q and Q are independent.

Also, what about the fact that, at least initially, it looks like the choice of q restricts the choice of Q?

But again, thanks for the responses so far. I'm still just trying to get a grip on what 'independent' means here.

 

[fzero]

For this system and transformation, you can explicitly compute that

$$\frac{dF}{dt} = p \dot{q} - P \dot{Q} .$$

We infer from this that $$F=F(q,Q)$$ with $$q$$ and $$Q$$ treated as independent variables, otherwise $$\partial Q/\partial q\neq 0$$ would mean that either $$\dot{q}$$ or $$\dot{Q}$$ would not appear in the total derivative above.

 

[homology]

For this system and transformation, you can explicitly compute that

$$\frac{dF}{dt} = p \dot{q} - P \dot{Q} .$$

We infer from this that $$F=F(q,Q)$$ with $$q$$ and $$Q$$ treated as independent variables, otherwise $$\partial Q/\partial q\neq 0$$ would mean that either $$\dot{q}$$ or $$\dot{Q}$$ would not appear in the total derivative above.

Okay I think I see where you're going with this. I've been focusing on the transformation where $$\partial Q/\partial q \neq 0$$ which throws me as it seems to indicate an explicit dependency on one another.