- #1

fluidistic

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## Homework Statement

I'm trying to find a generating function for the canonical transformation [itex]Q=\left ( \frac{\sin p}{q} \right )[/itex], [itex]P=q \cot p[/itex].

## Homework Equations

I am not really sure. I know there are 4 different types of generating function. I guess it's totally up to me to choose the type of the generating function I want, since it's not specified in the problem statement?

So let's say I want a type 1. Thus F_1 depends on q and Q.

[itex]\frac{\partial F_1}{\partial q}=p[/itex]

[itex]\frac{\partial F_1}{\partial Q}=-P[/itex].

## The Attempt at a Solution

From the relevant equations, [itex]\partial F_1=-P \partial Q[/itex] and [itex]\partial F_1=p \partial q[/itex].

I don't really know why, but apparently [itex]dF_1=pdq-PdQ[/itex]. It looks like [itex]dF_1=2 \partial F_1[/itex] for some reason unknown to me.

Now I think the main idea to get [itex]F_1[/itex] is to express [itex]dF_1[/itex] as a single differential and then integrate.

I have that [itex]p= \cot ^{-1} \left ( \frac{P}{q} \right )[/itex] and [itex]dQ=\frac{1}{\sin p } \left [ \frac{\cos p}{q}dp-\frac{\sin p}{q^2}dq \right ][/itex].

This gives me [itex]dF_1=(p+ \cot p)dq-q\cot ^2 p dp[/itex]. In a book I found that this is equal to [itex]d[q(p+\cot p)][/itex] (how?!), therefore [itex]F_1=q \cot ^{-1} \left ( \frac{P}{q} \right )+P[/itex]. But it seems that F depends on P and q instead of on q and Q. So I "found" a type 2 generating function instead of type 1... I think I know why, I found p instead of q. Anyway my problems remains.

1)Why is [itex]dF=pdq-PdQ[/itex]?

2)Why and how do you figure out that [itex](p+ \cot p)dq-q\cot ^2 p dp=d[q(p+\cot p)][/itex]?

I'd be very grateful if someone could answer these 2 questions.

Thanks!

Edit: I just answered the "why" of my question 2. I just don't know how do one figures this out.

Edit 2: I do not know why I found out a type 2, namely F(P,q) while I was looking for F(q,Q).

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