Finding the generator function for 4 variables

[LCSphysicist]

Homework Statement

Generating function

Relevant Equations

.

Prove directly that the transformation $$Q_{1} = q_{1}, P_{1} = p_{1} − 2p_{2}$$ $$Q_{2} = p_{2}, P_{2} = −2q_{1} − q_{2}$$ is canonical and find a generating functionSo the first part is easy and can be skipped here. I have some difficults regarding the second part, namely, the one that ask for the generator function.

My approach was to evaluate the object $\sum_i p_i dq_i - P_i dQ_i$, i have read at MG Calkin that if this object can be written as a total differential $dF$, this one is (can be) the generating function.

So, expandin the differential, i have got that $\sum_i p_i dq_i - P_i dQ_i = d(p_2q_2 + 2p_2q_1) \implies F = F(P_2,p_2) = -p_2P_2$.

What do you think? is it ok?

 

[Mark44]

Homework Statement:: Generating function
Relevant Equations:: .

Prove directly that the transformation $$Q_{1} = q_{1}, P_{1} = p_{1} − 2p_{2}$$ $$Q_{2} = p_{2}, P_{2} = −2q_{1} − q_{2}$$ is canonical and find a generating functionSo the first part is easy and can be skipped here. I have some difficults regarding the second part, namely, the one that ask for the generator function.

My approach was to evaluate the object $\sum_i p_i dq_i - P_i dQ_i$, i have read at MG Calkin that if this object can be written as a total differential $dF$, this one is (can be) the generating function.

So, expandin the differential, i have got that $\sum_i p_i dq_i - P_i dQ_i = d(p_2q_2 + 2p_2q_1) \implies F = F(P_2,p_2) = -p_2P_2$.

What do you think? is it ok?

It looks like you have omitted a lot of work.
Starting from $dF = \sum_i p_i dq_i - P_i dQ_i$, how did you arrive at $dF = d(p_2q_2 + 2p_2q_1)$ and then to $F(P_2,p_2) = -p_2P_2$?

 

[LCSphysicist]

It looks like you have omitted a lot of work.
Starting from $dF = \sum_i p_i dq_i - P_i dQ_i$, how did you arrive at $dF = d(p_2q_2 + 2p_2q_1)$ and then to $F(P_2,p_2) = -p_2P_2$?

Yes. Actually, i have just expressed all $dQ_i, dP_i$ in $dF = \sum_i p_i dq_i - P_i dQ_i$ in terms of $dq_i, dp_i$. Eventually some terms cut off and we can express the results as $dF = d(p_2q_2 + 2p_2q_1)$.
Now, $d(p_2q_2 + 2p_2q_1) = d(p_2(-P_2))$ from the summary, so i just concluded that $F(P_2,p_2) = -p_2P_2$

 

[Mark44]

Yes. Actually, i have just expressed all $dQ_i, dP_i$ in $dF = \sum_i p_i dq_i - P_i dQ_i$ in terms of $dq_i, dp_i$. Eventually some terms cut off and we can express the results as $dF = d(p_2q_2 + 2p_2q_1)$.

"Eventually" is what I'm asking you to elaborate on.

Now, $d(p_2q_2 + 2p_2q_1) = d(p_2(-P_2))$ from the summary, so i just concluded that $F(P_2,p_2) = -p_2P_2$