Just plain dont know what to do or how to even start

[Phymath]

Homework Statement

Well I've been working on these problems for the last 4 hours and still nothing...and out of despairation i turn to you all.

It has to do with Canonical transformations of the Hamiltonian:
1) Consider a type 1 generating function (F(q,Q,t)) where the following must be satisfied
$$p = \frac{\partial{F}}{\partial{q}} \ P = -\frac{\partial{F}}{\partial{Q}} \$$
show that for a single degree of freedom the possion bracket [Q,P] = 1 (aka canonical) now I've been able to show it is 0 obviously wrong and yes it makes use of many different forms of differentials any help on this would help

2) For Two particles interact via a central potential V(r1-r2) the H is
$$H= \frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+V(r_1-r_2)$$

$$Q = \frac{m_1 r_1+m_2 r_2}{M = m_1+m_2}, P = p_1+p_2$$

$$q= r_1 - r_2, \ p = \frac{m_2 p_1-m_1 p_2}{M}$$

this one I am really not sure about, however i think the transformation is type 3 so that may help doing that what does anyone think? any help is always awesome.show that the transformation is canonical

Homework Equations

$$[Q,P] = \frac{\partial{Q}}{\partial{q}}\frac{\partial{P}}{\partial{p}}-\frac{\partial{P}}{\partial{q}}\frac{\partial{Q}}{\partial{p}}$$

 

[Jhero]

The first problem involves canonical transformations and the second problem involves showing that a transformation is canonical. I can help you with both problems.

1) To show that [Q,P] = 1, we need to evaluate [Q,P] using the definition given:

[Q,P] = \frac{\partial{Q}}{\partial{q}}\frac{\partial{P}}{\partial{p}}-\frac{\partial{P}}{\partial{q}}\frac{\partial{Q}}{\partial{p}}

First, let's evaluate \frac{\partial{Q}}{\partial{q}}:

\frac{\partial{Q}}{\partial{q}} = \frac{\partial}{\partial{q}}\left(\frac{m_1 r_1+m_2 r_2}{M}\right) = \frac{m_1}{M}

Next, let's evaluate \frac{\partial{P}}{\partial{p}}:

\frac{\partial{P}}{\partial{p}} = \frac{\partial}{\partial{p}}(p_1+p_2) = 1

Now, let's evaluate \frac{\partial{P}}{\partial{q}}:

\frac{\partial{P}}{\partial{q}} = \frac{\partial}{\partial{q}}\left(\frac{m_2 p_1-m_1 p_2}{M}\right) = 0

Lastly, let's evaluate \frac{\partial{Q}}{\partial{p}}:

\frac{\partial{Q}}{\partial{p}} = \frac{\partial}{\partial{p}}(r_1-r_2) = 0

Plugging these values into the definition of the Poisson bracket, we get:

[Q,P] = \frac{m_1}{M}\cdot1 - 0\cdot0 = \frac{m_1}{M} = 1

Therefore, [Q,P] = 1 and the transformation is canonical.

2) To show that the transformation is canonical, we need to show that the Poisson bracket [q,p] = 1 using the definition of the Poisson bracket:

[q,p] = \frac{\partial{q}}{\partial{r_1}}\frac{\partial{p}}{\partial{p_1}}-\frac{\partial{p}}{\partial{r_1}}