Showing infinitesimal transformation is canonical

  • #1

Homework Statement


So we have infinitesimal transformations from ##q_i## to ##\bar{q_i}## and ##p_i## to ##\bar{p_i}## ( where ##p_i## represents the canonical momentum conjugate of ##q_i##) given by $$\bar{q_i} = q_i + \epsilon \frac{\partial g}{\partial p_i}$$ $$\bar{p_i} = p_i - \epsilon \frac{\partial g}{\partial q_i}$$ where ##g(q,p)## is any dynamical variable. We must show that this is a conanical transformation. The hint is to work in first order in ##\epsilon##.

Homework Equations

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For canonical transformations we have ##\{ \bar{q_i} , \bar{q_j} \} = 0, \{ \bar{p_i}, \bar{p_j} \} = 0, \{ \bar{q_i}, \bar{p_j} \} = \delta_{ij} ##


The Attempt at a Solution



Summing over ##j##, we have

$$\{ q_i,q_k \} = \left ( \frac{\partial q_i}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_i} \right ) \left ( \frac{\partial q_k}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_k} \right ) - \left ( \frac{\partial q_i}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_i} \right ) \left ( \frac{\partial q_k}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_k} \right ).$$

Any ##\epsilon^2## terms will be disregarded, so only the following remains:

$$\frac{\partial q_i}{\partial q_j} \frac{\partial q_k}{\partial p_j} - \frac{\partial q_i}{\partial p_j} \frac{\partial q_k}{\partial q_j} + \left( \epsilon \frac{\partial^2 g}{\partial q_j \partial p_i} \frac{\partial q_k}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_k} \frac{\partial q_i}{\partial q_j} \right ) - \left ( \epsilon \frac{\partial^2 g}{\partial p_j \partial p_i} \frac{\partial q_k}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_k} \frac{\partial q_i}{\partial p_j} \right ).$$

However, it does not look like these cancel to me. What have I done wrong?
 

Answers and Replies

  • #2
fzero
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The new Poisson bracket should be

$$ \{ A,B\} =\sum_j \left( \frac{\partial A}{\partial \bar{q}_j} \frac{\partial B}{\partial \bar{p}_j} -\frac{\partial B}{\partial \bar{q}_j} \frac{\partial A}{\partial \bar{p}_j} \right),$$

so you have to expand the deriviatives in ##\epsilon## using the chain rule.
 
  • #3
The new Poisson bracket should be

$$ \{ A,B\} =\sum_j \left( \frac{\partial A}{\partial \bar{q}_j} \frac{\partial B}{\partial \bar{p}_j} -\frac{\partial B}{\partial \bar{q}_j} \frac{\partial A}{\partial \bar{p}_j} \right),$$

so you have to expand the deriviatives in ##\epsilon## using the chain rule.
Sorry I should have written ##\{ \bar{q_i} , \bar{q_k} \}.## Since ##\bar{q_i}(q, p),## the definition of the Poisson bracket is $$\{ A, B \} = \left ( \frac{\partial A}{\partial q_j} \frac{\partial B}{\partial p_j} - \frac{\partial A}{\partial p_j} \frac{\partial B}{\partial q_j} \right)$$

where it is summed over ##j##.

I assumed ##\epsilon## was just a number?
 
  • #4
fzero
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Sorry about that, you're totally correct that we have to show that the Poisson bracket computed wrt to the old variables vanishes. So the missing ingredient is then that the old variables are canonical and satisfy

$$ \frac{\partial q_i}{\partial q_j} = \delta_i^j,~~~~\frac{\partial p_i}{\partial p_j} = \delta_i^j, ~~~~\frac{\partial q_i}{\partial p_j} = 0.$$
 
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  • #5
Sorry about that, you're totally correct that we have to show that the Poisson bracket computed wrt to the old variables vanishes. So the missing ingredient is then that the old variables are canonical and satisfy

$$ \frac{\partial q_i}{\partial q_j} = \delta_i^j,~~~~\frac{\partial p_i}{\partial p_j} = \delta_i^j, ~~~~\frac{\partial q_i}{\partial p_j} = 0.$$
Aha. Now everything cancels out correctly. Thanks! Now I just have to calculate the other Poisson brackets...
 

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