Showing infinitesimal transformation is canonical

In summary, the Homework Equations state that canonical transformations between momentum conjugates (##q_i##, ##p_i##) result in zero net momentum. To solve for the transformation, one must expand the derivatives in the infinitesimal parameter ##\epsilon## using the chain rule.
  • #1
Dazed&Confused
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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?
 
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  • #2
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
fzero said:
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
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
fzero said:
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...
 

FAQ: Showing infinitesimal transformation is canonical

What is an infinitesimal transformation?

An infinitesimal transformation is a small, incremental change in a system or process. It is often used in mathematics and physics to describe how a system changes over time.

Why is it important to show that an infinitesimal transformation is canonical?

Showing that an infinitesimal transformation is canonical is important because it ensures that the transformation preserves the fundamental properties of the system, such as energy and momentum. This is essential in understanding the behavior of a system and making accurate predictions.

What does it mean for an infinitesimal transformation to be canonical?

An infinitesimal transformation is canonical if it satisfies the canonical equations of motion, which describe how the positions and momenta of particles in a system change over time. This indicates that the transformation is mathematically consistent and preserves the fundamental properties of the system.

How is an infinitesimal transformation proven to be canonical?

An infinitesimal transformation can be proven to be canonical by showing that it satisfies the canonical equations of motion or by using the Hamiltonian formalism, which involves the use of Hamilton's equations of motion. Both methods involve mathematical proofs and calculations to demonstrate that the transformation is canonical.

What are some examples of systems where infinitesimal transformations are used?

Infinitesimal transformations are used in various fields of science, such as physics, mathematics, and engineering. Some examples include studying the motion of particles in a gravitational field, analyzing the behavior of electromagnetic waves, and understanding the dynamics of chemical reactions.

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