Symmetries in Lagrangian Mechanics

[sophiatev]

In Classical Mechanics by Kibble and Berkshire, in chapter 12.4 which focuses on symmetries and conservation laws (starting on page 291 here), the authors introduce the concept of a generator function G, where the transformation generated by G is given by (equation 12.29 on page 292 in the text)

$\delta q_\alpha = \partial G / \partial p_\alpha \ \delta \lambda$
$\delta p_\alpha = -\partial G / \partial q_\alpha \ \delta \lambda$

They seem to introduce G as a "general function of the coordinates, momenta, and time, G(q,p,t)", where q and p range over all n generalized coordinates $q_\alpha$ and $p_\alpha$. But if the above equations are true, they imply that

$\partial G / \partial p_\alpha \ \delta p_\alpha = -\partial G / \partial q_\alpha \ \delta q_\alpha$

This property does not seem generally true at all, and so I don't see why it would apply to a "general function" G. Am I missing something?

 

[vanhees71]

Why do you think the authors imply the last equation?

 

[Delta2]

Why do you think the authors imply the last equation?

I don't have that book and I am no good at Lagrangian mechanics, but isn't the last equation an algebraic consequence of the first two equations? Solve both equations for $\delta \lambda$ and equate what you get.

P.S I have no idea what $\delta \lambda$ is so I don't know if I can treat it algebraically.

 

[stevendaryl]

Hi. Yes, $G$ is a perfectly general function.

The book is discussing transformations of the state. A transformation is just a map from one configuration of coordinates and momenta to another configuration. For example, in the usual cartesian coordinates $x,y,z$, if you shift everything by $\delta \lambda$ in the x-direction, then the transformation is given by:

$x \rightarrow x + \delta \lambda$
$y \rightarrow y$
$z \rightarrow z$
$p^x \rightarrow p^x$
$p^y \rightarrow p^y$
$p^y \rightarrow p^y$

This transformation changes the x coordinate but nothing else.

Absolutely any mapping counts as a transformation, although the ones we are interested in for physics have some physical significance.

In general a transformation function would require one function for each coordinate and momenta:

$\delta q^j = Q^j(q,p,t) \delta \lambda$
$\delta p_j = P_j(q,p,t)\delta \lambda$

But an interesting type of transformation is one given by a generating function $G(q,p, t)$, which defines the transformations via

$\delta q^j = \dfrac{\partial G}{\partial p_j} \delta \lambda$

$\delta p_j = - \dfrac{\partial G}{\partial q^j} \delta \lambda$

$G$ is any function at all. It is just a way of generating a transformation.

For example, in one dimension, the transformation

$x \rightarrow x + \delta \lambda$
$p \rightarrow p$

is given by the generating function

$G = p$

Then $\dfrac{\partial G}{\partial x} = 0$ so $\delta p = 0$. $\dfrac{\partial G}{\partial p} = 1$, so $\delta x = \delta \lambda$

The function $G$ is the x-component of momentum, and the effect is to shift $x$. This is what is meant when they say that $p$ is the generator of translations.

A more interesting case is rotations. In two dimensions, a rotation is the transformation

$x \rightarrow x - y \delta \lambda$
$y \rightarrow y + x \delta \lambda$
$p_x \rightarrow p_x -p_y \delta \lambda$
$p_y \rightarrow p_y + p_x \delta \lambda$

(Note: this is an infinitesimal rotation, where we are allowed to approximate $sin(\delta \lambda)$ by $\delta \lambda$ and $cos(\delta \lambda)$ by 1. A real rotation is made up by summing many infinitesimal rotations.)

The generator for this transformation is:
$G = x p_y - y p_x$, which is just the angular momentum.

[Edit: was $G = x p_y - z p_x$]

These sorts of transformations are interesting because if the transformation leaves the system unchanged, then the corresponding generator is a constant.

 

[sophiatev]

Hi. Yes, $G$ is a perfectly general function.

The book is discussing transformations of the state. A transformation is just a map from one configuration of coordinates and momenta to another configuration. For example, in the usual cartesian coordinates $x,y,z$, if you shift everything by $\delta \lambda$ in the x-direction, then the transformation is given by:

$x \rightarrow x + \delta \lambda$
$y \rightarrow y$
$z \rightarrow z$
$p^x \rightarrow p^x$
$p^y \rightarrow p^y$
$p^y \rightarrow p^y$

This transformation changes the x coordinate but nothing else.

Absolutely any mapping counts as a transformation, although the ones we are interested in for physics have some physical significance.

In general a transformation function would require one function for each coordinate and momenta:

$\delta q^j = Q^j(q,p,t) \delta \lambda$
$\delta p_j = P_j(q,p,t)\delta \lambda$

But an interesting type of transformation is one given by a generating function $G(q,p, t)$, which defines the transformations via

$\delta q^j = \dfrac{\partial G}{\partial p_j} \delta \lambda$

$\delta p_j = - \dfrac{\partial G}{\partial q^j} \delta \lambda$

$G$ is any function at all. It is just a way of generating a transformation.

For example, in one dimension, the transformation

$x \rightarrow x + \delta \lambda$
$p \rightarrow p$

is given by the generating function

$G = p$

Then $\dfrac{\partial G}{\partial x} = 0$ so $\delta p = 0$. $\dfrac{\partial G}{\partial p} = 1$, so $\delta x = \delta \lambda$

The function $G$ is the x-component of momentum, and the effect is to shift $x$. This is what is meant when they say that $p$ is the generator of translations.

A more interesting case is rotations. In two dimensions, a rotation is the transformation

$x \rightarrow x - y \delta \lambda$
$y \rightarrow y + x \delta \lambda$
$p_x \rightarrow p_x -p_y \delta \lambda$
$p_y \rightarrow p_y + p_x \delta \lambda$

(Note: this is an infinitesimal rotation, where we are allowed to approximate $sin(\delta \lambda)$ by $\delta \lambda$ and $cos(\delta \lambda)$ by 1. A real rotation is made up by summing many infinitesimal rotations.)

The generator for this transformation is:
$G = x p_y - z p_x$, which is just the angular momentum.

These sorts of transformations are interesting because if the transformation leaves the system unchanged, then the corresponding generator is a constant.

Was that last equation meant to be $G = x p_y - y p_x$? But otherwise, I think that makes sense, thank you. Sounds like I was interpreting the $\delta q^j$ and $\delta p_j$ as the differential changes in the respective variables when taking a derivative (i.e. $\delta G = \sum \partial G / \partial q^j \ \delta q^j + \partial G / \partial p_j \ \delta p_j$) rather than the differential changes produced by the transformation.

 

[stevendaryl]

Was that last equation meant to be $G = x p_y - y p_x$?

Yes, you're right.