3
$\begingroup$

I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104, $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.

Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.

In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?

Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive).

$\endgroup$
  • $\begingroup$ I'm also having real trouble understanding this section! Seeing as Noether's theorem is so important, I'd really like to know what's going on so have added a bounty. $\endgroup$ – Arthur Morris Jun 12 at 23:58
1
+100
$\begingroup$

It seems to me that some of the notation in the mentioned textbook is implicit. Let's start with the phase space where coordinates are $(q,p)$. We have a canonical transformation to coordinates $(q',p')$. A point in the phase space will be denoted by $A$, and some other point by $B$.

$\quad \bullet\quad$ In the passive viewpoint, all quantities at a point $A$ have the same value in all coordinate systems, but possibly their functional dependence changes: $$f(q,p)\Big|_{A} = f'(q',p')\Big|_{A}$$

where I deliberately put a dash to the function $f$ to denote that the form of the function might change.

$\quad \bullet\quad$ In the active viewpoint, we "move" our points, say $A \to B$, and we want to know how our functions change. Our original function was (denoting by $q_{A}$ the coordinates of the point $A$ and similarly for the momentum $$f(q_A,p_{A})$$ $\quad \bullet\quad$ The change in our functions then corresponds to $$f(q_A,p_{A}) \to f(q_B,p_{B})$$

Note how this time in the right hand side $f$ does not have the dash. One important point is that from the knowledge of the passive transformation we can say something about the active transformation - the coordinates of the point $B$ in the old coordinate system $(q,p)$ are the same as the coordinates of the point $A$ in the new coordinate system $(q',p')$

Now the textbook says that we are interested in the change of our functions under an active transformation, and we denote this change by $\partial$. $$\partial f = f(q_B,p_{B}) - f(q_A,p_{A})$$ However, we cannot say at which point $\partial f$ is evaluated - is it at the point $A$ or at the point $B$?

To make this consistent, we must actually obtain a result that can be evaluated at a single point, so, as implicitly done in Goldstein, let's take this point to be $B$. This means that implicitly we actually have $$\partial f = \lim_{A\to B} \left(f(q_B,p_{B}) - f(q_A,p_{A})\right)$$

Technically, we must now evaluate $f(q_A, p_A)$ in terms of quantities that exist at the point $B$. We use our known relation $$f(q_A,p_{A}) = f'(q'_A,p'_A)$$ and we use our trick mentioned above - the coordinates of the point $B$ in the old coordinate system $(q,p)$ are the same as the coordinates of the point $A$ in the new coordinate system $(q',p')$ $\Rightarrow q'_A = q_B$, etc.

This actually means that $$f'(q'_A,p'_A) = f'(q_B,p_B)$$ And we have $$\partial f = \lim_{A\to B} \left(f(q_B,p_{B}) - f(q_A,p_{A})\right) = f(q_B,p_{B}) - f'(q_B,p_{B})$$ This is the meaning of "...where of course A and B will be infinitesimally close." in Goldstein under eq. (9.102). The point of it all is to apply this to examine the change in the Hamiltonian, now using our consistent definition $$\partial H = H(q_B, p_B) - H'(q_B, p_B)$$ And the good thing is, we know from earlier how to use the passive transformation properties, and also how to transform the Hamiltonian to a new pair of canonical variables: \begin{equation} H(q_A,p_A) = H'(q'_A,p'_A) = H'(q_B, p_B) = K(q_B, p_B) = H(q_B, p_B) + \frac{\partial F}{\partial t} \end{equation}

to get $$\partial H = H(q_B, p_B) - H(q_A,p_A) - \frac{\partial F}{\partial t}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.