# Generating function of infinitesimal translation - classical mechanics

I am reading Sakurai's Modern quantum mechanics and at some point it's trying to draw a parallel between classical and quantum mechanics.

It says

An infinitesimal translation in classical mechanics can be regarded as a canonical transformation, $$\mathbf{x}_{\mathrm{new}} \equiv \mathbf{X} = \mathbf{x} + d\mathbf{x}, \quad \mathbf{p}_{\mathrm{new}} \equiv \mathbf{P} = \mathbf{p}, \tag{1.6.28}$$ obtainable from the generating function $$F_2(\mathbf{x}, \mathbf{P}) = \mathbf{x}\cdot \mathbf{P} + \mathbf{p}\cdot d\mathbf{x}. \tag{1.6.29}$$

From the wikipedia page it seems that a generating function is something that one can differentiate to obtain the equation of motion of the system.

I mean I assume I need to differentiate $F$ with respect to $\mathbf{X}$ and $\mathbf{P}$? What equation of motion am I supposed to get from here?

• You just read the instructions to differentiate w.r.t. the arguments, so x and P to get p and X, the canonical tfmation displayed. – Cosmas Zachos Feb 12 '18 at 23:22
• Where do I even get $F$ from? I can see that somehow I need to have $\mathbf{X} \cdot$ something ? – SuperCiocia Feb 12 '18 at 23:24
• If it were not trivial to guess, a basic analytical mechanics book describes the methods.... Did you read up on cts? – Cosmas Zachos Feb 12 '18 at 23:27

Hints:

1. The infinitesimal change in position $d{\bf x}=\varepsilon~ {\bf f}({\bf x})$ can be thought of as an infinitesimal parameter $\varepsilon$ times a vector-valued function of the old position ${\bf x}$.

2. A type-2 generating function $F_2({\bf x},{\bf P},t)$ for a canonical transformation depends on the old position ${\bf x}$, the new momentum ${\bf P}$, and possibly time $t$.

3. If the above does not make sense, then you should study Hamiltonian mechanics and canonical transformations, as Cosmas Zachos suggests in above comment, e.g. Ref. 1.

References:

1. H. Goldstein, Classical Mechanics; Section 9.
• Is there a reason you changed $\mathrm{d}{\bf x}$ to $d{\bf x}$? – SuperCiocia Feb 14 '18 at 10:43
• Yes. $d{\bf x}$ is the notation used by Sakurai. Personally I would prefer $\delta{\bf x}$ for an infinitesimal change. $\mathrm{d}{\bf x}$ looks like a one-form. – Qmechanic Feb 14 '18 at 11:33