# How is the Poisson bracket $\{\mathbf{c},\mathbf{l}\cdot\hat{n}\}=(\hat{n}\times \mathbf{c})$, for constant $\mathbf{c}$, and not zero?

The Poissonian formulation of mechanics tells us that for a generating function $g(q,p,t)$, the Poisson bracket of some function/variable $f(q,p,t)$ with the generating function corresponds with an infinitesimal change in $f$ along the transformation or "motion" generated by $g$.

$$\delta f = \epsilon \left\{f,g \right\}$$

An example of this is momentum conservation due to invariance under infinitesimal translations. To show this, take $f$ to be the Hamiltonian and $g$ to be $\mathbf{p}\cdot\hat{n}$, where $\mathbf{p}$ is the momentum $p_x \hat{x}+p_y\hat{y}+p_z\hat{z}$ and $\hat{n}$ is an arbitrary unit vector. The canonical transformation generated by $\mathbf{p}\cdot \hat{n}$ is an infinitesimal translation along the $\hat{n}$ direction of the system variables with which the Hamiltonian is evaluated.

\begin{align*} \epsilon\left\{H,\mathbf{p}\cdot\hat{n}\right\}&=\epsilon\left(\sum_i \frac{\partial H}{\partial q_i}\frac{\partial\,(\mathbf{p}\cdot\hat{n})}{\partial p_i}-\frac{\partial H}{\partial p_i}\frac{\partial\,(\mathbf{p}\cdot\hat{n})}{\partial q_i}\right)\\ &=\epsilon\left(\sum_i \frac{\partial H}{\partial q_i}(\hat{n})_i\right)\\ &=\epsilon (\nabla_q H)\cdot \hat{n}\\ &\\ &\implies \left\{H,\mathbf{p}\cdot\hat{n}\right\}=(\nabla_q H)\cdot \hat{n} \end{align*}

Now, if we were to take an polar angle $\theta$ about some axis $\hat{n}$ to be a coordinate, the above procedure with $\mathbf{l}$, the angular momentum, in place of $\mathbf{p}$ would then translate as an infinitesimal "translation" of the $\theta$ variable - i.e. a rotation about the $\hat{n}$ axis. An example of this is given in Landau & Lifshitz, Goldstein, and many other mechanics textbooks - the rotation of a constant vector $\mathbf{c}$ about a specified axis.

$$\left\{\mathbf{c},\mathbf{l}\cdot\hat{n}\right\}=\hat{n}\times\mathbf{c}$$

In terms of the interpretation of the Poisson brackets through generating functions (which I just gave), I can see why this would be true. The vector $\mathbf{c}$ changes by an amount $d\theta(\hat{n}\times\mathbf{c})$ when rotated by an infinitesimal angle $d\theta$ about an axis $\hat{n}$, and that result can be reached by simple analytical geometry. However, by direct evaluation of the Poisson bracket, I can't see why this isn't zero (as $\mathbf{c}$ is a constant). The angular momentum operator (vector-valued function in terms of phase space variables) is given by

\begin{align*} \mathbf{l}&=\mathbf{r}\times\mathbf{p}\\ &=(yp_z-zp_y)\hat{x}+(zp_x-xp_z)\hat{y}+(xp_y-yp_x)\hat{z} \end{align*}

Note that this, assuming a typical classical Hamiltonian, entirely in terms of phase space variables. Now, the Poisson bracket of this with a constant vector is

\begin{align*} \left\{\mathbf{c},\mathbf{l}\cdot\hat{n}\right\}&=\sum_i\left(\frac{\partial \mathbf{c}}{\partial q_i}\frac{\partial (\mathbf{l}\cdot\hat{n})}{\partial p_i}-\frac{\partial \mathbf{c}}{\partial p_i}\frac{\partial (\mathbf{l}\cdot\hat{n})}{\partial q_i}\right)\\ &=0\,\,\,\,(\mathbf{c}\textrm{ doesn't depend on phase space variables)} \end{align*}

P.s: I originally wrote this question extremely briefly because I thought somebody would certainly know what I'm talking about.

• In the definition of the Poisson bracket derivatives wrt. the canonical coordinates and momenta are taken. The derivatives of $c$ vanish and hence the Poisson bracket as well. Feb 22 '16 at 9:42
• @Urgje $\vec c$ is a vector, that implies a certain transformation behaviour of the components on changes of the coordinate system, therefore a constant vector $\vec c$ does not necessarily have constant components (consider, for example spherical coordinates where the vector basis changes from point to point). Feb 22 '16 at 14:22
• @SebastianRiese That $c$ is a vector does not imply that it depends on the canonical variables. Feb 22 '16 at 22:38
• @ArturodonJuan (3 comments up) for the record, when a question is closed (technically, put on hold), that's not necessarily such a bad thing. The whole point of putting on hold is to keep answers away until you can edit the question to be good. Feb 23 '16 at 3:46
• I have a strong feeling this has to do with the fact that $\mathbf{l}\cdot\hat{n}$ is the momentum conjugate to the polar angle about the $\hat{n}$ axis, and so the angular momentum operator/function that I wrote down in terms of the standard $x$, $y$, and $z$ doesn't apply - i.e. I need to express \mathbf{c} in terms of coordinates of which at least one needs to be the angle $\theta$ about the \hat{n} axis. Feb 23 '16 at 16:02

The starting point is that the $3$-vector $\vec{\bf c}$ transforms in the $3$-dimensional irreducible vector representation of the rotation group $SO(3)$,

$$\tag{1} \{ \vec{\bf c}, \vec{\bf L}\cdot \hat{\bf n} \}_{PB}~=~ \hat{\bf n}\times \vec{\bf c},$$

where $\hat{\bf n}$ is an arbitrary unit vector, whose Poisson bracket (PB) with anything vanishes

$$\tag{2} \{ \hat{\bf n}, \cdot \}_{PB}~=~0.$$

We assume that $\vec{\bf c}$ is not identically zero. Since the PB with $\vec{\bf c}$ does not vanish, the $3$-vector $\vec{\bf c}$ cannot be a constant. It must be a function of the phase space variables $\vec{\bf r}$ and $\vec{\bf p}$. It can be thought of as being of the form

$$\tag{3} \vec{\bf c}~=~\vec{\bf r}f+ \vec{\bf p}g+ \vec{\bf L}h,$$

where

$$\tag{4} f~=~f(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2), \quad g~=~g(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2), \quad\text{and}\quad h~=~h(r^2,p^2,\vec{\bf r}\cdot\vec{\bf p},L^2),$$ are three suitable functions of the phase space $SO(3)$ scalars

$$\tag{5} r^2,\quad p^2,\quad \vec{\bf r}\cdot\vec{\bf p},\quad\text{and}\quad L^2.$$

References:

1. H. Goldstein, Classical Mechanics; Section 9-6 in 2nd edition or Section 9.7 in 3rd edition.
• The question specifies $\mathbf c$ as a constant, though. Are you saying this is impossible? Feb 23 '16 at 18:30
• $\uparrow$ Yes. I updated the answer. Feb 23 '16 at 19:23
• That makes very little sense to me. $f:\mathrm{PS}\to\mathbb R$, $f\equiv 1$ is a perfectly valid constant function on phase space. Why is $f:\mathrm{PS}\to\mathbb R^3$, $f\equiv (1,0,0)$ suddenly impossible? Feb 23 '16 at 19:42
• Because then $\{ f, \cdot \}_{PB}=0$. Feb 23 '16 at 20:03
• Yeah, that's sort of what the OP is asking - why is that not allowed? Heck, you could even rephrase the PB as $$\left\{\hat{e}\cdot \mathbf{c},\mathbf{l}\cdot\hat{n}\right\}_\mathrm{PB}=\hat{e} \cdot (\hat{n}\times\mathbf{c}),$$ and then choosing $\mathbf c=\hat n=(1,0,0)$ you reduce to the real-valued case $f\equiv 1$. Is $f\equiv 1$ allowed but $f\equiv \hat e \cdot \hat e$ forbidden? There are obviously issues with rotational invariance and active vs passive transformations, but it doesn't help to not talk about them. Feb 23 '16 at 20:19

Let's just do it for a simple example. By $\vec{c}$ I imagine you mean the location of the particle relative to some origin, so $\vec{c}=\vec{r}$. Later on for simplicity we'll suppose further the particle is located on the x-axis (but it is important to do this only after differentiating as we will see).

We'll also suppose we are rotating around the z axis so that $\hat{n}=\hat{z}$.

Then we have \begin{equation} \{\vec{c} , \vec{l}\cdot\vec{n}\}=\{\vec{r},xp_y - y p_x\} =\{x\hat{x}+y\hat{y}+z\hat{z},xp_y - y p_x\}= -y \hat{x}+x\hat{y}. \end{equation}

Now that we have differentiated (meaning, evaluated the brackets) we can set $y=0$ and $x=R$ (that is, we can suppose our particle started on the $x$-axis at the position $R$). Then \begin{equation} \{R \hat{x},\vec{l}\cdot\hat{z}\}=R \hat{y}=\hat{z}\times(R\hat{x}) \end{equation} which is consistent with your formula.

Incidentally, you might be worried that I started off by setting $\vec{c}=\vec{r}$. I think in the framework you are working in--particle mechanics--the vectors should all start from the same origin. If you want to start taking poisson brackets of vectors with different origins, I think you really need to generalize this discussion to field theory (which will complicate the story a bit because in addition to rotating the direction of the vector you need to rotate the origin, so you will end up with an additional term). So I think that may be what you have in mind but that is a more complicated story.

• I'm not sure field theory really needs to be introduced to handle multiple origins. Field theory may be where that comes in withing most physics curricula, but logically speaking I don't see the need. Feb 23 '16 at 18:46