Commutator of vector fields $x∂_x + y∂_y + z∂_z$ and $∂_z$ The commutator of the vector fields $x∂_x + y∂_y + z∂_z$ and $∂_z$ is given by:
$[x∂_x + y∂_y + z∂_z,∂_z] = -∂_z$.
I'm having trouble interpreting this result. The commutator is the vector that connects p and p': p being the point one ends in by first following an integral curve of the first vector field over a distance $\delta t$ and then the second, starting in $(x_0,y_0,z_0)$, and p' vice versa, starting from the same point, over the same distance.
Here's my problem. I believe the integral curves of the first vector field are straight lines through the origin. The integral curves of the second field are lines parallel to the z-axis. If this is indeed the case, why would the order in which we follow the integral curves matter? Looking at both transformations:
$(x_0,y_0,z_0) \rightarrow_1 (x_0+\delta t,y_0+\delta t,z_0+\delta t) \rightarrow_2 (x_0+\delta t,y_0+\delta t,z_0+2\delta t)$
$(x_0,y_0,z_0) \rightarrow_2 (x_0,y_0,z_0+\delta t) \rightarrow_2 (x_0+\delta t,y_0+\delta t,z_0+2\delta t)$
Therefore we end up in the same point. Obviously my question is this: why is the commutator not zero, where did I go wrong?
 A: The first transformation should take
$$\newcommand{\de}{\delta}(x_0,y_0,z_0)\to
(x_0+x_0\de t,y_0+y_0\de t,z_0+z_0\de t).$$
Follow that by the second takes that to
$$(x_0+x_0\de t,y_0+y_0\de t,z_0+z_0\de t+\de t).$$
Doing it the other way round gives
$$(x_0+x_0\de t,y_0+y_0\de t,z_0+\de t+z_0\de t +\de t^2)$$
a difference of $(0,0,-\de t^2)$.
A: You can use them as partial derivatives applied to a function $f(x,y,z)$ and it should be clear:
\begin{align*}
&[x\partial_x + y\partial_y + z\partial_z,\partial_z]f(x,y,z) \\
&\quad= (x\partial_x + y\partial_y + z\partial_z)\bigl(\partial_z f(x,y,z)\bigr) - \partial_z\bigl((x\partial_x + y\partial_y + z\partial_z)f(x,y,z)\bigr) \\
&\quad = (x\partial_{xz} + y\partial_{yz} + z\partial_{zz}) f(x,y,z) - (x\partial_{xz} + y\partial_{yz} + \partial_z + z\partial_{zz}) f(x,y,z)\\
&\quad= - \partial_z f(x,y,z),
\end{align*}
because we have to use the product rule: $$\partial_z(z\partial_z) = \underbrace{\partial_z(z)}_{=1} \cdot\partial_z + z\cdot \underbrace{\partial_z(\partial_z)}_{\partial_{zz}}.$$
A: $ [ x \partial_x+y \partial_y+z \partial_z,  \partial_z  ]= - \partial_z $ is an operator equation. It needs to operate on something.
\begin{eqnarray*}
[ x \partial_x+y \partial_y+z \partial_z,  \partial_z  ] \phi &=&( x \partial_x+y \partial_y+z \partial_z)  \partial_z   \phi - \partial_z ( x \partial_x \phi+y \partial_y \phi+z \partial_z \phi \\
&=& x \partial_x \partial_z   \phi -x \partial_z \partial_x   \phi +y \partial_y \partial_z   \phi -y \partial_z \partial_y   \phi +z \partial_z \partial_z   \phi - \partial_z(z \partial_z   \phi)\\
&=&- \partial_z \phi.
\end{eqnarray*}
