Commutator of vector fields $x∂_x + y∂_y + z∂_z$ and $∂_z$

Question

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?

Answer 1

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 $(0,0,-\de t^2)$.

Answer 2

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: $$\partial_z(z\partial_z) = \underbrace{\partial_z(z)}_{=1} \cdot\partial_z + z\cdot \underbrace{\partial_z(\partial_z)}_{\partial_{zz}}.$$

Answer 3

$ [ 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*}