Non-commutativity of unit polar bases

[chartery]

I'm having trouble(s) showing that unit polar bases do not commute.

Adapting <https://math.stackexchange.com/questions/3288981>

taking: $\hat{\theta} = \frac{1}{r}\frac{\partial }{\partial \theta} ( =\frac{1}{r}\overrightarrow{e}_{\theta})$
then $\hat{r}\hat{\theta} = \frac{\partial }{\partial r}\left( \frac{1}{r} \frac{\partial f}{\partial \theta}\right)$ by chain rule will differ from $\hat{\theta}\hat{r} = \frac{1}{r}\frac{\partial }{\partial \theta}\left( \frac{\partial f}{\partial r}\right)$
so far so good, but...in <https://physics.stackexchange.com/questions/198280> (first answer)
(via $\frac{\partial ^{2}x}{\partial \hat{r}\partial \hat{\theta}} vs \frac{\partial ^{2}x}{\partial \hat{\theta}\partial \hat{r}}$)

$\overrightarrow{e_{\hat{r}}}=\overrightarrow{e_{r}}$ seems to justify $(\frac{\partial }{\partial \hat{r}}\left( -sin\theta \right)= )$, $\frac{\partial }{\partial \hat{r}}\left( -\frac{y}{r} \right)=\frac{\partial }{\partial r}\left( -\frac{y}{r} \right)$
but I don't see how to apply $\frac{\partial }{\partial \hat{\theta}}$ in $\frac{\partial }{\partial \hat{\theta}}\left( cos\theta \right)$Finally, when just multiplying out the commutator of Cartesian versions, everything seems to cancel:
$[\hat{r},\hat{\theta}] = \left( cos\theta\overrightarrow{x}+sin\theta\overrightarrow{y} \right)\left( -sin\theta\overrightarrow{x}+cos\theta\overrightarrow{y} \right)-\left( -sin\theta\overrightarrow{x}+cos\theta\overrightarrow{y} \right)\left( cos\theta\overrightarrow{x}+sin\theta\overrightarrow{y} \right)$ =0??Evidently, my grasp of the underlying logic of the various manipulations is flimsy. As Jim Hacker said, can you tell me what I don't know :-)?

 

[strangerep]

There are various shortcuts you can use. I'll show some of them via an explicit worked example:
$$[\hat e_\theta \,, \hat e_\phi] ~\equiv~ \Big[ \frac1r \, \partial_\theta ~,~ \frac{1}{r \sin\theta} \, \partial_\phi \Big] ~=~ \frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \, \partial_\phi \Big] ~.$$In the above, I've moved all the $r$ terms out the front of the commutator because $r$ is independent of $\theta$ and $\phi$, hence $r$ can pass straight through $\partial_\theta$ and $\partial_\phi$. (To further check my righthand side above, expand the commutator using the Leibniz product rule $[A,BC] = B[A,C] + [A,B]C$.)

The next step is $$\frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \, \partial_\phi \Big] ~=~ \frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \Big] \, \partial_\phi ~,$$where I've moved $\partial_\phi$ outside of the commutator because $\theta$ and $\phi$ are independent variables. That leaves $$\frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \Big] \partial_\phi ~=~ \frac{1}{r^2} \;\left[ \partial_\theta \left( \frac{1}{ \sin\theta} \right) \right] \partial_\phi~=~ -\,\frac{\cot\theta}{r^2 \sin\theta}\; \partial_\phi ~=~ -\,\frac{\cot\theta}{r}\; \hat e_\phi ~.$$
If you're having trouble with commutators involving $\partial$ apparently acting on nothing, write the commutator like $[A,B]f$, where $f$ is an arbitrary function, and expand the commutator explicitly as $A(B(f)) - B(A(f))$.

HTH.

 

[PeterDonis]

I'll show some of them via an explicit worked example

Shouldn't it be $1 / r$ outside the commutator?

 

[PeroK]

There are various shortcuts you can use. I'll show some of them via an explicit worked example:
$$[\hat e_\theta \,, \hat e_\phi] ~\equiv~ \Big[ \frac1r \, \partial_\theta ~,~ \frac{1}{r \sin\theta} \, \partial_\phi \Big] ~=~ \frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \, \partial_\phi \Big] ~.$$In the above, I've moved all the $r$ terms out the front of the commutator because $r$ is independent of $\theta$ and $\phi$, hence $r$ can pass straight through $\partial_\theta$ and $\partial_\phi$. (To further check my righthand side above, expand the commutator using the Leibniz product rule $[A,BC] = B[A,C] + [A,B]C$.)

The next step is $$\frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \, \partial_\phi \Big] ~=~ \frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \Big] \, \partial_\phi ~,$$where I've moved $\partial_\phi$ outside of the commutator because $\theta$ and $\phi$ are independent variables. That leaves $$\frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \Big] \partial_\phi ~=~ \frac{1}{r^2} \; \partial_\theta \left( \frac{1}{ \sin\theta} \right) ~=~ -\,\frac{\cot\theta}{r^2 \sin\theta}\; \partial_\phi ~=~ -\,\frac{\cot\theta}{r}\; \hat e_\phi ~.$$
If you're having trouble with commutators involving $\partial$ apparently acting on nothing, write the commutator like $[A,B]f$, where $f$ is an arbitrary function, and expand the commutator explicitly as $A(B(f)) - B(A(f))$.

HTH.

More generally, you can see directly that:
$$[\partial_{\theta}, f(\theta)\partial_{\phi}] = f'(\theta) \partial_{\phi} + f(\theta)\partial_{\theta}\partial_{\phi} - f(\theta)\partial_{\phi}\partial_{\theta} = f'(\theta) \partial_{\phi}$$

 

[chartery]

Thanks, that does help fix the chain rule version in my head. But I am uneasy starting on tetrads when the other two (via $\frac{\partial }{\partial \hat{\theta}}$, or especially the Cartesians) imply I am missing some basic understanding of the manipulations of even the simplest 2D unit polars.

 

[PeroK]

Thanks, that does help fix the chain rule version in my head. But I am uneasy starting on tetrads when the other two (via $\frac{\partial }{\partial \hat{\theta}}$, or especially the Cartesians) imply I am missing some basic understanding of the manipulations of even the simplest 2D unit polars.

If you mean this:

Finally, when just multiplying out the commutator of Cartesian versions, everything seems to cancel:
$[\hat{r},\hat{\theta}] = \left( cos\theta\overrightarrow{x}+sin\theta\overrightarrow{y} \right)\left( -sin\theta\overrightarrow{x}+cos\theta\overrightarrow{y} \right)-\left( -sin\theta\overrightarrow{x}+cos\theta\overrightarrow{y} \right)\left( cos\theta\overrightarrow{x}+sin\theta\overrightarrow{y} \right)$ =0??Evidently, my grasp of the underlying logic of the various manipulations is flimsy. As Jim Hacker said, can you tell me what I don't know :-)?

That makes little sense. The commutator applies to operators, not to vectors.

 

[chartery]

If you mean this:That makes little sense. The commutator applies to operators, not to vectors.

Yes, I got a bit mixed up between the various e's, thetas, hats, and arrows. It should have been

$[\hat{r},\hat{\theta}] = \left( cos\theta\partial_{x}+sin\theta\partial_{y} \right)\left( -sin\theta\partial_{x}+cos\theta\partial_{y} \right)-\left( -sin\theta\partial_{x}+cos\theta\partial_{y} \right)\left( cos\theta\partial_{x}+sin\theta\partial_{y} \right)$

but the result still seems zero, given commutativity of $\partial_{x}\partial_{y}$ ?

 

[PeroK]

Yes, I got a bit mixed up between the various e's, thetas, hats, and arrows. It should have been

$[\hat{r},\hat{\theta}] = \left( cos\theta\partial_{x}+sin\theta\partial_{y} \right)\left( -sin\theta\partial_{x}+cos\theta\partial_{y} \right)-\left( -sin\theta\partial_{x}+cos\theta\partial_{y} \right)\left( cos\theta\partial_{x}+sin\theta\partial_{y} \right)$

but the result still seems zero, given commutativity of $\partial_{x}\partial_{y}$ ?

$\theta$ is a function of $x$ and $y$. For example:
$$\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$$

 

[chartery]

Many thanks again. I hope you were not startled by the sound of distant forehead slapping.

 

[chartery]

That leaves $$\frac{1}{r^2} \Big[ \partial_\theta ~,~ \frac{1}{ \sin\theta} \Big] \partial_\phi ~=~ \frac{1}{r^2} \; \partial_\theta \left( \frac{1}{ \sin\theta} \right) ~=~ -\,\frac{\cot\theta}{r^2 \sin\theta}\; \partial_\phi ~=~ -\,\frac{\cot\theta}{r}\; \hat e_\phi ~.$$
HTH.

@strangerep, sorry, I missed you were separate, so forgot to thank you. Now I am working through your reply, I have a question. In the first equality just quoted, what happened to the $~ \left( \frac{1}{ \sin\theta} \right)\partial_\theta~$ term from the commutator? (Assuming the missing $\partial_\phi$ on right is a typo?)

 

[PeroK]

@strangerep, sorry, I missed you were separate, so forgot to thank you. Now I am working through your reply, I have a question. In the first equality just quoted, what happened to the $~ \left( \frac{1}{ \sin\theta} \right)\partial_\theta~$ term from the commutator? (Assuming the missing $\partial_\phi$ on right is a typo?)

You should work through this yourself. At this level, you need to move away from needing every line in a proof to be fully explained. You need to be able, if necessary, to fill in the blanks yourself.

This seems to be quite a common problem when students move up to advanced undergraduate or graduate level. They are used to being spoon-fed every detail. At a certain level it becomes impossible to give every algebraic step, because there are too many. You need to develop self-sufficiency in this respect.

There's a further clue in my post #2, which you "liked". Hopefully you also understood it!

 

[strangerep]

@strangerep, sorry, I missed you were separate, so forgot to thank you. Now I am working through your reply, I have a question. In the first equality just quoted, what happened to the $~ \left( \frac{1}{ \sin\theta} \right)\partial_\theta~$ term from the commutator?

$\partial_\theta$ acting on nothing can be "shortcut" to zero. This is actually a frequently-used trick. To understand it the first time, make the commutator act on an arbitrary function $g$, like this: $$[\partial_\theta \,, f(\theta)] g~=~ \partial_\theta fg ~-~ f \partial_\theta g ~=~ (\partial_\theta f) g + f \partial_\theta g ~-~ f \partial_\theta g ~=~ (\partial_\theta f) g ~,$$ where we have used the Leibniz product rule.

Therefore, we can shortcut this to say $$[\partial_\theta \,, f(\theta)] = \partial_\theta f ~.$$

(Assuming the missing $\partial_\phi$ on right is a typo?)

Yes, that was a typo. Now corrected.

 

[strangerep]

Shouldn't it be $1 / r$ outside the commutator?

I don't think so, unless I'm missing something. Where precisely?

 

[chartery]

$\partial_\theta$ acting on nothing can be "shortcut" to zero. This is actually a frequently-used trick. To understand it the first time, make the commutator act on an arbitrary function $g$, like this: $$[\partial_\theta \,, f(\theta)] g~=~ \partial_\theta fg ~-~ f \partial_\theta g ~=~ (\partial_\theta f) g + f \partial_\theta g ~-~ f \partial_\theta g ~=~ (\partial_\theta f) g ~,$$ where we have used the Leibniz product rule.

Thanks very much for the spoon-feeding. My knowledge is patchy, and being 'woolly' on the flurry of notations, it helps me avoid overlooking subtleties and errors applying generalisations, which of course are obvious in hindsight. (Sorry @PeroK, was deficiency in grounding rather than inclination.). Otherwise I can find later that I didn't understand as well as I thought.

 

[PeterDonis]

I don't think so, unless I'm missing something. Where precisely?

Never mind, I saw what I was missing. You factored $1 / r^2$ out of the commutator because there are two factors of $1 / r$ inside it.