B Solving for the Nth divergence in any coordinate system

[Vanilla Gorilla]

TL;DR Summary

Suppose we have an operator that, in Cartesian Coordinates, represents the sum of all the UNMIXED Nth partial derivatives. How would we write a more general formula for this operator that works in any coordinate system?

Preface​

We know that, in Cartesian Coordinates, $$\nabla f= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}$$ and $$\nabla^2 f= \frac{\partial^2 f}{\partial^2 x} + \frac{\partial^2 f}{\partial^2 y} + \frac{\partial^2 f}{\partial^2 z}$$
Generalizing these formulas, let's construct an operator $\nabla^{n}$, such that $$\nabla^{n}=\frac{\partial^n f}{\partial^n x} + \frac{\partial^n f}{\partial^n y} + \frac{\partial^n f}{\partial^n z}$$ even more generally $$\nabla^{n} =\sum_{i=1}^{m} \frac{\partial^n f}{\partial^n x_{i}}$$ when there are $m$ dimensions.

My Question​

How would we write a more general formula that works in any coordinate system? I.e., just as we have

and $$\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)$$, is there a more general (I.e., applies in any coordinate system) formula for the $\nabla^{n}$ operator?
My first guess would probably be $$\nabla^{n} = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k^{n-1} F \right)$$ but past that, I'm not sure.

Any help is much appreciated!
P.S., I'm not always great at articulating my thoughts, so my apologies if this question isn't clear. Also, I know this isn't high school material, but I am currently in high school, which is why I made my level "Basic/high school level."

Note to moderators: I think that is the most appropriate forum for this post. However, if not, apologies, and please feel free to move at your own discretion :)

 

[anuttarasammyak]

TL;DR Summary: Suppose we have an operator that, in Cartesian Coordinates, represents the sum of all the UNMIXED Nth partial derivatives. How would we write a more general formula for this operator that works in any coordinate system?

We know that, in Cartesian Coordinates, ∇f=∂f∂x+∂f∂y+∂f∂z and

\nabla f=[\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}]f
, a vector. Its innerproduct with itself is
\nabla \cdot \nabla f= [\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2} ]f
, a scalar. Considering $\nabla$ is a vector, I am not sure how we should interpret your $\nabla^n$.

 

[Vanilla Gorilla]

I meant it as a scalar. Sorry for any confusion! I’m not sure if divergence is the right name, but I meant for both to be scalar quantities that are just sums of unmixed partial derivatives.

Hopefully that clears things up? :)

 

[pasmith]

In cartesians, your operator is A^{i_1 \dots i_n}\partial_{i_1} \cdots \partial_{i_n} for a particular choice of A. That is the expression you need to generalize.

You may need to build it out of repeated application of the external derivative and the Hodge star.

 

[Vanilla Gorilla]

How would I do that exactly?
I am familiar with the exterior derivative, but am unsure if that’s the same thing as the external derivative; likewise, I have no knowledge of the Hodge Star Operator, which I assume is what you mean when you say Hodge Star.

 

[anuttarasammyak]

If you have already proved the formula in OP of
\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)
I think you have a good reason of recurrence formula
\Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k\Delta^{n-1} F\right)

 

[Vanilla Gorilla]

That's kind of along the lines of what I was thinking, but I don't know how I'd prove it

 

[anuttarasammyak]

If
\Delta G=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kG\right),
for any G, then choose G as
G= \Delta^{n-1} F

I cannot find meaning of odd power of nabla applying scalar function. For example can you show where we meet
(\mathbf{i}+\mathbf{j}+\mathbf{k})\cdot\nabla F =\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}
in physics ?

 

[Vanilla Gorilla]

How could we turn that into an explicit, rather than a recursive formula?
Also, thank you for your help, it is much appreciated! :)

 

[anuttarasammyak]

How could we turn that into an explicit, rather than a recursive formula?

Explicitly
\Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_{i_1}(\sqrt{\vert g\vert} g^{i_1k_1}\partial_{k_1}<br /> \frac{1}{\sqrt{\vert g\vert}}\partial_{i_2}(\sqrt{\vert g\vert} g^{i_2k_2}\partial_{k_2}<br /> ...<br /> \frac{1}{\sqrt{\vert g\vert}}\partial_{i_n}(\sqrt{\vert g\vert} g^{i_nk_n}\partial_{k_n}<br /> F )...))
It seems not easy to perform calculation.

 

[Vanilla Gorilla]

That is incredibly tedious-looking, but the proof of concept is amazing. Thank you so much!

 

[Vanilla Gorilla]

OK, so I've taken a few days to mull this over, and I think this should be correct, but I'm not sure. My main concern is if $m$ is the appropriate number to sum to here $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}}$$ However, I am unsure all around as to the correctness of my reasoning and conclusions. Any further criticism is appreciated! :)

Let's construct an operator, such that $$\nabla^{n} =\sum_{i=1}^{m} \frac{\partial^n f}{\partial^n x_{i}}$$ in $m$ dimenions, when written in the Cartesian Basis. Then $$\nabla^n F = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k\nabla^{n-1} F\right) = \frac{1}{\sqrt{\vert g\vert}}\partial_{i_1}(\sqrt{\vert g\vert} g^{i_1k_1}\partial_{k_1} \frac{1}{\sqrt{\vert g\vert}}\partial_{i_2}(\sqrt{\vert g\vert} g^{i_2k_2}\partial_{k_2} ... \frac{1}{\sqrt{\vert g\vert}}\partial_{i_n}(\sqrt{\vert g\vert} g^{i_nk_n}\partial_{k_n} F )...))$$ by This. Let's convert this out of Einstein notation for our use in regular PDEs. $$\nabla^n F = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k\nabla^{n-1} F\right) = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k (\nabla^{n-1} F)\right) = \frac{\partial_i \left(\sqrt {\vert g \vert} g^{ik} \partial_k (\nabla^{n-1} F)\right)} {\sqrt{\vert g \vert}} =$$ $$\sum^{m}_{\begin{align} {i = 1} \nonumber \\ {k = 1} \nonumber \\ \end{align}} \frac{\partial_i \left(\sqrt {\vert g \vert} g^{ik} \partial_k (\nabla^{n-1} F)\right)} {\sqrt{\vert g \vert}} =$$ $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ Expanding this twice to get the gist of the pattern: $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-2 } F)\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}} = $$ $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-3 } F)\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}}$$ However, this is highly cluttered, so let's replace it again with the following $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ For example, $$\nabla^3 F = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ where $$\Delta F = \nabla^2 F = \nabla^{ n-1 } F = \frac{1} {\sqrt{\vert g \vert}} \partial_i \left( \sqrt { \vert g \vert } g^{ik} \partial_k F\right) = \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} $$

Summary​

So, in summary, my final results are that, 1, $$\nabla^3 F = \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}}$$ in any general basis, while in the Cartesian Basis, it is $$\nabla^3 F = \sum_{i = 1}^{ m } \frac {\partial^3 F } {\partial x_i^3} $$
Likewise, $$\nabla^n F = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ in any general basis, while in the Cartesian Basis, it is $$\nabla^{n} F = \sum_{i=1}^{m} \frac{\partial^n F}{\partial^n x_{i}}$$

 

[anuttarasammyak]

Then ∇nF=1|g|∂i(|g|gik∂k∇n−1F)=1|g|∂i1(|g|gi1k1∂k11|g|∂i2(|g|gi2k2∂k2...1|g|∂in(|g|ginkn∂knF)...)) by This.

Just a quick notice that $\nabla$ here should be Laplacian $\Delta$.

 

[Vanilla Gorilla]

Just in that instance? Id everything else OK then?

 

[anuttarasammyak]

Why don't you calculate
\Delta^2 F
by hand and compare it with your formula of n=2 for confirmation ?
In notation
\partial_j F = F_{,j}
\Delta F=|g|^{-1/2}(g^{ik}|g|^{1/2}F_{,k})_{,i}=|g|^{-1/2}(g^{ik}_{,i}|g|^{1/2}F_{,k})+|g|^{-1/2}(g^{ik}|g|^{1/2}_{,i}F_{,k})+|g|^{-1/2}(g^{ik}|g|^{1/2}F_{,k,i})
Only the last term survives in Euclid space. Partial derivative is
(\Delta F)_{,j}=[|g|^{-1/2}(g^{ik}|g|^{1/2}F_{,k})_{,i}]_{,j}=|g|^{-1/2}_{,j}(g^{ik}|g|^{1/2}F_{,k})_{,i}+|g|^{-1/2}(g^{ik}|g|^{1/2}F_{,k})_{,i,j}
and so on.

..Ah, I am sorry to say, in 3d Euclid space
\Delta^2=(\partial_x^2+\partial_y^2+\partial_z^2)^2
so
\partial^4_x+\partial^4_y+\partial^4_z=\Delta^2-2(\partial_x^2\partial_y^2+\partial_y^2\partial_z^2+\partial_z^2\partial_x^2)
What you want is not power of $\Delta$ itself.

 

[Vanilla Gorilla]

I’m afraid I don’t understand what you’re getting at

 

[Vanilla Gorilla]

OK, I'm getting back to this now, and I think I see what you meant, I.e., that the formula I constructed was for $\Delta^n$ and not $\nabla^n$, which was what I was looking for. How could I fix this?

 

[Vanilla Gorilla]

I don't know if the following is right, but it was an attempt to generalize, equations (89)-(96) from this link, found on pages 13-14.

My Attempt​

Note that my notation of $\left[\prod\limits^{\cdot}_{n} \nabla \right]$ is just meant to denote $n$ $\nabla$s dotted together (I didn't want to put subscripts in the big operator's argument to not confuse with later covariant derivative subscripts).
My operator $$\nabla^{n} f = \sum_{q=1}^{m} \frac{\partial^n f}{\partial^n x_{q}} = \left[\prod\limits^{\cdot}_{n} \nabla \right] \cdot f= \nabla \cdot \nabla \cdot \nabla ... \cdot \nabla \cdot f = \nabla \cdot \nabla \cdot \nabla... \cdot \nabla f = g_{ c_{1} c_{2} } \left [\prod\limits_{k = 1}^{n} g^{ c_{k} d_{k} } \nabla_{d_{k}} \right] f =$$ $$g_{ c_{1} c_{2} } g^{ c_{1} d_{1} } ... g^{ c_{n} d_{n} } \nabla_{d_{1}} ... \nabla_{d_{n}} f = \delta^{d_{1} }_{c_{2}} g^{ c_{2} d_{2} } ... g^{ c_{n} d_{n} } \nabla_{d_{1}} ... \nabla_{d_{n}} f = g^{ d_{1} d_{2} } ... g^{ c_{n} d_{n} } \nabla_{d_{1}} ... \nabla_{d_{n}} f$$ where $\nabla_{r}$ represents a covariant derivative. So, rewriting using semicolon covariant derivative notation $\vec {V}_{\alpha;\beta \gamma} = \nabla_{\gamma} \nabla_{\beta} \vec {V}_{\alpha}$
$$\nabla^{n} f = g^{ d_{1} d_{2} } ... g^{ c_{n} d_{n} } \nabla_{d_{1}} ... \nabla_{d_{n}} f = g^{ d_{1} d_{2} } ... g^{ c_{n} d_{n} } f_{;d_{n}...d_{1}} = g^{ d_{1} d_{2} } g^{ c_{3} d_{3} } ... g^{ c_{n} d_{n} } f_{;d_{n}...d_{1}} = g^{ d_{1} d_{2} } \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] f_{;d_{n}...d_{1}} =$$ $$g^{ d_{1} d_{2} } ~ \bar {k} ~ f_{;d_{n}...d_{1}}$$ where $$\bar {k} = \begin{cases} 1 & \text{if } n < 3 \\ \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] & \text{if } n \geq 3 \end{cases}$$ when $n \in \mathbb{N}$. Even more generally, $$\nabla^{n} f = g^{ d_{1} d_{2} } ~ \bar {k} ~ f_{;d_{n}...d_{1}}$$ where $$\bar {k} = \begin{cases} \frac {1} { g^{ d_{1} d_{2} } } & \text{if } n=1 \\ 1 & \text{if } n = 2 \\ \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] & \text{if } n \geq 3 \end{cases}$$

Summary​

So, assuming I did it right, we have found that:
The operator defined to be $\nabla^{n} f = \sum\limits_{q=1}^{m} \frac{\partial^n f}{\partial^n x_{q}}$ in Cartesian Coordinates, should, in general, tensorial coordinates be the following, when $n \in \mathbb{N}$, $$\Large {\nabla^{n} f = \begin{cases} { f_{;d_{1}} } & \text{if } n=1 \\ { g^{ d_{1} d_{2} } ~ f_{ ; d_{2} d_{1} } } & \text{if } n = 2 \\ { g^{ d_{1} d_{2} } ~ \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] ~ f_{;d_{n}...d_{1}} } & \text{if } n \geq 3 \end{cases}}$$

 

[anuttarasammyak]

For verification of your effort, let us see 2d polar coordinates
x^2+y^2=r^2
\frac{y}{x}=tan\phi
xdx+ydy=rdr
xdy-ydx=r^2d\phi
\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}
=\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
similarly
\frac{\partial}{\partial y}=\sin \phi\frac{\partial}{\partial r}+\frac{\cos \phi}{r}\frac{\partial}{\partial \phi}
if I do it right. Does the sum of these match with your $\nabla$ ?

 

[Vanilla Gorilla]

Hmmm, I don't think it worked. That's unfortunate. I'm not really sure where to go from here now.

 

[anuttarasammyak]

continuing my previous post, as
\frac{\partial}{\partial x}=\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
\frac{\partial}{\partial y}=\sin \phi\frac{\partial}{\partial r}+\frac{\cos \phi}{r}\frac{\partial}{\partial \phi},
\frac{\partial^n}{\partial x^n}=(\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})^n
\frac{\partial^n}{\partial y^n}=(\sin \phi\frac{\partial}{\partial r}+\frac{\cos \phi}{r}\frac{\partial}{\partial \phi})^n
You may get it step by step from 1 to n. i.e. for n=2
\frac{\partial^2}{\partial x^2}=(\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})^2
=\cos \phi\frac{\partial}{\partial r}\cos \phi\frac{\partial}{\partial r}-\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}\cos \phi\frac{\partial}{\partial r}+\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
=\cos^2 \phi \frac{\partial^2}{\partial r^2 }+2\frac{\cos\phi \sin\phi}{r^2}\frac{\partial}{\partial \phi}-2\frac{\cos\phi \sin\phi}{r}\frac{\partial}{\partial \phi}\frac{\partial}{\partial r}+\frac{\sin^2 \phi}{r}\frac{\partial}{\partial r}+\frac{\sin^2 \phi}{r^2}\frac{\partial^2}{\partial \phi^2}
For n=3
\frac{\partial^3}{\partial x^3}=(\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})^3 =(\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})(\cos^2 \phi \frac{\partial^2}{\partial r^2 }+2\frac{\cos\phi \sin\phi}{r^2}\frac{\partial}{\partial \phi}-2\frac{\cos\phi \sin\phi}{r}\frac{\partial}{\partial \phi}\frac{\partial}{\partial r}+\frac{\sin^2 \phi}{r}\frac{\partial}{\partial r}+\frac{\sin^2 \phi}{r^2}\frac{\partial^2}{\partial \phi^2})=...
and so on. You may be able to find a general rule for calculation.

 

[Vanilla Gorilla]

So the nth derivative is just the 1st derivative to the nth power? I’m not sure I’m understanding correctly.

 

[anuttarasammyak]

\frac{d^2 F}{dx^2}=\frac{d}{dx}(\frac{dF}{dx})=(\frac{d}{dx})^2 F
But the trouble is
(\cos \phi\frac{\partial}{\partial r}-\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})^2 \neq<br /> (\cos \phi\frac{\partial}{\partial r})^2-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}+(\frac{\sin \phi}{r}\frac{\partial}{\partial \phi})^2
The true result is as written in my previous post. A familiar relation of
(a+b)^2=a^2+2ab+b^2
is not applicable here due to uncommutable character of differentiation operator whose position or order in the formula matters.

 

[Vanilla Gorilla]

Yea that’s why I was confused. Is there a simple rule analogous to the comma to semicolon rule in GR to turn Cartesian sums of partial derivatives into general invariant form?

 

[anuttarasammyak]

Generalization of
\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}
to coordinate change of $x^i \rightarrow q^i$ is
\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}
The operator you want is
\delta^{i_1 i_2 i_3 ... i_n}\prod_{r=1}^n\frac{\partial}{\partial x^{i_r}}=\delta^{i_1 i_2 i_3 ... i_n}\prod_{r=1}^n(\frac{\partial q^{j_r}}{\partial x^{i_r}}\frac{\partial }{\partial q^{j_r}})
where $\delta^{i_1 i_2 i_3 ... i_n}=1$ for $i_1=i_2=i_3=...=i_n$ , 0 otherwise. .
$\frac{\partial q^{j_r}}{\partial x^{i_r}}$ are functions of $\{ q^i \}$, transformed coordinates. They do not commute with $\{ \frac{\partial }{\partial q^i } \}$ so products should be treated carefully.

 

[Vanilla Gorilla]

Suppose $\mu$ is a function of $y$, $\mu (y)$ $$\frac {\partial \mu} {\partial x_{ \theta }} = \frac{\partial y_{i}}{\partial x_{ \theta }}\frac{\partial \mu}{\partial y_{i}} $$ $$\therefore, ~ \text {by the Product Rule,} ~ \partial^{n}_{x_{ \theta }} \left[\frac {\partial \mu} {\partial x_{ \theta }}\right] = \partial^{n}_{x_{ \theta }} \left[ \frac{\partial y_{i}}{\partial x_{ \theta }}\frac{\partial \mu}{\partial y_{i}} \right] = \sum^{n}_{k=0} \binom{n}{k} * d^{(n-k)} (\frac {\partial y_{i}} {\partial x_{ \theta }}) * d^{(k)} (\frac{\partial \mu} {\partial y_{i}} ) =$$ $$\sum^{n}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial^{n-k}_{x_{ \theta }} \left [\frac {\partial y_{i}} {\partial x_{ \theta }} \right ] ~ \partial^{k}_{x_{ \theta }} \left [\frac{\partial \mu} {\partial y_{i}} \right ] \right ] $$ So $$\partial^{n}_{x_{ \theta }} \left[\frac {\partial \mu} {\partial x_{ \theta }}\right] = \partial^{n+1}_{x_{ \theta }} \left[\mu\right] = \sum^{n}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial^{n-k}_{x_{ \theta }} \left [\frac {\partial y_{i}} {\partial x_{ \theta }} \right ] ~ \partial^{k}_{x_{ \theta }} \left [\frac{\partial \mu} {\partial y_{i}} \right ] \right ] $$ Setting $n=(n-1)$ $$\therefore ~ \partial^{n}_{x_{ \theta }} \left[\mu\right] = \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial^{n-k}_{x_{ \theta }} \left [\frac {\partial y_{i}} {\partial x_{ \theta }} \right ] ~ \partial^{k}_{x_{ \theta }} \left [\frac{\partial \mu} {\partial y_{i}} \right ] \right ] = \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial^{n-k+1}_{x_{ \theta }} \left [y_{i} \right ] ~ \partial^{k}_{x_{ \theta }} \left [\partial^{}_{y_{i}} \mu \right ] \right ] = $$ $$\sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial^{n-k+1}_{x_{ \theta }} \left [y_{i} \right ] ~ \partial^{k}_{x_{ \theta }} \partial^{}_{y_{i}} \mu \right ] $$ So $$ \partial^{n}_{x_{ \theta }} \left[ \mu \right] = \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial_{ x_{ \theta }^{n-k+1} } y_{i} ~ \partial_{x_{ \theta }^{k} y_{i}^{} } \mu \right ] $$ $$ \sum_{\theta = 1}^{m} \partial^{n}_{x_{ \theta }} \left[ \mu \right] = \sum_{\theta = 1}^{m} \left [ \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial_{ x_{ \theta }^{n-k+1} } y_{i} ~ \partial_{x_{ \theta }^{k} y_{i}^{} } \mu \right ] \right ] $$ By my definition of the $\nabla^{n}$ operator, $$ \nabla^{n} \mu = \sum_{\theta = 1}^{m} \partial^{n}_{x_{ \theta }} \left[ \mu \right] = \partial^{n}_{x_{ 1 }} \left[ \mu \right] + \partial^{n}_{x_{ 2 }} \left[ \mu \right] + ... \partial^{n}_{x_{ m }} \left[ \mu \right] = \sum_{\theta = 1}^{m} \left [ \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial_{ x_{ \theta }^{n-k+1} } y_{i} ~ \partial_{x_{ \theta }^{k} y_{i}^{} } \mu \right ] \right ] $$ In conlusion, assuming I did this right, $$\LARGE \nabla^{n} \mu = \sum_{\theta = 1}^{m} \left [ \sum^{n-1}_{k=0} \left [ \frac{n!} {k! (n-k)!} ~ \partial_{ x_{ \theta }^{n-k+1} } y_{i} ~ \partial_{x_{ \theta }^{k} y_{i}^{} } \mu \right ] \right ] $$ P.S., thank you so much for dealing with all of my misunderstandings through all of this, I really appreciate it :)

 

[anuttarasammyak]

I am afraid that keeping to apply
\partial_{x_\theta}
or d as it is without transferring to new coordinates, to
\frac{\partial y_i}{\partial_{x_\theta}} and
\frac{\partial \mu}{\partial y_i}
, which are functions of new coordnates { y } not of original Cartesian { x }, is incomplete in the sense that everything is expressed by new coordinates { y }. In my simple example #19, they are functions of r, $\phi$ not of x,y. I think \partial_{x_\theta} should be replaced by expression in new coordinates with
\frac{\partial}{\partial x^\theta}=\frac{\partial q^j}{\partial x^\theta}\frac{\partial }{\partial q^j}

BTW have we gotten to the same result for this simple example of #19?

 

[Vanilla Gorilla]

BTW have we gotten to the same result for this simple example of #19?

No, it didn't work, unfortunately. I'm not sure what I'm missing.

 

[Vanilla Gorilla]

Let's try my 1st guess which was $$\nabla^{n} = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k^{n-1} F \right)$$ In Polar, we have $$g_{ij}=\left(\begin{matrix}1 && 0 \\ 0 && r^2\end{matrix}\right)$$ So $$g^{ij}=\begin{pmatrix}1 && 0 \\ 0 && r^{-2} \end{pmatrix}$$ and $$\sqrt {|g|}=\sqrt {ad-bc} = \sqrt {ad-bc} = \sqrt {r^2} = r$$ $$\nabla^{n} = \frac { \partial_i \left(\sqrt {\vert g \vert} g^{ik} \partial_k^{n-1} F \right) } {\sqrt {\vert g \vert} } $$ $$\nabla^{n} = \frac { \partial_i \left( r g^{ik} \partial_{k}^{n-1} F \right) } { r } $$ $$\nabla^{n} = \frac { \partial_{ \mu } \left( r g^{ \mu \beta } \partial_{ \beta }^{ n-1 } F \right) } { r } $$ expanding the Einstein Notation $$\nabla^{n} F = \frac { \partial_{ r } \left( r g^{ r r } \partial_{ r }^{ n-1 } F \right) } { r } + \frac { \partial_{ r } \left( r g^{ r \theta } \partial_{ \theta }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( r g^{ \theta \theta } \partial_{ \theta }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( r g^{ \theta r } \partial_{ r }^{ n-1 } F \right) } { r } $$ Recall that $$g^{ij} = \begin{matrix} r\\ \theta\\ \end{matrix} ~~ \overset {\LARGE {\begin{matrix} r & \theta \\ \end{matrix}}} { \begin{pmatrix}1 & 0 \\ 0 & r^{-2} \end{pmatrix} } $$ So we're now left with $$\nabla^{n} F = \frac { \partial_{ r } \left( r \cdot 1 \cdot \partial_{ r }^{ n-1 } F \right) } { r } + \frac { \partial_{ r } \left( r \cdot 0 \cdot \partial_{ \theta }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( r \cdot \frac {1} {r^{2}} \cdot \partial_{ \theta }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( r \cdot 0 \cdot \partial_{ r }^{ n-1 } F \right) } { r } $$ $$\nabla^{n} F = \frac { \partial_{ r } \left( r \cdot 1 \cdot \partial_{ r }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( r \cdot \frac {1} {r^{2}} \cdot \partial_{ \theta }^{ n-1 } F \right) } { r } $$ $$\nabla^{n} F = \frac { \partial_{ r } \left( r \cdot \partial_{ r }^{ n-1 } F \right) } { r } + \frac { \partial_{ \theta } \left( \frac {1} {r} \cdot \partial_{ \theta }^{ n-1 } F \right) } { r } = \frac { 1} {r} \left [ \partial_{ r } \left( r \cdot \partial_{ r }^{ n-1 } F \right) + \partial_{ \theta } \left( \frac {1} {r} \cdot \partial_{ \theta }^{ n-1 } F \right ) \right ]=$$ $$\frac {1} {r} \left [ \frac {\partial} {\partial r } \left( r \cdot \frac {\partial^{ n-1 }} {\partial r^{ n-1 } } F \right) + \frac {\partial} {\partial \theta } \left( \frac {1} {r} \cdot \frac {\partial^{ n-1 }} {\partial \theta^{ n-1 } } F \right ) \right ]= \frac {1} {r} \left [ \frac {\partial} {\partial r } \left( r \cdot \frac {\partial^{ n-1 }} {\partial r^{ n-1 } } F \right) + \frac {1} {r} \frac {\partial} {\partial \theta } \left( \frac {\partial^{ n-1 }} {\partial \theta^{ n-1 } } F \right ) \right ]=$$ $$\frac {1} {r} \left [ \frac {\partial} {\partial r } \left( r \cdot \frac {\partial^{ n-1 }} {\partial r^{ n-1 } } \left [ F \right ] \right) + \frac {1} {r} \frac {\partial^{ n }} {\partial \theta^{ n } } \left [ F \right ] \right ] = \frac {1} {r} \frac {\partial} {\partial r } \left( r \cdot \frac {\partial^{ n-1 }} {\partial r^{ n-1 } } \left [ F \right ] \right) + \frac {1} {r^{2}} \frac {\partial^{ n }} {\partial \theta^{ n } } \left [ F \right ] =$$ $$\frac {1} {r} \left( 1 \cdot \frac {\partial^{ n-1 }} {\partial r^{ n-1 } } \left [ F \right ] + r \frac {\partial^{ n }} {\partial r^{ n } } \left [ F \right ] \right) + \frac {1} {r^{2}} \frac {\partial^{ n }} {\partial \theta^{ n } } \left [ F \right ] = \frac {1} {r} \left( \frac {\partial^{ n-1 } F} {\partial r^{ n-1 } } + r \frac {\partial^{ n } F} {\partial r^{ n } } \right) + \frac {1} {r^{2}} \frac {\partial^{ n } F } {\partial \theta^{ n } } = \frac {1} {r} \frac {\partial^{ n-1 } F} {\partial r^{ n-1 } } + \frac {\partial^{ n } F} {\partial r^{ n } } + \frac {1} {r^{2}} \frac {\partial^{ n } F } {\partial \theta^{ n } } $$ Rewritten in a new notation, $$\nabla^{n} F = \frac {1} {r} \partial_{r^{ n-1 }} F + \partial_{r^{ n } } F + \frac {1} {r^{2}} \partial_{\theta^{ n } } F = \frac {1} {r} F_{r^{ n-1 }} + F_{r^{ n } } + \frac {1} {r^{2}} F_{\theta^{ n } } = F_{r^{ n } } + \frac {1} {r} F_{r^{ n-1 }} + \frac {1} {r^{2}} F_{\theta^{ n } } $$ In the case of $n=2$, this simplifies to $$\nabla^{2} F = \Delta F = F_{r^{ 2 } } + \frac {1} {r} F_{r^{ 2-1 }} + \frac {1} {r^{2}} F_{\theta^{ 2 } } = F_{rr } + \frac {1} {r} F_{r} + \frac {1} {r^{2}} F_{\theta \theta} $$ Giving $$\Delta F = F_{rr } + \frac {1} {r} F_{r} + \frac {1} {r^{2}} F_{\theta \theta}$$ as our final result, which I believe is correct, per Laplace's Equation in Polar Coordinates: $$\Delta u = 0 \rightarrow u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0$$ which would imply that my first guess, $$\nabla^{n} = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k^{n-1} F \right)$$ was correct

 

[anuttarasammyak]

which would imply that my first guess, ∇n=1|g|∂i(|g|gik∂kn−1F) was correct

In OP you wrote

　
We already know n=2, Laplacian satisfies that relation. Investigation of n=1, 3 or more is necessary for verification of your guess. Your guess

for n=1
\nabla^1F=\frac{1}{\sqrt{|g|}}\frac{\partial( \sqrt{|g|}g^{ik}e_{k})}{\partial y^i}\ F+\frac{1}{\sqrt{|g|}}\sqrt{|g|}g^{ik}e_{k}\ \frac{\partial F}{\partial y^i}
where $e_k=(\frac{\partial}{\partial y^k})^0=1$ for any k. {y} is new coordinate transformed from Cartesian. If I did it right you see RHS first tem has F with no partial derivative operated. Obviously it does not match with calculation in #19.

 

[Vanilla Gorilla]

Ok, let's try $n=5$ $$\nabla^{n} F = F_{r^{ n } } + \frac {1} {r} F_{r^{ n-1 }} + \frac {1} {r^{2}} F_{\theta^{ n } } = F_{r^{ 5 } } + \frac {1} {r} F_{r^{ 5-1 }} + \frac {1} {r^{2}} F_{\theta^{ 5 } } = F_{r^{ 5 } } + \frac {1} {r} F_{r^{ 4 }} + \frac {1} {r^{2}} F_{\theta^{ 5 } } $$

 

[Vanilla Gorilla]

I want to try and calculate if this formula is correct $$g^{ d_{1} d_{2} } \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] f_{;d_{n}...d_{1}}$$ but I'm unsure of how to quickly calculate iterated covariant derivatives, like we have with $f_{;d_{n}...d_{1}}$

 

[anuttarasammyak]

Your $f_{;d_{n}...d_{1}}$ are all n-th derivative. As you calculated in #29

$\Delta$ or $\nabla^2$ contains RHS 2nd term, first derivative. How do you concilitate it ?

 

[Vanilla Gorilla]

I'm sorry, I don't know what you mean by "concilitate"

 

[anuttarasammyak]

My bad typo, conciliate. $\Delta$ has first derivative of r. I wonder your #32 goes with it.

 

[Vanilla Gorilla]

Ok, so I was rereading through your posts 19, 21, 23, and 25, and I think I understand them better now. I also think I MIGHT have found what I was looking for $$\nabla^{n} f = \sum_{i=1}^{m} \left [ \partial_{x_{i}}^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \partial_{x_{i}} \right ]^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \frac {\partial } {\partial {x_{i}}} \right ] ^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \left [ f \right ] \right )$$ and the $\nabla^{n}$ operator itself $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ where $x$ represents the Cartesian Basis, and $q$ represents the arbitrary Basis. There are $m_x$ dimensions in the Cartesian Basis and $m_q$ dimensions in the arbitrary Basis.
I think this holds, since, in Einstein Notation, we have the multivariable chain rule as $$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}$$

 

[Vanilla Gorilla]

Is the like meant to indicate that that is correct? :D

 

[anuttarasammyak]

I observe my #25 and your #36 coincide.

 

[Vanilla Gorilla]

Also, would it be necessary for $m_x=m_q$ for that equation $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ to work?

 

[anuttarasammyak]

We need same number of parameters to describe same space.

 

[Vanilla Gorilla]

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as $$-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}$$ in my calculation? Do I just discard those?

 

[anuttarasammyak]

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as −2cos⁡ϕ∂∂rsin⁡ϕr∂∂ϕ in my calculation? Do I just discard those?

-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
We have no problem on the leftest
-2\cos \phi. Applying produclt rule of differentiaion,
\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}
=[\frac{\partial}{\partial r}\frac{\sin \phi}{r}]\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}
=-\frac{\sin \phi}{r^2}\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}

As for change of order of operators for an example
\frac{d}{dx}xA=A+x\frac{d}{dx}A
As operator we may delete A as
\frac{d}{dx}x=1+x\frac{d}{dx}
1=\frac{d}{dx}x-x\frac{d}{dx}=[\frac{d}{dx},x]
In general
[\frac{d}{dx},f(x)]=f&#039;(x)