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
$$\Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_{i_1}(\sqrt{\vert g\vert} g^{i_1k_1}\partial_{k_1} \frac{1}{\sqrt{\vert g\vert}}\partial_{i_2}(\sqrt{\vert g\vert} g^{i_2k_2}\partial_{k_2} ... \frac{1}{\sqrt{\vert g\vert}}\partial_{i_n}(\sqrt{\vert g\vert} g^{i_nk_n}\partial_{k_n} F )...))$$
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 $n \in \mathbb{N}$, $$\Large {\nabla^{n} f = \begin{cases} { f_{;d_{1}} } & \text{if } n=1 \\ { g^{ d_{1} d_{2} } ~ f_{ ; d_{2} d_{1} } } & \text{if } n = 2 \\ { g^{ d_{1} d_{2} } ~ \left[\prod\limits_{w = 3}^{n} g^{ c_{w} d_{w} } \right] ~ f_{;d_{n}...d_{1}} } & \text{if } n \geq 3 \end{cases}}$$