# Covariant derivative

1. May 7, 2005

### Nemesis_one

Hello!
I am trying t solution Navier-Stokes equation and I cannot find something about Laplacian. I would like to solution Laplace’a equation for each component.I am trying to transform cylindrical coordinate. I would like to search equation for covariant derivative. For divergence of a contravariant tensor I found:

Aj;k=1/gkk(dAj/dxk)-ri;jkAi

where r is christoffel symbol

I do not know what I have to change that obtain the same for Laplace’a equation? If Christoffel symbols do not change?
I will be grateful for any directions and instructions.

2. May 7, 2005

### dextercioby

What you've written there doesn't make any sense.You may have meant

$$A^{j}{}_{;k}=A^{j}{}_{,k}+\Gamma^{j}{}_{lk} A^{l}$$

There's no big deal,really,u also need the laplacian...

$$A^{j}{}_{;k}{}^{;k}$$

which can be gotten by applying the divergence on the object in the RHS of the gradient written above.

Daniel.

3. May 8, 2005

4. May 8, 2005

### dextercioby

You got the laplacian in formulas (91-92).What else do u need...?

Daniel.

5. May 8, 2005

### Nemesis_one

Yes but I have to solution for each components. I have to have laplacian for r,teta and z seperately. Formulas (91 -92) are showed finaly equation.

Kamil

6. May 8, 2005

### dextercioby

Hmm,i don't understand,what laplacian for "r","theta" & "z",there's only one operator for all 3 of them at the same time,if you're considering the laplacian in 3D.

Daniel.

7. May 8, 2005

### Nemesis_one

At page http://mathworld.wolfram.com/CylindricalCoordinates.html formulas (47-49). If I add this I will obtain derivative for r component in cylindrical coordinate( [v*delatV]r) I have to obtain the same for Laplacian equation for each components ( [delta2V]r) I have this solution but I do not know how to solution step by step. For derivative is equation (46) now I tam trying to find equation for laplacian. (How I can insert equation???)

Kamil

8. May 8, 2005

### dextercioby

I'm really sorry,but i'll just tell u what to do.I won't be doin' any calculations.

Here http://mathworld.wolfram.com/ScaleFactor.html u lear about scale factor (a.k.a.Lamé parameters).U can apply formula #2 from that page to compute the 3 scale factors for the cylidrical coordinates.

Then u go here http://mathworld.wolfram.com/Laplacian.html where in formula #1 u have to use the scale factors computed before in order to get the expression for the laplacian in cylidrical coordinates...

Is there something else to it...?

Daniel.

9. May 8, 2005

### dextercioby

Here's the deal.U need to put the N-S equation for incompressible flow in a generally covariant form.

$$\frac{d\vec{u}}{dt}=\vec{f}+\frac{1}{\rho}\nabla p+\frac{\mu}{\rho}\Delta \vec{u}$$ (1)

I won't use the column,semicolumn notation,since it won't be that easy to follow,because many indices will come up.

Take the convective derivative of the vector field 'convective velocity'

$$\frac{d\vec{u}}{dt}=\left(\partial_{t}u^{i}+u^{j}\nabla_{j}u^{i}\right)\vec{e}_{i}$$ (2)

$$\nabla_{j}u^{i}=\partial_{j}u^{i}+\Gamma^{i}{}_{jk}u^{k}$$ (3)

So u can couple (2) & (3) to get

$$\frac{d\vec{u}}{dt}=\left(\partial_{t}u^{i}+u^{j}\partial_{j}u^{i}+u^{j}\Gamma_{jk}u^{k}\right)\vec{e}_{i}$$ (4)

Next term

$$\vec{f}=f^{i}\vec{e}_{i}$$ (5)

Next

$$\nabla p=\left(g^{ij}\nabla_{j}p\right)\vec{e}_{i}$$ (6)

The last & the most interesting by far is the vector laplacian.U'll see why i used the other notation for the derivative & the covariant derivative

$$\Delta\vec{u}=\left(\nabla^{j}\nabla_{j}u^{i}\right)\vec{e}_{i}=\left(g^{jk}\nabla_{k}\nabla_{j}u^{i}\right)\vec{e}_{i}$$ (7)

To be continued.

Daniel.

Last edited: May 8, 2005
10. May 8, 2005

### dextercioby

Alright.U need to compute this animal $\nabla_{k}\nabla_{j}u^{i}$

$$\nabla_{k}\nabla_{j}u^{i} =\nabla_{k}\left(\partial_{j}u^{i}+\Gamma^{i}{}_{jl}u^{l}\right)$$ (8)

Imagine that tensor $\nabla_{j}u^{i}$ as being what it is,namely a (1,1) tensor.Denote it generically as $T_{j}{}^{i}$.Then

$$\nabla_{k}T_{j}{}^{i}=\partial_{k}T_{j}{}^{i}+\Gamma^{i}{}_{km}T_{j}{}^{m}-\Gamma^{n}{}_{kj}T_{n}{}^{i}$$ (9)

Applying (9) to our case we get

$$\nabla_{k}\nabla_{j}u^{i}=\partial_{k}\left(\partial_{j}u^{i}+\Gamma^{i}{}_{jl}u^{l}\right)+\Gamma^{i}{}_{km}\left(\partial_{j}u^{m}+\Gamma^{m}{}_{jl}u^{l}\right)-\Gamma^{n}{}_{kj}\left(\partial_{n}u^{i}+\Gamma^{i}{}_{nl}u^{l}\right)$$ (10)

Contract (10) with the inverse metric tensor $g^{jk}$ and u have the last term of N-S equations,expressed in the basis $\vec{e}_{i}$.

Now u'll have to apply this abstract tensor formalism to the orthonormal case of cylidrical coordinates.U'll need both the metric & its inverse (check Mathworld page),the gradient of a scalar in cylidrical coords.(for the pressure term,check the Mathworld page) and the second kind Christoffel symbols (check the Mathworld page).

I think you're all set...

Daniel.

Last edited: May 8, 2005
11. May 8, 2005

### Nemesis_one

Thank you very much:) Now I try to understand this:)

12. May 8, 2005

### Nemesis_one

No good:( I do not understand index:( Therefore I cannot solute this:( Can you do one example??? For r components?

Kamil

13. May 8, 2005

### dextercioby

All,those indices (subscripts and superscripts) take 3 values.Each time 2 get repeated in diagonal direction (NE-SW or NW-SE),a summation over all three possible values is understood.

E.g.the derivative of a vector

$$\partial_{i}u^{i}=\partial_{r}u^{\rho}+\partial_{\varphi}u^{\varphi}+\partial_{z}u^{z}$$

And similar to everything else...I'm sorry,but that's what you have to do to prove the formula you're already been given.

Daniel.

14. Jun 10, 2011

### rieuk

Thanks for this posting Dexter, it's helping my understanding. I'm trying to fully understand the derivation of the formulation for the vector laplacian in transformed coordinates - do you know anything on the web that also describes this (basically what you've written) but in a more explanatory way?