Covariant derivative

  • #1
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.
 

Answers and Replies

  • #2
dextercioby
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What you've written there doesn't make any sense.You may have meant

[tex] A^{j}{}_{;k}=A^{j}{}_{,k}+\Gamma^{j}{}_{lk} A^{l} [/tex]

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

[tex] A^{j}{}_{;k}{}^{;k} [/tex]

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

Daniel.
 
  • #4
dextercioby
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You got the laplacian in formulas (91-92).What else do u need...?

Daniel.
 
  • #5
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
dextercioby
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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
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
dextercioby
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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
dextercioby
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Here's the deal.U need to put the N-S equation for incompressible flow in a generally covariant form.

Start with the vector form

[tex] \frac{d\vec{u}}{dt}=\vec{f}+\frac{1}{\rho}\nabla p+\frac{\mu}{\rho}\Delta \vec{u} [/tex] (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'

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

[tex] \nabla_{j}u^{i}=\partial_{j}u^{i}+\Gamma^{i}{}_{jk}u^{k} [/tex] (3)

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

[tex]\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} [/tex] (4)

Next term

[tex]\vec{f}=f^{i}\vec{e}_{i} [/tex] (5)

Next

[tex] \nabla p=\left(g^{ij}\nabla_{j}p\right)\vec{e}_{i} [/tex] (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

[tex] \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} [/tex] (7)

To be continued.

Daniel.
 
Last edited:
  • #10
dextercioby
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Alright.U need to compute this animal [itex] \nabla_{k}\nabla_{j}u^{i} [/itex]

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

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

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

Applying (9) to our case we get

[tex] \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) [/tex] (10)

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

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...:wink:

Daniel.
 
Last edited:
  • #11
Thank you very much:) Now I try to understand this:)
 
  • #12
No good:( I do not understand index:( Therefore I cannot solute this:( Can you do one example??? For r components?

Kamil
 
  • #13
dextercioby
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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

[tex] \partial_{i}u^{i}=\partial_{r}u^{\rho}+\partial_{\varphi}u^{\varphi}+\partial_{z}u^{z} [/tex]

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
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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?
 

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