# Solving Laplace's Equation with Covariant Derivative

• Nemesis_one
In summary, you need to solve a system of equations in a generally covariant form in order to obtain the laplacian for each component of a tensor.
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.

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.

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

Daniel.

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

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.

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

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.

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

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

Kamil

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.

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?

## 1. What is Laplace's Equation?

Laplace's Equation is a partial differential equation that describes the relationship between the potential and the sources of a conservative field. It is commonly used in physics and engineering to solve problems related to electric and gravitational fields.

## 2. What is the role of the covariant derivative in solving Laplace's Equation?

In the context of solving Laplace's Equation, the covariant derivative is used to generalize the concept of differentiation to curved spaces. By accounting for the curvature of space, the covariant derivative allows for a more accurate and comprehensive solution to Laplace's Equation.

## 3. What are the key steps involved in solving Laplace's Equation with covariant derivative?

The key steps in solving Laplace's Equation with covariant derivative include:

• Defining the problem and identifying the sources of the conservative field
• Applying the covariant derivative to the potential function
• Simplifying the resulting equation using the metric tensor
• Solving for the potential function using boundary conditions

## 4. What are the advantages of using covariant derivative in solving Laplace's Equation?

Using covariant derivative in solving Laplace's Equation allows for a more accurate and comprehensive solution, as it takes into account the curvature of space. This is particularly useful in problems involving non-uniform gravitational or electric fields. Additionally, the use of covariant derivative can help simplify the equation and make it more manageable to solve.

## 5. Are there any practical applications of solving Laplace's Equation with covariant derivative?

Yes, there are several practical applications of solving Laplace's Equation with covariant derivative. Some examples include calculating the electric potential in a non-uniform electric field, predicting the flow of fluids in a curved space, and determining the trajectory of a particle under the influence of a non-uniform gravitational field.

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