- #1

- 708

- 20

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- A
- Thread starter LagrangeEuler
- Start date

- #1

- 708

- 20

- #2

Gold Member

- 765

- 258

See Arfken on this. For example, in spherical polar coordinates here is the Laplacian acting on a scalar field (first scan) and components, with respect to spherical polar coordinate unit vectors, of the Laplacian acting on a vector field (second scan). Arfken says on the vector Laplacian that "It is best obtained by using the vector identity (Eq. 1.80)". This vector identity is

##

\nabla \times (\nabla \times \vec{V}) = \nabla \nabla \cdot \vec{V} - \nabla \cdot \nabla \vec{V} .

##

- #3

Gold Member

- 2,003

- 1,064

[tex]g^{jk}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^k} A^i[/tex] or

[tex]g^{jk}A^i_{,j,k}[/tex]

- #4

- 20,004

- 10,653

Yes and no. It depends on what you mean by ”act with the same operator on the vector field”. If you mean actually acting on the vector field including taking the fact that the basis vectors are not constant then fine. If you mean applying the operator to the individual components, then no.

The best way is however to work with the covariant derivative, which will always work.

No. This expression is not generally covariant. The result is a vector field and as such its components must show the correct transformation properties. This expression does not.

[tex]g^{jk}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^k} A^i[/tex] or

[tex]g^{jk}A^i_{,j,k}[/tex]

- #5

- 20,004

- 10,653

I must also disagree with Arfken here. That involves taking the curl twice, gradient once, and divergence once, which is a mouthful. Better would be to learn about the covariant derivative or use knowledge about how basis vectors in curvilinear coordinates are affected by derivatives.See Arfken on this. For example, in spherical polar coordinates here is the Laplacian acting on a scalar field (first scan) and components, with respect to spherical polar coordinate unit vectors, of the Laplacian acting on a vector field (second scan). Arfken says on the vector Laplacian that "It is best obtained by using the vector identity (Eq. 1.80)". This vector identity is

##

\nabla \times (\nabla \times \vec{V}) = \nabla \nabla \cdot \vec{V} - \nabla \cdot \nabla \vec{V} .

##

- #6

Gold Member

- 2,003

- 1,064

Do you mean derivatives should be covariant derivatives, i.e.No. This expression is not generally covariant.

[tex]g^{jk}A^i_{:j:k}[/tex]

to include the general cases that ##\Gamma##s are not zero ? I was not sure whether OP deals with such general cases.

Last edited:

- #7

- 20,004

- 10,653

OP has stated that the question is about general curvilinear coordinates. In such coordinates the basis vectors change and therefore the connection coefficients are non-zero.Do you mean derivatives should be covariant derivatives, i.e.

[tex]g^{jk}A^i_{:j:k}[/tex]

to include the general cases that ##\Gamma##s are not zero ? I was not sure whether OP deals with such general cases.

- #8

Gold Member

- 2,003

- 1,064

Vector Laplacian https://mathworld.wolfram.com/VectorLaplacian.html (2)-(4) and

Tensor Laplacian https://mathworld.wolfram.com/TensorLaplacian.html (1)-(5).

Share:

- Replies
- 2

- Views
- 674

- Replies
- 20

- Views
- 2K

- Replies
- 11

- Views
- 681

- Replies
- 5

- Views
- 578

- Replies
- 1

- Views
- 491

- Replies
- 6

- Views
- 736

- Replies
- 11

- Views
- 864

- Replies
- 2

- Views
- 740

- Replies
- 4

- Views
- 562

- Replies
- 13

- Views
- 1K