# Vector analysis question. Laplacian of scalar and vector field

• A
• LagrangeEuler

#### LagrangeEuler

If we define Laplacian of scalar field in some curvilinear coordinates ## \Delta U## could we then just say what ##\Delta## is in that orthogonal coordinates and then act with the same operator on the vector field ## \Delta \vec{A}##?

If we define Laplacian of scalar field in some curvilinear coordinates ## \Delta U## could we then just say what ##\Delta## is in that orthogonal coordinates and then act with the same operator on the vector field ## \Delta \vec{A}##?

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} .
##

#### Attachments

• anuttarasammyak
It is expressed in tensor analysis
$$g^{jk}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^k} A^i$$ or
$$g^{jk}A^i_{,j,k}$$

If we define Laplacian of scalar field in some curvilinear coordinates ## \Delta U## could we then just say what ##\Delta## is in that orthogonal coordinates and then act with the same operator on the vector field ## \Delta \vec{A}##?
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.

It is expressed in tensor analysis
$$g^{jk}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^k} A^i$$ or
$$g^{jk}A^i_{,j,k}$$
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.

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} .
##
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.

No. This expression is not generally covariant.
Do you mean derivatives should be covariant derivatives, i.e.
$$g^{jk}A^i_{:j:k}$$
to include the general cases that ##\Gamma##s are not zero ? I was not sure whether OP deals with such general cases.

Last edited:
Do you mean derivatives should be covariant derivatives, i.e.
$$g^{jk}A^i_{:j:k}$$
to include the general cases that ##\Gamma##s are not zero ? I was not sure whether OP deals with such general cases.
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.

• anuttarasammyak