# Vector analysis question. Laplacian of scalar and vector field

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• LagrangeEuler
In summary: 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.
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}##?

LagrangeEuler said:
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} .
##

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

LagrangeEuler said:
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.
anuttarasammyak said:
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.

julian said:
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.

Orodruin said:
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:
anuttarasammyak said:
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

## 1. What is vector analysis?

Vector analysis is a branch of mathematics that deals with the study of vectors, which are quantities that have both magnitude and direction. It involves the use of mathematical operations such as addition, subtraction, and multiplication to analyze and manipulate vectors.

## 2. What is the Laplacian of a scalar field?

The Laplacian of a scalar field is a mathematical operator that measures the rate of change of a scalar quantity at a given point in space. It is represented by the symbol ∇² and is calculated by taking the sum of the second derivatives of the scalar field with respect to each coordinate axis.

## 3. What is the Laplacian of a vector field?

The Laplacian of a vector field is a mathematical operator that measures the divergence of a vector field at a given point in space. It is represented by the symbol ∇² and is calculated by taking the sum of the second derivatives of each component of the vector field with respect to each coordinate axis.

## 4. What is the significance of the Laplacian in vector analysis?

The Laplacian is an important tool in vector analysis as it allows us to determine the behavior of scalar and vector fields at a given point in space. It is used in various applications, such as fluid dynamics, electromagnetism, and quantum mechanics, to analyze the behavior of physical quantities.

## 5. How is the Laplacian of a scalar and vector field calculated?

The Laplacian of a scalar or vector field can be calculated using the appropriate formula, depending on the type of field. For a scalar field, it is calculated by taking the sum of the second derivatives of the field with respect to each coordinate axis. For a vector field, it is calculated by taking the sum of the second derivatives of each component of the field with respect to each coordinate axis. Both of these calculations involve the use of the Laplacian operator, represented by the symbol ∇².

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