Vector Laplacian!

1. Apr 4, 2007

Swapnil

They say that vector laplacian is defined as the following:

$$\nabla^2 \vec{A} = \nabla(\nabla\cdot\vec{A}) - \nabla\times(\nabla \times\vec{A})$$

Is the above definition true for all coordinate systems or just for cartesian coordinate system?
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Also, wikipedia say the following can be used to evaluate the laplacian of a vector:

$$\nabla^2 \vec{A} = (\nabla^2 A_x, \nabla^2 A_y, \nabla^2 A_z)$$

Is this only true of in cartesian coordinates or can a similar form also be used to evaluate the laplacian of a vector in other coordinate systems? For example, would the following be correct in spherical coordinate system?:

$$\nabla^2 \vec{A} = (\nabla^2 A_r(r,\theta,\phi), \nabla^2 A_{\theta}(r,\theta,\phi), \nabla^2 A_{\phi}(r,\theta,\phi))$$

where the scalar laplacian operator is given in spherical coordinates(i.e. it is calculated by taking the divergence of the gradient in spherical coordiantes).
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Also, take a look at the following link:
http://www.uic.edu/classes/eecs/eecs520/textbook/node23.html#curl_curl

How does the following equation make sense?

$$\nabla\times\nabla = \nabla\nabla\circ -\nabla\circ\nabla$$

The expression on the left gives you a vector, the first expression to the right also gives you a vector but then the second expression on the right gives you a scalar!? How can two vectors add up to a scalar?

Last edited by a moderator: Apr 22, 2017
2. Apr 4, 2007

StatMechGuy

If the Laplacian is to make sense at all, any identity proven for cartesian coordinates should definitely hold for any other coordinate system.

3. Apr 4, 2007

Manchot

The first definition is independent of coordinate system. The second definition (the Laplacian of each component) is most definitely coordinate-dependent.

4. Apr 5, 2007

lpfr

When I learned what a laplacian was, it was still a scalar operator and was defined as:
$$\nabla^2=\nabla\cdot\nabla$$
It is a scalar operator and can be applied to a scalar giving a scalar or to a vector giving a vector.
I do not think there is such thing as "vector laplacian"

5. Apr 5, 2007

Swapnil

Well, there is. And it crops up in a lot of different places like Maxwell's wave equation for the E-field for instance.
http://en.wikipedia.org/wiki/Vector_laplacian

BTW, I have also seen gradient of a vector!

Last edited: Apr 5, 2007
6. Apr 5, 2007

7. Apr 5, 2007

masudr

*spoiler*
Indeed. Further verification in the 7th row of the table featured in this Wikipedia article.

EDIT: The article contains the solution to the exercise in Ida's Electromagnetic Engineering, pg. 113. So the above is a spoiler in that sense.

Last edited: Apr 5, 2007
8. Apr 6, 2007

lpfr

First: you forgot a parenthesis:
$$\nabla\times\nabla = \nabla(\nabla\circ\ )-\nabla\circ\nabla$$
Second: it is not an equation. It is a recipe to use an operator. At this stage you do not have vectors, just "vector operators" at left and right sides. You must apply the operators to obtain vectors or scalars.

9. Apr 6, 2007

Swapnil

I know, but the rightmost operator has two weird things going on: First, if we assume that it operates on a vector, then it is taking the gradient of a vector!, and second, if we assume that the output/result of the operator is a vector then it still doesn't make any sense because divergence takes a vector and gives you a scalar not a vector!

10. Apr 6, 2007

lpfr

$$\nabla\times(\nabla\times\ \ ) = \nabla(\nabla\circ\ )-(\nabla\circ\nabla)$$
Better now?

11. Apr 6, 2007

Swapnil

I don't know how this makes things better. We still have to operate all the operators from right to left and you would still have the problem I mentioned in the previous post.

12. Apr 6, 2007

robphy

It may be helpful to look at this using tensorial-notation.

13. Dec 2, 2008

bcorso1

You can see what is meant by the gradient of a vector by expanding everything out in it's basis form
(this is less elegant, however, since the laplacian changes in different coordinate systems).

The easiest example is Cartesian Coordinates:

The gradient of a scalar is a vector:
$$\nabla {f}(x,y,z)= (\hat{e_{x}} \frac{\partial }{\partial x}+ \hat{e_{y}} \frac{\partial }{\partial y}+ \hat{e_{z}} \frac{\partial }{\partial z}) f(x,y,z)$$

$$=\hat{e_{x}}\frac{\partial f}{\partial x}+ \hat{e_{y}}\frac{\partial f}{\partial y}+ \hat{e_{z}}\frac{\partial f}{\partial z}$$

Likewise the gradient of a vector is a matrix/tensor:
$$\nabla\overline{u}(x,y,z)= (\hat{e_{x}} \frac{\partial }{\partial x}+ \hat{e_{y}} \frac{\partial }{\partial y}+ \hat{e_{z}} \frac{\partial }{\partial z}) (\hat{e_{x}}u_{x}+ \hat{e_{y}}u_{y}+ \hat{e_{z}}u_{z})$$

$$=\hat{e_{x}} \hat{e_{x}}\frac{\partial u_{x}}{\partial x}+ \hat{e_{x}} \hat{e_{y}}\frac{\partial u_{y}}{\partial x}+ \hat{e_{x}} \hat{e_{z}}\frac{\partial u_{z}}{\partial x}+ \hat{e_{y}} \hat{e_{x}}\frac{\partial u_{x}}{\partial y}+ \hat{e_{y}} \hat{e_{y}}\frac{\partial u_{y}}{\partial y}+ \hat{e_{y}} \hat{e_{z}}\frac{\partial u_{z}}{\partial y}+ \hat{e_{z}} \hat{e_{x}}\frac{\partial u_{x}}{\partial z}+ \hat{e_{z}} \hat{e_{y}}\frac{\partial u_{y}}{\partial z}+ \hat{e_{z}} \hat{e_{z}}\frac{\partial u_{z}}{\partial z}$$

The coefficient in front of each $$\hat{e_{i}} \hat{e_{j}}$$ can be thought of as representing the matrix component $$\nabla\overline{u}_{i,j}$$ in the cartesian basis.

The same thing can be done in other coordinate systems to get the matrix elements in that basis. However, be careful because:
(1) $$\nabla$$ changes is each coordinate system
(2) the basis vectors in other coordinate systems can be functions of the variables
(e.g. in cylindrical coordinates $$\hat{e_{r}}=cos(\theta)\hat{e_{x}}+sin(\theta)\hat{e_{y}}$$ , so $$\frac{\partial e_{r}}{\partial \theta}\neq 0$$)

Hope this helps!