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They say that vector laplacian is defined as the following:

[tex]\nabla^2 \vec{A} = \nabla(\nabla\cdot\vec{A}) - \nabla\times(\nabla \times\vec{A})[/tex]

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:

[tex]\nabla^2 \vec{A} = (\nabla^2 A_x, \nabla^2 A_y, \nabla^2 A_z)[/tex]

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?:

[tex]\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))[/tex]

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?

[tex]\nabla\times\nabla = \nabla\nabla\circ -\nabla\circ\nabla[/tex]

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?

[tex]\nabla^2 \vec{A} = \nabla(\nabla\cdot\vec{A}) - \nabla\times(\nabla \times\vec{A})[/tex]

Is the above definition true for all coordinate systems or just for cartesian coordinate system?

--- --- ---

Also, wikipedia say the following can be used to evaluate the laplacian of a vector:

[tex]\nabla^2 \vec{A} = (\nabla^2 A_x, \nabla^2 A_y, \nabla^2 A_z)[/tex]

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?:

[tex]\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))[/tex]

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).

--- --- ---

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?

[tex]\nabla\times\nabla = \nabla\nabla\circ -\nabla\circ\nabla[/tex]

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?

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