Vector identity question

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Does ∇⋅A = A ⋅∇? If not then, what does the latter actually equal?
 

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  • #2
Delta2
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if [itex]\vec{A}=(A_x,A_y,A_z)[/itex] then [itex]\vec{A}\cdot\nabla=A_x\frac{\partial}{\partial x}+A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}[/itex] that is A⋅∇ is still an operator and can be applied to a scalar or a vector. For example applied to scalar φ, [itex](\vec{A}\cdot\nabla)\phi=A_x\frac{\partial\phi}{\partial x}+A_y\frac{\partial\phi}{\partial y}+A_z\frac{\partial\phi}{\partial z}[/itex]
 
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if [itex]\vec{A}=(A_x,A_y,A_z)[/itex] then [itex]\vec{A}\cdot\nabla=A_x\frac{\partial}{\partial x}+A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}[/itex] that is A⋅∇ is still an operator and can be applied to a scalar or a vector. For example applied to scalar φ, [itex](\vec{A}\cdot\nabla)\phi=A_x\frac{\partial\phi}{\partial x}+A_y\frac{\partial\phi}{\partial y}+A_z\frac{\partial\phi}{\partial z}[/itex]
So they are equal? Since it is multiplication, they are the same then?
 
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No they are not equal, [itex]\nabla\cdot\vec{A}[/itex] is not an operator itself , its the result of the operator [itex]\nabla[/itex]applied to vector [itex]\vec{A}[/itex] via the dot product and it is a scalar quantity refered as the divergence of vector A. It is [itex]\nabla\cdot\vec{A}=\frac{\partial{A_x}}{\partial x}+\frac{\partial{A_y}}{\partial y}+\frac{\partial{A_z}}{\partial z}[/itex].
 
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No they are not equal, [itex]\nabla\cdot\vec{A}[/itex] is not an operator itself , its the result of the operator [itex]\nabla[/itex]applied to vector [itex]\vec{A}[/itex] via the dot product and it is a scalar quantity refered as the divergence of vector A. It is [itex]\nabla\cdot\vec{A}=\frac{\partial{A_x}}{\partial x}+\frac{\partial{A_y}}{\partial y}+\frac{\partial{A_z}}{\partial z}[/itex].
Nice thank you! Are you familiar with vector potentials in physics? I am working on proving the following but I have been running into trouble. Trying to prove that ∇xA=B!
 

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Fredrik
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I assume that what you want to prove is the second equality in the second line. The proof is fairly long and tedious. You have to use a definition of the cross product, and the product rule. What definition of the cross product are you using? Are you familiar with this one (using the Levi-Civita symbol and Einstein's summation convention)?
$$(\vec A\times\vec B)_i=\varepsilon_{ijk} A_j B_k.$$
 
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For that you can also use the definition of the operator ∇ in spherical coordinates and the identity [itex]\nabla\times(A\times B)=A(\nabla\cdot B)-B(\nabla\cdot A)+(B\cdot\nabla)A-(A\cdot\nabla)B[/itex] which can be simplified in this case where [itex]A=m[/itex] is a constant vector, by setting the 2 medium terms to zero.
 
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I found out this is not so long afterall if one knows how to do it in spherical coordinate system. However extra caution is needed when working with operators like [itex]\vec{m}\cdot\nabla[/itex] in spherical coordinates. The derivatives of the spherical unit vectors are not always zero.
 
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