# Vector identity question

1. Apr 15, 2015

### majormuss

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

2. Apr 15, 2015

### Delta²

if $\vec{A}=(A_x,A_y,A_z)$ then $\vec{A}\cdot\nabla=A_x\frac{\partial}{\partial x}+A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}$ that is A⋅∇ is still an operator and can be applied to a scalar or a vector. For example applied to scalar φ, $(\vec{A}\cdot\nabla)\phi=A_x\frac{\partial\phi}{\partial x}+A_y\frac{\partial\phi}{\partial y}+A_z\frac{\partial\phi}{\partial z}$

3. Apr 16, 2015

### majormuss

So they are equal? Since it is multiplication, they are the same then?

4. Apr 16, 2015

### Delta²

No they are not equal, $\nabla\cdot\vec{A}$ is not an operator itself , its the result of the operator $\nabla$applied to vector $\vec{A}$ via the dot product and it is a scalar quantity refered as the divergence of vector A. It is $\nabla\cdot\vec{A}=\frac{\partial{A_x}}{\partial x}+\frac{\partial{A_y}}{\partial y}+\frac{\partial{A_z}}{\partial z}$.

5. Apr 16, 2015

### majormuss

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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6. Apr 16, 2015

### Fredrik

Staff Emeritus
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)?
$$(\vec A\times\vec B)_i=\varepsilon_{ijk} A_j B_k.$$

7. Apr 16, 2015

### Delta²

For that you can also use the definition of the operator ∇ in spherical coordinates and the identity $\nabla\times(A\times B)=A(\nabla\cdot B)-B(\nabla\cdot A)+(B\cdot\nabla)A-(A\cdot\nabla)B$ which can be simplified in this case where $A=m$ is a constant vector, by setting the 2 medium terms to zero.

Last edited: Apr 16, 2015
8. Apr 17, 2015

### Delta²

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 $\vec{m}\cdot\nabla$ in spherical coordinates. The derivatives of the spherical unit vectors are not always zero.