What is the difference between $(\phi\vec{\nabla}) \cdot \vec{A}$ and $\phi(\vec{\nabla}\cdot\vec{A})$?

Question

$\vec{\nabla} \cdot (\phi\vec{A}) = (\phi\vec{\nabla}) \cdot \vec{A} + \phi(\vec{\nabla}\cdot\vec{A})$

But what is the difference between $(\phi\vec{\nabla}) \cdot \vec{A}$ and $\phi(\vec{\nabla}\cdot\vec{A})$ ?

I cannot get that. They are different, but when I try to work it out, they both becomes the same.

Please state the difference between the above two quantities and please provide an example.

I am in my First Semester of Physics Undergrad course.

Answer 1

I am not familiar with the notation you are using, but it is clear to me that $\phi \nabla$ is meant to be the derivative of $\phi$. If one uses einstein notation, we can see that the given equation should follow from just the 1D product rule you are familiar with, $$ \partial_i (\phi A_i) = \partial_i\phi A_i + \phi(\partial_i A_i) $$
Since clearly $\partial_iA_i = \nabla\cdot A$, we are forced to conclude that $\partial_i \phi = \phi \nabla$.

Note that this is distinct from the term in (say) the material derivative, which in coordinates is

$$ [(u\cdot \nabla) u]_i = u_j\partial_ju_i$$

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user470992 has kindly requested a computed example with the functions $$A = \begin{pmatrix} 3xyz^2 \\ + 2xy^3\\ - x^2yz \end{pmatrix}$$ and $\phi = 3x^2 - yz$ ; calculating $\nabla \cdot ( \phi \vec{A}) $ at $p=(1,-1,1)^T$ : $$\nabla \phi|_p = \left.\begin{pmatrix} 6x \\-z\\-y \end{pmatrix}\right|_p = \begin{pmatrix} 6 \\-1\\1 \end{pmatrix} \\ \nabla \cdot A|_p = (3yz^2+6xy^2-x^2y)|_p = -3+6+1=4$$ This gives $ 6(-3) + (-1)(-2) + 1(1) + 4(4)=-15+16 = 1$.

Answer 2

It should be $${\nabla} \cdot (\phi\vec{A}) = ({\nabla_\phi \phi}) \cdot \vec{A} + \phi({\nabla}_A\cdot\vec{A})$$

where the frist $\nabla_\phi$ works on $\phi$, and the second $\nabla_A$ works on $\vec{A}$.

Answer 3

The equation

$\vec \nabla \cdot (\phi \vec A) = (\phi \vec \nabla)\cdot \vec A + \phi \vec \nabla \cdot \vec A \tag 1$

is wrong as written. The correct expression for $\vec \nabla \cdot (\phi \vec A)$ is

$\vec \nabla \cdot (\phi \vec A) = (\vec \nabla \phi )\cdot \vec A + \phi \vec \nabla \cdot \vec A. \tag 2$

(2) may be verified by evaluating the terms in a cartesian coordinate system. If

$A = (A_x, A_y, A_z), \tag 3$

then

$\dfrac{\partial}{\partial x}(\phi A_x) = \dfrac{\partial \phi}{\partial x} A_x + \phi \dfrac{\partial A_x}{\partial x}, \tag 4$

$\dfrac{\partial}{\partial y}(\phi A_y) = \dfrac{\partial \phi}{\partial y} A_y + \phi \dfrac{\partial A_y}{\partial y}, \tag 5$

$\dfrac{\partial}{\partial z}(\phi A_z) = \dfrac{\partial \phi}{\partial z} A_z + \phi \dfrac{\partial A_z}{\partial z}; \tag 6$

summing (4)-(6) yields

$\vec \nabla \cdot (\phi A) = \dfrac{\partial}{\partial x}(\phi A_x) + \dfrac{\partial}{\partial y}(\phi A_y) + \dfrac{\partial}{\partial z}(\phi A_z)$ $= \dfrac{\partial \phi}{\partial x} A_x + \dfrac{\partial \phi}{\partial y} A_y + \dfrac{\partial \phi}{\partial z} A_z + \phi \dfrac{\partial A_x}{\partial x} + \phi \dfrac{\partial A_y}{\partial y} + \phi \dfrac{\partial A_z}{\partial z} = \vec \nabla \phi \cdot \vec A + \phi \vec \nabla \cdot \vec A, \tag 7$

validating (2).

The usual notation is that operators act from left to right. Therefore,

$(\phi \vec \nabla)\cdot \vec A = \phi \vec \nabla \cdot \vec A, \tag 8$

as my easily seen via operator associativity or direct evaluation in a cartesian frame as was done above in the case of (2); there is in fact no essential difference 'twixt $(\phi \vec \nabla) \cdot \vec A$ and $\phi \vec \nabla \cdot \vec A$. Thus, (1) also may read

$\vec \nabla \cdot (\phi \vec A) = 2\phi \vec \nabla \cdot \vec A, \tag 9$

which omits the derivatives of $\phi$ required by the Leibniz rule for differentiating products.

For example, if we take

$\vec A = (x, y, z) \tag{10}$

and

$\phi = y, \tag{11}$

we obtain

$\vec \nabla \cdot (\phi \vec A) = \vec \nabla \cdot (xy, y^2, yz) = y + 2y + y = 4y, \tag{12}$

$\vec \nabla \phi \cdot \vec A = (0, 1, 0) \cdot (x, y, z) = y, \tag{13}$

$\phi \vec \nabla \cdot A = y \vec \nabla \cdot (x, y, z) = 3y; \tag{14}$

one sees that (2) holds here. Evaluating (1) or (9) however yields

$4y = 2y(3) = 6y, \tag{15}$

which is not, in general, the case.