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

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.

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  • $\begingroup$ Perhaps there is a typo. Assuming some form of product rule, I would guess the equality $\vec{\nabla} \cdot (\phi\vec{A}) = (\vec{\nabla}\phi) \cdot \vec{A} + \phi(\vec{\nabla}\cdot\vec{A})$ instead. $\endgroup$
    – edm
    Oct 28, 2017 at 17:01

3 Answers 3

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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 $\vec{A}$.

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  • $\begingroup$ Yeah... I get that... thanks! But please give me an example of these. Also, I don't thing \nabla will work on \phi. \phi is written at left of the grad operator. Operators don't work left-ward. Only in the direction of right. Correct me if I am wrong. $\endgroup$
    – physrito
    Oct 28, 2017 at 17:08
  • $\begingroup$ $\nabla$ Operators is different from vector, It is meaningful only when they work on some symbols. In fact it is a derivative operation,When the product is derivatived, It calculates the derivative of the two items separately, and then sums. So the derivative of $\phi $ is expressed $\nabla \phi$ ,rather than $ \phi \nabla$ $\endgroup$ Oct 28, 2017 at 17:25
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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

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


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$.

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  • $\begingroup$ No, \phi\nabla is not meant to be a derivative of \phi. \nabla is the gradient operator. $\endgroup$
    – physrito
    Oct 28, 2017 at 17:05
  • $\begingroup$ @user470992 I'm aware what a gradient operator is, but if we agree that a divergence is written $\nabla \cdot F = \partial_i F_i$ then I have done nothing wrong and it is only a matter of notation. $\endgroup$ Oct 28, 2017 at 17:08
  • $\begingroup$ I knew you knew. I just gave it to be clear. $\endgroup$
    – physrito
    Oct 28, 2017 at 17:13
  • $\begingroup$ @user470992 Also, the gradient is the derivative. At least, it is, if you don't care about how things behave under coordinate transforms, so maybe this is to fix some sort of notational inconsistency elsewhere? $\endgroup$ Oct 28, 2017 at 17:15
  • $\begingroup$ I am doing exercise from Spiegel's Vector Analysis. I don't think there will be a notational anomaly. $\endgroup$
    – physrito
    Oct 28, 2017 at 17:16
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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.

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