In summary, using the convective operator gets you a different answer than using the material derivative.f

Homework Statement
$$A = (x^2, y^2, z^2)$$
$$B = (z, y, x)$$

Calculate $$grad(A \cdot B)$$
Relevant Equations
$$\nabla (A \cdot B) = (B \cdot \nabla)A + (A \cdot \nabla)B + B \times (\nabla \times A) + A \times (\nabla \times B)$$
Calculating dot product then doing gradient on it gets you:
$$(2xz + z^2, 3y^2, x^2 + 2xz)$$

Using the formula, which you're required to do, gets a whole different answer.
Lets do each term individually.
##(B \cdot \nabla)A##
$$(B \cdot \nabla) = 1$$

## (A \cdot \nabla)B##
$$(A \cdot \nabla) = 2(x + y + z)$$

For the cross product terms, ##(\nabla \times A)## and ##(\nabla \times B)## both gets you the zero vector, which cross with anything still just gives zero.

So you're left with
$$1A + 2(x + y + z)B = (x^2 + 2xz + 2yz +2z^2, 2xy + 3y^2 + 2yz, 2x^2 + 2xy + 2xz + z^2)$$

Which is nothing like the answer. I've recalculated every single piece of this equation 10 times and I can testify that the equation given in Relevant Equations is false.

What makes you think that ##\vec B\cdot\nabla=1##? It should be a differential operator.

Edit: Note that it is not ##\nabla
\cdot \vec B##, which is equal to one.

• topsquark
##(B \cdot \nabla)A##
$$(B \cdot \nabla) = 1$$

$(B \cdot \nabla)A$ means $$B_x \frac{\partial A}{\partial x} + B_y\frac{\partial A}{\partial y} + B_z \frac{\partial A}{\partial z}.$$

• topsquark, PhDeezNutz, Addez123 and 1 other person
$(B \cdot \nabla)A$ means $$B_x \frac{\partial A}{\partial x} + B_y\frac{\partial A}{\partial y} + B_z \frac{\partial A}{\partial z}.$$
well that's a confusing use of parentheses.

well that's a confusing use of parentheses.

We’ll get used to it. If you are going into physics you’re going to need it.

For instance the force on an electric dipole ##\vec{p}## is ## \vec{F} = \left( \vec{p} \cdot \nabla \right) \vec{E}## where ##\vec{E}## is the external field that the dipole is immersed in.

But yeah it is perplexing when you first come across it. You’re unlikely to come across it in a first multivariable calculus class. I’d bet money you’re doing Griffiths E&M now.

• topsquark
You apply the expression ## \vec{p} \cdot \nabla## to each component of ##\vec{E}##

I believe this expression in general is called the “Material Derivative”

$(B \cdot \nabla)A$ means $$B_x \frac{\partial A}{\partial x} + B_y\frac{\partial A}{\partial y} + B_z \frac{\partial A}{\partial z}.$$
are you sure its not
$$(B \cdot \nabla)A = (B_x \frac{\partial A}{\partial x} , B_y\frac{\partial A}{\partial y} , B_z \frac{\partial A}{\partial z})$$
?
Because the answer is suppose to be a vector.

@PhDeezNutz not doing Griffiths rn, but good guess! :p
Funny fact I've completed electro magnetic course with almost an A, but it was years ago and I never actually passed the vector.

Because the answer is suppose to be a vector.
If ##A## is a vector then so is ##\partial A/\partial x## etc.

well that's a confusing use of parentheses.
Not as confusing as having differential operators acting to the left …

I believe this expression in general is called the “Material Derivative”
It is not. The material derivative is the derivative of some quantity along the flow of some velocity field ##\vec u## and is given by
$$\frac D{Dt} = \partial_t + \vec u \cdot \nabla$$

• PhDeezNutz

The full expression for ##\left( \vec{A} \cdot \nabla \right) \vec{B} ## is the following

##\left( \vec{A} \cdot \nabla \right) \vec{B} = \left( A_x \frac{\partial }{\partial x} + A_y \frac{\partial }{\partial y} + A_z \frac{\partial }{\partial z} \right) \left( B_x, B_y, B_z \right)##

Distributing that entire operation over each of the components

##\left( \vec{A} \cdot \nabla \right) \vec{B} = \left( A_x \frac{\partial B_x}{\partial x} + A_y \frac{\partial B_x }{\partial y} + A_z \frac{\partial B_x}{\partial z} , A_x \frac{\partial B_y}{\partial x} + A_y \frac{\partial B_y }{\partial y} + A_z \frac{\partial B_y}{\partial z} , A_x \frac{\partial B_z}{\partial x} + A_y \frac{\partial B_z }{\partial y} + A_z \frac{\partial B_z}{\partial z} \right)##

• Thanks a ton, just figured that out and now I get the correct answer.
Thanks! :)

• PhDeezNutz
It is not. The material derivative is the derivative of some quantity along the flow of some velocity field ##\vec u## and is given by
$$\frac D{Dt} = \partial_t + \vec u \cdot \nabla$$

I guess someone on wiki messed up

https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates Yup they did, it's called the convective operator apparently.

o,i calculate this way:
p =A \cdot B =x^2z+y^3+z^2x,