Calculate grad(A dot B)

[Addez123]

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:
$$(2xz + z^2, 3y^2, x^2 + 2xz)$$
which is the correct answer.

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.

 

[Orodruin]

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.

 

[pasmith]

$(B \cdot \nabla)A$
$$(B \cdot \nabla) = 1$$

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

 

[Addez123]

$(B \cdot \nabla)A$ 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.

 

[PhDeezNutz]

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.

 

[PhDeezNutz]

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”

 

[Addez123]

$(B \cdot \nabla)A$ 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.

 

[Orodruin]

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 …

 

[Orodruin]

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]

@Addez123

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

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

 

[Addez123]

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.

 

[alex_Hou]

o,i calculate this way:
p =A \cdot B =x^2z+y^3+z^2x,
the answer should be Jacobian
(\frac{\partial p}{\partial x},\frac{\partial p}{\partial y},\frac{\partial p}{\partial z})
=(2xz+z^2,3y^2,x^2+2xz)

 

[Orodruin]

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

This was done in the OP already.