Question in the curl of a cross product.

[yungman]

The last equation in the previous post should be:
$$\nabla(V_1\cdot V_2)= V_1\times(\nabla\times V_2)<br /> +(V_1\cdot\nabla)V_2$$.

So

$$(\vec V_1\cdot\nabla)\vec V_2 = \nabla(\vec V_1\cdot \vec V_2) - \vec V_1\times(\nabla\times \vec V_2)$$

I coordinates independent?


How about

$$( \vec A \cdot \nabla) B \;\hbox { where B is a scalar function.}$$

 

[Rap]

How about

$$( \vec A \cdot \nabla) B \;\hbox { where B is a scalar function.}$$

It is coordinate independent.

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

 

[yungman]

It is coordinate independent.

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

Just the gradiant of B?

Thanks

 

[Rap]

Just the gradiant of B?

Thanks

Yes, np.

 

[yungman]

Once you know the method, the derivation amounts to just writing the formula: $$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)<br /> -{\vec V}\times(\nabla\times{\vec V})$$.
It uses the algebraic identity
$$a\times(b\times c)=b(a\cdot c)-c(a\cdot b).$$
Letting $$b=\nabla$$, $$a=V_1$$, $$c=V_2$$,
and considering $$V_1$$ to be a constant vector for the mathematical moment, you get:
$$\nabla(V_1\cdot V_2)=V_1\times(\nabla\times V_2)<br /> +(V_1\cdot\nabla)V_2$$.
Note: Since V_1 is constant, it must always be to the left of
$$\nabla$$ on the RHS. Do this again, considering V_2 constant. This gives the final answer.
I have tried to describe a method that takes several pages in the book, but I hope you can get the point.

I cannot verify this equation compare to Wolfram site.

Using your equation:

$$(\vec A \cdot \nabla) \vec B = \nabla ( \vec A \cdot \vec B) -\vec A \times ( \nabla \times \vec B)$$

$$\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z \;\Rightarrow \; \nabla (\vec A \cdot \vec B) = \hat x ( \frac {\partial A_x B_x} {\partial x} + \frac {\partial A_y B_y}{\partial x} + \frac {\partial A_z B_z}{\partial x}) + ...$$

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

$$\nabla \times \vec B = \hat x \left ( \frac {\partial B_z}{\partial y} - \frac {\partial B_y}{\partial z} \right ) + \hat y \left ( \frac {\partial B_x}{\partial z} - \frac {\partial B_z}{\partial x} \right ) + \hat z \left ( \frac {\partial B_y}{\partial x} - \frac {\partial B_x}{\partial y} \right )$$

$$\vec A \times (\nabla \times \vec B) = \hat x \left ( A_y\frac {\partial B_y}{\partial x} - A_y \frac {\partial B_x}{\partial y} + A_z \frac {\partial B_z}{\partial x} - A_z \frac {\partial B_x}{\partial z} \right ) + \hat y(...) + \hat z (...)$$

I only use the $\hat x$ only because that's enough to show the error already. I just show the total of the $\hat x$ components below:

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


Compare to Wolfram

http://mathworld.wolfram.com/ConvectiveOperator.html?affilliate=1


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

As you see, they don't match. Feel free to check my work shown above.

 

[Rap]

I cannot verify this equation compare to Wolfram site. Using your equation:

$$(\vec A \cdot \nabla) \vec B = \nabla ( \vec A \cdot \vec B) -\vec A \times ( \nabla \times \vec B)$$

Your work is good - you can see right here that the starting equation is not right. The del operator must never operate on A, as it is doing on the first term to the right of the equal sign. When A, B, and C are vectors, there is a sum term which is rewritten:

$$\sum_j A_j B_i C_j = B_i\sum_j A_j C_j$$

and that is correct, but this does not work when B is the del operator:

$$\sum_j A_j \partial_i C_j \ne \partial_i\sum_j A_j C_j$$

In other words, you cannot just substitute the del operator into what is a vector identity. The correct equation is

$$(\vec A \cdot \nabla) \vec B = (\nabla \vec B)\cdot A -\vec A \times ( \nabla \times \vec B)$$

where $\nabla \vec B$ is a tensor: $(\nabla \vec B)_{ij}=\partial_i B_j$

The incorrect step was in assuming $(\nabla \vec B)\cdot A=\nabla(\vec B \cdot \vec A)$ which is fine when $\nabla$ is replaced by a vector, but not when it is not.

I don't like messing around with these kinds of equations, it seems like there are a hundred different theorems, like recipes in a cookbook and, you try to solve problems by flipping through a bunch of recipes, hoping to find the right one. Using abstract index notation, you need only a few theorems, the notation is cleaner, and the core of the problem is isolated from the particular vectors or vector operators you are using. The problem is, its something new and different.

In abstract index notation (which is ultimately coordinate-system free!), the dot product is written

$$(\vec A \cdot \vec B)=A_i B_i$$

where double indices imply summation. The cross product is written:

$$(\vec A \times \vec B)_i=\epsilon_{ijk}A_j B_k$$

where $\epsilon_{ijk}$ is the permutation pseudotensor (1 for even permutations of 123, -1 for odd permutations of 123 and zero otherwise). The above theorem is written:

$$A_i \partial_i B_j=A_j\partial_i B_j-\epsilon_{ijk}A_i(\epsilon_{kmn}\partial_m B_n)=A_j\partial_i B_j-\epsilon_{ijk}\epsilon_{kmn}A_i\partial_m B_n$$

The theorem solves itself when you realize one important theorem, that has only to do with the permutation pseudotensor, not the components of the vectors or operators involved:

$$\epsilon_{ijk}\epsilon_{kmn}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$$

where $\delta_{ij}$ is the Kronecker delta, 1 when i=j, zero otherwise. Of course, when the del operator $(\partial_i)$ is involved you have to be careful to keep track of what it is operating on. $\partial_i A_j B_k$ is nebulous, it should be written $\partial_i (A_j) B_k$ or $B_k \partial_i A_j$ when the del operator is operating only on A. When operating on both it should be written $\partial_i(A_j B_k)$

 

[yungman]

Thanks for the detail reply. I have not studied tensors yet, so I guess at this point I wasted enough time on this, I should just use the Wolfram formulas and be done with it.

Yes the reason I spend like two hours work through the proof, triple check my work was because I didn't feel good in replacing a vector with the $\nabla$.

Do you have any easy derivation of:

$$(\nabla \cdot \vec A)f = \vec A \cdot \nabla f \;\hbox { where f is a scalar function. }$$

Thanks

 

[Rap]

Thanks for the detail reply. I have not studied tensors yet, so I guess at this point I wasted enough time on this, I should just use the Wolfram formulas and be done with it.

Yes the reason I spend like two hours work through the proof, triple check my work was because I didn't feel good in replacing a vector with the $\nabla$.

Do you have any easy derivation of:

$$(\nabla \cdot \vec A)f = \vec A \cdot \nabla f \;\hbox { where f is a scalar function. }$$

Thanks

I think you mean

$$(\vec A \cdot \nabla)f=\vec A \cdot \nabla f$$

The above symbolic statement is true by definition - the notation $A \cdot \nabla$ operating on f is defined to mean $A \cdot \nabla f$. In abstract index notation, its true by the rules:

$$(A_i\partial_i)f=A_i(\partial_i f) = A_i \partial_i f$$

More explicitly, in Cartesian coordinates:

$$\left(\sum_i A_i \frac{\partial}{\partial x_i}\right)f=\sum_i A_i \frac{\partial f}{\partial x_i}$$

 

[yungman]

Thanks.