Question in the curl of a cross product.

[yungman]

This might be math problem, but I only see it in EM books.

$$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$$.

What is $$\vec A \cdot \nabla$$?

 

[Rap]

This might be math problem, but I only see it in EM books.

$$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$$.

What is $$\vec A \cdot \nabla$$?

In Cartesian coordinates:

$$\vec A \cdot \nabla = A_x\frac{\partial }{\partial x}+A_y\frac{\partial }{\partial y}+A_z\frac{\partial }{\partial z}$$

It has to be applied to something. For example, a scalar function f:

$$(\vec A \cdot \nabla) f = A_x\frac{\partial f }{\partial x}+A_y\frac{\partial f}{\partial y}+A_z\frac{\partial f}{\partial z}$$

Thats the same as $$\vec A \cdot (\nabla f)$$ but when applied to a vector, its not.

 

[yungman]

In Cartesian coordinates:

$$\vec A \cdot \nabla = A_x\frac{\partial }{\partial x}+A_y\frac{\partial }{\partial y}+A_z\frac{\partial }{\partial z}$$

It has to be applied to something. For example, a scalar function f:

$$(\vec A \cdot \nabla) f = A_x\frac{\partial f }{\partial x}+A_y\frac{\partial f}{\partial y}+A_z\frac{\partial f}{\partial z}$$

Thats the same as $$\vec A \cdot (\nabla f)$$ but when applied to a vector, its not.

Thanks for the reply.

But in this case, it is applied to a vector! What does this mean?

Thanks

 

[lurflurf]

Ignore that it works fine applied to a vector. You can see by the chain rule that if A is a rate of change the convective operator is a rate of change as well. It arises often not only in EM but also in fluids, heat, and many other areas.

http://mathworld.wolfram.com/ConvectiveOperator.html

The four product derivatives are related by an identity so one only needs to deal with three, though yours is not often chosen to be avoided.

$$\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}$$

 

[netheril96]

In Cartesian coordinates:

$$\vec A \cdot \nabla = A_x\frac{\partial }{\partial x}+A_y\frac{\partial }{\partial y}+A_z\frac{\partial }{\partial z}$$

It has to be applied to something. For example, a scalar function f:

$$(\vec A \cdot \nabla) f = A_x\frac{\partial f }{\partial x}+A_y\frac{\partial f}{\partial y}+A_z\frac{\partial f}{\partial z}$$

Thats the same as $$\vec A \cdot (\nabla f)$$ but when applied to a vector, its not.

No, you are wrong.

$$(\mathbf{A}\cdot\nabla) \mathbf{B}= \mathbf{A}\cdot(\nabla\mathbf{B})$$

$$\nabla\mathbf{B}$$ is a second-order tensor.

 

[Meir Achuz]

No, you are wrong.

$$(\mathbf{A}\cdot\nabla) \mathbf{B}= \mathbf{A}\cdot(\nabla\mathbf{B})$$

$$\nabla\mathbf{B}$$ is a second-order tensor.

They are equivalent forms.
$$(\mathbf{\hat A}\cdot\nabla)$$
is called the "directional derivative". It is the rate of cchange of anything (scalar, vector, ...) in the A direction.
It is most useful acting on the vector r, where it gives
$$(\mathbf{A}\cdot\nabla) \mathbf{r}=\mathbf{A}$$.

 

[Rap]

No, you are wrong.

$$(\mathbf{A}\cdot\nabla) \mathbf{B}= \mathbf{A}\cdot(\nabla\mathbf{B})$$

$$\nabla\mathbf{B}$$ is a second-order tensor.

Absolutely right. I should have said it was more complicated.

 

[yungman]

OK, this is part of an exercise, now I read the Wolfram link and I "peek" at the solution manual:


Given by the solution manual:

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

Which is in agreement with Mathworld Wolfram (2) in the link.

If you look at the equation again, it is like:

$$(\vec A \cdot \nabla) \;=\; (\hat x A_x + \hat y A_y + \hat z A_z ) \cdot (\hat x \frac {\partial}{\partial x} +\hat y \frac {\partial}{\partial y} +\hat z \frac {\partial}{\partial z})$$

But I cannot figure out this in cylindrical and spherical coordinates.

 

[Rap]

OK, this is part of an exercise, now I read the Wolfram link and I "peek" at the solution manual:Given by the solution manual:

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

Which is in agreement with Mathworld Wolfram (2) in the link.

If you look at the equation again, it is like:

$$(\vec A \cdot \nabla) \;=\; (\hat x A_x + \hat y A_y + \hat z A_z ) \cdot (\hat x \frac {\partial}{\partial x} +\hat y \frac {\partial}{\partial y} +\hat z \frac {\partial}{\partial z})$$

But I cannot figure out this in cylindrical and spherical coordinates.

I need $$(A \cdot \nabla)B$$ in cylindrical coordinates too, if anyone can find it. I am calculating it, but its very complicated, and I will need a check against my calculations.

 

[yungman]

I need $$(A \cdot \nabla)B$$ in cylindrical coordinates too, if anyone can find it. I am calculating it, but its very complicated, and I will need a check against my calculations.

go to post #4 and follow the link.

 

[Meir Achuz]

I need $$(A \cdot \nabla)B$$ in cylindrical coordinates too, if anyone can find it. I am calculating it, but its very complicated, and I will need a check against my calculations.

Are you sure you do need it in cylindrical coordinates?
$$(A \cdot \nabla)B$$ can usually be evaluated without a coordinate system and then the system introduced.
What is the equation you need it for?

 

[yungman]

Are you sure you do need it in cylindrical coordinates?
$$(A \cdot \nabla)B$$ can usually be evaluated without a coordinate system and then the system introduced.
What is the equation you need it for?

Can you show us? That's exactly what I am looking for.

Thanks

 

[Rap]

Are you sure you do need it in cylindrical coordinates?
$$(A \cdot \nabla)B$$ can usually be evaluated without a coordinate system and then the system introduced.
What is the equation you need it for?

I need to calculate $$\rho (\vec V \cdot \nabla) \vec V$$ in the Boltzmann fluid equation for conservation of momentum. I am sure I want it in cylindrical coordinates. I also want to have the general statement for future reference.

 

[Rap]

go to post #4 and follow the link.

Bingo. That's it, thanks.

Edit - and it matches my calculations! This is rare. LOL - makes me worry about Wolfram.

 

[Meir Achuz]

I need to calculate $$\rho (\vec V \cdot \nabla) \vec V$$ in the Boltzmann fluid equation for conservation of momentum. I am sure I want it in cylindrical coordinates. I also want to have the general statement for future reference.

$$(\vec V \cdot \nabla) \vec V=\vec V$$ in any coordinate system.

 

[yungman]

$$(\vec V \cdot \nabla) \vec V=\vec V$$ in any coordinate system.

Any magic formula for my case?

 

[Meir Achuz]

Any magic formula for my case?

What does A.del act on in your case?

 

[yungman]

What does A.del act on in your case?

Acting on B. I am hopping to get the general formula so I don't have to rely blindly on the Mathworld Wolfram formulas for different coordinate systems. I have not manage to find any definition of Convection Operator so for now, the only way to solve any problem is to blindly using the formulas.

Thanks

 

[Rap]

$$(\vec V \cdot \nabla) \vec V=\vec V$$ in any coordinate system.

I don't see how that's possible. If V is constant, $$(\vec V \cdot \nabla) \vec V=0$$ for example, because any spatial derivative of V is zero.

 

[Meir Achuz]

I don't see how that's possible. If V is constant, $$(\vec V \cdot \nabla) \vec V=0$$ for example, because any spatial derivative of V is zero.

I'm sorry. I thought he meant Del with respect to V.
You can write $$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)<br /> -{\vec V}\times(\nabla\times{\vec V})$$, if that helps.
If a specific form of V is given,
$$(\vec V \cdot \nabla) \vec V$$ can usually be worked out directly.

 

[Rap]

I'm sorry. I thought he meant Del with respect to V.
You can write $$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)<br /> -{\vec V}\times(\nabla\times{\vec V})$$, if that helps.
If a specific form of V is given,
$$(\vec V \cdot \nabla) \vec V$$ can usually be worked out directly.

Thanks, that does help, because I am dealing with irrotational flow $$\nabla \times \vec V=0$$.

 

[yungman]

I'm sorry. I thought he meant Del with respect to V.
You can write $$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)<br /> -{\vec V}\times(\nabla\times{\vec V})$$, if that helps.
If a specific form of V is given,
$$(\vec V \cdot \nabla) \vec V$$ can usually be worked out directly.

Yes, this is what I am looking for. Can you kindly put down the derivation also? I want to learn how it come about.

Many thanks.

 

[Rap]

Yes, this is what I am looking for. Can you kindly put down the derivation also? I want to learn how it come about.

I don't think you want my calculations - they are very inelegant and messy. I know there is a general way to calculate these things that is very elegant and much more informative. Check out http://en.wikipedia.org/wiki/Curvilinear_coordinates

The case of cylindrical coordinates is probably the simplest case possible. The fundamental thing is the Jacobian.

 

[yungman]

I don't think you want my calculations - they are very inelegant and messy. I know there is a general way to calculate these things that is very elegant and much more informative. Check out http://en.wikipedia.org/wiki/Curvilinear_coordinates

The case of cylindrical coordinates is probably the simplest case possible. The fundamental thing is the Jacobian.

What I meant is the derivation of:

$$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)-{\vec V}\times(\nabla\times{\vec V})$$

I believe Meir Achuz said it is independent to coordinate systems.

 

[Rap]

What I meant is the derivation of:

$$(\vec V \cdot \nabla) \vec V=(1/2)\nabla(V^2)-{\vec V}\times(\nabla\times{\vec V})$$

I believe Meir Achuz said it is independent to coordinate systems.

I would do it using Einstein notation. It boils everything down a bunch of easy steps and one hard step. Einstein notation means sum over every repeated index (they only show up alone or in pairs) $\partial_i$ means $\partial/\partial x_i$ where $x_1,x_2,x_3=x,y,z$. The components of a vector A are given by $A_i$

Dot product
$$\vec A \cdot \vec B \rightarrow A_i B_i$$

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

where $\epsilon_{ijk}$ is the permutation pseudotensor. Its equal to 1 for even permutations of 123 (e.g. 231, 312, etc.) and -1 for odd permutations (132, 321, etc.)

So what we want to prove is

$$V \times (\nabla \times V)=\frac{1}{2}\nabla V^2-(V\cdot \nabla) V$$

In Einstein notation, thats

$$\epsilon_{ijk}V_j(\epsilon_{k\ell m}\partial_\ell V_m) = \frac{1}{2}\partial_i (V_j V_j) - V_j\partial_j V_i$$

Collecting some terms and realizing that $\partial_i (V_j V_j)=2V_j\partial_i V_j$ we get:

$$\epsilon_{ijk}\epsilon_{k\ell m}V_j \partial_\ell V_m = V_j\partial_i V_j - V_j\partial_j V_i$$

Now comes the hard part. We have to realize that:

$$\epsilon_{ijk}\epsilon_{k\ell m}=\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}$$

where $\delta_{ij}$ is 1 when i=j, zero otherwise. The rest is easy. For example

$$\delta_{i\ell}V_j\partial_\ell V_m=V_j\partial_i V_m$$

I know that last step is not straightforward, but if you think about it, its true, and its independent of all the vectors and partial derivatives, its the core of the proof.

 

[Meir Achuz]

Yes, this is what I am looking for. Can you kindly put down the derivation also? I want to learn how it come about.
Many thanks.

There is a derivation of this and similar vector differential identities without introducing coordinate systems in the first chapter of Franklin's "Classical Electromagnetism".

 

[yungman]

There is a derivation of this and similar vector differential identities without introducing coordinate systems in the first chapter of Franklin's "Classical Electromagnetism".

Can you please write them out here? I don't have the book.

Is it a good book? I have another question in another thread that I am not happy with all the books that I have on the coordiantes transformation. Even as good as Griffiths don't get into that.

Thanks

Alan

 

[Meir Achuz]

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.

 

[Meir Achuz]

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

 

[Rap]

There is a derivation of this and similar vector differential identities without introducing coordinate systems in the first chapter of Franklin's "Classical Electromagnetism".

Coordinate systems are not introduced using this notation. See http://en.wikipedia.org/wiki/Abstract_index_notation