Question in the curl of a cross product.

AI Thread Summary
The discussion revolves around the mathematical expression of the curl of a cross product in electromagnetism, specifically the operator \(\vec A \cdot \nabla\). Participants clarify that \(\vec A \cdot \nabla\) acts as a directional derivative and can be applied to both scalar and vector fields, with different implications. The conversation also touches on the complexity of calculating \((\vec A \cdot \nabla) \vec B\) in cylindrical coordinates, with requests for general formulas and derivations. A specific example related to fluid dynamics is discussed, emphasizing the need for accurate calculations in the context of the Boltzmann fluid equation. The importance of understanding these concepts across different coordinate systems is highlighted throughout the thread.
yungman
Messages
5,741
Reaction score
294
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?
 
Physics news on Phys.org
yungman said:
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.
 
Rap said:
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
 
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}
 
Rap said:
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.
 
netheril96 said:
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}.
 
Last edited:
netheril96 said:
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.
 
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.
 
Last edited:
yungman said:
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.
 
  • #10
Rap said:
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.
 
  • #11
Rap said:
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?
 
  • #12
Meir Achuz said:
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
 
  • #13
Meir Achuz said:
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.
 
  • #14
yungman said:
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.
 
Last edited:
  • #15
Rap said:
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.
 
Last edited:
  • #16
Meir Achuz said:
(\vec V \cdot \nabla) \vec V=\vec V in any coordinate system.

Any magic formula for my case?
 
  • #17
yungman said:
Any magic formula for my case?
What does A.del act on in your case?
 
  • #18
Meir Achuz said:
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
 
  • #19
Meir Achuz said:
(\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.
 
  • #20
Rap said:
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.
 
  • #21
Meir Achuz said:
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.
 
  • #22
Meir Achuz said:
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.
 
  • #23
yungman said:
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.
 
  • #24
Rap said:
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.
 
  • #25
yungman said:
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.
 
  • #26
yungman said:
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".
 
  • #27
Meir Achuz said:
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
 
  • #28
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.
 
Last edited:
  • #29
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.
 
  • #30
Meir Achuz said:
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
 
  • #31
Meir Achuz said:
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.}
 
  • #32
yungman said:
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
 
  • #33
Rap said:
It is coordinate independent.

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

Just the gradiant of B?

Thanks
 
  • #34
yungman said:
Just the gradiant of B?

Thanks

Yes, np.
 
  • #35
Meir Achuz said:
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.
 
  • #36
yungman said:
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)
 
Last edited:
  • #37
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
 
  • #38
yungman said:
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}
 
Last edited:
  • #39
Thanks.
 
Back
Top