How to approach vector calculus identities?

[fatpotato]

Homework Statement

Prove the relation $\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A\cdot (\nabla \times B)$

Relevant Equations

Definition of the vector product, definition of the divergence.

Majoring in electrical engineering imply studying Griffiths book on electrodynamics, so I have begun reading its first chapter, which is a review of vector calculus. A list of vector calculus identities is given, and I would like to derive each one, with one of them being $\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A\cdot (\nabla \times B)$.

Now, I have two questions about this :

1. Methodology question : I have seen a lot of threads here on PF about deriving these identitites, and Einstein's summation notation and Levi-Civita symbol are mentioned every time. Given that I will be majoring in electrical engineering and not in physics, can a mere engineer learn these notations or does this involve higher unreachable mathematical concepts?
2. Actual question about the identity : I have seen an identity $A \cdot (B \times C)$ involving a determinant, but since $\nabla$ is an operator and not an actual vector, does it even make sense to use the identity?

I would consider using $A \cdot (B \times C) = det \left(\begin{array}{ccc} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{array} \right)$, which would yield in this case $\nabla \cdot (A \times B) = det \left(\begin{array}{ccc} \partial_x & \partial_y & \partial_z \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array} \right)$, am I allowed to put $\nabla$ in this matrix?

Thank you.

 

[MidgetDwarf]

What you are referring to A*(B X C) is called the scalar triple product. The keyword here is scaler.

Its a simple derivation (mostly calculation) using the definition of the dot product and cross product.

If you want to study Griffiths, then take a look at Vector Calculus, Linear Algebra, and Differential Forms by Hubbard and Hubbard.

I like it better than Mardsen: Vector Calculus.

 

[pasmith]

You cna prove these identities by expanding everything in cartesian coordinates. The summation convention allows you to use less paper.

 

[PeroK]

Now, I have two questions about this :

1. Methodology question : I have seen a lot of threads here on PF about deriving these identitites, and Einstein's summation notation and Levi-Civita symbol are mentioned every time. Given that I will be majoring in electrical engineering and not in physics, can a mere engineer learn these notations or does this involve higher unreachable mathematical concepts?
2. Actual question about the identity : I have seen an identity $A \cdot (B \times C)$ involving a determinant, but since $\nabla$ is an operator and not an actual vector, does it even make sense to use the identity?

I would say:

Don't worry about Levi-Civita for now. But, do (certainly) use the determinant form of the cross product and curl.

$nabla$ has some vector properties, so in principle vector identities translate to $\nabla$ identities. You should be able to check that one you quoted is valid.

If you have a vector identity, then do it component by component. Or, do it for the x-component and appeal to the right-handed symmetry of x-y-z coordinates.

Stick with Griffiths - there's all you need in those first 58 pages!

 

[fatpotato]

If you want to study Griffiths, then take a look at Vector Calculus, Linear Algebra, and Differential Forms by Hubbard and Hubbard.

Thank you, though I don't feel I will ever reach this level, it looks like a good reference.

The summation convention allows you to use less paper.

Since I have to steal paper sheets from my school's printer, I will definitely give it a try!

I would say:

Don't worry about Levi-Civita for now. But, do (certainly) use the determinant form of the cross product and curl.

has some vector properties, so in principle vector identities translate to identities. You should be able to check that one you quoted is valid.

If you have a vector identity, then do it component by component. Or, do it for the x-component and appeal to the right-handed symmetry of x-y-z coordinates.

Stick with Griffiths - there's all you need in those first 58 pages!

Thank you PeroK, your messages always cheer me up when my stupidity brings me down.

For anyone interested, here is the method used. The first term obtained by developing the determinant expression gives a partial derivative which can be expanded using the product rule :
$$\frac{\partial}{\partial x}A_yB_z = A_y\frac{\partial}{\partial x}B_z + B_z\frac{\partial}{\partial x}A_y$$
Then, one can find each matching term from $B\cdot (\nabla \times A)$ and $- A\cdot(\nabla \times B)$ and conclude that the expression does in fact hold, which is maybe less strenuous than expanding everything.

Edit : missing cross product symbol

 

[SammyS]

Edit : missing cross product symbol

The cross product symbol is simply " \times ", which I see you have found by now.

##B\cdot (\nabla \times A)##

gives

$B\cdot (\nabla \times A)$

 

[PhDeezNutz]

https://web.iitd.ac.in/~pmvs/courses/mcl702/notation.pdf

the above link is a pretty good primer to teach you the basics of summation notation.

 

[robphy]

A helpful hint with such identities...

To start, write everything out in components.
Be consistent with the algebra and use the cyclic nature x->y->z->x with the same sign.
Swapping exactly two will introduce a minus sign.

It might help to arrange your calculation into sections (and not just write it out on a single line).
After a while, you see the patterns and then appreciate the index notations... and maybe differential forms.

 

[fatpotato]

Thank you for the subsequent replies to this post.

the above link is a pretty good primer to teach you the basics of summation notation.

Thank you very much for the reference, I will try to get accustomed while studying EM.

A helpful hint with such identities...

To start, write everything out in components.
Be consistent with the algebra and use the cyclic nature x->y->z->x with the same sign.
Swapping exactly two will introduce a minus sign.

I believe you are mentioning tensor notation? Thank you for the advice, I will keep that in mind.

 

[Dr Transport]

@Orodruin wrote this for our site

https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/