Vector differential identity proof

[_jo_]

Hi,
I am a engineering student and I am currently upgrading my maths level on my own to follow physics courses. While reading a book, I came across a vector differential identity that I don't manage to prove using index notation.
The identity is the following:
$$\nabla(\vec{A}\cdot\vec{B}) =<br /> \vec{A} \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B}<br /> + \vec{B} \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}$$
Could you please give me a hint on how to prove this ?
Thank you for your time.

 

[lurflurf]

Why do you want to use index notation? In any case what makes this formula a bit strange is that it can be cast into many forms and that is in some sense not the natural one.

For example the identity

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

shows we can express each of those four terms in terms of the others
also we can chose to use or not use dyatic terms

two links
https://www.physicsforums.com/showthread.php?t=297027
https://www.physicsforums.com/showthread.php?t=273630

 

[_jo_]

Thank you, I wanted to use index notation because it proved really usefull for developping vector identities. But if this identity's proof does not lend itself to this notation, I don't see the necessity to use it.

In one of your links, you posted this:

$$\nabla (\vec{A} \cdot \vec{B}) = \nabla_{\vec{A}}(\vec{A} \cdot \vec{B}) + \nabla_{\vec{B}}(\vec{A} \cdot \vec{B})$$
$$\nabla_{\vec{A}}(\vec{A} \cdot \vec{B}) = B \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}$$
$$\nabla_{\vec{B}}(\vec{A} \cdot \vec{B}) = A \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B}$$
$$\nabla(\vec{A}\cdot\vec{B}) = <br /> \vec{A} \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B} + \vec{B} \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}$$

But how do you develop the 2^nd and 3^rd lines from the partial gradient ?

 

[lurflurf]

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

follows from the vector algebra identity

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

or in index form the identity (not guaranteed typo free)

δ_ilδ_jk=ε_mjlε_imk+δ_ijδ_kl=ε_ijmε_mlk+δ_ikδ_jl

can be applied to

δ_ilδ_jka_jb_k,l

Care must be taken when using algebra identities to deduce calculus identities, mainly because by normal convention differential operators are right acting so we must not change multiplication order. We would not like to make an error like

$$\nabla\cdot b=b\cdot\nabla$$

It is also possible and convenient to change to bidirectional differential operators and back so that manipulations can be unrestricted.

 

[_jo_]

Thanks a lot ! That was really helpful !