Vector Calculus Identities: Proving v · ∇v = ∇(0.5v2 + c × v)

[MaxManus]

Homework Statement

Show that v\nablav = \nablaxvxv
v · ∇v = ∇(0.5v² + c × v

c=∇ × v

My attempt
∇(A · B)= B · ∇A + A · ∇B + B×(∇×A) + A×(∇×B)

Replace A and B with V

∇(v · v)= v · ∇v + v · ∇v + v×(∇×v) + v×(∇×v)

v · ∇v = ∇(0.5v² - v×(∇×v)

Is v×(∇×v) = =∇ × v × v?
And am I on the right track?

I can't fint such a rule in my textbook

 

[lanedance]

doesn't vxv = 0?

 

[gabbagabbahey]

Homework Statement

Show that v\nablav = \nablaxvxv
v · ∇v = ∇(0.5v² + c × v

c=∇ × v

What you've written doesn't make much sense...Do you mean v\matbf{\nabla}v=(\mathbf{\nabla}\times \textbf{v})\times\textbf{v}+(\textbf{v}\cdot\mathbf{\nabla})\textbf{v} ?

My attempt
∇(A · B)= B · ∇A + A · ∇B + B×(∇×A) + A×(∇×B)

Replace A and B with V

∇(v · v)= v · ∇v + v · ∇v + v×(∇×v) + v×(∇×v)

v · ∇v = ∇(0.5v² - v×(∇×v)

Is v×(∇×v) = =∇ × v × v?
And am I on the right track?

I can't fint such a rule in my textbook

Well, \textbf{a}\times\textbf{b}=-\textbf{b}\times\textbf{a}, so you tell us whether or not (\mathbf{\nabla}\times \textbf{v})\times\textbf{v}=\textbf{v}\times(\mathbf{\nabla}\times \textbf{v})

 

[MaxManus]

Thanks to both of you.
The first line is wrong

I was suppose to show that
v · \nablav = ∇\nabla(0.5v²) + c × v

c=∇\nabla× v.
I then usend
∇\nabla(A · B)=B · \nablaA + A · \nablaB + B×(\nabla×A) + A×(\nabla×B)

Replaced A and B with v

And got

v\nablav = \nabla(0.5v²) - v×(\nabla×v)

Using A×B = - B×A

- v×(\nabla×v) = (\nabla×v)×v

And I have:

v · \nablav = ∇\nabla(0.5v²) + c × v

c=∇\nabla× v.

Am I right?

Thanks again for all the help

 

[gabbagabbahey]

Looks good to me

 

[MaxManus]

Thanks.