Vector differential identity proof (using triple product)

[Earthland]

Homework Statement

Need to prove that:

(v⋅∇)v=(1/2)∇(v⋅v)+(∇×v)×v

Homework Equations

Vector triple product

(a×b)×c=-(c⋅b)a+(c⋅a)b

The Attempt at a Solution

I know I could prove that simply by applying definitions directly to both sides. I haven't done that because that is tedious, and I strive for a more elegant proof and thought that triple product would give it to me.

Since we are dealing with operators the order of the vectors is important and applying triple product would give me

(v⋅∇)v=(v⋅v)∇+(∇×v)×v

But it seems to me that (v⋅v)∇≠ (1/2)∇(v⋅v). It can't be right, because (v⋅v)∇still needs to be applied to something, while the left hand of the equation is something definite. And I can't see where the 1/2 should come from.

Is it wrong to apply triple product to this problem? How else could I prove this identity? Any help?

 

[Orodruin]

It is wrong to apply the triple product in the way you have done, you need to keep track of which v the differential operator is acting on, which should be the same before and after applying the identity.

 

[Earthland]

I can't quite understand what you mean. I kept track of the order according to triple product definition, but what does it even mean to keep track on the v that ∇ is acting on? I can't just make it act on v where it isn't acting on v.

 

[Orodruin]

You cannot simply treat the nabla as if it was a vector. You can do this for the vector structure, but not for what function the differential operator is acting on. The differential operator is always going to act on the v that corresponds to the v that was standing to the right of the nabla in your original expression. Thus the term with both v to the left of the nabla has no meaning unless you realize that the differential operators in this nabla are actually acting on the components of one of the vs.

 

[Ray Vickson]

I can't quite understand what you mean. I kept track of the order according to triple product definition, but what does it even mean to keep track on the v that ∇ is acting on? I can't just make it act on v where it isn't acting on v.

No, but you have made it not act on v where it should!

 

[Earthland]

You cannot simply treat the nabla as if it was a vector. You can do this for the vector structure, but not for what function the differential operator is acting on. The differential operator is always going to act on the v that corresponds to the v that was standing to the right of the nabla in your original expression. Thus the term with both v to the left of the nabla has no meaning unless you realize that the differential operators in this nabla are actually acting on the components of one of the vs.

That's what I thought also, but I blindly followed the triple product structure provided by wikipedia. So should I just change it to

∇(v⋅v)

But still, 1/2 is missing

 

[Orodruin]

Yes, in there the nabla is acting on both vs. You have to relate this to the case where it is only acting on one of them.

This is why I clearly prefer tensor/index notation for deriving vector relations. You get rid of these ambiguities that appear from trying to apply vector relations to nabla operators.

 

[Earthland]

Both v-s produce a scalar field and thus ∇ acting on it produces a vector field, which it must be because another vector is added to it. ∇ acting on only one v means acting on vector field which would produce a tensor field. So there is some kind of relation I should use?

 

[Orodruin]

You can either note that the gradient of v^2 is going to be symmetric in terms of which v the nabla acts, or you could simply apply index notation.