Proving vector identity using levi-civita tensor help

[ATOMICJOCK]

Using the fact that we can write the vector cross-product in the form: (A× B)i =ε ijk Aj Bk ,
where εijk is the Levi-Civita tensor show that:

∇×( fA) = f ∇× A− A×∇f ,

where A is a vector function and f a scalar function.


Could you please be as descriptive as possible; as I'm not sure how to even start this problem.

 

[Pengwuino]

Well you can rewrite the left hand side as $$\epsilon_{ijk}\partial_j(fA_k)$$ and act with the derivative. The Levi-Civita tensor is a constant so the partial does not act on it.

 

[ATOMICJOCK]

PENGUINO,

Thanks for the reply; but can you provide this answer as a step by step solution please? I'm a visual person, and prefer to see everything so that i can learn.

 

[Pengwuino]

It's a two step problem. The derivative is acting on a product of two functions, what does a derivative do when it acts $${{\partial} \over {\partial x}}(f(x)*g(x))$$?

 

[ATOMICJOCK]

but A is a vector function, and f is a scalar function...I'm sorry i haven't had much sleep, could you please write everything out.

 

[Pengwuino]

but A is a vector function, and f is a scalar function...I'm sorry i haven't had much sleep, could you please write everything out.

This will only be helpful if you do the work yourself. The beauty of index notation is that $$A_i$$ means the scalar part of the i-th component.

So if you have a vector $$A = 3xy{\bf{\hat x}} + 9zx{\bf{\hat y}}$$, the $$A_x$$ signifies the x-component of the vector A, 3xy.

 

[ATOMICJOCK]

Lol I still don't understand. I mean i know the basic definition of the Levi-Civita Tensor, but that's it...realy need detailed step for this question to understand it.

 

[hunt_mat]

Okay, you know that you were told that:

$$\nabla\times (f\mathbf{A})=\epsilon_{ijk}\partial_j(fA_k)$$

So just use the product rule to show that:

$$\epsilon_{ijk}\partial_j(fA_k)=\epsilon_{ijk}f\partial_{j}A_{k}+\epsilon_{ijk}A_{k}\partial_{j}f$$

and you should also know that:

$$(\mathbf{A}\times\mathbf{B})_{i}=\epsilon_{ijk}A_{j}B_{k}$$

Right? can you continue from here?

 

[ATOMICJOCK]

It makes sense, but would be nice to see the completed end result, brain just isn't working at the moment.

 

[hunt_mat]

One more hint and then you're on your own:

$$(\nabla\times\mathbf{A})_{i}=\epsilon_{ijk}\partial_{j}A_{k}$$

 

[ATOMICJOCK]

Thanks very much for the help; although more lines would help.

 

[hunt_mat]

I won't just give you the answer, that is against the board rules and I have been stung once before. The tips I have given you are enough for you to quite easily do the problem. My suggestion is that you try and put these lines together and we can correct you on your ideas. We want to help but we won't do the problems for you.