Levi-Civita proofs for divergence of curls, etc

[theuserman]

I've also posted this in the Physics forum as it applies to some physical aspects as well.
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I want to know if I'm on the right track here. I'm asked to prove the following.

a) \nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})
b) \nabla \times (f \vec{A}) = f(\nabla \times \vec{A}) - \vec{A} \times (\nabla f) (where f is a scalar function)

And want (read: need to, due to a professor's insistence) to prove these using Levi-Civita notation. I've used the following for reference:
http://www.uoguelph.ca/~thopman/246/indicial.pdf and http://folk.uio.no/patricg/teaching/a112/levi-civita/

Here's my attempts - I need to see if I have this notation down correctly...

a) \nabla \cdot (\vec{A} \times \vec{B})

= \partial_i \hat{u}_i \cdot \epsilon_{jkl} \vec{A}_j \vec{B}_k \hat{u}_l

= \partial_i \vec{A}_j \vec{B}_k \hat{u}_i \cdot \hat{u}_l \epsilon_{jkl}

Now I thought it'd be wise to use the identity that \hat{u}_i \cdot \hat{u}_l = \delta_{il}.

= \partial_i \vec{A}_j \vec{B}_k \delta_{il} \epsilon_{jkl}

In which we make i = l (and the \delta_{il} goes to 1).

= \partial_i \vec{A}_j \vec{B}_k \epsilon_{jki}

Then using 'scalar derivative product rules' we get two terms. Now, here's where I get a little mixed up. I'm wondering if we rearrange the terms and then modify the epsilon to go in order the the terms.

= \vec{B}_k \partial_i \vec{A}_j \epsilon_{kij} + \vec{A}_j \partial_i \vec{B}_k \epsilon_{jik}

Now since the first epsilon is 'even' it remains positive, the other epsilon is 'odd' so that term becomes negative and we end up with the required result.

= \vec{B} (\nabla \times \vec{A}) - \vec{A} (\nabla \times \vec{B})

b) \nabla \times (f \vec{A}) (where f is a scalar function)

= \partial_i f \vec{A}_j \hat{u}_k \epsilon_{ijk}

= f \partial_i \vec{A}_j \hat{u}_k \epsilon_{ijk}+ \vec{A}_j \partial_i f \hat{u}_k \epsilon_{jik}

Once again, the first epsilon is the positive ('even') while the other is negative ('odd').

= f (\nabla \times \vec{A}) - \vec{A}(\nabla f)

Man, my hands hurt from all that tex work :P Been awhile for me.
Since my teacher refuses to tell me if this is the correct method (he's only willing to show the concepts, and while I can appreciate that I don't want my mark to go to hell), can anyone help me out?

[theuserman]

Some new equations I need to prove and have no idea how to go about it... I'll show what I've attempted so far.

a) \nabla (\vec{A} \cdot \vec{B}) = \vec{A} \times (\nabla \times \vec{B}) + \vec{B} \times (\nabla \times \vec{A}) + (\vec{A} \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla) \vec{A}

b) \nabla \times (\vec{A} \times \vec{B}) = (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B} + \vec{A}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{A})

Attempts:

a) \nabla (\vec{A} \cdot \vec{B})

= \delta_i (\vec{A}_j \hat{u}_j \cdot \vec{B}_k \hat{u}_k)

= \delta_i (\vec{A}_j \vec{B}_j)

= \vec{B}_j \delta_i \vec{A}_j + \vec{A}_j \delta_i \vec{B}_j

Now I'm thinking we do the other side of the equation and see if they add up??
1st term :
\vec{A} \times (\nabla \times \vec{B})

= \epsilon_{ijk} \vec{A}_j (\nabla \times \vec{B})_k

= \epsilon_{ijk} \vec{A}_j \epsilon_{klm} \nabla_l \vec{B}_m

= \epsilon_{kij} \epsilon_{klm} \vec{A}_j \nabla_l \vec{B}_m

= (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \vec{A}_j \nabla_l \vec{B}_m

So we make i=l, j=m and for the other case i=m and j=l

= \vec{A}_j \nabla_l \vec{B}_j - \vec{A}_l \nabla_l \vec{B}_m

= \vec{A}_j \partial_l \vec{B}_j - \vec{A}_l \partial_l \vec{B}_m

Which seems to resemble what I got when I did this normally...

2nd term : (I'm guessing we just switch the terms)
\vec{B} \times (\nabla \times \vec{A})

= \vec{B}_j \partial_l \vec{A}_j - \vec{B}_l \partial_l \vec{A}_m

3rd term:
(\vec{A} \cdot \nabla)\vec{B}

= (\vec{A}_i \cdot \nabla_j)\vec{B}_k

= (\vec{A}_i \nabla_i) \vec{B}_k

= \vec{A}_i \nabla_i \vec{B}_k

4th term:
(\vec{B} \cdot \nabla)\vec{A}

= \vec{B}_i \nabla_i \vec{A}_k

Adding them all up gets us:

\vec{A}_j \partial_l \vec{B}_j - \vec{A}_l \partial_l \vec{B}_m + \vec{B}_j \partial_l \vec{A}_j - \vec{B}_l \partial_l \vec{A}_m + \vec{A}_i \partial_i \vec{B}_k + \vec{B}_i \partial_i \vec{A}_k

Since most of these are dummy variables I can change them to...


\vec{A}_j \partial_l \vec{B}_j - \vec{A}_l \partial_l \vec{B}_m + \vec{B}_j \partial_l \vec{A}_j - \vec{B}_l \partial_l \vec{A}_m + \vec{A}_l \partial_l \vec{B}_m + \vec{B}_l \partial_l \vec{A}_m

Which is \vec{B}_j \delta_i \vec{A}_j + \vec{A}_j \delta_i \vec{B}_j !!!
HAH!

b) \nabla \times (\vec{A} \times \vec{B}) I'm pretty sure I have this down.

= \partial_l \hat{u}_l \times (\vec{A}_i \vec{B}_j \hat{u}_k \epsilon_{ijk})

=\partial_l \vec{A}_i \vec{B}_j \epsilon_{ijk} (\hat{u}_l \times \hat{u}_k)

[(\hat{u}_l \times \hat{u}_k) = \hat{u}_m \epsilon_{lkm}]

=\partial_l \vec{A}_i \vec{B}_j \hat{u}_m \epsilon_{ijk} \epsilon_{mlk}

=\partial_l \vec{A}_i \vec{B}_j \hat{u}_m (\delta_{im} \delta_{jl} - \delta_{il} \delta_{jm})

=\partial_j \vec{A}_i \vec{B}_j \hat{u}_i- \partial_i \vec{A}_i \vec{B}_j \hat{u}_j

Using scalar derivative scalar product rule:

= \vec{A}_i \partial_j \vec{B}_j \hat{u}_i + \partial_j \vec{B}_j \vec{A}_i \hat{u}_i - (\vec{A}_i \partial_i \vec{B}_j \hat{u}_j + \vec{B}_j \vec{A}_i \partial_i \hat{u}_j)

= \vec{A}(\nabla \cdot \vec{B}) + (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B} - \vec{B}(\nabla \cdot \vec{A})

Alright, so it took me 2 hours to type it but I managed to figure it out.
Thanks anyway.

[Phrak]

I've also posted this in the Physics forum as it applies to some physical aspects as well.
---

I want to know if I'm on the right track here. I'm asked to prove the following.

a) \nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})
...
a) \nabla \cdot (\vec{A} \times \vec{B})

= \partial_i \hat{u}_i \cdot \epsilon_{jkl} \vec{A}_j \vec{B}_k \hat{u}_l

= \partial_i \vec{A}_j \vec{B}_k \hat{u}_i \cdot \hat{u}_l \epsilon_{jkl}

Now I thought it'd be wise to use the identity that \hat{u}_i \cdot \hat{u}_l = \delta_{il}.


I haven't got a lot of wherewithall at the moment to put a lot into this, but I can help a little with equation a)

Change your equation to:

a) \nabla \cdot (\vec{A} \times \vec{B}) = \partial_i \hat{u}_i \cdot \epsilon_{jki} \vec{A}_j \vec{B}_k \hat{u}_i


Then using 'scalar derivative product rules' we get two terms. Now, here's where I get a little mixed up. I'm wondering if we rearrange the terms and then modify the epsilon to go in order the the terms.

= \vec{B}_k \partial_i \vec{A}_j \epsilon_{kij} + \vec{A}_j \partial_i \vec{B}_k \epsilon_{jik}

Now since the first epsilon is 'even' it remains positive, the other epsilon is 'odd' so that term becomes negative and we end up with the required result.

= \vec{B} (\nabla \times \vec{A}) - \vec{A} (\nabla \times \vec{B})


For my own work, when things get a little dicy, I define e.

\ \ e_{ijk} = 1 where i,j,k are odd (cyclic) permutations of (1,2,3) and otherwise the permutations are are 0.

\ e_{123} = e_{231} = e_{312} = 1

To make things a bit transparent, define \ \overline{e}

\ \overline{e}_{ijk} = -1 where i,j,k are odd permutations of (1,2,3) and otherwise are 0.

\ \overline{e}_{321} = \overline{e}_{213} = \overline{e}_{132} = -1

So you get

\epsilon_{ijk} = e_{ijk} + \overline{e}_{ijk}

[theuserman]

Phrak,
Thanks for the tip with \epsilon , as for your first suggestion it makes sense to make it i... Since it is a dummy variable.

Thanks.

[theuserman]

Just completeness, I'm going to prove that:


\nabla \times (\nabla \times \vec{v}) = \nabla(\nabla \cdot \vec{v}) - \nabla^2\vec{v}


So we have

\nabla \times (\nabla \times \vec{v})



= \partial_i(\nabla \times \vec{v})_j \epsilon_{ijk}



= \partial_i(\partial_l v_m \epsilon_{jlm})_j \epsilon_{ijk}



=\partial_i \partial_l v_m \epsilon_{jlm} \epsilon_{ijk}



= \partial_i \partial_l v_m (\delta_{lk}\delta_{mi} - \delta_{li}\delta_{mk})



= \delta_{lk}\delta_{mi} \partial_i \partial_l v_m - \delta_{li}\delta_{mk}\partial_i \partial_l v_m



= \partial_m \partial_l v_m - \partial_i \partial_i v_m



= \partial_l \partial_m v_m - \partial_i^2 v_m



= \nabla(\nabla \cdot \vec{v}) - \nabla^2 \vec{v}


Viola.