- 11

- 0

I've also posted this in the Math forum as it is math as well.

---

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

a) [tex]\nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})[/tex]

b) [tex]\nabla \times (f \vec{A}) = f(\nabla \times \vec{A}) - \vec{A} \times (\nabla f)[/tex] (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 [Broken] 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) [tex]\nabla \cdot (\vec{A} \times \vec{B}) [/tex]

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

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

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

= [tex]\partial_i \vec{A}_j \vec{B}_k \delta_{il} \epsilon_{jkl}[/tex]

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

= [tex]\partial_i \vec{A}_j \vec{B}_k \epsilon_{jki} [/tex]

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.

= [tex] \vec{B}_k \partial_i \vec{A}_j \epsilon_{kij} + \vec{A}_j \partial_i \vec{B}_k \epsilon_{jik}[/tex]

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.

= [tex] \vec{B} (\nabla \times \vec{A}) - \vec{A} (\nabla \times \vec{B}) [/tex]

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

= [tex] \partial_i f \vec{A}_j \hat{u}_k \epsilon_{ijk}[/tex]

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

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

= [tex] f (\nabla \times \vec{A}) - \vec{A}(\nabla f)[/tex]

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?

---

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

a) [tex]\nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})[/tex]

b) [tex]\nabla \times (f \vec{A}) = f(\nabla \times \vec{A}) - \vec{A} \times (\nabla f)[/tex] (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 [Broken] 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) [tex]\nabla \cdot (\vec{A} \times \vec{B}) [/tex]

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

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

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

= [tex]\partial_i \vec{A}_j \vec{B}_k \delta_{il} \epsilon_{jkl}[/tex]

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

= [tex]\partial_i \vec{A}_j \vec{B}_k \epsilon_{jki} [/tex]

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.

= [tex] \vec{B}_k \partial_i \vec{A}_j \epsilon_{kij} + \vec{A}_j \partial_i \vec{B}_k \epsilon_{jik}[/tex]

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.

= [tex] \vec{B} (\nabla \times \vec{A}) - \vec{A} (\nabla \times \vec{B}) [/tex]

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

= [tex] \partial_i f \vec{A}_j \hat{u}_k \epsilon_{ijk}[/tex]

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

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

= [tex] f (\nabla \times \vec{A}) - \vec{A}(\nabla f)[/tex]

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?

Last edited by a moderator: