MHB Vector Index Notation Proof: Solved by Sam

[SamJohannes]

Hi everyone!

I've got a vector index notation proof that I'm struggling with.

View attachment 3457

(sorry ignore the c, that's the question number)
I've simplified it u * (del X del)
and from there I've sort of assumed del X del = 0. Is that right and if so could somebody please explain it? Else any help on the correct proof would help.

Cheers, Sam

 

[Ackbach]

Hi everyone!

I've got a vector index notation proof that I'm struggling with.

View attachment 3457

(sorry ignore the c, that's the question number)
I've simplified it u * (del X del)
and from there I've sort of assumed del X del = 0. Is that right and if so could somebody please explain it? Else any help on the correct proof would help.

Cheers, Sam

Your proof seems somewhat incoherent. Let $\vec{u}=\langle u_1,u_2,u_3 \rangle$. Then:
\begin{align*}
\nabla\times\vec{u}&=\langle \partial_y u_3-\partial_z u_2, -\partial_x u_3+\partial_z u_1,
\partial_x u_2-\partial_y u_1 \rangle \\
\nabla\cdot(\nabla\times\vec{u})&=\partial_x(\partial_y u_3-\partial_z u_2)
+\partial_y(-\partial_x u_3+\partial_z u_1)+\partial_z(\partial_x u_2-\partial_y u_1).
\end{align*}
Can you finish? (You might need Clairaut's Theorem.)

 

[SamJohannes]

Thanks Ackback!

You showed clear steps and I understand the rest just cancels out via Clairaut's thm.

I didn't make it clear that the question asks for the solution in tensor notation. So is this a perfectly viable way to prove it in this manner?

Cheers

 

[Ackbach]

Thanks Ackback!

You showed clear steps and I understand the rest just cancels out via Clairaut's thm.

I didn't make it clear that the question asks for the solution in tensor notation. So is this a perfectly viable way to prove it in this manner?

Cheers

This is definitely not tensor notation. You should be using the Levi-Civita symbol for the cross product and the Einstein Summation Convention thus:
$$(\nabla\times\vec{u})_i=\varepsilon_{ijk}\partial_ju_k.$$
That gives you the $i$th component of the cross product. Then, taking the divergence yields
$$\nabla\cdot(\nabla\times\vec{u})=\varepsilon_{ijk}\partial_i\partial_ju_k.$$
At this point, the contracted epsilon identity is not available to you, because you don't have two Levi-Civita symbols. But, you can work with Clairaut's Theorem and the properties of the Levi-Civita symbol to get the final result.

 

[SamJohannes]

Sorry about any lack of clarity. Since the start of this post I learned a little bit about Latex so here's my attempt.

If got to here, which is what I had at that beginning but I poorly explained it.
$$\varepsilon_{ijk}\partial_i\partial_ju_k=u_k\varepsilon_{kij}\partial_i\partial_j$$
I guess what I'm asking is that this notation implies I am crossing two del operators and if you following that through it does what you said and cancels out via clairaut's thm. I'm just unsure as to whether this cross operation is legal, or even makes sense in maths terms?

EDIT:
The above question still stands but I think I the following is the answer:
$$\partial_i\varepsilon_{ijk}\partial_ju_k= -\partial_i\varepsilon_{jik}\partial_ju_k = -\partial_j\varepsilon_{jik}\partial_iu_k $$
Which implies the negative equals the positive and therefore the only solution is 0. Sound okay?

 

[Ackbach]

Sorry about any lack of clarity. Since the start of this post I learned a little bit about Latex so here's my attempt.

If got to here, which is what I had at that beginning but I poorly explained it.
$$\varepsilon_{ijk}\partial_i\partial_ju_k=u_k\varepsilon_{kij}\partial_i\partial_j$$

That last expression is definitely not equal to the first. Both partial derivative signs MUST stay on the left of the $u_k$, since they are operators.

I guess what I'm asking is that this notation implies I am crossing two del operators and if you following that through it does what you said and cancels out via clairaut's thm. I'm just unsure as to whether this cross operation is legal, or even makes sense in maths terms?

EDIT:
The above question still stands but I think I the following is the answer:
$$\partial_i\varepsilon_{ijk}\partial_ju_k= -\partial_i\varepsilon_{jik}\partial_ju_k = -\partial_j\varepsilon_{jik}\partial_iu_k $$
Which implies the negative equals the positive and therefore the only solution is 0. Sound okay?

You're very close, I think. You've got the right idea. Just a clean-up:
\begin{align*}
\nabla\cdot(\nabla\times\vec{u})&=\varepsilon_{ijk}\partial_i \partial_j u_k \\
&=-\varepsilon_{jik}\partial_i \partial_j u_k \\
&=-\varepsilon_{jik}\partial_j \partial_i u_k \\
&=-\nabla\cdot(\nabla\times\vec{u}).
\end{align*}
I think you do need to relate it back to the original expression.