please help me sci not hmwk

>>7727445
Nobody here has taken calc 2.

>>7727445

Do you know how to use index notation, proving vector identities with tensors is a lot easier.

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>>7727445
OP here i'll post what I got to, I know the curl of the gradient of a scalar is 0 what next?

>>7727445
The i-th component of your expression is
[eqn] \epsilon_{ijk}\partial_j(\phi\partial_k\phi) = \epsilon_{ijk}[(\partial_j\phi)(\partial_i\phi) + \phi\partial_{ij}\phi][/eqn]

The [math]\epsilon[/math] is anti-symmetric in j,k and the bracket is symmetric.

>>7727456
I have used a bit of index notation but I'm not super comfortable with it

>>7727478
Hey! The fuck is this! Where's my cool looking math?!

Let's try again
[eqn] \epsilon_{ijk}\partial_j(\phi\partial_k\phi) = \epsilon_{ijk}\[(\partial_j\phi)(\partial_i\phi) + \phi\partial_{ij}\phi\][/eqn]

>>7727490
Fuck it, you are on your own OP

[eqn] \epsilon_{ijk}\partial_j(\phi\partial_k\phi) = \epsilon_{ijk}[(\partial_j\phi)(\partial_i\phi) + \phi\partial_{ij}\phi][/eqn]

[eqn] \epsilon_{ijk}\partial_j(\phi\partial_k\phi) = \epsilon_{ijk}[(\partial_j\phi)(\partial_i\phi) + \phi\partial_{ij}\phi][/eqn]

test

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>>7727495
>>7727490
ty for trying

>>7727445
Use [math]\nabla \times ( \psi \mathbf{A} ) = \psi ( \nabla \times \mathbf{A}) + ( \nabla \psi ) \times \mathbf{A}[/math]
then
[math]\mathbf{A}\times \mathbf{A}=0[/math]
and
[math]\nabla \times ( \nabla \psi )=0[/math].

>>7727672
[math]\nabla \times ( \phi \nabla \phi)
= \phi (\nabla \times \nabla \phi) + (\nabla \phi) \times (\nabla \phi)
= 0 + 0=0[/math]

These problems wont matter now that Christ Lagrange has perfected our theory of the universe.

>>7727510
stupid frogposter

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>>7727672
>>7727679
Thank you my friend, it was so obvious

>>7727679
Set [math]\psi=\frac{1}{2}\phi^2[/math]. Then [math]\nabla \times ( \phi \nabla \phi) = \nabla \times ( \nabla \psi) = 0[/math]

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> (ϕ∇ϕ)

brother, what did they do to you?

>>7727991
nice

>[eqn]
[math]

Use tensor notation you stupid faggot
[math]
\epsilon_{ijk}\partial_{j}(\phi \partial_{k}\phi)
= \epsilon_{ijk}(\partial_{j}\phi \partial_{k} \phi + \phi \partial_{j}\partial_{k} \phi)
[/math]
You're contracting an antisymmetric tensor [math]\epsilon_{ijk}[/math] with a symmetric tensor, so the result is zero.

>>7727445
what the fuck is this
upside down delta cross product golden ratio upside down delta golden ratio equals zero comma?

>>7728532
Both work.

>>7727490
[eqn] \epsilon_{ijk} \partial_j (\phi \partial_k \phi) = \epsilon_{ijk} [( \partial_j \phi )( \partial_i \phi ) + \phi \partial_{ij} \phi ] [/eqn]

Test

stoke's theorem baby
[eqn] \int_\Omega \nabla \times (\phi \nabla \phi) =\int_{\partial \Omega} (\phi \nabla \phi) = 0[/eqn]
Where the last step comes from that it's a conservative vector field.

>>7729623
Fun fact: All of vector calc comes from the abstract Stokes' Theorem applied in the right way.