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.
>>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
>>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
>>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]
> (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.
