Ordering

[ornate hull]

Hi, I am struggling to complete this question, particularly evaluating the RHS of the identity, as all I try I get the answer as a scalar.

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[wooden brook]

what is atled?

i get that it's a vector, but is it some special vector? what are its coordinates?

[ornate hull]

it’s grad no?

so grad f is the derivative of f

and doing the dot or cross product gives you curl and div

[wooden brook]

gradient of what?

this is a cross product of two vectors

what is the left vector?

[ornate hull]

it’s not really a vector more of a function?

defined by partial differentiation

[ornate hull]

this is also known as curl(v)

you take the cross product with the derivative?

i’ll be honest i’m not too sure

[wooden brook]

oh, overloaded notation

are you saying x stands for two different operations here?

[ornate hull]

no it doesn’t they’re both the cross product, and everything is a vector i was just lazy with my latex,

it’s v cross product of curl(w)

it’s on the left hand side where i think it is implying regular multiplication i’m getting confused do i expand or cross product??

[wooden brook]

curl is defined for vector fields

w is not a vector field

the usual notation is

$$ \nabla \times F $$

[hearty grove]

aL

[wooden brook]

curl of vector field F

[ornate hull]

i’ll grab the og question hold on

my apologies for the confusion

i even have the check sheet i’m just very uncertain of how they got there

[wooden brook]

that's a lot better

[ornate hull]

(i’ll take this as hint to continue practicing my latex lmao)

[wooden brook]

you should use appropriate terminology

not vector

but vector field

[ornate hull]

what’s the different again? so i have full understanding

[wooden brook]

vector is an element of a vector space

vector field is a vector valued function

now that's cleared up, what exactly are you stuck with?

can you do the arithmetic?

[ornate hull]

nope that’s the bit i’m stuck with and this

[ornate hull]

thank u :)

i thought the inner bracket would be 2 so it evaluates to 2w but that is not the case

[wooden brook]

do you denote

$$ \mathrm{div}v = v\cdot \nabla $$

[hearty grove]

aL

[ornate hull]

no and in my notes it specifically says the two are not equal

[wooden brook]

ok so what is the operator dot?

you are applying it here for example

binary operation of two vector fields

[ornate hull]

a dot product between the vector fields v and w

[wooden brook]

the notation is like super confusing

does this mean take curl of v dot w?

[ornate hull]

i believe so (i got that part right)

[wooden brook]

so why wouldn't it be

[ornate hull]

i also don’t know what it’s telling me

[wooden brook]

$$ \nabla \times (v\cdot w) $$

[hearty grove]

aL

[ornate hull]

wait no it’s the gradient of it not the cross product beacuse the dot product becomes a scalar

[wooden brook]

ok now we're getting somewhere

almost..

are you saying the nabla stands for two different things in these expressions?

[ornate hull]

yes

[wooden brook]

christ almighty..

why are they confusing new students like this i'll never know

[ornate hull]

the nabla with a cross or not or nothing is all different things

[wooden brook]

are you doing physics?

[ornate hull]

I KNOW RIGHT googling online did show me lots of different notations

nope mathematics

[wooden brook]

what does v dot nabla mean

[ornate hull]

that’s what i’ve found online

hold on

[wooden brook]

ok so it's a vector field

[ornate hull]

wait how?

also i’m not as confident on the first part as ive made an arithmatic error somewhere ?

[wooden brook]

no wait

[ornate hull]

i think i’m miss reading it then, is it not the x part of a derived by x and so on?

[wooden brook]

you apply it to a vector field

[ornate hull]

using the given vector fields how would you do that?

i’ve done this which i know is wrong but i’m just not sure

[wooden brook]

this notation is cursed

[ornate hull]

i’ll be sure to tell mike when i see him

[wooden brook]

$$ (a\cdot \nabla)w = a\cdot (\nabla w) $$

[hearty grove]

aL

[wooden brook]

if I'm reading this right

[tough delta]

You apply the operator to each coordinate function

[ornate hull]

WAIT IS THAT RLLY IT

no that ends as a scalar

[tough delta]

It doesnt

You have a vector of functions being differentiated

I can TeX it out when I get to campus

[ornate hull]

that would be amazing thank you

and thank you @wooden brook ok still stuck but moving forward slowly :)

[tough delta]

$v=(v_1,v_2,v_3),w=(w_1,w_2,w_3)$ (where ofc these are all vector fields)

$v\cdot\nabla=v_1\partial_x+v_2\partial_y+v_3\partial_z$, which 'is a scalar', so acts on $w$ like how any other scalar would.

$(v\cdot\nabla)w=(v_1\partial_xw_1+v_2\partial_yw_1+v_3\partial_zw_1,...)$, likewise for the rest of the components

[hearty grove]

Omegabet_

[tough delta]

(which is the same as $v\cdot(\nabla w)$, where $\nabla w$ is the vector of gradients of w's components)

[hearty grove]

Omegabet_

[ornate hull]

that all makes sense i’m just working through it thank you!

[tough delta]

Yeah, the notation is standard but takes a second to get use to. By all means it is mnemonic more than anything

especially when courses typically just cover curl, divergence, and laplacians, and not the more exotic operators that can come up

[ornate hull]

hell yeah i did it and got the correct answer thank you guys! i’m not sure my working is the clearest thing in the world but i understand it now :)

[gentle birch]

@ornate hull

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[ornate hull]

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