How to calculate the flux of a vector field? (differential geometry)

[MathNugget]

Say, $X=a\frac{\partial}{\partial y}-\frac{\partial}{\partial x} \in \chi(\mathbb{R}^3)$ .
How would I (practically) calculate the flux ? I've been looking over the wikipedia page on the topic...I fail to see which would be the supposed answer to such question.

I also checked this out https://tutorial.math.lamar.edu/classes/calciii/surfintvectorfield.aspx . I suppose it applies here (the partial derivatives would be the basis of the space, instead of i j k, but I don't think have an object here, so I cannot apply what I saw on the last website)...

 

[blamocur]

What is $\chi (\mathbb R^3)$ ?

It seems that to compute flux you need both a vector field and a surface.

 

[MathNugget]

What is $\chi (\mathbb R^3)$ ?

It seems that to compute flux you need both a vector field and a surface.

Basis of the space of vector fields on smooth manifolds

Let $M$ be a smooth ($C^\infty$) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$. $\mathfrak{X}(M)$ is a ...

math.stackexchange.com

The $\chi(M)$ is basically the weird X appearing on this post. The set of all vector fields on M.

And there's no surface given in this exercise...unfortunately. I guess I'll put this exercise on the back burner.

 

[blamocur]

And there's no surface given in this exercise...unfortunately. I guess I'll put this exercise on the back burner.

Why don't you post the exact text of the exercise?

 

[MathNugget]

Why don't you post the exact text of the exercise?

Let $a\in \mathbb{R}$, $X=a\frac{\partial}{\partial y}-\frac{\partial}{\partial x} \in \chi(\mathbb{R}^3)$, $\alpha=yzcos(x)dx\wedge dy \in \Omega^2(\mathbb{R}^3)$.
a) Calculate the flux of X. Is X complete $\forall a \in \mathbb{R}$ ?
b) Calculate $\mathcal{L}_X\omega$ .

This was the subject. I suppose there was a typo, and $\alpha = \omega$ (I am 99% certain of it). I think differential geometry is 1 of those topics were less context is better, but this is the (exam) context.

 

[blamocur]

Since there is no surface involved I am guessing by "flux" they mean "exterior derivative", but I still don't really understand the problem statement. In particular, is $a$ related to $\alpha$ ? And what is $\mathcal L_X\omega$?

 

[MathNugget]

Since there is no surface involved I am guessing by "flux" they mean "exterior derivative", but I still don't really understand the problem statement. In particular, is $a$ related to $\alpha$ ? And what is $\mathcal L_X\omega$?

I'll start with the last part. According to wikipedia ,

$\mathcal L_X\omega$ is the lie derivative of $\omega$ along X.

I've been to this class (but I cannot solve any problem). $a$ and $\alpha$ are totally unrelated. I am also fairly certain a) uses just the formula of X, and $a$ being an arbitrary real number.

I checked out the exterior derivative thing. It's likely what this is about (I've had half of this course involving wedges)

 

[MathNugget]

Lie derivative - Wikipedia

en.wikipedia.org

If you scroll down, you'll see this

I don't know what it is, but it describes my circumstances perfectly...

Sorry I put so much pressure on you. These things are so absurd, I am just trying to find a solving pattern and then replicate it on every similar problem... I don't plan on continuing my studies after finishing (bachelor's degree?), so I am attempting to worm my way through the exams left...

 

[blamocur]

You've reached the

Sorry I put so much pressure on you.

Not at all; but you have reached the boundaries of what I remember o these subjects.

These things are so absurd

In my recollection they are super elegant but not easy

As for the flux of $X$, I am guessing it is 0 according to Poincare's lemma mentioned in this section. But don't take my word for it -- I don't really know the subject.

 

[MathNugget]

In my recollection they are super elegant but not easy

Well, I know if I see Lie brackets, they're supposed to simplify at some point, and if I see wedges, there's supposed to be some 0's out there. The rest is a mis(t)ery. Despite watching such wizardry, it was more of a speed-writing 3rd grade competition than what I'd call a math class.

 

[MathNugget]

Found the solution: (this probably has a different name in english)
I was supposed to do (something like) this: (solve a system or ordinary differential equations):

$x'(t)=-1 \\ y'(t)= a \\ z'(t)=0$
Now that I look at it, this one looks stupidly easy. Usually there are coefficients depending on x, y and z.
I also have to add some (initial knowledge?)
$x(0)=x_0\\ y(0)=y_0\\ z(0)=z_0$

$x(t)=-t\\ y(t)=at\\ z(t)=0$

The initial conditions don't really do much. And the answer would be something like (-t, at, 0)

 

[blamocur]

The initial conditions don't really do much. And the answer would be something like (-t, at, 0)

Does not sound right. What is $x(0)$ in your solution?

 

[MathNugget]

Does not sound right. What is $x(0)$ in your solution?

Well, to solve it, as far as I know, we need some extra knowledge (which we don't have), so we suppose we have that extra data to complete the Cauchy problem and make it have a unique solution.

Here we usually suppose the value in 0 is something. The site I mentioned prefers to say $x(t_0)=x_0$, to be more general, but we can simplify it a little more by supposing we know the value in 0...
Maybe it's something that happens only at my university, I don't know, but I am fairly certain of this.

 

[blamocur]

Well, to solve it, as far as I know, we need some extra knowledge (which we don't have),

Your post #11 has all the knowledge you need:
A general solution for $x^\prime(t) = -1$ is $x(t) = -t + C$ where $C$ is an arbitrary constant unless there are initial conditions.

 

[MathNugget]

Your post #11 has all the knowledge you need:
A general solution for $x^\prime(t) = -1$ is $x(t) = -t + C$ where $C$ is an arbitrary constant unless there are initial conditions.

True.

Just curious, what would be the name of this thing in english? (the one where we solve the differential equations linked to the coefficients of the vector field)

local flow? We often use flux and local flow interchangeably here... but I don't study math in english, so I attempt to the words by looking on wikipedia.

 

[blamocur]

You might be talking about curl and divergence, both of which are special cases of exterior derivatives.