Commutator of vector fields $x∂_x + y∂_y + z∂_z$ and $∂_z$

The commutator of the vector fields $x∂_x + y∂_y + z∂_z$ and $∂_z$ is given by:$[x∂_x + y∂_y + z∂_z,∂_z] = -∂_z$.I'm having trouble interpreting this result. The commutator is the vector that connects p and p': p being the point one ends in by first following an integral curve of the first vector field over a distance $\delta t$ and then the second, starting in $(x_0,y_0,z_0)$, and p' vice versa, starting from the same point, over the same distance.Here's my problem. I believe the integral curves of the first vector field are straight lines thr...Read more

vector fields - Surface Tension of Floating Object

I was reading this thesis about surface tension and its role in floating bodies. I couldn't quite understand at page 10 how the author applied the 2D divergence theorem outside the region bounded by the contour. Shouldn't you use the divergence theorem for a region inside the boundary?The line integral is evaluated around a boundary C \pi, but the double integral is evaluated outside the region bounded by C\piHere is my thinking. The divergence theorem in 2D is:$\iint_S \nabla \cdot \textbf{F}\hspace{0.2cm} dA = \int_{\partial S} \textbf{F}\cdo...Read more

vector fields - Why are line integrals not always path independent?

The text I am reading says that a line integral,$$\int_{C}{\mathbf{F}\cdot\textrm{ d}\mathbf{r}}$$is path-independent whenever $\mathbf{F}$ is a gradient field (or in the realm of physics, a conservative field).My question is, when is a line integral path-dependent? Under the characterization of path-independence I described above, it seems like every line integral must be path-independent because every vector field is the gradient field of some potential function. Or is this false?...Read more

Explicit derivation of the interior product between the Hamiltonian Vector Field and the symplectic two-form

I am trying to understand Example 5.12 of the book Geometry, Topology and Physics by Nakahara. In particular, consider the two-form $\omega=dp_{\mu} \wedge dq^{\mu}$ (1) and the Hamiltonian vector field $X_f=\frac{\partial f}{\partial p_{\mu}}\frac{\partial }{\partial q^{\mu}}-\frac{\partial f}{\partial q^{\mu}}\frac{\partial }{\partial p_{\mu}}$ (2)In the book it says that "it is easy to verify" that:$i_{X_f} \omega = -\frac{\partial f}{\partial p_{\mu}}dp^{\mu}-\frac{\partial f}{\partial q^{\mu}} dq^{\mu}=-df$ (3).But I am a bit confused abo...Read more

surface integrals - Stokes' Theorem - Vector Field

I am having problems trying to verify Stokes' theorem (below) as part of a question. $$\iint_{S} \text{curl} \vec F \cdot d\vec S=\oint_{c} \vec F \cdot d\vec r$$The vector field in question is $\vec F=(xy,z^3,xz^2)$ and I am looking over a triangle plane that joins $(0,0,1)$ , $(0,4,1)$ and $(4,0,1). $Computing the surface integral (over the triangle plane) does not seem to yield the same answer as the closed line integral (taking around the edge of the triangle) so I know I am making a mistake somewhere!My working is below:Surface Integral:Fu...Read more

Irrotational fields and divergence

Let $F,G$ be $C^1$ vector fields from $\mathbb R^n$ in itself. The condition$$\int_{\partial A}F\cdot \nu_A\ d\sigma=\int_{\partial A} G\cdot \nu_A\ d\sigma$$for every bounded domain $A$ whose boundery is $C^1$ does not imply that $F-G$ is constant. Prove that if $F$ and $G$ are irrotational and bounded then $F-G$ must be constant.Until now I've obtained that $\int_A div (F-G)=0$, so $div (F-G)=0$ (correct?).General suggestions about properties of irrotational and zero-divergence vector fields are welcome too.Thank you in advance!...Read more

vector fields - Divergence in spherical coordinates vs. cartesian coordinates

The divergence in spherical coordinates is derived as$$\nabla\vec A=\frac{1}{r^2}\frac{\partial(r^2A_r)}{\partial r}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(A_\theta\sin\theta)+\frac{1}{r\sin\theta}\frac{\partial A_\varphi}{\partial\varphi}$$However, this seems to fail at a simple example. Let's take the identity $\vec F(\vec r):=\vec r$ for example. The divergence of the vector field $\vec F(\vec r)=\begin{pmatrix}x\\y\\z\end{pmatrix}$ in cartesian coordinates is$$\nabla\vec F(\vec r)=\frac{\partial}{\partial x}F_x(\vec r)+\frac{\...Read more

vector fields - Identifying a scalar function

We know that a scalar is invariant under rotations. What about a scalar function? Should it also be invariant under rotations? Therefore, under rotation $\phi(x,y,z)$ must be equal to $\phi^\prime(x^\prime,y^\prime , z^\prime)$. Where $ (x^\prime,y^\prime , z^\prime)$ is the rotated coordinate system. Does it imply that $$\phi(x,y,z)=x^2+y^2+z^2$$ is the only possible scalar function in three dimension? Can $\phi(x,y,z)=x^2+yz$ be a scalar function?...Read more

vector fields - Corollary of Poincaré-Lemma

In the script of my professor, there is the following corollary of the Poincaré Lemma: If $\mathcal F$ is a vector field with rot$(\mathcal F)=0$ and $U\subset\mathbb R^n$ is an open ball, then there is a function $f:U\to\mathbb R$ with grad$(f)=\mathcal F$.Could anyone make clear how this follows directly from the Poincaré Lemma? I don't really see it. Any help is greatly appreciated!...Read more

How to find a line field in a vector field

I have a vector field described by $$\mathbf{F}(x,y) = y \hat{x} -x\hat y$$I am trying to find the field lines for it, ideally by choosing a starting point and then having a parameter t that will lead me through all points that are on the field line, starting with the starting point.Now, I do know that the field lines are circles, but that's only from drawing them, which wouldn't work with every field.I have also tried saying that $$ \frac {dy}{dx}=\frac{-x}{y}$$$$y \cdot dy=-x \cdot dx$$$$y^2 = - x^2 + C $$$$C=x^2+y^2$$Which is true, but now I...Read more

Finding a potential function using integration (vector fields)

For the method of finding a potential function of a conservative vector field, usually I would let $\nabla \phi$ be equal to the vector field, and do some integration (indefinite) and differentiating whilst equating it to other components of $\nabla \phi$. However, in this youtube video (at 11:47), the technique was to do a definite integral twice, over some random path. Where could I learn this method and why does this work? The question is to find the potential scalar field for the conservative vector field:$$F = \begin{pmatrix} x \\ y \end...Read more

Intuition on why a field is not conservative

I know the definition of a conservative vector field is that the integral $\int_{C} F \cdot dr$ is path independent, or it is zero if $C$ is a closed loop, or the curl of $F$ is zero (in a simply connected region), or there exists a scalar field $\phi$ such that $\nabla \phi = F$. But what is the intuition behind this example?$F = (x^2 - xy)i + (y^2 - xy)j$. The curl is obviously non-zero, but what is the intuition, if any, for the result that this vector field is not conservative? Perhaps the vector field isn't """"evenly distributed""" (as in...Read more

Are most vector fields conservative?

I started learning line integrals and vector fields, and my question is: are the majority of "nice" functions (like polynomials, trig etc) conservative?I ask this because of the definition:$\vec{f}$ is conservative $\iff$ there exists a scalar field $\vec{F}$ s.t. $\nabla \vec{F} = \vec{f}$.When equating the components to solve for $F$, it seems that almost all $\vec{f}$ can be found as the gradient of some scalar field $\vec{F}$ (assuming that we can integrate the functions of course). Also, I see some sources verify that the curl of $f$ is ...Read more

Determining if a field $f$ is conservative

So there is a question that I did where it made me find the integral of $f \cdot dr$ under the path $C$ (which is a semicircle when $x=0$ so $y^2 + z^2 = $) for the arc that connects $(0,0,-1)$ to $(0,0,1)$. For clarity, the function $f$ is $(x^2 + y^2 + z^2 , -z, y+1)$. I evaluated this to be $\pi + 2$. Now, it asks whether $f$ is conservative, and the answer says : "No, the line integral is dependent on the path". Now I know this is exactly the definition of a conservative field (that it is independent of its path), but how can we come to thi...Read more