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

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

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

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

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

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

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

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

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

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

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

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

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

Given is the vector field $\vec{w} = f \left( |\vec{x}| \right)\vec{x} $.How do I find the scalar function $f$ so that $\vec{\nabla} \cdot \vec{w} = 0$ ??...Read more

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