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0 votes

Interpretation of "inverse" curl

$ \newcommand\R{\mathbb R} $Yes, the expression $v\times\nabla$ makes sense. It is vector-like, and you have to decide what it's going to act on and how. If $f : \R^n \to \R$ is a scalar function then ...
4 votes

The necessity of the Lie bracket of vector fields in the definition of the Riemann curvature tensor

The two definitions of the coefficients ${R^t}_{kij}$ coincide when $[X,Y]=0$ and this is the case for $X= \frac{\partial}{\partial x^i}$ and $Y= \frac{\partial}{\partial x^j}$. However the definition ...
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3 votes

The necessity of the Lie bracket of vector fields in the definition of the Riemann curvature tensor

Without the $\nabla_{[X,Y]}$ the expression for $R(X,Y)Z$ would not be a tensor because it would depend on derivatives of the $X,Y$ fields. A tensor $T(X,Y)$ depends only on the components of $X$, $Y$ ...
1 vote

Integration by parts and divergence of curl

If boundary terms are $0$ then: $$ \int (\nabla \times A_2(x))\cdot\nabla f(x) \ dx^3 =- \int \big[\nabla\cdot(\nabla \times A_2(x))\big] f(x) \ dx^3 $$ by integration by parts which is what you are ...
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0 votes

Multivariable calculus: Can two trajectories of a vector field ever cross each other?

are you also in UIUC's NetMath for MVC? It sure sounds like it. I'm working on L6 right now; great to meet a fellow student outside my school. Are you at UIUC or also a high schooler? How did the ...
2 votes
Accepted

Why do these results not contradict Green's Theorem?

By definition, a conservative vector field is one which is the gradient of some scalar function. In $\mathbb{R}^2$, vector fields which have zero curl are only conservative on domains without holes (...
0 votes

How to approximate a vector-valued function?

Have you tried solving the following least-squares problem? $$ \underset{{\bf X} \in \Bbb R^{2 \times 2}}{\text{minimize}} \quad \| {\bf X} {\bf A} - {\bf B} \|_{\text{F}}^2 $$ If $\bf A$ has full row ...
2 votes

Questions about the definitions of r

It seems to me that here $i$ and $j$ represent respectively the following unit vectors : $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. And $x,y$ are simply real ...
3 votes
Accepted

Questions about the definitions of r

These represent different things. Written in more common and clear mathematical notation, you have $$r = \sqrt{x^2 + y^2} \qquad \vec r = x \hat\imath + y \hat\jmath$$ respectively. The former ...
2 votes

Left Invariant Vector field of matrix Lie group

Okay here is an attempt at a more full answer: You aren't pushing forward a matrix, you are pushing forward the map $L_A : G \to G; B \mapsto AB $. This is clearly a diffeomorphism of $G$ (by ...
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0 votes

What is the Jacobian matrix?

The Jacobian matrix finds multiple applications in robotics: a detailed explanation can be found in the book "Modern Robotics: Mechanics, Planning, and Control," by Kevin M. Lynch and Frank ...
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1 vote

$N$-dimensional Anti-Curl Operator

$ \newcommand\R{\mathbb R} \newcommand\dd{\mathrm d} $ I think what you're looking for is given by geometric calculus. This is essentially calculus done over the Clifford algebra associated to the ...
0 votes

Vector fields on a sphere: equivalence of two definitions

From $\Phi(x)$ we can construct a vector field in $\mathbb R^{n+1}$ (where we indentify $T_x\mathbb R^{n+1}$ with $\mathbb R^{n+1}$ ): $$X(x)=|x|\Phi\left(\frac{x}{|x|}\right). $$ Now i use the ...
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1 vote

Lie algebras of vector fields over $\mathbb{R}$ and over $S^1$

For any Lie algebra $L$, denote the $q$-th cohomology of $L$ by $H^q(L)$. By De-Rham theorem we know that: $$H^1(V(\mathbb{S}^1))=H_{De-Rham}^1(\mathbb{S}^1)\cong H_{Singular}^1(\mathbb{S}^1; \mathbb{...
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0 votes

Vector potential of magnetic dipole

Ok, as I mention above \begin{equation*} \oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} = \oint\limits_L \vec{r}'\times [d\vec{l} \times \vec{R}] . \end{equation*} Let's consider first integral $\...
  • 165
2 votes

Proof of differentiability for F-related vector fields

Suppose that $Df:TM\rightarrow TN$ is the derivative of $f$, that is $Df(p, v):= \big(Df(p)\big)(v)$ (or $df_p(v)$ in your notation) where $p \in M$ and $v \in T_pM$. First show that $Df$ is smooth ...
3 votes
Accepted

Flow of a vector field $f(x,y)=(x,-y+x^2)$

Given a differential equation (possibly a system) \begin{align} X'(t) &= F(X(t)), & X(0) &= X_{0} \end{align} the flow of the equation set is defined by $\varphi(X_{0}, t) = X(t)$. This ...
1 vote
Accepted

Variable-coefficient Laplacian identity proof

Using the product rule to expand the RHS in either index notation $$\eqalign{ \def\BR#1{\left(#1\right)} \def\LR#1{\Big(#1\Big)} \def\a{\mu}\def\b{{\bf v}}\def\n{\nabla}\def\p{\partial} \p_k\LR{\a\:\...
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1 vote
Accepted

Do the points lie on the same plane?

Yes, that is correct. Equivalently, you may check directly that $$\begin{align}\det\left(\overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{PS}\right) &=\begin{vmatrix}2&0&4\\3&1&...
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1 vote
Accepted

Lie bracket of vector fields - what have I done wrong?

Informally, we have $$ \frac{\partial~~}{\partial y} = \frac{\partial ~~}{\partial (2t)} = \frac{1}{2}\frac{\partial~~}{\partial t} = \frac{1}{2}\frac{\partial~~}{\partial x}, $$ and \begin{align} v &...
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