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0 votes
0 answers
34 views

Changing coordinates of $2$nd order partial operators

Let's work in $\mathbf R^n$. If we want to change coordinates $\mathbf x\to\mathbf r$, with them related like $$ \mathbf x=\mathbf x(\mathbf r) $$ Then, the second order generic partial operator in ...
Conreu's user avatar
  • 2,638
0 votes
0 answers
115 views

tensor identity

The tensor $t$ is defined in terms of a scalar $\varrho$ and the two vectors $v$ and $u$ (and derivatives) as follows: $$t_{i j}:=\varrho v_i u_j-\frac{1}{2} \varrho \varepsilon_{i lm} v_l (\...
reverendjamesm's user avatar
0 votes
1 answer
57 views

$\nabla\cdot(f\vec{g})=f\nabla\vec{g}+\vec{g}\cdot\nabla f$ using Levi-Civita

I need to prove the following equality: $\nabla\cdot(f\vec{g})=f\nabla\vec{g}+\vec{g}\cdot\nabla f$. I know that proving this equality using the properties of $\nabla$ and $(\cdot)$ is easy, what the ...
user19872448's user avatar
2 votes
1 answer
75 views

geometric interpretation on covariant derivateve in curvilinear coordinates.

I'm having trouble understanding what does the covariant derivative do in a coordinate system where we have changing basis vectors. I always thought it was giving us the change in coordinates while ...
Krum Kutsarov's user avatar
25 votes
7 answers
11k views

How would you explain a tensor to a computer scientist?

How would you explain a tensor to a computer scientist? My friend, who studies computer science, recently asked me what a tensor was. I study physics, and I tried my best to explain what a tensor is, ...
closedvolumeintegral's user avatar
2 votes
1 answer
67 views

Differential surface element and nabla operator

If we have the vector field $\vec{u}=\vec{A}\times \vec{v}$, where $\vec{A}=\text{const.}$ and we integrate over some closed curve, by using Stokes' theorem we get: $$ \begin{align} \oint_{\partial S}...
Krum Kutsarov's user avatar
0 votes
1 answer
44 views

If $S$ is a symmetric matrix, then rewriting $\int{S:\nabla \phi} dx$

I am trying to prove that: If $S$ is a symmetric matrix, then one can rewrite $\int{S:\nabla \phi} \text{dx}$ as $\int{S:D(\phi)} \text{dx}$, where $D(\phi)$ is the symmetric gradient of $\phi$. Any ...
ali's user avatar
  • 194
1 vote
1 answer
104 views

Exercise 5.22 in the book Geometrical Methods of Mathematical Physics (by Bernard Schutz)

I can't figure it out about the Exercise 5.22 in the book Geometrical Methods of Mathematical Physics (by Bernard Schutz). Could any one give me a help? Thanks. ($\bar{V}$ means vector V. ) Exercise ...
yo-yos's user avatar
  • 63
0 votes
2 answers
119 views

What is the gradient in this skewed coordinate system?

Consider two bases $e_i$ and $f_i$ of $\mathbb{R}^2$ defined by: $$\begin{aligned} (f_1,f_2) &= (e_1,e_2)\cdot F\\ \begin{pmatrix}1&-1\\1&1\end{pmatrix} &= \begin{pmatrix}1&...
l'étudiant's user avatar
0 votes
0 answers
52 views

Intuition about the divergence of a vector field in non-orthogonal basis

My textbook defines the divergence of a vector field in a non orthogonal constant basis the following way: $$div(\vec{u})=\vec{a}^i\cdot\frac{\partial \vec{a}_ku^k}{\partial x^i}=\frac{\partial u^i}{\...
Krum Kutsarov's user avatar
1 vote
0 answers
45 views

How to integrate by parts the vectorial product between a vector and a gradient [closed]

I am having a problem trying to check if the identity below should be positive or negative. $$ \int\; \boldsymbol{A} \times (\boldsymbol{\nabla} a ) = \pm \int (\boldsymbol{\nabla} \times \boldsymbol{...
Nathan's user avatar
  • 11
0 votes
0 answers
51 views

Differential operators of tensor fields. Hamilton operator

The very first thing my textbook says is that the Hamilton operator is defined as: $$\vec{\nabla}=\vec{a}^i\nabla_i$$ Where $\nabla_i$ is the covariant derivative and " $\vec{a}^i$ is the ...
Krum Kutsarov's user avatar
2 votes
2 answers
93 views

Derivation or intuition on the covariant derivative for higher rank tensors

So the derivation in my textbook for the covariant derivative of a vector field $\vec{u}$ in curvilinear coordinates $\xi^k$ is the following: $$\frac{\partial \vec{u}}{\partial\xi^j}=\frac{\partial (...
Krum Kutsarov's user avatar
0 votes
0 answers
25 views

Why does the invariant 1-tensor integral gives 0 for any volume?

I was going through David Tong's vectors calculus notes. In Chapter 6.1.3 (Invariant Integrals) he gives the example Here are some examples. First, suppose that we have a 3d integral over the ...
vueenx's user avatar
  • 1
0 votes
0 answers
92 views

Averaging the components of a unit vector over a circle

I have a following problem: "Calculate the average values of the products of the components of the unit vectors: $$\langle n_i \rangle, \langle n_i n_j \rangle, \langle n_i n_j n_k \rangle, \...
intex2dx's user avatar
0 votes
1 answer
59 views

Metric Tensor Grid

Let, we are in a 2d metric where $g_{xx}=1, g_{yy}=x^2$, therefore $|e_x|=1$, $|e_y|=x$. If we try to draw the metric in a grid - it looks something like the image I uploaded. Note that, along the X ...
Nayeem1's user avatar
  • 109
1 vote
1 answer
140 views

Covariant and partial derivative of a vector field (not component)

Is the covariant derivative of a vector field (not the components of a vector) same as the partial derivative? I am adding a screenshot from page 69 from General Relativity: An introduction for ...
Nayeem1's user avatar
  • 109
0 votes
0 answers
61 views

Confusion between covariant and partial derivatives

Let, we are in 1d cartesian space with metric $g_{xx} = x^2$. Let we have a vector $v = 1/x e_x$. Since the vector is designed to shrink its components as the basis grows - its total length will ...
Nayeem1's user avatar
  • 109
0 votes
2 answers
101 views

The difference between two indices suffix notation

Recently reading a set of lecture notes on vector calculus, which is a topic I am already familiar with. However during this I came across this representation of the gradient vector... $$\frac{\...
Maximus's user avatar
  • 80
7 votes
2 answers
289 views

What is $\left ( \vec{\nabla} \times \vec{A} \right ) \cdot \left ( \vec{\nabla} \times \vec{A} \right )$?

I'm trying to rewrite $\left ( \vec{\nabla} \times \vec{A} \right ) \cdot \left ( \vec{\nabla} \times \vec{A} \right )$ in some other way. I tried using Levi-Civita symbol and Kronecker delta, but I'm ...
Bemciu's user avatar
  • 120
5 votes
3 answers
259 views

What are the linear transformations that preserves the cross product, i.e. $ R(v\times w) = (Rv) \times (Rw), \forall v,w \in \mathbb{R}^3 $

Let us just focus on $\mathbb{R}^3$ currently. We study the set of all $3\times 3$ matrices $R$ satisfying $$ R(v\times w) = (Rv) \times (Rw), \forall v,w \in \mathbb{R}^3 $$ where $\times$ is the ...
Mr. Egg's user avatar
  • 658
1 vote
0 answers
40 views

Whats the significance of $g^{-1}$ (the inverse metric) appearing in tangential projection?

Let $M \subseteq (\mathbb{R}^n,g_E)$ be an embedded submanifold, with the embedding $F : M \to \mathbb{R}^n$. It is well known (c.f. Lee, doCarmo) that the covariant derivative on $M$ with respect to ...
colossal's user avatar
  • 198
4 votes
2 answers
413 views

Proving $A\times(\nabla\times B)+B\times(\nabla\times A)=\nabla(A\cdot B)-(A\cdot\nabla)B -(B\cdot\nabla)A$ with Einstein summation

So, I'm seeking to prove the below identity, for $A,B$ vectors fields in $\mathbb{R}^3$: $$A \times (\nabla \times B) + B \times (\nabla \times A) = \nabla (A \cdot B) - (A \cdot \nabla)B - (B \cdot \...
PrincessEev's user avatar
  • 45.9k
0 votes
1 answer
241 views

Divergence of a Tensor Field

Given a tensor field $\hat{\tau}$, I wish to calculate $\nabla\cdot\hat{\tau}$. My first question: is this actually the divergence of the tensor field? Wikipedia seems to differentiate between div$(\...
Mjoseph's user avatar
  • 1,019
0 votes
0 answers
126 views

Tensor and Gauss divergence theorem

I am trying to see whether, in spherical coordinates, $$\int \left( \boldsymbol{r} \times \boldsymbol{\nabla} \cdot \boldsymbol{T} \right) \cdot \boldsymbol{e}_z dV$$ where $T$ is a 2D symmetric ...
Tetrapoil's user avatar
0 votes
0 answers
120 views

How do I simplify $\delta_{ij} \delta^{jk}$?

How do I simplify $\delta_{ij} \delta^{jk}$? I know that $\delta_{ij} \delta_{jk}=\delta_{ik}$, but what do I do if the there's a Kronecker Delta symbol with upper indices and one with lower indices?
math's user avatar
  • 93
0 votes
0 answers
34 views

Can you define a tensor by integrating one vector with respect to another?

I was reading this question, simply I was wondering about integrating a vector with respect to another vector field. In the question, the OP asks if the following quantity has any sensible meaning: $$\...
FizzKicks's user avatar
  • 211
6 votes
0 answers
207 views

'Tensor Calculus' by J.L. Synge and A. Schild (1979 Dover publication) . Exercise 12 page 110.

I'm solving all the exercises of 'Tensor Calculus' by J.L. Synge and A. Schild (1979 Dover publication) . Till now everything went smooth, but now I'm stuck at exercise 12. page 110 of the third ...
Bufo Viridis's user avatar
0 votes
1 answer
143 views

Tensor calculus - product of metric tensor and second covariant derivative of a scalar (Laplace-Beltrami operator)

I am trying to prove the following. Suppose we have a scalar function $\phi$ (sufficiently differentiable), the metric tensor $g_{ij} = \dfrac{\partial y^\alpha}{\partial x^i}\dfrac{\partial y^\alpha}{...
gibtskrummung's user avatar
0 votes
1 answer
444 views

Tensor calculus - gradient of the Jacobian determinant

Given an invertible coordinate transform between a set of coordinates $\{y^1, ..., y^n \}$ and $\{x^1, ..., x^n \}$ where $y^i = y^i(x^1,...,x^n)$ and $x^i = x^i(y^1,...,y^n)$ for each $i \in \{1,...,...
gibtskrummung's user avatar

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