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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
0 votes
1 answer
210 views

Vector Laplacian in Curved Spaces

The vector gradient, $\mathbb{L}$, is defined as $$ (\mathbb{L} W)^{ij} \equiv \nabla^{i} W^{j} + \nabla^{j} W^{i} - \frac{2}{3} g^{ij} \nabla_{k} W^{k} \,, $$ where $\nabla_{i}$ is the covariant ...
Thiago's user avatar
  • 698
0 votes
2 answers
236 views

Power of a second-order tensor

I have the following equation: $$ \dfrac{\mathbf{V}^T \cdot \mathbf{V}}{\mathbf{V}^T : \mathbf{V}} + \dfrac{\mathbf{V} \cdot \mathbf{V}^T}{\mathbf{V} : \mathbf{V}^T} = \dfrac{\mathbf{D}^2}{\left \| \...
user avatar
1 vote
2 answers
116 views

Why is $\delta\mathbf{u}\cdot \mathrm{div}(\mathbf{\sigma}) = -\mathrm{grad}(\delta\mathbf{u}) \mathbf{:} \mathbf{\sigma}$?

Context is from this deal.ii tutorial. Screenshot of the relevant part is below. I don't get the transformation from $\delta\mathbf{u}\cdot \mathrm{div}(\mathbf{\sigma})$ to $-\mathrm{grad}(\delta\...
MaxD's user avatar
  • 866
3 votes
1 answer
6k views

Calculating the Divergence of a Tensor

I am working through a fluid dynamics paper and came across this equation: $$ \frac{\partial \vec{v}}{\partial t} + \vec{v}\cdot\nabla\vec{v}=\nabla\cdot T - \frac{1}{\rho}\nabla \phi\tag1$$ where T ...
Mjoseph's user avatar
  • 1,019
2 votes
1 answer
133 views

What do these tensor partial derivatives mean?

In the Wikipedia page on Ricci calculus the following tensor derivative equation is given: $$A_{\alpha \beta ..., \gamma}:= \frac \partial {\partial x^\gamma}A_{\alpha \beta ...}.$$ However, what does ...
user56834's user avatar
  • 13.4k
0 votes
0 answers
62 views

Transformations of curvilinear coordinate systems

I'm reading a book covering multiple topics in mathematics and physics. There's a chapter on tensors that had a very curious statement. It says "all curvilinear coordinate systems can relate to ...
EM_1's user avatar
  • 259
2 votes
1 answer
149 views

Extremely complex vector-matrix expression and its differentiation by vector

Given: $Q=R_z(\psi)R_y(\xi)R_x(\phi)$ - rotation matrix $\boldsymbol{\theta}=\left[\begin{array}{@{}c@{}} \phi \\ \xi \\ \psi \end{array} \right]$ - vector of angles $p=Q\left[\begin{array}{@{}...
ayr's user avatar
  • 731
0 votes
3 answers
362 views

Deriving product rule for divergence of a product of scalar and vector function in tensor notation

On page-94 of the 4th edition in the international version of Griffith's Electrodynamic, the following identity is used: $$ \int \left[ V(\nabla \cdot \vec{E} ) + \vec{E} \cdot \nabla V \right]dV= \...
Cathartic Encephalopathy's user avatar
0 votes
1 answer
379 views

$C_{ij}=T_{ijklmn} D_{kl} D_{mn}$ where $T_{ijklmn}$ is a rank 6 isotropic tensor, $C_{ij}$ is symmetric and $D_{ij}$ is antisymmetric

I was doing a question on Tensors and hit a roadblock The Question: Suppose that $C_{ij}$ and $D_{ij}$ satisfy the quadratic relationship $C_{ij} = T_{ijklmn} D_{kl}D_{mn}$, where $T_{ijklmn}$ is an ...
Swapna's user avatar
  • 23
3 votes
1 answer
148 views

Proving $(\nabla \times \mathbf{v}) \cdot \mathbf{c} = \nabla \cdot (\mathbf{v} \times \mathbf{c})$ using cylindrical coordinates

Assuming the form of divergence in polar coordinates is known, I am attempting to use the following definition of the curl of a vector field to determine the form of the curl in cylindrical ...
Jacob Wilson's user avatar
0 votes
1 answer
51 views

How do I reexpress the equation $\nabla \times (\nabla \times gs)) \times (\nabla \times \nabla(f\nabla \cdot t))$?

How do I go about reexpressing $\nabla \times (\nabla \times bs)) \times (\nabla \times \nabla(c\nabla \cdot t))$ where s and t are vector properties and b and c are scalar. I don't know where to even ...
ellen_230920000's user avatar
2 votes
1 answer
98 views

Expression for Rank 2 Tensor in Vector Notation

How does one write the following expression $D_{jk} (r_k \delta_{ij} - r_{i}\delta_{jk} - r_j \delta_{ik})$ in matrix notation? Is this just $\textbf{D} (\textbf{r} \times \textbf{I})$?
Tom's user avatar
  • 3,005
0 votes
1 answer
167 views

Gradient in tensor form

I found a problem which had $$\partial_i (A_i \vec{G})= (\vec{\nabla} .\vec{ A} )\vec{G}+ (\vec{A}.\nabla) \vec{G} $$ but my problem is what does $$\partial_i (A_i \vec{B})$$ even mean? it doesn't ...
SHIN101's user avatar
  • 11
3 votes
2 answers
404 views

Help with the gradient in different co-ordinate systems

Let $L(x,y)$ be the linear Taylor series expansion of some function $f(x,y)$. This can be written as $$L(x,y)=f(x_0,y_0)+f_x(x-x_0)+f_y(y-y_0)$$ Or in more compact form as $$L(x,y)=f(x_0,y_0)+\nabla f ...
boigadendro's user avatar
0 votes
1 answer
42 views

Levi-Civta symbol question

$$\delta_{kl}\epsilon_{ijk}\epsilon_{jki} = \delta_{kl} (\delta_{jl}\delta_{ki} - \delta_{ji}\delta_{kl})$$ $$\delta_{kl}\epsilon_{ijk}\epsilon_{jki} = \delta_{kj}\delta_{ki} -\delta_{ii}\delta_{ji} = ...
Z. Huang's user avatar
1 vote
1 answer
1k views

Partial derivatives in tensor/index notation!

For my continuum mechanics class, I'm tasked with finding $\nabla u$, $u$ being $u = b\frac{x}{|x|^3}$. Here, $b$ is a scalar constant. Attempt at the solution: I rewrite $\frac{1}{|x|^3}$ as $(x_i^2)^...
amber's user avatar
  • 11
2 votes
1 answer
427 views

Proof of expression for Christoffel symbols of the first kind, $[i , j k] = {\bf e}_i \cdot \frac{\partial {\bf e}_j}{\partial x^k}$

On page 155 of Vector and Tensor Analysis with Applications, by A.I Borishenko and I.E. Tarapov, the authors assert that, $$\frac{\partial {\bf e}_j}{\partial x^k} = \left\{ i \atop j \; k \right\} {\...
Olumide's user avatar
  • 1,261
1 vote
0 answers
80 views

Calculate the angle between vectors in equation. When does the conductivity tensor component take the form $\sigma_{ab} = \bar \sigmaδ_{ab}$?

In a certain anisotropic conductive material, the relationship between the current density $\vec j$ and the electric field $\vec E$ is given by: $$\vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\...
kpopgirl's user avatar
1 vote
1 answer
1k views

How should I calculate fourth order tensor times second order tensor?

Let's say I have two second-order tensors ${\mathbf{S}} = {S_{ij}}{{\mathbf{e}}_i} \otimes {{\mathbf{e}}_j}$ and ${\mathbf{T}} = {T_{ij}}{{\mathbf{e}}_i} \otimes {{\mathbf{e}}_j}$ . Then, I know ${\...
ENHorse's user avatar
  • 11
0 votes
1 answer
149 views

Tensor Calculus Notation

In tensor notation, we know the following is true for general vectors: $$ \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = A_i\epsilon_{ijk}B_jC_k = -B_j\epsilon_{jik}A_iC_k $$ However, if we try and ...
Jacob Wilson's user avatar

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