All Questions
25
questions
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 ...
- 2,638
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 ...
- 194
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.
$$ \int\; \boldsymbol{A} \times (\boldsymbol{\nabla} a ) = \pm \int (\boldsymbol{\nabla} \times \boldsymbol{...
- 11
1
vote
1
answer
140
views
Covariant and partial derivative of a vector field (not component)
Is the covariant derivative of a vector field (not the components of a vector) same as the partial derivative?
I am adding a screenshot from page 69 from General Relativity: An introduction for ...
- 109
0
votes
2
answers
101
views
The difference between two indices suffix notation
Recently reading a set of lecture notes on vector calculus, which is a topic I am already familiar with. However during this I came across this representation of the gradient vector...
$$\frac{\...
- 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 ...
- 120
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$(\...
- 1,019
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:
$$\...
- 211
0
votes
1
answer
143
views
Tensor calculus - product of metric tensor and second covariant derivative of a scalar (Laplace-Beltrami operator)
I am trying to prove the following.
Suppose we have a scalar function $\phi$ (sufficiently differentiable), the metric tensor $g_{ij} = \dfrac{\partial y^\alpha}{\partial x^i}\dfrac{\partial y^\alpha}{...
0
votes
1
answer
444
views
Tensor calculus - gradient of the Jacobian determinant
Given an invertible coordinate transform between a set of coordinates $\{y^1, ..., y^n \}$ and $\{x^1, ..., x^n \}$ where $y^i = y^i(x^1,...,x^n)$ and $x^i = x^i(y^1,...,y^n)$ for each $i \in \{1,...,...
0
votes
3
answers
362
views
Deriving product rule for divergence of a product of scalar and vector function in tensor notation
On page-94 of the 4th edition in the international version of Griffith's Electrodynamic, the following identity is used:
$$ \int \left[ V(\nabla \cdot \vec{E} ) + \vec{E} \cdot \nabla V \right]dV= \...
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 $$L(x,y)=f(x_0,y_0)+\nabla f ...
- 33
1
vote
1
answer
1k
views
How should I calculate fourth order tensor times second order tensor?
Let's say I have two second-order tensors
${\mathbf{S}} = {S_{ij}}{{\mathbf{e}}_i} \otimes {{\mathbf{e}}_j}$
and
${\mathbf{T}} = {T_{ij}}{{\mathbf{e}}_i} \otimes {{\mathbf{e}}_j}$
. Then, I know
${\...
- 11
4
votes
0
answers
101
views
What is a neat way to solve $\nabla\mathbf{u}+\nabla\mathbf{u}^T=\mathbf{\mathbf{C}}$?
Let $\mathbf{u}:\mathbb{R}^3\to\mathbb{R}^3$ be a smooth enough vector field that satisfies the following equation
$$\nabla\mathbf{u}+\nabla\mathbf{u}^T=\mathbf{C},\tag{1}$$
where $\nabla\mathbf{u}$ ...
- 15k
1
vote
0
answers
58
views
Prove that if $V=\text{constant}$ then the second part in the paratheses after the integral sign is equal to $0$
$$\frac{\mathrm d}{\mathrm dt^i} \underset{\large V(t)}{\iiint} \Psi \,\mathrm dV= \underset{\large V(t)}{\iiint} \left({\frac{\partial \Psi}{\partial t}+\nabla\cdot\Psi\mathbf v_i}\right)\, \mathrm ...
- 8,382