Movatterモバイル変換


[0]ホーム

URL:


Sorry, we no longer support your browser
Please upgrade toMicrosoft Edge,Google Chrome, orFirefox. Learn more about ourbrowser support.
Skip to main content

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities includingStack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange
Loading…
Physics

Questions tagged [tensor-calculus]

Ask Question

Tensor calculus (tensor analysis) is a systematic extension of vector calculus to multivector and tensor fields in a form that is independent of the choice of coordinates on the relevant manifold, but which accounts for respective sub-spaces, their symmetries, and their connections.

2,665 questions
Filter by
Sorted by
Tagged with
1vote
1answer
123views

Does anyone have a clear image relating the concepts of vectors, covectors, tensors and metric? Sure I can do computations with them, but I still have limited image of how all of these relate to each ...
0votes
1answer
136views

If that's the case, how is the Einstein field equation satisfied for different coordinate systems of our choosing? I'm asking in context of my previous question.
2votes
0answers
75views

According to Weinberg' QFT vol I, (2.4.8), the generator of angular momentum $J^{\rho\sigma}$ transform as$$U(\Lambda, a) J^{\rho \sigma} U^{-1}(\Lambda,a) =\Lambda_\mu^{\,\,\rho} \Lambda_\nu^{\,\,\...
0votes
1answer
141views

How from these two equations:$$\nabla^\mu R_{\rho\mu}=\frac{1}{2}\nabla_\rho R$$and$$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$$follows$$\nabla^\mu G_{\mu\nu}=0~?$$Also, what is $\nabla^\mu$ ...
1vote
0answers
131views

In a 5-dimensional spacetime written as a warped product,$$ds^2 = e^{2A(\rho)}\,dt^2 - d\rho^2 - e^{2B(\rho)}\,d\Omega_3^2,$$where $d\Omega_3^2$ is the metric on the unit 3-sphere $(R[S^3]=6)$ and ...
3votes
2answers
168views

I’m learning general relativity from "Covariant Physics: From Classical Mechanics to General Relativity and Beyond" by Moataz H. Emam.He uses pseudo-Riemannian manifolds and index notation, ...
-1votes
1answer
84views

In tensor notation, a vector $\vec v$ can be expanded as $ \vec v = e_i\, v^i$,where $e_i$ are the basis vectors and $v^i$ are the components.Under a coordinate transformation $x^i \to x^{i'}$, the ...
-3votes
1answer
112views

When we consider no variations of the field configuration, we have:$$T^{\mu\nu}= \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial^\nu\phi - g^{\mu\nu}\mathcal{L}.$$How can one prove ...
1vote
2answers
117views

Well, from precisely mathematicians, i am not sure if scalar and vector can be view as i said. But i notice thatScalar describes as 1 aspect quantity, whether it is magnitude or amount of something....
2votes
0answers
183views

The algebraic Bianchi identity for the Riemann tensor is$R_{a[bcd]}=0. \tag{1}$A dual Riemann tensor, defined as $*R_{abcd}=\tfrac{1}{2}\epsilon_{ab}{}^{ef}R_{efcd}$, obeys a corresponding algebraic ...
8votes
5answers
2kviews

In many books on Physics, I read this:(i) Some quantities are represented in terms of numbers; they are called scalars.(ii) Some quantities are expressed in terms of numbers and directions; they are ...
1vote
0answers
101views

If $|g|$ is the absolute value of the determinant of the metric, the Levi-Civita Tensor $\epsilon$ in terms of its tensor density $\epsilon'$is defined as:$$\epsilon_{\mu_1..\mu_n}=\sqrt{|g|}\epsilon'...
4votes
3answers
346views

I'm trying to understand the connection between the general (vector) form of Fourier’s law, related, for example to this YouTube videofor heat conduction:$$\mathbf{J} = -k \boldsymbol{\nabla} T \...
1vote
0answers
62views

I am reading Schwarz's book "Quantum Field Theory and Standard Model", chap 17, anomalous magnetic moment. In 17.2, page 319, when simplifying the integral, the book says "Using $k^\mu ...
0votes
0answers
120views

I am reding Introduction to General Relativity Book by Maurice Bazin, Menahem Max Schiffer, and Ronald Adler.1st page$$\Im_{\alpha \beta}^\gamma=T_{\alpha \beta}^\gamma \sqrt{-g}$$is a tensor ...

153050per page
1
2345
178

Hot Network Questions

more hot questions
Newest tensor-calculus questions feed

[8]ページ先頭

©2009-2025 Movatter.jp