Tag Info

New answers tagged

3

For the benefit of brevity I'll give a straight answer: the second Weyl scalar is not related to electromagnetic radiation in any way. As per my previous comment, a simple proof is the Schwarzschild (or Kerr) spacetime. It describes a stationary black hole, without electromagnetic (or even gravitational) radiation, yet there is a basis in which all Weyl ...


0

“Always a polytope” – definitely not. Moreover, in certain situation $Ω$, if a closed set, may not change at all; I mean product with the 0-dimensional set of states $Ω_{\rm id} = \{1\}$ (one point), considered as a subset of 1-dimensional vector space $V_{\rm id} = {\mathbb R}$. It has the only effect, the unit effect, and corresponds to 1-state quantum ...


0

I have an answer here describing the general concept of true physical vectors and tensors which might be useful. (It also might be things you already know very well.) This turned out to be superlong, I hope you don't die of age before finishing reading. I at least added some useful titles of sections so you can skip some parts. The conclusion is that it is ...


1

There are two kinds of mathematical situations considering vectors and covectors. They are those: In the space without scalar product (dot product) they are very different entities and one should never confuse them. In the space with scalar product they are the same thing viewed from two perspectives. One can freely convert one entity to another with ...


0

Most of the answers posted here are incorrect. The Wikipedia page for the gradient says The gradient of $f$ is defined as the unique vector field whose dot product with any vector $v$ at each point $x$ is the directional derivative of $f$ along $v$. A look at Theodore Frankel's The Geometry of Physics confirms this. Other posters have said that the ...


4

I understand force to be a 1-form, through the following reasoning. Given a time-independent, conservative lagrangian $L$, its differential (a 1-form in the purest sense) is $$ \mathrm{d}L = p_a ~\mathrm{d}\dot{x}^a + f_a~\mathrm{d} x^a $$ where $$ p_a = \frac{\partial L}{\partial \dot{x}^a},~f_a = \frac{\partial L}{\partial x^a}. $$ So the components of ...


1

You're not really correct about what a tensor is. Here's a more precise perspective that might help clear things up. First remember that a vector $\mathbf v$ can be written $$\mathbf v = \sum_i v_i \mathbf e_i$$ where $v_i$ are numbers called components and $\mathbf e_i$ are basis vectors. A tensor is a generalization of a vector. By definition a rank ...


0

I'm not thinking in dimensions in this answer, so perhaps there will be a flaw: Force as gradient On any Riemnannian manifold $\mathcal{M}$ (think of $\mathbb{R}^n$ with the dot product), the gradient of a scalar field $U : \mathcal{M} \to \mathbb{R}$ is the vector field constructed by the dual of its exterior derivative $\mathrm{d}f$, so it is a vector ...


7

Since torque is defined as the rate of change of angular momentum, the more fundamental question would be whether angular momentum is a vector in SR. The answer is no, because there is no vector cross product in four dimensions. Angular momentum is a rank-2 tensor.


1

In these particular cases, the authors are interested in the conformal structure, i.e. lightcone structure, of the manifold. A conformal structure can be defined by an equivalence class of metrics, all of which are related to each other by a conformal transformation, $$g_{ab}\sim e^{\omega(x)}\bar{g}_{ab}$$ A nice way to characterize a conformal structure ...


1

The word "pseudotensor" is used in the sense that Emilio Pisanty mentioned, but it also has a completely different and fairly common meaning in general relativity: a multidimensional array of numbers indexed by spacetime coordinates that doesn't transform as a tensor. Energy pseudotensors are an example. Both of these meanings are mentioned in the Wikipedia ...


1

From the Wikipedia page on pseudotensors, a pseudotensor is usually a quantity that transforms like a tensor under an orientation-preserving coordinate transformation (e.g., a proper rotation), but additionally changes sign under an orientation reversing coordinate transformation (e.g., an improper rotation, which is a transformation that can be ...


1

The algebraic rule that if $xy=0$ then $x=0$ or $y=0$ does hold for tensor products: if you had $a^x b_y = 0$ you could conclude that $a^x = 0$ or $b_y = 0$. But there's no such rule for contractions like $a^x b_x$. For example suppose $a$ and $b$ written out in components were $(1,0)$ and $(0,1)$ respectively. You can't just move a variable to the other ...



Top 50 recent answers are included