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You can create a bilinear mapping $M : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ using the standard inner product and your matrix $\xi$ above: $$M(x, y) = \langle y, \xi x \rangle$$ Likewise, each bilinear mapping defines a matrix using that relationship.


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What you have here is a simple tensor. We may write this tensor as $F \otimes F \otimes F$.


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Try to formalize what internal stress state should actually mean. What science has come up with is the notion of sectional forces and moments, together with a way to relate them to internal stress state. You need to do a Gedankenexperiment: Imagine a smooth cut passing through the material, including the point $p$. Remove the material on one side of the cut ...


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Take a look at Couple stress theory for solids


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To understand the trace, it is good to spell out the isomorphism between $End(V)$ and $V\otimes V^*$: $$V\otimes V^*\to End(V): v\otimes\omega \mapsto (x\mapsto \omega(x)v).$$ Under this isomorphism, composition of endomorphisms becomes $$(V\otimes V^*) \times (V\otimes V^*) \to V\otimes V^*: (v_2\otimes\omega_2, v_1\otimes \omega_1)\mapsto ...


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$\newcommand{\d}[1]{\;\mathrm{d} #1} \newcommand{\c}[3]{\Gamma^#3_{#1 #2}}\newcommand{\riem}[4]{\mathcal R^{#4}_{\; #1 #2 #3}}$As it turns out setting $r=0$ and calculating the Riemann tensor as an attempt to cut some corners, which is what I did earlierĀ¹, doesn't do the job. I'll start by taking the original metric $$\d s^2 = ...


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If we assume the sphere has a constant density, and we can ignore contributions from the pressure, then the geometry inside a solid sphere is described by the Schwarzschild interior metric: $$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 d\Omega^2 $$ ...


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This question is better suited for PHYS.SE, but here's the outline of an answer. It is mathematically the easiest way I know. With no symmetry, there are clearly 21 (7*6/2) independent components in the Voigt elasticity matrix, but symmetry always reduces this number. By the definition of $[C]$, $$\{\sigma\}_{6 \times 1} = [C]_{6 \times 6} ...


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$M$ looks like this: $M(u) = (F_{12}(u),\dots,F_{1n}(u),F_{23}(u),\dots F_{2n}(u)),\dots ,F_{nn}(u))$ $M(u) = (F_{12}(u),F_{21}(u),\dots) = (F_{12}(u),-F_{12}(u),\dots)$ doesn't make much sense. then the first variation: $\delta M(u) = \frac{d}{d\epsilon} M(F_{12}(u+\epsilon v),\dots,F_{nn}(u+\epsilon v))\big|_{\epsilon = 0}$ using the chain rule: ...


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Let's start from the top. If you have any programming background, then the following might make some more sense to you: A tensor is merely a function that is linear on each of its arguments and produces a number. For instance, let $f: \mathbb R^n \to \mathbb R$. That is, for any vector $v \in \mathbb R^n$, $f(v)$ is some real number. If $f$ is ...


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By whatever arguments, you yourself arrived at the correct answer: $dx_i$ computes at each point $p\in{\mathbb R}^n$ the $i$th coordinate of any tangent vector $v$ attached at $p$. A simple way to see this is as follows: $x_i(\cdot)$ can be viewed as a real-valued function on ${\mathbb R}^n$. In order to compute the differential $dx_i(p)$ we have to look at ...


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To remove notational confusion, let $\lambda^1,\ldots,\lambda^n$ be the dual basis of $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$. Then, your goal is to show that $$d(x^i)=\lambda^i.$$ i.e. $d(x^i)=dx^i$, where the later is the formal symbol usually used for $\lambda^i$. Note that this justifies the notation $dx^i$. Now, compute: ...


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As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate. But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...


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Given a $p$-form $\theta \in \bigwedge^p E^*$, we can define an alternating multilinear map $h \colon E^{n-p} \to \bigwedge^n E$ by $$h(u_1, \ldots, u_{n-p}) = \theta \wedge \tilde{u}_1 \wedge \ldots \wedge \tilde{u}_{n-p}.$$ Let $b \colon \mathbb{R} \to \bigwedge^n E$ be the linear map $$b(t) = t \omega.$$ Because $\bigwedge^n E$ is one-dimensional and ...


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The equation you wish to prove is linear in each of the vectors $u_j$ and $v_j$. Therefore it suffices to show the identity when these vectors are basis vectors. There are $n$ basis vectors from which we now want to choose the $n$ vectors $u_1,\dots,u_n$. If we choose any two to be the same, then both sides of the identity vanish (and the identity is true), ...


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Hint : (1) Let $$ U=[u_1\cdots u_n],\ V=[v_1\cdots v_n]$$ Then $$ (U^TV)_{ij} = u_i\cdot v_j =g(u_i,v_j) $$ (2) ${\rm det} (U^TV)={\rm det}\ U {\rm det}\ V$


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I don't know how to generalize for different sizes of the matrices, but here goes my answer: Try to show the following lemma: Let $T_1,\ldots,T_m \in L(V,V)$ and $A_1,\ldots,A_m$ be square matrices $n \times n$. Show that i) $(T_1 \otimes \cdots \otimes T_m)P(\sigma)=P(\sigma)(T_{\sigma(1)} \otimes \cdots \otimes T_{\sigma(m)})$ ii) There exists a ...


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The oriented volume spanned by three vectors is a multilinear alternating map of the vectors themselves. That's the starting point. See, e.g., the book by S.Winitzki.



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