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2

$\newcommand{\G}{\mathbf{G}} \newcommand{\H}{\mathbf{H}} \newcommand{\A}{\mathbf{A}} \newcommand{\B}{\mathbf{B}} \newcommand{\tH}{\tilde{\H}} \newcommand{\tG}{\tilde{\G}} \newcommand{\Hp}{\H^+} \newcommand{\Gp}{\G^+}$I will prove the answer two ways. The first way is the way you were "supposed" to do it, and the second way is another way of doing it. The ...


2

$$ \delta^{[\mu_1\mu_2\ldots \mu_n]}_{\nu_1\nu_2\ldots \nu_n}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} \prod_{i=1}^n \delta^{\mu_{\pi(i)}}_{\nu_i}. $$ More generally, $$ T^{[\mu_1\mu_2\ldots \mu_n]}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} T^{\mu_{\pi(1)}\mu_{\pi(2)}\ldots \mu_{\pi(n)}}. $$ Here $S_n$ is the symmetric group of permutations, ...


13

For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source) Since it is not necessarily the case that $\sigma_{11}=\sigma_{31}=\sigma_{21}$ (all pointing the in the same $\mathbf{e}_1$ direction)--or any of the other $\sigma_{ij}$ combinations, we need to have 9 components describing it, ...


19

Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor. Pressure is part of the stress tensor. The diagonal elements ...


0

Contract both sides of the first equation with $e^{ijk}$. In three dimensions, we have the identity for the Levi-Civita, $$\epsilon_{ijk} \epsilon^{imn} = \delta_j^m \delta_k^n - \delta_j^n \delta_k^m.$$ Using the antisymmetry of the $M_{jk}$, you obtain the second equation.


1

It's a well known identity that $\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ Therefore, if $$J_{i} = -\frac{1}{2}\epsilon_{ijk}M_{jk}$$ then we have: $$\begin{align} \epsilon_{lmi}J_{i} &= -\frac{1}{2}\epsilon_{lmi}\epsilon_{ijk}M_{jk}\\ ...


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Let's do it in 1D for simplicity: you consider a portion of thread of length $dL$ and section $S$, with a net body force density $f$, say $f=\rho S g$ where $\rho S$ is the lineic mass density. On $dL$, you also have stress from the rest of the thread, which are $T_+ = \sigma_{zz} (z+dL) S$ at the $z+dL$ end and $T_- = -\sigma_{zz}(z) S$ at the other end. ...


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Hyperplanes can be defined by (1) a direction normal to them and (2) one point within them, so a train of parallel hyperplanes certainly encodes direction information. A good concrete model for an one-form is a wavevector $k$ that defines a linear plane wave solution of D'Alembert's equation $(\nabla^2 - c^{-2})\,\psi=0$, when the wavevector is thought of ...


3

You could make an argument that the four-current is most naturally defined as a vector density. This is because a vector density uniquely defines a three-form (i.e. a totally antisymmetric tensor $J_{\alpha\beta\gamma}$), and three-forms can be integrated on hypersurfaces without any reference to a metric. So if you think of a current as an object which, ...


2

Clearly MTW's definition of $J^\alpha$ is a vector field due to the argument given. Note however that in the Wikipedia article you linked, under Summary, $J^\alpha$ is defined with an additional factor $\sqrt{-g}$, making it a vector density. You can easily see this by replacing the ordinary derivative in the definition of $J^\mu$ by a covariant one (which ...


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In the context of e.g. a pseudo-orthogonal Lie group $$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$ of pseudo-orthogonal matrices $\Lambda$ for the metric $$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{1,\ldots,1}_{p~\text{times}},\underbrace{-1,\ldots -1}_{q~\text{times}}), \qquad n~=~p+q,$$ ...


3

The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more. The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical ...


0

Yes, he defined the vector as behaving that way (a vector rotation is equivalent to a change of basis), otherwise it would not be a vector. A tensor is a different kind of object, it has at least two indexes and behaves different that a vector under transformation of coordinates (as defined in your book). You might probably read more about linear algebra up ...


2

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z ...


1

The covariant derivative for a general tensor of the form $T^{a_1\dots a_n}_{b_1 \dots b_n}$ is given by, $$\nabla_c T^{a_1\dots a_n}_{b_1 \dots b_n} = \partial_c T^{a_1\dots a_n}_{b_1 \dots b_n} + \Gamma^{a_1}_{cd}T^{d\dots a_n}_{b_1 \dots b_n} + \dots - \Gamma^d_{c b_1}T^{a_1\dots a_n}_{d \dots b_n} - \dots$$ Taking the covariant derivative of a ...


1

A covariant derivative of a tensor is itself a tensor. Actually, when we say something is covariant (or invariant under coordinate transformation), we mean that thing is a tensor. So, in this case $\nabla_\mu V^\nu\equiv T_\mu{}^\nu$. Now calculate $\nabla_\alpha T_\mu{}^\nu$ easily. \begin{equation} \nabla_\alpha T_\mu{}^\nu=\partial_\alpha ...



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