6

There can be different notions of "symmetric". Also, keep in mind that $\delta$ is just the index form of the identity operator. The first one pertains to the transpose of an operator, which is just the adjoint for a real case. Generally, for the transpose of a tensor, we have $$\left<v,M w\right>=\left<M^T v,w\right>\,,$$ or in index ...


5

From Wikipedia: $$ R_{abcd}=-R_{abdc} $$ $$ \Rightarrow g^{ad}R_{abcd} \\=-g^{ad}R_{abdc} \\=-R_{bc} $$ Here's a table of all the possible contractions, all derived using three facts: The definition of the Ricci tensor: $R_{bd}=g^{ac}R_{abcd}$ The antisymmetry of the Riemann tensor in its first two and last two indices: $R_{abcd}=-R_{bacd}=-R_{abdc}$ When ...


4

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via $$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{...


3

The quick answer is that yes, it does make sense, if we interpret symmetric to mean self-adjoint. Symmetry of a bilinear form $g$ (or higher rank tensor) means $g(X,Y)=g(Y,X)$, but symmetry of a linear operator $A$ means $\langle X,A(Y)\rangle = \langle A(X),Y\rangle $ where the brackets denote some chosen inner product. In that sense, symmetry of a linear ...


3

$$\omega^{ab}(\theta_{ab}-\theta_{ba})=\omega^{ab}(\theta_{ab})-\omega^{ab}(\theta_{ba})=\omega^{ab}(\theta_{ab})+\omega^{ba}(\theta_{ba})=2\omega^{ab}(\theta_{ab})$$ where the last but one equality holds since $w$ is antisymmetric and the last equality is due only to rename the dummy indices (so the indices summed over) as suggested in a comment; in ...


3

TL;DR - I suspect your confusion lies in the Physics 101 example that e.g. the ordered pair ("temperature","pressure") does not define a vector because when we change our coordinates, temperature and pressure don't transform. However, if we are working in cartesian coordinates, the object (temperature)$\hat x$ + (pressure)$\hat y$ is a ...


3

Provided that the connection is torsion-free, $\Gamma^{\sigma}_{[\mu\nu]}=0$, there's a neat cancellation that happens. I'll work out a few examples to get a feel for this: $$ \require{cancel} \newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} (dA)_{\mu_1\mu_2\dots\mu_{p+1}}\sim\nabla_{[\mu_1}A_{\mu_2\dots\mu_{p+1}]} $$ $$ \\(dA)_{\mu\nu}\...


2

Nihar Karve provided an explanation why the two formulas which are denoted by $(\text{2.76})$ in the original post are identical in a curved spacetime with no torsion, i.e. with a symmetric connection. Here is the missing part to his post (which is useful in GR extensions with torsion): Let $\nabla$ be an operator which generalizes the $d$ from curved ...


2

Exactly as @Jacob1729 says. Strange though it may seem, positions are not vectors. Positioning in this or that coordinate system is just tagging points on a manifold. The definition of vectors essentially depends on a concept of tangency. And for that you need a derivative. Stop worrying about $x^i$ and $x'^i$. Those are just parametrisations of your ...


2

UPDATE: For clarity, here are some references relevant to the OP's question. There may be differences in conventions among the references. (Readers unfamiliar with this spacetime viewpoint may misinterpret the title of the OP's question.) Robert Geroch's General Relativity 1972 Lecture Notes (p. 53-54): An electromagnetic field is a (smooth) antisymmetric ...


2

In mathematics you can study manifolds that have non-zero torsion, in that case the connection in local coordinates is not symmetric in both lower indices, in this case torsion is not zero. In physics and particular in General Relativity it is assumed that the torsion is zero, so the connection (called in this case "Levi-Civita") of such a manifold ...


2

Turns out after some algebra, $$ \begin{align*} X^{(\mu\nu)}Y_{(\mu\nu)} &= \frac{1}{4}\left(X^{\mu\nu}Y_{\mu\nu} + X^{\mu\nu}Y_{\nu\mu} + X^{\nu\mu}Y_{\mu\nu} + X^{\nu\mu}Y_{\nu\mu}\right) \\ &= \frac{1}{4}\left(2X^{\mu\nu}Y_{\mu\nu} + 2X^{\nu\mu}Y_{\mu\nu}\right) & \text{Renaming dummy variables} \\ &= \frac{1}{2}\left(X^{\mu\nu} + ...


2

In general, a $(p,q)$-tensor eats $p$ covectors and $q$ vectors and returns a real number; such an object has $p$ indices upstairs and $q$ indices downstairs. The components of the metric tensor are written $g_{ij}$ because the metric $\mathbf g$ is a $(0,2)$-tensor which eats two vectors (and no covectors) and returns their inner product. If you see the ...


2

Those two expressions are equivalent. When contracting tensor indices, it doesn’t matter which one is up and which is down. So it doesn’t matter which one you contact with the (inverse) metric first.


1

Suppose you have an arbitrary tensor $T_{\mu\nu}$. It can be separated into the symmetric part and the antisymmetric part, i.e. $T_{\mu\nu}=T_{(\mu\nu)}+T_{[\mu\nu]}$. If you contract its indices with an antisymmetric tensor $B^{\mu\nu}$, you will have $$T_{\mu\nu}B^{\mu\nu}=T_{[\mu\nu]}B^{\mu\nu}\tag{1},$$ but this does not mean that $T_{\mu\nu}=T_{[\mu\nu]}...


1

Given a curve $\gamma: \mathbb R \rightarrow \mathcal M$ where $\mathcal M$ is the manifold under consideration (in your case, presumably $\mathbb R^2$), we define the tangent vector to $\gamma$ at the point $p=\gamma(0)$ to be a map $\mathcal V_{\gamma,p}$ which eats smooth functions $f:\mathcal M\rightarrow \mathbb R$ and spits out the following number: $$\...


1

$s_0$ is just the unit matrix. It is not explicitly written in the Hamiltonian - every entry in the $4\times4$ matrix is a $2\times2$ unit matrix itself, but since it's just a unit matrix, a simplified notation is used.


1

Equation (1.98) reduces to $0=0$ if any two of the free indices are the same. So consider how they can all be different: either they are three different spatial indices, or two are different spatial indices and the third is temporal. The former gives the fourth 3D Maxwell equation, and the latter gives the third. In more detail, when the three indices are ...


1

So, when you applied the equation $B_{ij}=X_{ijkl}A_{kl}$ to get $X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}X_{pqkl}A_{kl}$, note that in the expression $X_{ijkl}A_{kl}$, $k$ and $l$ are essentially dummy indices, and can be renamed to whatever we want. In particular, the next step is much clearer if we rename them to $x$ and $y$! Then we instead get $$X_{...


1

For clarification, write $\vec{u} = u_x \hat{x} + u_y \hat{y} + u_z\hat{z}$ in components, and $d\vec{S} = \hat{r} R^2 \sin\theta d\theta d\phi$ $$ \vec{I} \equiv \oint \oint \vec{\nabla} \vec{u} \cdot d\vec{S} $$ \begin{align} I_\alpha = & \oint \oint \vec{\nabla} u_\alpha(\vec{r})\cdot d\vec{S}\\ =& R^2 \int_0^\pi \sin\theta d\theta \int_0^{...


1

Use $R_{[abc]d} = 0$ and other symmetries of Riemann tensor.


1

If $\:\mathbf{f}\:$ is a pure force 3-vector applied on a particle of velocity 3-vector $\:\mathbf{u}\:$, then the 4-dimensional vector \begin{equation} \mathbf{F}\boldsymbol{=}\gamma_{\mathrm u}\left(\mathbf{f}\:,\:\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c} \right) \tag{01}\label{01} \end{equation} is a Lorentz 4-vector. We use the term $''$pure ...


1

Since $V$ is a Killing vector field, $\nabla_iV_j=-\nabla_jV_i$. Using this in equation 2, the desired relation follows.


1

There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so the action of the connection coefficient part of the covariant derivative is undefined on it.


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