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1

For this to make sense, we have to talk about $F$ as a 2-form, i.e. $\mathbf{F}=\frac{1}{2}F_{\mu\nu} \: dx^\mu \wedge dx^\nu$ Then $\star\mathbf{F}$ is its dual. In 4d space: $\star \left(dx^\mu\wedge dx^\nu\right) = \frac{1}{2}g^{\mu\kappa}g^{\nu\sigma}\epsilon_{\kappa\sigma\rho\phi} dx^\rho\wedge dx^\phi$ Thus: $\star \mathbf{F} = \frac{1}{4}F_{\mu\...


1

Every Lorentz invariant is a Lorentz scalar. That is just true by definition. The trick here is to specify what an object is a scalar, vector, tensor, spinor, etc with respect to. For instance, the electric charge density is a scalar with respect to spatial rotations, but not with respect to boosts. Likewise, the four-momentum of a neutral particle can ...


2

“Scalar” and “Lorentz invariant” are synonyms in the context of Special and General Relativity. However, it is possible to have constant tensors whose components don’t actually change when transformed, such as the Minkowski metric tensor $\eta_{\mu\nu}$ in Special Relativity. We don’t call these “invariant”. Some people call these tensors “isotropic”; ...


0

Thanks to @Brick I could finish my calculation using: \begin{equation} (g^{ip} A_p)_{\mid k}=A^i_{\mid k} \end{equation} \begin{equation} g^{ip}_{\mid k} g_{pd} A^d + g^{ip} A_{p \mid k} \end{equation} Since $(g^{ip} g_{pd})_{\mid k} = 0$ we can conclude: \begin{equation} A^i_{\mid k} = -g_{pd\mid k} g^{ip} A^d + g^{ip} A_{p \mid k} \end{equation} Which ...


1

You made your mistake in the first line. Between the second and the third equal signs you raised the index through a partial derivative, but $$ g^{ip} A_{p|k} \neq A^{i}_{|k} .$$ (It would be true for the covariant derivative.) I think you'll find this is different by exactly the term that you're trying to cancel at the end, although you may need to simply ...


1

This is an arbitrary convention that was fixed historically around the time that Einstein published the general theory of relativity. It's similar to the right-hand rule for defining torque, or the convention that the charge of the electron is negative. Although it's arbitrary, it's fixed, so different authors do not use different conventions as a matter of ...


2

There are two objects in relativity that contain the same information, but are of different mathematical nature - vectors and covectors. Covector is dual vector to the original vector. That is, it is member of different vector space, but (in space endowed with metric) there exists natural one-to-one correspondence between vectors and covectors. Now, since ...


2

Your intuition is correct in my opinion. The relativistic equivalent of $\rho$ would be $j^{\mu} = (\rho c, j^i)$ which leads to a vector equation. However you should not think of masses being the source of the gravitational field but more about energy in general. This leads to the stress-energy-tensor. \begin{equation} T^{\mu \nu} = \rho \gamma^2 c^2\left( ...


0

The 3D rotation group $SO(3)$ has double-cover $SU(2)$, which is a subgroup of $SL(2,\mathbb{C})$. TL;DR: A spinor index of a 2-component spinor is raised and lowered with the 2D Levi-Civita symbol, which can informally be viewed as a "symplectic metric" for the symplectic group $Sp(2,\mathbb{C})\cong SL(2,\mathbb{C})$. In more detail, in case of the Lie ...


5

Let $F\in \Omega^p(\mathbb{R}^d)$ be an arbitrary $p$-form in $d$ dimensions, where $1\leq p\leq d$. Clearly any form $F$ squares to zero: $F\wedge F=0$ if $p$ is odd or if $2p>d$. Clearly any simple/decomposable $F$ squares to zero: $F\wedge F=0$. Proposition 1: The opposite implication $$\forall F\in \Omega^p(\mathbb{R}^d):~~F\wedge F=0~\...


1

No, for an Euclidean 3D space the rotations (and translations) are maps between reference frames, while tensors are independent of reference frames. See also my related Phys.SE answer here in the context of SR.


2

Why ???, the coordinate system is arbitrary, can't be polar or spherical ?, i understand that locally the inertial frame has behaviour as SR but in SR i can have polar, spherical and some other's coordinate system's that derivative of basis are not equals to zero. All of those statements of Schultz revolve around the idea of a coordinate system whose origin ...


0

I don't think you need metric compatibility to prove this although you can use it. There is a much simpler way with repeated use of (any) metric to lower the index. $ U^\alpha\nabla_\alpha V^\beta=W^\beta $ $ \Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}V_\gamma $ $ \Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha ...


3

Let me add an argument that is much less general than the answer by Mike Stone, but somewhat more elementary. It is valid for 2-forms regardless of the dimension of the base space the forms act on. So, a general 2-form $F$ can be written in an arbitrarily chosen basis of 1-forms $e^i$ as $F=\frac12f_{ij}e^i\wedge e^j$. The matrix of coefficients $f_{ij}$ is ...


2

The general condition for an $n$-form with components $T^{\mu_1,\mu_2,\ldots \mu_n}$ to be writable as a single wedge product $$ T=U_1\wedge U_2,\wedge\ldots \wedge U_n $$ is $$ T^{\mu_1\mu_2\ldots\mu_{n-1}[ \nu_1}T^{\nu_2\nu_3\ldots \nu_{n+1}]}=0 $$ for all possible choices of the indices. where $[\ldots]$ means antisymmetrize. Thus a three index ...


0

The inverse metric tensor $g^{\mu\nu}$ (or with any other choice of index naming) is the inverse of $g_{\mu\nu}$, in the matrix sense. Since it's usually $4\times 4$ or larger, I would shug it into Mathematica or Wolfram Alpha to get $g^{\mu\nu}$. If $g_{\mu\nu}$ is diagonal, then $g^{\mu\nu}$ is element-wise the reciprocal of each element of $g_{\mu\nu}$. ...


1

I'll give another proof, just for fun. We know that in a pseudo-Riemannian manifold $(M,g)$, geodesics are critical points of the energy functional $$E[\gamma] = \frac{1}{2}\int_I g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))\,{\rm d}t$$Since $K$ is Killing, the flow of $K$ (consisting of isometries) leaves the Lagrangian we're integrating invariant. So ...


1

I think you've pretty much answered the question yourself. Notations like $x^\alpha$ and $\Lambda^\alpha{}_\beta$ can be read either as a notation for a whole tensor or as a notation for one of its components. If you like, you can imagine this as two different notations where the translation between the two notations is trivial. One thing that seems missing ...


0

I don't think there is an easy way to do this. Assuming that $U$ is the underlying vector space and $U^*$ is it's dual, you are staring with $\mathbf{T} \in V=U\otimes U\otimes U^*$ and $\mathbf{S} \in W=U\otimes U^*\otimes U^*$ and then seeking to define a unique linear map: $\alpha: V\times W\to Q,\quad Q=U \times U^*$ The problem, I think, is ...


1

I would think of it like this, using the fact that covariant derivative commutes with contractions (use C for contraction) and Liebniz rule $\nabla_X (T(\omega,Y)) =\nabla_X (C C ( T\otimes \omega \otimes Y)) = CC \nabla_X(T\otimes \omega \otimes Y) = CC((\nabla_X T)\otimes \omega \otimes Y+ T\otimes (\nabla_X\omega) \otimes Y+ T\otimes \omega \otimes( \...


1

The covariant derivative is defined to obey the Leibnitz rule. If the ${\bf e}_i$ are a vielbein basis then We define the action of $\nabla_X$ on any function $f(x)$ by $$ \nabla_Xf= Xf = X^\mu \partial_\mu f, $$ and on the elements ${\bf e}_i$ of a vielbein basis by $$ \nabla_X {\bf e}_i = {\bf e}_j {\omega^j}_{i\mu}X^\mu. $$ We extend to any other ...


1

This longer answer is a supplement to tparker's good concise answer. The notation $dx^a$ is used for a couple of different-but-related things: First perspective: Intuitively, we often think of it as an infinitesimal displacement. In this view, $ds^2=g_{ab}dx^a dx^b$ is giving us the infinitesimal change $ds$ in proper time or proper distance (depending on ...


3

The metric $\eta$ is a bilinear function that inputs two four-vectors (call them $A$ and $B$) and outputs the scalar $A \cdot B = A^\mu B_\mu = A^\mu \eta_{\mu \nu} B^\nu$. It is clearly bilinear (i.e. separately linear in both the first argument $A$ and the second argument $B$) by the distributive property.


3

The standard way to obtain the coefficients of a metric compatible connection goes as follows. First, we demand that $$\nabla_\mu g_{ab} = 0$$ for all $\mu,a,b$. Plugging in the connection coefficients gives us that $$\nabla_\mu g_{ab} = \partial_\mu g_{ab} - \Gamma^i_{a\mu} g_{ib} - \Gamma^i_{b\mu} g_{ai}=0$$ Next we permute the indices to get two ...


2

The index $\alpha$ is actually a dummy index. You are using the Einstein summation convention, this means you need to expand the terms, for example $$g^{\alpha \theta} g_{\alpha r ,r} = g^{r \theta} g_{r r,r} + g^{\theta \theta} g_{\theta r,r}.$$ If you do this for all three terms and plug in the metric you should get the right result.


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