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0

After some more thought, I think I've realized what's going on. The short answer is that the above result is correct, and is just a specific case of a more general construction. Here's more explanation: the point is that the original tangent field $k^a$ is only defined on the geodesic $\gamma$, but the expression $\nabla_a k^b$ is only sensible if the ...


3

Before going further, I would suggest you to read Chapter 13 ("Spinors") of R.Wald's book "General Relativity". In that chapter, you will see that 2-spinors are simply vectors living in a two-dimensional complex vector space. The capital letters in the indices are simply the abstract index notation for these vectors (see Section 2.4 in Chapter 2 of the same ...


0

Assume a path $X:\mathbb{R}\to\mathcal{M}$ in your manifold $\mathcal{M}$ parameterized by $s\in\mathbb{R}$ linking points $X(0),\,X(1)\in\mathcal{M}$. Write the length of this path as the functional $\mathcal{L}(X)=\int_0^1\sqrt{g(\mathrm{d}_s\,X,\,\mathrm{d}_s\,X)}\,\mathrm{d} s$. Now use the calculus of variations, i.e. apply Euler-Lagrange equation to ...


2

The geodesic equation can be derived by extremizing the length ("proper time" in the case of general relativity) of a path connecting two fixed points. One requires that, after introducing a parameter $\lambda$ so that for the geodesic $x^{\mu} = x^{\mu}(\lambda)$ connecting points $A$ and $B$: ...


2

When we vary $F^{ab}F_{ab}$ with respect to the metric, we must also specify what we are holding fixed. Assuming that the context is that of electromagnetism, we consider the four-potential $A_b$ as an independent variable, and therefore under variations of other variables (such as the metric), it is held fixed, as is $F_{ab} = \partial_a A_b - \partial_b ...


1

Using the Dolbeault bigrading, the (2,0) and (0,2) components of the Kähler metric $g_{zz}=0$ and $g_{\bar{z}\bar{z}}=0$ do indeed vanish, respectively. In particular, the formula $$g_{z\bar{z}}=\partial_{z}\partial_{\bar{z}}K$$ for the mixed (1,1) components does not generalize to the (2,0) and (0,2) components.


0

This may not be quite what you're after, but "Relativistic Hydrodynamics" by Rezzolla and Zanotti covers (relativistic) hydrodynamics in the language of differential geometry. This is a graduate level textbook on hydrodynamics in the context of general relativity (hence the differential geometry). It covers kinetic theory (including quantum effects), ...


3

This is called Helmholtz theorem, which states that for any vector field $\vec{F}$ that is twice continuously differentiable in a bounded domain, we can perform the decomposition $$ \vec{F} = - \vec{\nabla} \Phi + \vec{\nabla}\times\vec{A} $$ See http://en.wikipedia.org/wiki/Helmholtz_decomposition for a derivation


5

I) Disclaimer: In this answer we will use the (traditional) physicist's definition of tensors using indices and their transformation properties under coordinate transformations. Moreover, let us suppress time dependence $t$ for simplicity. II) Let the manifold $Q$ be the configuration space. The Lagrangian $L:TQ\to \mathbb{R}$ transforms as a scalar ...


0

In a rather general approach you can consider the Poisson bracket ${g,f}$ as expressing the rate of change of $g$ as a consequence of a flow induced by $f$. As mentioned by AngusTheMan in the comments, you get the time variation of $g$ if $f=H$ (assuming that quantities are not explicitly time-dependent). Here $g$ and $f$ are any (smooth) functions on the ...


3

First, they do not transform in an actual "representation" in the sense of a linear representation of the group of coordinate transformation since their behaviour under a coordinate transformations $x\mapsto y(x)$ is given as $$ {\Gamma^\alpha}_{\beta\gamma} \overset{y(x)}\mapsto \frac{\partial x^\mu}{\partial y^\beta}\frac{\partial x^\nu}{\partial ...


4

The Christoffel symbols do not transform under any representation. The reason for this is that they do not transform linearly, which puts them out of the game altogether. The transformation law is $$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde ...


0

Ref. 1 defines a local translation on spacetime $M$ as a diffeomorphism. The word local means here point-dependent ($x$-dependent). An infinitesimal diffeomorphisms $\xi^{\mu}(x)$ can be identified with a vector field. Note that in flat Minkowski spacetime we called affine transformations for global Poincare transformations. The word global means here ...


2

I'm not completely sure what you want, but honestly the entirety of Spivak's Calculus on manifolds is devoted to exactly that. If you want something that feels familiar, you can simply find $\nabla$ in various coordinate systems in Wikipedia, but if you want a less coordinate-centric view then you're probably going to need to step outside of your comfort ...


1

I would say that the Wikipedia page on curvilinear coordinates and the article Mathematical Physics Lessons - Gradient, Divergence and Curl in Curvilinear Coordinates by James Foadi are enough to understand what is going on.


0

$$ \nabla \epsilon_{...}=\nabla \sqrt{g}[...]=gg^{\mu\nu}\nabla g_{\mu\nu}[...]=0$$ In the last step we use Jacobi's formula for differentiating a determinant and $\nabla g_{\mu\nu}=0$. Here $[...]$ describes permutations and consists of $-1,0,1$ as you see from this article.


3

If $Q$ is configuration space, then the Lagrangian is a function $L: TQ\times \mathbb{R}\to \mathbb{R}$. Let the cotangent bundle $M:=T^{\ast}Q$ be the corresponding phase space. The Hamiltonian/phase space Lagrangian is a function $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


2

Let $X$ be the phase space. Then $L_\text{ph}(q,p,\dot{q},\dot{p},t)$ is a function on $TX\times \mathbb{R}$1, since the coordinates of $TX\times\mathbb{R}$ are precisely the coordinates of $X$, i.e. $(q,p)$ and their derivatives $(\dot{q},\dot{p})$ (and time $t$). If Hamilton's equations are fulfilled, there are relations among $q,\dot{q},p,\dot{p}$ (the ...


1

The Ricci tensor obviously a tensor that accepts two vectors and outputs a number. This number represents in some sense the "average" sectional curvature at a given location on the manifold $M$. In GR, we usually use spacetime manifolds. In his book Riemannian Geometry, Manfredo Do Carmo states the following on page 97: Let $x = z_n$ be a unit vector ...


6

At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric: $\Gamma^i_{jk} = g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk})$. Plugging this into the right-hand side of your expression will yield the left-hand side. However, one can obtain your expression directly from one of the ...


1

Comments to the question (v1): Let there be given an $n$-dimensional manifold $M$ with a smooth vector field $X\in \Gamma(TM)$. If $(x^1, \ldots, x^n)$ is some local coordinates on $M$, then the vector field takes the form $$\tag{A} X~=~X^i(x)\frac{\partial}{\partial x^i},$$ and one may study the autonomous first-order ODE $$\tag{B} ...


-1

there is NO curvature of spacetime. To prove this claim we will first assume that there IS curvature of spacetime. From that we derive the following: spacetime must have properties. The truth of the matter is that spacetime does not have properties. Therefore, our initial assumption was invalid, i.e. opposite must be true: there is no curvature of spacetime. ...


4

When people study continuum mechanics they usually do so at first in $\mathbb{R}^3$ where we have usually implied the usual metric tensor $(g_{ij}) = \operatorname{diag}(1,1,1)$ and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert ...


1

If four vector notation is less intuitive then refer back to three vectors \begin{align*} \vec{E} &= - \vec{\nabla}\phi - \frac{\partial\vec{A}}{\partial t} \\ \vec{B} &= \vec{\nabla}\times\vec{A} \end{align*} For a static point particle \begin{align*} \vec{E} &= \frac{e}{r}\hat{r}\\ \vec{B} &= 0 \end{align*} The solution up to gauge ...


3

Stokes' theorem needs no physical reason to be true. However, there is a nice intuitive description of the two-dimensional case. Tesselate the surface with little (infinitesimal) oriented squares and consider the integral as the sum of the curl on all these little squares: The inner sides of the squares have no contribution to this sum at all, because ...


2

When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. With this understanding, we call $g(.,.):=\langle.,.\rangle$ an inner product as usual. This lack of positive definiteness has many consequences. It is the ...


2

Comments to the question (v2): On one hand, let there be given a configuration space $(Q,g)$ endowed with a metric $g$. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric $g$ and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form $\omega$ on the cotangent bundle $T^{\ast}Q$ ...


1

No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the $\dot q$ variables. You then get a metric if such kinetic term is quadratic in $\dot q$ (and of course sensible kinetic energy is positive-definite). The metric and symplectic structures on a manifold are ...


0

Velocities and Spatial Accelerations are twists and Forces and Momenta are wrenches. Both are screws (two-vectors) with one vector free and the other a spatial field. All of them transform with the same laws and their interactions have many dual properties. NOTE: See "A treatise on the theory of screws", Stawell R Ball, ...


0

Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the ...


0

Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat $d\Omega^2$ more carefully: $r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$ so ...


3

I'll prove a formula that is probably easier to use for this. \begin{equation} \begin{split} \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\ &= ...


1

All timelike geodesics in Minkowski spacetime start at past timelike infinity and end at future timelike infinity. The worldlines of Rindler observers are not geodesics, whereas the worldlines of Minkowski observers are. Heuristically think of a flat Euclidean plane. There are plenty of inextendible curves that don't go to infinity, but all geodesics start ...


0

If the observer is not in free-fall, the metric-tensor $g_{\mu,\nu}(s)$ at the observer's position, expressed in local coordinates around the observer, will not be $\eta_{\mu,\nu}$. Your first assumption about the path $(\gamma)$ is wrong. I guess what you are aiming at is the notion of the space of coordinates around a point, which is indeed a flat space ...


3

Let $M$ be your spacetime, a smooth manifold equipped with (pseudo) Riemannian metric (for example $\mathbb{R}^{(1,3)}$ for special relativity). The set of reference frames is the frame bundle over $M$, usually denoted $FM$. Explicitly a frame at point $p$ in $M$ can be viewed as an ordered orthonormal basis (with respect to the the inner product defined ...


0

If you are working with the Newman Penrose null tetrad, i.e. with respect to which the metric looks like $g=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right)$ then the standard notation is $\lbrace l,n,m,\overline{m}\rbrace$, where $l,n$ distinguish the ...



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