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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 ...


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 ...


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 ...


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 ...


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 $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


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 ...


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) \\ &= ...


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 ...


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 ...


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


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 ...


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 ...


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

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 $T^{\ast}Q$ ...


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

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 ...


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.


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

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 ...


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 ...


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 ...


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 ...


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.



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