Tag Info

Hot answers tagged

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 ...


3

The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at $r=2m$. There is no physical connection and the two pieces can be viewed as separate solutions. The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different ...


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) \\ &= ...


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

The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order ...


2

Relativists tend to use the proper time, $d\tau$, and the proper distance, $ds$, interchangably. If you're working with proper time you'd expect the equation for it to look like: $$ d\tau^2 = dt^2 + \text{other terms} $$ while if you're working with proper distance you expect: $$ ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms} $$ The sign problem comes ...


1

Over the real numbers, any non-degenerate quadratic form is determined (up to a change of basis) by its signature, which consists entirely of $1$s and $-1$s.


1

Our model for spacetime is that of a manifold, which is the mathematical term for something that looks like $\mathbb{R}^n$ in any zoomed-in patch, and where all these patches are stitched together in a sensible way. On our manifold we have $n$ coordinates -- real numbers that describe each point and vary smoothly from point to point. We also add to our ...


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.


1

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

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

One thing you can use the curvature tensor for is to detect singularities in the spacetime. For the Schwarzschild solution, the simpler curvature scalars formed from the Ricci tensor, $R$ and $R_{ab} R^{ab}$ vanish everywhere due to the fact that this is a vacuum solution to the Einstein equations. But since you have the full Riemann tensor at your ...


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 ...



Only top voted, non community-wiki answers of a minimum length are eligible