# Tag Info

1

Suppose you're in a coordinate system where the Christoffels don't vanish at some point. To choose a coordinate system where the Christoffel symbols vanish at a given point $p$, you must apply a Christoffel symbol change of variables: $$0={\bar\Gamma}^k{}_{ij} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r{}_{pq}\, ... 1 My answer to How does the Hubble parameter change with the age of the universe? (which is itself adapted from Equation for Hubble Value as a function of time) explains how to calculate the scale factor. In fact we calculate the time as a function of the scale factor rather than the other way around. The equation we use is:$$ t(a) = \frac{1}{H_0}\int_0^a ...

0

The metric is telling you how to calculate the proper time along a path of your choosing. If you select a path where the time is everywhere constant then as you integrate along that path $dt = 0$ and any terms involving $dt$ disappear. It is as simple as that.

-1

The question is why you would want to do this in the first place. The equations of motion that you obtain one a compact region still aren't the real equations of motion, since the compact region is a mathematical choice to simplify the formulation of the problem. It is implicitly understood that the true equations of motion are only obtained on the limit of ...

1

The boundary of a subset of a topological space is abstractly defined as the set-theoretic difference between its closure and its interior. Since topological spaces in general have neither coordinates nor metrics, this notion is independent of the metric. Since the spacetime manifold is a manifold, it is a topological space (locally homeomorphic to ...

0

There is no problem with saying that we have a region with boundary as far as the underlying manifold goes. The problem is that you don't know the metric on the boundary or how to integrate. You should rephrase your question accordingly and in that case yes you need to be careful what boundary conditions you take as already said.

0

It does not make sense to have $\delta g_{\mu \nu}\neq 0$. Actually, for finite region, even if the variation of metric is zero at boundary, the derivative of metric is not zero, it will contribute a surface term to the variation of actio, thus we have to add another surface term to cancel this contribution, which consists of exterior curvature of the ...

3

Take a trace of Einstein equations (trace of $g_{\mu \nu}$ is $D$), you obtain $$R - \frac{D}{2} R + D \Lambda = 0$$ Or $$R=\frac{D \Lambda}{D/2-1}$$ Then substitute this expression for $R$ into full Einstein equations and you obtain trivially $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

0

Manifolds are defined such that locally they look like Euclidean space; this is why we call them smooth manifolds. A riemannian manifold is a manifold that locally has some inner product structure, ie a way of measuring length and angles. Lengths and angles are invariants, hence will have an invariant expression in terms of a local coordinate basis; and ...

0

In Riemannian geometry there is a beautiful theorem which states that a manifold with a symmetric connection is locally flat everywhere if and only if the curvature tensor vanishes. Therefore, in a locally flat coordinates such that $\Gamma_{jk}^i=0$, $g_{ij}$ is constant throughout the chart and a linear transformation can be used to diagonalize the metric ...

0

If $ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta}$ were true for all points of space, we would have no curvature, hence no gravity! Take for example a sphere (the Earth), locally we can measure distances by $ds^2=dx^2+dy^2$, but this can't hold for two arbitrary points on the sphere. In fact, this coordinate system changes from point to point ...

0

Let $\mathcal{M}$ be the space time manifold, whose local charts (open sets) are described by $U_i$. A local coordinate frame $S_i$ is a map $\xi\colon U_i\mapsto \mathbb{R}^N$ such that $\xi(m) = (x_1,\ldots,x_N) \in \mathbb{R}^N, m\,\in U_i$. Let, moreover, $g$ be a $(0,2)$ rank tensor (the metric). A change of coordinates is any smooth invertible map ...

0

$g_{\mu \nu}(x)$ means that $g$ is a function of location ($x$) --- so it varies across the manifold, which is the problem. I think that if $g \ne g(x)$, then necessarily $g = \eta$ ... Hopefully someone else can chime in on that.

2

The values of individual entries in the metric tensor depend on the coordinate system you choose. In the case of the FLRW metric there is a natural choice of coordinates called the comoving coordinates. In particular the comoving time has a very simple interpretation because it is equal to the proper time of a stationary observer, which obviously means ...

0

I am assuming the final equality $-\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}/2$ is a known result you are trying to check against your calculation of $-\sqrt{g}\ \text{tr}(g_{\mu\nu}\delta g^{\mu\nu})/2$. If so, you have essentially arrived at the same result, but notice that the expression $g_{\mu\nu}\delta g^{\mu\nu}$ in your final equality is already a ...

1

A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity. The equivalence principle ...

0

but keep getting confused if I should sum over each term separately. It's not particularly clear what you mean here. Your expression uses the Einstein summation convention, which means that every repeated index is summed over. In principle this should be done for each term separately, so your expression reads $${R_{00}}= ... 1 Let's suppress some dimensions to simplify:$$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 $$This quantity$$\Delta s^2$$is preserved by changes of reference frame, just as in Galilean physics the quantity$$\Delta r^2 = \Delta x^2 + \Delta y^2 $$is preserved by rotations. Notice it is also the equation of a hyperbola. Thus, the effect of a frame shift is ... 0 The answer to my question is simpler than I suspected. It is fairly easy to describe the movements of both soldiers mathematically. The first soldier's spear is being transported along X_1 = [y^2+z^2, -x^2, 0]. The second soldier's spear along X_2 = [y^2+z^2, 0, -x^2]. Both are valid parallel transports relative to the tangent vector field V=[0, -z, ... -8 The FLRW metric starts with "the assumption of homogeneity and isotropy of space". But Einstein described a gravitational field as space that is "neither homogeneous nor isotropic" : Hence setting expansion aside, for the universe as a whole there's no overall gravitational field, and light goes straight. Because of this we say the universe is flat, as ... 3 Definition 1. A spacetime is said to be spatially homogeneous if there is a one-parameter family of spacelike hypersurfaces \Sigma_t foliating the spacetime such that for each t and for any points p,q\in\Sigma_t there is an isometry of the spacetime metric g which takes p to q. Definition 2. A spacetime is said to be isotropic if at each point ... 3 Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric:$$T^a T_a = \sum_{a,b} g_{ab} T^a T^b You have terms like ...

Top 50 recent answers are included