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22

The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not ...


15

A homogeneous cosmology is one in which there are no "special" places in the universe: at a given instant in time, the universe appears the same at every location (on large enough spatial scales). An isotropic cosmology is one in which there are no "special" directions: at a given instant in time, the universe appears the same in every direction (again, on ...


15

The significance of the metric: $$ d\tau^2 = dt^2 - dx^2 $$ is that $d\tau^2$ is an invarient i.e. every observer in every frame, even accelerated frames, will agree on the value of $d\tau^2$. In contrast $dt$ and $dx$ are coordinate dependant and different observers will disagree about the relative values of $dt$ and $dx$. So while it is certainly true ...


15

Firstly, the Equivalence Principle reduces to the statement that in a freefall frame, the spacetime manifold is locally exactly as it is for special relativity (see my answer here for a fuller explanation of why this is so). So you can swiftly reduce your question to "Why does Minkowski spacetime have a nontrivial, non-Euclidean signature?". Since we're ...


14

It means that the laws of physics are the same everywhere and the same in every direction. It is of fundamental importance as these symmetries give rise to conservation laws. The isotropy of the universe means that angular momentum is conserved; its homogeneity means that momentum is conserved. A similar symmetry, that the laws of physics are the same for ...


13

as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: ...


12

In hindsight, here is a short proof. The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols $$ \Gamma^{\lambda}_{\mu\nu}~=~0$$ are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a ...


12

Gödel's rotating universe solution does allow for time travel (closed timelike curves), but it has nothing to do with wormholes--in such a universe one could travel into one's own past just by taking a rocket on a long-term looping path through space, from any starting location. This page discussing Gödel's solution includes a spacetime diagram showing how ...


11

You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor. It is not necessarily obvious whether a given metric is ...


11

First, there is no mechanical algorithm to solve a general differential equation. Einstein's equations are obviously no exception – in fact, they belong among the more complicated and less "solvable" equations among those one may learn about. Analytically writable solutions only exist in very special, simple, and/or symmetric cases (simple enough equations ...


11

To calculate the Hubble constant we need the a scale factor, $a(t)$. This is a measure of how much the universe has expanded. We take the scale factor to be unity at the current moment, so if $a = 2$ that means the universe has expanded twice as much as it has right now. Likewise $a = 0.5$ means the universe had expanded only half as much as it has right ...


10

By defintion, Minkowski space $\mathbb{R}^{p,q}$ must have signature $(p,q)=(1,d-1)$, with metric, $$ds^2 = -dt^2 +dx_1^2 + dx^2_2 + \dots$$ The signature $(+,+,\dots)$ corresponds to Euclidean space, which is obtained by a Wick rotation, $$t\to -i\tau$$ to imaginary time $\tau$, and the metric is modified, in the case of Wick rotated Minkowski space, to ...


10

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.] II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing $$\tag{1} V \otimes ...


10

First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices. In this parameterization, the four velocity takes the form \begin{equation} ...


9

I think it might help to think about the spacetime interval $\text{d}s^2$ as a measure of movement in spacetime relative to the speed of light. Let's say that you want to move from a point $p=(0,0,0,0)$ to another point $p'=(t,x,0,0)$. The quantity $\text{d}s^2 = c^2\text{d}t^2-\text{d}x^2$ is then: Positive if $x<ct$, which means that you traversed the ...


9

This is true - in fact you could define $\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$. I assume this meant to say $$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$ Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch $g^{\sigma\rho}$ with $\nabla_\nu$, which you ...


8

Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange). The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ...


8

The covariant derivative is metric compatible, so $\nabla_{\alpha} g_{\beta \gamma} = 0$. This is the condition that the inner product is preserved under parallel transport.


8

An interesting question indeed :-) Yes, you can flip the overall sign of the Minkowski metric, and in fact a lot of physicists do this! The sign choice $\operatorname{diag}(-1, 1, 1, 1)$ is conventional in fundamental quantum field theory and in quantum gravity, if I remember correctly, whereas $\operatorname{diag}(1, -1, -1, -1)$ is conventional in particle ...


8

On any manifold we can define the differential $df$ of a scalar $f$. The differential is a 1-form: something that eats vectors and spits out scalars, or even less formally, something with one down index. We have the following formula for the differential, $$df = \frac{\partial f}{\partial x^i} dx^i$$ (sum over $i$ implied). You can write it in index notation ...


8

and Δs is the space-time interval. Actually, many (most?) will say that the spacetime interval is $\Delta s^2$. In other words, $\Delta s^2$ is not the squared interval; it is the symbol for the interval. Since this has been questioned in a comment, I provide some references below: Bernard Schutz writes in Gravity from the Ground up: An Introductory ...


8

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows: Option one: One defines ...


8

Let's separate out some definitions: metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ such that the following axioms hold for all $x,y,z \in X$: $d(x,y) \geq 0$, $d(x,y) = 0 \Leftrightarrow x = y$, $d(x,y) = d(y,x)$, and $d(x,z) \leq d(x,y) + d(y,z)$. pseudo-metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ ...


8

A metric on a manifold $M$ is, by definition, a symmetric 2-tensor field $g$ with the property that $g_x$ is positive-definite for every $x\in M$ (plus some smoothness/continuity requirements if $M$ is smooth/topological). This ensures that the norm of a vector in a fibre of the tangent bundle to $M$ is a non-negative number, and that the angle between ...


7

From the little I've read, it seems that spacetime is Lorentzian. Unfortunately, the need for a metric that isn't positive-definite escapes my understanding. Could someone explain the reasoning? The short answer: - because of the attractiveness of the geometrical language and because of the Lorentz transformations, which do not preserve any positive ...


7

The Lagrangian you wrote is $$L=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\sigma}\frac{dx^\nu}{d\sigma}}$$ I'm sure you also know that $$d\tau^2=-g_{\mu\nu}dx^\mu dx^\nu$$ Plugging this into the first equation, we get $$L=\sqrt{\frac{d\tau^2}{d\sigma^2}}=\frac{d\tau}{d\sigma}$$ I'm not sure how much more help you want. Do you want to know why you have been told to use ...


7

Why is the scalar product of four-velocity with itself -1 The scalar product is invariant In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component. See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have $$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 ...


7

Those Greek letters are indices indexing the components of $g$. Generally if one expresses a rank-2 tensor like $g$ as a matrix, the first index indexes the rows, the second the columns. In your example, we have $g_{rr} \equiv g_{11} = 1$, $g_{\theta\theta} \equiv g_{22} = r^2$, $g_{r\theta} \equiv g_{12} = 0$, etc. As you can see, we sometimes use numbers ...


7

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...



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