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11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...


9

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here. We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions ...


8

I think the problem in understanding this is the idea of "space being sucked into a black hole." The reality is matter is "sucked" into a black hole. Space is warped around the black whole, but space is not "sucked" into anything. Here's the issue. What is space? You can't touch space (or better, the space-time continuum). So, one view is that space is ...


7

Assume $\eta_{\alpha\beta} \neq \eta_{\beta\alpha}$. Because it's irrelevant what letter we use for our indices, $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}.$$ Then $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = ...


5

If you look at the formula for the Christoffel symbol in terms of the metric tensor you see that you need to take derivatives. So ... Potential $\overset{\text{take derivative}}\rightarrow$ Force $\rightarrow$ Acceleration. Is like: Metric $\overset{\text{take derivative}}\rightarrow$ Christoffel Symbol $\rightarrow$ Geodesic Deviation. I'm not sure any ...


4

Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric, whereas $\nabla_Xg = 0$ says something about the connection. Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a ...


4

If you stick to one convention out of many other conventions, you should have same results regardless of signature of metric. Here I follow Carroll's conventions: http://amzn.com/0805387323. For Christoffel symbol, we have \begin{equation} \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda\rho}(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho ...


4

The metric tensor is created from the spacetime interval equation. On top of that, $[dx^\alpha,dx^\beta]=0$. Suppose we have a 1+1 dimensional spacetime, if you are given an interval equation resembling: $$ds^2=-a\,dx_0^2+b\,dx_1^2+c\,dx_0dx_1$$ Obviously, $\eta_{0\,0}=-a$ and $\eta_{1\,1}=b$. Now, you can assume that, say, $\eta_{0\,1}=c/3$ and ...


4

When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe. It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are ...


3

Schutz's A First Course in General Relativity explains it pretty well, I think. The pertinent page is here, but if international copyright law won't let you read it, the basic gist of it is the following: If the gravitational acceleration was the same everywhere in space, then we could shift into a freely falling reference frame, and freely falling ...


3

As one way, consider the parallel transport of two vectors $p^\mu$ and $q^\nu$ along the same curve $C$ such that the same angle between the two, $$ \cos\phi=\frac{p^\alpha q_\alpha}{\sqrt{p^\beta q_\beta}\sqrt{p^\gamma q_\gamma}} $$ is preserved. Parallel transport tells us that for a general vector $f^\alpha$, $$ ...


2

The contravariant metric tensor is the inverse metric tensor. If you have a general $g_{ab}$ you can find $g^{ab}$ by matrix inversion (which can usually be done in Mathematica or any other program of the kind). In the special case of a diagonal metric tensor you can verify that $g^{ii} = 1/g_{ii}$.


2

Let $T$ be some tensor field, $V$ a vector field, intuitively: Covariant derivative $\triangledown_V T$ measures how far away a tensor is from being parallel transported along a vector field $V$. Lie derivative $\mathcal L _V T$ measures how much a tensor changes under the one-parameter group of transformations generated by vector field $V$. Parallel ...


2

I think this a way of understanding the situation that is both intuitive and correct. The more experienced GRers will no doubt shout at me if this is wrong or misleading. If we can write the metric so that the timelike Killing vector is everywhere orthogonal to the spacelike surfaces then there will be no cross terms between the time and spatial elements in ...


2

To decide if two metrics are related by a change of frame and/or coordinate transformation is called the equivalence problem. It can be solved using the Cartan-Karlhede algorithm. Given a metric $g$ expressed in some coordinates $x_i$, the algorithm computes a set of invariantly defined curvature invariants expressed as functions of $x_i$. For example, the ...


2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...


2

You never have to make energy come from nowhere, the fact that the Einstein tensor has zero divergence means thavyiu can write any spacetime and the corresponding stress-energy tensor will have zero divergence. And zero divergence means that changes in energy (or momentum) are effected only by energy (or momentum respectively) net flowing in or out of a ...


2

The formula you have for the metric is not quite right. The curvature tensor is skew in the first pair of indices and the last pair but your metric has it being symmetric. Noneltheless, I agree with you calculation of the Christoffel symbol. I note that our formula for ${\Gamma^\lambda}_{\mu\nu}$ is symmetric in the last two indices as befits a ...


2

The correct way is to define the reparametrization-invariant action $$ S[X] = \int d\tau \sqrt{g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau} }. $$ Note that the choice of $\tau$ is arbitrary. The system has a large group of gauge symmetries - those are reparametrizations of the worldline (different choices of $\tau$). One way ...


2

The Schwarzschild metric as you've written it is only one particular coordinate system and the fact that $r$ and $t$ switch roles at the evnt horizon is an artifact of that coordinate system. There are other coordinate systems which make certain properties of the metric more intuitive. The ones that might be most useful for you are ones which can be drawn as ...


1

What the author means is as follows. Consider the (un-normalized) vector field $\partial_r$ where $r$ is the radial coordinate; $\partial_r$ is thus just the vector field orthogonal to the level sets $r = \text{const.}$ or, equivalently, it is the vector field foliating said level sets. As an aside, note that Schwarzschild coordinates are perfectly valid ...


1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...


1

If you plot a continuous curve in spacetime, it could be a path of a body if the tangent to the curve exists and has positive squared interval. General relativity is a geometrical theory, so everything is written in a geometrical way and the geometrical generalization is the predictions the theory makes. So outside the event horizon your curve has to have ...


1

It appears that all you have to do is integrate (6) to obtain: $$\alpha = \int dx \frac{2GM}{c^2}\frac{y}{(x^2+y^2)^\frac{3}{2}} = \frac{2GM}{c^2}\frac{xy}{y^2\sqrt{x^2+y^2}}=\frac{2GM}{c^2}\frac{x}{y\sqrt{x^2+y^2}} \equiv \frac{4GM}{c^2R}$$ So we obtain: $$R=\frac{2y\sqrt{x^2+y^2}}{x}$$


1

If it's diagonal, you can just focus on the diagonal elements, that is : $g_{aa} g^{aa} = 1$


1

A normal Killing vector is obtanied by solving the usual Killing equation $\nabla_{(\mu} \xi_{\nu)} = 0$ A conformal Killing vector is obtanied by solving a slightly different equation, the conformal Killing equation: $\nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = 2 \alpha g_{\mu \nu}$ where $\alpha$ is obtanied by taking the trace of the equation above. ...


1

You just need to plug your expression for $dT^2$ back into the original metric, with the substitution $\psi' = -C/A$. This gets you \begin{align*} ds^2 &= -A(r) \left[ dT^2 - {\psi'}^2 dr^2 + 2 \frac{C}{A} dr dt \right] + B(r) dr^2 + 2 C(r) \, dr \, dt + D(r) r^2 \, d\Omega^2 \\ &= -A(r) dT^2 + \left[ B(r) + A(r) \left( \psi'(r) \right)^2 ...


1

The signature is one convention (both in special relativity and in general relativity). But in general relativity there are many different conventions besides just the signature. The front inside cover of Misner Thorne and Wheeler lists conventions for signature, for the Riemann Tensor, for the Einstein Tensor, and for the use of Greek and Latin indices and ...


1

It is not clear what OP is asking, but consider the following chain of reasoning: Scalar curvature $R$ is an invariant independent of choice of coordinates. In GR, regions of matter, e.g., a star, is modelled with non-zero curvature. The Minkowski space has zero curvature. Hence, if a spacetime $(M,g)$ has non-zero scalar curvature $R$ at a point $p\in M$, ...


1

According to what I have read, we have measured the universe to be flat More or less. I'm happy enough with the WMAP results that indicate that the universe is flat. To be blunt I never thought it could be anything other than flat. the shape of the universe is directly related to the mass-energy density. That's what they say. But IMHO two out of three ...



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