# Tag Info

22

Black holes are solutions of vacuum EFE on a space-time with singularities. EFE in the vacuum are: $$G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 0.$$ These are sometimes written as $$R_{\mu \nu} = 0,$$ which is completely equivalent in spacetimes of all dimensionalities except for $d = 2$. This is easy to see. Write down the contraction of ...

17

You can see this in many different ways. Here is one explanation: The field equations of electrodynamics in harmonic gauge reads \begin{align} \Box A_\mu=J_\mu \end{align} This is a linear theory and the solution can be determined in terms of the sources, i.e. \begin{align} A_\mu(x)=\int d^3x' G(x,x')J_\mu(x') \end{align} where $G$ is the appropriate Green's ...

16

The short answer is no. The components of the Riemann tensor, which is an object measuring the curvature of the manifold at a point, can take any real number value. In classical GR, there's nothing which bounds these numbers, but when they diverge to infinity at a point it indicates a problem with our classical description (see discussions of singularities ...

7

I'm trying to motivate the general relativity theory. Concretely upon which arguments is deduced that spacetime must have a structure of curved Riemannian manifold. First, you consider spacetime without gravity. You notice that inertial objects have worldlines which are straight lines in spacetime and accelerometers measure how much a worldline bends in ...

6

The manifestly covariant equation is $\Sigma F^{\mu}=D_{\tau}P^{\mu}$ where $F^{\mu}$ is the four force, $D_{\tau}$ is the covariant derivative, and $P^{\mu}$ is the four momentum. Curvature is a rank four tensor, so it cannot be described as simply positive or negative. However, if you wish to include gravity or fictitious forces, that can be done by ...

5

The only thing I would consider to be a "limit" to spacetime curving in general relativity is a gravitational singularity: A gravitational singularity, spacetime singularity or simply singularity is a location in spacetime where the mass and gravitational field of a celestial body is predicted to become infinite by general relativity in a way that ...

3

The video is just nonsense. Early on, he starts talking about "curvature in time" and saying "time can curve." He just has no idea what he's talking about. Curvature is curvature of spacetime, not time or space separately. Gravitational time dilation isn't generically even something that can be defined in general relativity. It can only ...

3

The Friedmann metric with $k=0$ is spatially flat. It is consistent with current cosmological observations.

3

The theory of general relativity is a metric theory of gravity, with a Lorentzian metric (ie a metric with signature $-+++$). This means that any manifold with such a metric can be analyzed using the theory of general relativity. As far as topology goes, there are a few requirements for this to be the case. Most famously, the manifold has to be paracompact ...

3

You can correctly answer either yes or no to the question of whether gravity is a source of gravity. If you want to say no, then you can say that gravitational fields don't contribute to the stress-energy tensor, and that this has to be so because of the equivalence principle -- you can't make a tensor out of the gravitational field, because it vanishes for ...

3

When you derive the EFE from the Einstein-Hilbert action, $$S_{EH}= \frac{1}{2 \kappa} \int R \sqrt{-g} \, d^d x \ ,$$ at no point do you need to restrict the number of dimenions $d$ to 4 (but remember that the coupling constant $\kappa$ depends on $d$). Variation with respect to the metric leads to the usual field equations, $$G_{\mu \nu} = \kappa T_{\mu ... 3 We perceive gravity as a fictitious force as we are pushed off our geodesic by the surface of the Earth, according to the geodesic equation (the version that uses time):$$ \ddot x^{\mu} = -\Gamma^{\mu}_{\alpha\beta} \dot x^{\alpha}\dot x^{\beta} + -\Gamma^0_{\alpha\beta} \dot x^{\alpha}\dot x^{\beta}\dot x^{\mu}$$The point of this equation, here, is to ... 3 How much of the attractive "force" (well, it is not actually a force) felt by an object in the gravitational field of another object like a planet or star is caused by clocks closer to the object runnning slower (compared to clocks further away) compared to the object moving toward the heavier object due to curved geodesics? The geodesics are ... 3 Motion vs space expansion The two body situation cannot separated the two hypothesis you offer. Either can be assumed. Actually in order to establish with data that the space in our universe is expanding, astrophysicists needed data from the galaxies all around the earth In 1912, Vesto Slipher discovered that light from remote galaxies was redshifted. ... 3 What Landman writes is just nonsense. The components of the metric are not measures of curvature. Measures of curvature involve second derivatives of the metric's components. Spacetime is locally flat, so we can always choose local coordinates such that the components of the metric are the same as in flat spacetime. There is no such thing as purely temporal ... 2 A mode k is associated with a wavelength \lambda\propto 1/k so when a mode is k < H it is said to be superhorizon meaning that the wavelength is larger than the Hubble radius, because Hubble radius is proportional to 1/H. Be careful, the Hubble radius is no horizon but it's common to say that if something is larger than the Hubble radius then it'... 2 The spacetime of the string worldsheet is what is fundamental in perturbative string theory, not the target spacetime. You can define the worldsheet and its dynamics in an intrinsic way, without making reference to a "container space". In fact, that was a Gauss great achievement, "manifolds exist with independence of whether or not they accept ... 2 The object will only contract by a factor of$$V'=\frac{V}{\gamma}=V\sqrt{1-\frac{v^2}{c^2}}$$where V' is the measurement* made from our "stationary frame" (from which the object's frame is moving at relative velocity v) and V is the measurement made from the object's rest frame. * While V usually indicated Volume, the same relationship ... 2 is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved? is there a any explicit metric by this property? Sure, it's called the Friedman-Robertson-Walker metric and it describes the large-scale evolution of our Universe:$$ ds^2 = -dt^2 + a^2(t) (dx^2 + dy^2 + dz^2) $$On any surface of constant t, the metric is that ... 2 The Lorentz contraction is a statement about how a given object, which should be regarded as a region of spacetime (not just space since the object continues to exist over time), will be determined to have different spatial dimensions depending on which inertial reference frame is being adopted in order to specify how spacetime is to be divided into "... 2 There is no a mathematical compelling reason: curvature is permitted by local observations about free falling bodies and by the experimental identification of gravitational and inertial mass. The point is that the consequences of the assumption are experimentally confirmed. If we assume that curvature enters the picture and geodesics (with curvature) ... 2 We don't have any better way to define straightness than by saying that the world-line of an inertially moving test particle is straight. By the equivalence principle, we have to count free-falling particles as inertial. (We can't say they're acted on by a gravitational force, because there is always a frame in which that force is zero.) Once you accept this,... 2 No, it isn't only time curvature that can cause gravitational effects, but the temporal curvature terms (i.e. ones that include time and space, as opposed to just spatial directions) are the only ones we typically notice in everyday life. The Earth curves space and time together in comparable amounts (see this for the exact values) and in theory all of these ... 1 As @G. Smith said, The Friedmann metric with k=0 is spatially flat, but it is worth noting that this is an idealisation which depends on a perfectly uniform matter distribution. As such, we know it is an approximation to reality. It is only flat on a slice of constant cosmic time, which means it is only flat for observers with constant position in ... 1 Your question is ambiguous. You ask whether the shape will "change as predicted by the Lorentz transformation". But for an object in uniform motion, the Lorentz transformation does not predict any change in shape over time; it predicts that the shape will be different (but unchanging) in different frames. So there are two ways to interpret your ... 1 Length contraction gives L' = \frac{L}{\gamma}. Say the system S' moves at a certain velocity v_0\hat{x}, and lets look at the distance between two points:$$d=\left| p_1 - p_2 \right|d' = \left|\frac{p_1}{\gamma} - \frac{p_2}{\gamma}\right|=\left| \frac{p_1-p_2}{\gamma}\right| So the relation between two points can't switch with regards to the ...

1

Contrary to popular belief, there is no "space expansion" effect in general relativity. I covered this in detail in another answer recently. Friedmann coordinates are just a Lorentzian analog of polar coordinates (latitude and longitude). You can put them on any manifold that has the appropriate symmetries. That you can do so tells you only that ...

1

The Planck length is $l_p = \sqrt{\frac{\hbar G}{c^3}}$, so unless any of these constants evolve it will keep its value.

1

No, time-orientability is not enough. An oriented pseudo-Riemannian manifold needs to be time-oriented (or space-oriented) and have vanishing second Stiefel-Whitney class to admit a spin structure. It's a bit hard to find good references for this in the pseudo-Riemannian and not just the Riemannian setting, but most citations of this fact (e.g. in "Spin ...

1

The external field effect (EFE) is about whether the gravitational force of the external gravity sources can be effectively cancelled out (via equivalence principle) by changing to the free-fall reference frame. This hinges on the fact that the Einstein field equation of gravity is linear in the Riemann curvature tensor $R$ (for brevity I omitted the indices ...

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