# Tag Info

11

Great question. 1) There is indirect (and circumstantial) evidence that they do merge. While there are some famous examples of apparently 'binary' (or more accurately 'dual') AGN (e.g. Komossa+2003, or Rodriguez+2006) there seems to be a very conspicuous dearth of such systems --- suggesting that they don't spend very long at observable separations. Note ...

9

The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric. The exact form ...

5

Yes. I think Randall Munroe put it perfectly in this comic: The rubber sheet analogy does not tell you much about actual gravity.

4

Newtonian limit As a starting point, let us take Newtonian gravitation. There, we know two things: $$- \Phi_{,i} = \frac{d^2 x^i}{d t}$$ $$\Delta \Phi = 4 \pi G \rho$$ We must first investigate how does $\Phi$ correspond to the metric $g_{\mu \nu}$. For that, we assume that Newtonian gravity is valid for slow particles $v/c \ll 1$ on an almost Minkowski ...

4

Let us fix a reference frame $S$, where a particle of charge $q$ and velocity $v$ lies. It can be experimentally proven that, if another such particle $q'$ is present elsewhere in the universe, the initial one is subject to a force $\textbf{F}=q\textbf{E}$, where $\textbf{E}$ can be measured and addressed to the other body $q'$. Likewise, if a current $i$ ...

4

Now there is a light ray moving outward at the speed of light. I'm afraid that isn't the case; within the event horizon of a Schwarzschild black hole, the radial coordinate is timelike and so, moving 'outward' toward the horizon is as impossible as moving 'backward' in time. This plain to see in the Kruskal–Szekeres coordinates: Image credit See ...

4

Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation $$\frac{dx}{ds}= X(x(s))\tag{1}$$ with a generic initial condition $z$ at $s=0$ in the manifold $M$ (Minkowski spacetime in your example). $X$ is your vector field on ...

3

Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle $\theta$ of a light ray in the ...

3

In Feynman's Lectures on Physics (volume 2 chapter 42) he states that the field equation is equivalent to the following statement: For all local inertial observers, the scalar curvature of space at a point is proportional to the energy density at that point. Simple, right? By requiring the correct Newtonian limit the constant of proportionality can be ...

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 $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

3

The Guardian article is over dramatising a bit. GPS satellites normally orbit at an approximately fixed altitude and orbital speed so their gravitational time dilation is constant. Because Galileo 5 and 6 are in elliptical orbits their time dilation is constantly varying. The variation is partly due to changes in altitude and partly due to changes in the ...

2

I think a possible analogy would be to imagine that the singularity is a waterfall. By emitting light, you are trying to send a signal upstream using a tame fish. Outside the event horizon the fish is able to make headway against the current. But the river flows so fast within the event horizon as it approaches the waterfall, that your fish ends up going ...

2

Here's a few methods to find the Einstein Field Equations : 1) The classical route The classical method is to note the similarity between the geodesic equation $$\ddot x^\sigma + {\Gamma^\sigma }_{\mu\nu} \dot x^\mu \dot x^\nu = 0$$ And the classical equation of motion for particles in a gravitational field : ...

2

For better clarity, let's define the following: Axial direction = the direction the person & light beam are drawn into the BH. Radial direction = the direction perpendicular to the axial direction. If we, looking in the same direction as the person & light are being drawn into the BH, watch the light beam as it is drawn into the BH, we will see the ...

2

From the perspective of a photon: There is no such perspective. I was trying get an understanding of the universe from the photons perspective. There is no such perspective. Consider the following excerpt from I am driving my car at the speed of light and I turn on my headlights. What do I see?: Sometimes people persist: What would the ...

2

In this answer we assume a spherically symmetric spacetime, no cosmological constant $\Lambda=0$, and signature convention $(-,+,+,+)$ for the metric. I) Birkhoff's theorem (BT) only works for a vacuum branch of a spherically symmetric spacetime, i.e. in a radial interval $r_1<r<r_2$ without any matter, cf. e.g. this Phys.SE post. Therefore BT would ...

2

When you talk about a point in space, you're talking about a specific set of $(x, y, z)$ coordinates. Of course there's no use to talking about a point in space unless something is happening there, e.g. $(0, 6, 0)$ is the cannonball's starting location". An event is the same idea in $3+1D$ spacetime- it's a specific set of $(t, x, y, z)$ coordinates. ...

2

The basic problem with this is the so-called "cosmological principle". We expect the Universe to be 'homogeneous' (statistically the same no matter where you look, provided you zoom out far enough) and 'isotropic' (the same no matter how you rotate it). Failure of either condition would essentially mean that the Universe has a centre, which would be weird. ...

2

The problem is that no "intuitive" explanation can capture what gravity is actually about, because if it could, then general relativity should itself be intuitive. The rubber-sheet analogy is in my view not a totally misleading analogy for what it wants to show (namely that masses curve spacetime), but it tackles the wrong problem - the main problem being ...

2

I can't say that I have ever seen any attempts at simulating non-causal spacetimes (the closest I've seen is the simulation of fields upon such spacetimes). A few non-causal spacetimes do admit a time slicing, by the way, although by definition not all of these slices are achronal. Just solving it like any other PDE might be an avenue worth exploring, but ...

1

If you believe in "block time", that all past, present and future exist simultaneously like that famous loaf of bread in the Brian Greene video, and that our "present" is determined by our mind/neurons/consciousness tuning in to a specific slice of time, then it may be possible to train your mind/neurons/consciousness to tune in to a different slice of time. ...

1

I found the book that I mentioned above: H. Stephani, D. Kramer, M. MacCallum, C. Hoenselaers, E. Herlt, “Exact Solutions of Einstein’s Field Equations: 2nd Edition”, (2003), Cambridge University Press.

1

Goedel proved that circular time is possible. I.e., travelling forward in time eventually reaches the past and returns to the present. Not very practical,, but these circular time universes are solutions to Einstein's GR equations, so that settles it....

1

Photons do not have an inertial reference frame because photons travel at the speed of light in any inertial reference frame, and obviously you can not have a frame in which the "at rest" particle is not at rest. In fact, as far as velocities allowed in an inertial frame go, the $c$ is the asymptotic limit - "asymptotic" meaning of course that $c$ is not ...

1

The statement is wrong, though sort of true. Gravitational waves are exceedingly hard to create and significant energy is radiated as gravitational waves only for massive stars rotating rapidly at a short distance. In principle the Earth-Moon system radiates gravitational waves, but at such a ridiculously low intensity that it's fair to say it doesn't ...

1

Following David Z's answer, the proof for the last paragraph is: since $t$ is an affine parameter it satisfies: $$\frac{d^2x^a}{dt^2}-\Gamma^a_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}=0 \tag1$$ the parameter $t'$ must be related in some way to $t$, that is: $$t'=t'(t) \tag2$$ use the chain rule to get: ...

1

speed is $$\frac{\text{distance}}{\text{time interval}}$$ but at the event horizon of a black hole, time interval becomes $0$. Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, ...

1

$p=\frac{1}{3}\rho$ is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.$^1$ $p=-\rho$ follows from the fact that the energy momentum tensor of $\Lambda$-style dark energy is proportional to the metric. Thus, at a point and in the proper ...

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 ...

1

I know of two reasons for why we should consider gravity to be a force. The first is purely classical and Newtonian: tidal forces. Gravity is solely responsible for producing tidal forces, and they cannot be considered a fictitious force, whereas the usual acceleration due to gravity in some sense can always be thought of as fictitious. The way you know ...

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