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18

The accretion of matter onto a compact object cannot take place at an unlimited rate. There is a negative feedback caused by radiation pressure. If a source has a luminosity $L$, then there is a maximum luminosity - the Eddington luminosity - which is where the radiation pressure balances the inward gravitational forces. The size of the Eddington ...


9

A 12 billion Solar mass black hole sounds massive, but actually it's not all that big. The radius of the event horizon is given by: $$ r_s = \frac{GM}{c^2} $$ and for a 12 billion Solar mass black hole this works out to be about $1.8 \times 10^{13}$m. This seems big, but it's only about 0.002 light years. For comparison, the radius of the Milky way is ...


6

Strictly speaking the two well known uncharged black hole metrics (Schwarzschild and Kerr) are vacuum solutions. This means the stress-energy tensor is zero everywhere except at the singularity where it is undefined. This is what Alfred means when he says "A static (Schwarzschild) black hole 'contains' no matter". However this strikes me as a bit of a cheat ...


4

The "signed arc-length" is not used in relativity and I give reasons why. You are free to call and denote it in any way you like. $s$ and $\ell$ are interchangeably used to denote arc-length of space-like paths $g(\gamma',\gamma')>0$ in relativity and $\tau[\gamma]$ is used for "proper time" of $g(\gamma',\gamma')<0$ time-like paths but the notation ...


4

Let's consider a circular orbit in Schwarzschild coordinates, taken to be in the equatorial plane for simplicity. The test particle's position has components $x^\mu = (t, r, \pi/2, \phi)$, where $t$ and $\phi$ vary linearly with time/proper time and $r$ is constant. Then the $4$-velocity is $u^\mu = (\dot{t}, 0, 0, \dot{\phi})$, where dots denote derivatives ...


4

General relativity was written to describe all behaviour of spacetime. And expanding universe is one small subset of the whole general class of spactimes in the phase space of general relativity. So yes, relativity does apply to an expanding universe. At least general relativity does.


3

By the term black hole we normally mean one of four spacetime geometries, the Schwarzschild, Reissner–Nordström, Kerr or Kerr-Newman metrics. The universe is (we believe) approximately described by the Friedmann–Lemaître–Robertson–Walker metric, and it is not a black hole. The Big Bang is not the same as the singularity at the centre of a black hole. For ...


3

The fact that the energy-momentum tensor is called the source of curvature doesn't mean that there can't be any curvature where there is no energy-momentum. In fact, even if $T_{\mu\nu}=0$ across all spacetime, there are still nontrivial solutions of Einstein's equations, in the form of gravitational waves. You should remember that $T_{\mu\nu}$ is a ...


2

Quoting Sean Carrol's article linked by Симон Тыран, which makes the case for energy not being conserved: Having said all that, it would be irresponsible of me not to mention that plenty of experts in cosmology or GR would not put it in these terms. We all agree on the science; there are just divergent views on what words to attach to the science. In ...


2

I am going to answer the question 'When just considering GR without evaporation nor QM, is an empty (containing no matter) black hole possible?' and I will omit the 'anything' part, because is an ambiguous term. And the answer is yes, they are possible. As stated by another person here, Schwarzschild black holes and the rotating and charged versions are ...


2

In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = ...


2

1) Dirac operator Pseudo-Hermitian in flat and not pseudo-hermitian in curved space if you define it as in your first equation (without symmetrizing). because $(i\gamma^a e^\mu_a\partial_\mu)^\dagger = \gamma^0 (i\gamma^a e^\mu_a\partial_\mu)\gamma^0$, whereas $(\gamma^a \left[ \gamma^b,\gamma^c\right])^\dagger =\gamma^0 ( \left[ ...


1

Short answer: yes. But what do the original "straight lines" mean now? they cannot be defined in a nice way in the new metric, because the natural geometric entities now are geodesics, and a quadrilateral of four right angles does not exist (apart from possible special choice of corners). You must define your volume in a correct way (see below) Long answer: ...


1

There are a couple notions of time in GR. Typically when you write down a space you do so in some set of coordinates $(t,x,y,z)$. The $t$ there is the coordinate time. It marches forward forever, and doesn't care at all what the matter and energy content of the system may be. This seems very similar to the Aristotelian time. However, the fundamental ...


1

First, I highly suggest reading up on the concept of Locality. The issue is where you're measuring the speed from... Remember that it isn't so much that light can't escape due to the escape velocity, as it is that space itself is being dragged into the black hole (and anything residing in it), which happens to be falling in at the speed of light where you ...


1

Long story short: conservation of energy only holds locally where you can assume a static spacetime. On large scales the expansion of the universe gets relevant, so energy is said not to be conserved universally since the amount of dark energy per volume stays the same while the volume increases, see Sean Carroll's article, from which I quote: The famous ...


1

Whether energy is or isn't conserved in an expanding universe is a somewhat vexed issue. On the one hand you have an experienced physicist claiming that energy is conserved, and on the other hand you have an experienced physicist claiming that energy is not conserved. The problem is that accounting for energy in general relativity is a complicated business. ...


1

The problem is that the only gravity well to which we have easy access is the one that we are sitting in. While it's true that the combined gravity of the Earth and Sun will mean that CBM radiation reaching the Earth's surface is blue shifted, the blue shift is only a factor of about 1.00000002 and this is far less than the experimental errors in measuring ...


1

Singularities exist in theoretical 'perfect' solutions to General Relativity, but when you look at actual natural Kerr-like objects spinning in a noise filled background of GR waves and other incoming radiation and matter, its likely that no physical real singularities exist. Brandon Carter, referring to spinning black holes (all real black holes spin): ...


1

We have the frame $\{e_\mu\}_{\mu=0,\dotsc,3}$ in terms of which the velocity vector is $v=v^\mu e_\mu$. There are a few properties of the affine connection which I would like to summarize: $$\nabla_{fX}Y=f\nabla_XY$$ $$\nabla_X(fY)=f\nabla_XY+X(f)Y$$ $$\nabla_{e_\mu}e_\nu=\Gamma^\lambda{}_{\mu\nu}e_\lambda$$ Using this, let's get to work. We have ...


1

The story is the following. We start with the simplest Poincare-invariant action that does not depend on the parametrization $$ S=-m\int dl=-m\int\sqrt{-ds^2}$$ here $ds^2$ is the interval. We can rewrite it as $$S=-m\int\sqrt{dX^\mu dX^\nu \eta_{\mu\nu}}$$ here $\eta_{\mu\nu}$ is the Minkowski metric. Now if we suppose that $X^\mu$ depend only on $\tau$ we ...


1

Who's to say then, that we are not moving with the ether? Could we measure the difference? Yes that was the point of the experiment. They obviously didn't know which direction the ether was coming from so measured the difference through the year when we were at different positions around the sun and moving in a different direction. They also (IIRC) ...


1

You can imagine setting up two "Michelson & Morley" experiments, one stationary and one in a moving vehicle. Relativity says you would get the same results, so which experiment is stationary with respect to the ether?


1

The magnitude of the Ricci scalar places no restrictions on the Weyl tensor. After all, in the Schwarzschild metric the Ricci scalar is zero everywhere (except at the singularity where it's undefined). You may be able to link the Ricci scalar to the curvature in your specific case, but in general your assumption is not a safe one.


1

One of the biggest surprises that General Relativity has given us is that under certain circumstances the theory predicts its own limitations. There are two physical situations where we expect for General Relativity to break down. The first is the gravitational collapse of certain massive stars when their nuclear fuel is spent. The second one is the far past ...


1

No. You seem to be implying that the fact that the world-line of the satellite looks curvy, is what is meant by "curvature of space-time". No, that's wrong. The world-line of the satellite looks curvy in the picture, yes. But, it should be clear that you can have a curvy-looking world-line without gravity. Anything that orbits anything for any ...


1

"Or is it more correct to say that space is (almost) flat because that is what occurs without any mass at all? In this case, the answer might be: space is almost flat because there is so little mass in the universe." This is correct if you limit yourself to the vicinity of the solar system, spacetime is relatively flat because there is relatively small mass ...


1

If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines ...


1

We first note that the vanishing of the Ricci tensor does not imply the vanishing of the Riemann tensor. Thus the vacuum equations, $R_{\mu\nu}=0$, do not imply that spacetime is flat. The vacuum equations tell us that certain linear combination of components of the Riemann tensor vanishes. When solving differential equations, one usually has to worry ...


1

The formula for gravitational time dilation1 is $$\frac{t_0}{t_f}=\sqrt{1-\frac{2GM}{rc^2}}$$ For a sphere, $$M=V \rho = \frac{4}{3} \pi r^3 \rho$$ So $$\frac{t_0}{t_f}=\sqrt{1-\frac{8G \pi r^2 \rho}{3c^2}}$$ So the greater the density, the greater the time dilation. Has any relation been observed or postulated to exist between the energy-density (or the ...



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