# Tag Info

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The luminosity of the Galaxy is currently estimated to be around $5\times10^{36}$ W and thus an integrated "mass loss" in the form of radiation of of order $10^{-3} M_{\odot}$/yr. But how much radiation is present in the Galaxy? An order of magnitude estimate could be that the Galaxy (including the dark matter) is of order 100,000 light years in radius and ...

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According to Wald's GR [...] e.g. in a case where there are several massive bodies in relative motion--there exists no natural set of curves whose comparison with geodesics could be used to define gravitational force. Why not? Why not, indeed. What could be more "natural" than to make the required comparison ("with geodesics") for each participant ...

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Given $\bar{x}^i(x^j)$ transformation law of coordinates. The vector components $X^i$ and the covector $X_i$ are defined to transform like: $$\bar X^i = \frac{\partial\bar x_i}{\partial x_j} X^j,\quad\quad\quad \bar X_i = \frac{\partial x_i}{\partial\bar x_j} X_j$$ Both are simultaneously defined independent of the metric. And their definitions is ...

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This is discussed in section 4.3 in my 1984 edition. The quote supplied can't really be understood in isolation - you need to consider the whole section. Wald's point is that in general relativity there are no inertial observers because in general spacetime is nowhere flat. In Newtonian physics or special relativity acceleration can be measured relative to ...

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Since the metric $g_{\mu\nu}=g_{\nu\mu}$ is symmetric, we must demand that $$\tag{1} \delta g_{\mu\nu}~=~\delta g_{\nu\mu}~=~\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},$$ and therefore $$\tag{2} ... 1 The property you are referring to is called geodesic completeness. It is an important concept in the study of singularities in general relativity. There are somewhat trivial examples of geodesic incompleteness where you are just "missing" part of the spacetime, i.e. you could have Minkowski space with a point removed. In these cases you usually consider ... 1 It is not true. Sometimes the spacetime itself and therefore the geodesics can be extended. Consider for example the manifold described by (t,x,y,z) with x,y,z> 0 and Minkowski metric. This is nothing more than Minkowski spacetime truncated to a smaller region, but it is a perfectly valid manifold for all purposes. In the diagram we see schematically ... 1 I don't believe we need to invoke GR for this. So, to state your problem with variables, two objects A and B with identical rest masses m_{0} start at the origin at t=0 and then head in opposite directions along the X axis with equal speeds v, A moving in the positive direction and B in the negative direction. To an observer O stationary at ... 0 There is a relatively new theory (2012) called the firewall theory, that says that at the event horizon there is a huge "wall of fire" as such. This is because quantum entangled particles that cross the horizon (or one half of a pair of entangled particles) becomes tricky and starts breaking laws like the monogamy of entanglement. So a group of physicists ... 2 The singularity comes from the scale factor a(t):$$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$By solving the Friedmann equations for the scale factor we know that:$$a(t) = a_0 t^{\lambda}$$where \lambda is some positive number that depends on the matter-radiation ratio of the universe. At t=0 the scale factor becomes a(0)=0. So at ... 4 I think the following image sums up why your model, at least for our galaxy, is wrong rather nicely: These are the orbits for 6 stars in the inner region of the galaxy. The orbital period for S2, for instance, is 15 years for an orbit that is roughly twice the size of Sedna's orbit--which takes it 12 thousand years to complete its orbit. Using Kepler's ... 8 There are many problems with this line of reasoning. The most common galaxy types are elliptical galaxies and spiral galaxies, and there might be a parallel with star systems, where the most common types are systems with a single star, and binary systems with two stars in the middle. There is simply no justification for this. The dynamics of stellar ... -1 The stars in the galactic disk rotates with almost the same orbital velocity 200-230 km/s around the galactic center. Unlike a star system where the planets follow Kepler's third law. To explain the almost constant orbital speed of the stars in the galaxy, we have calculated dark matter, which is distributed in such a way, that it gives stars almost the same ... 2 This is not true. There are trivial counter-examples. For example, take \mathbb{R} with the trivial metric. Then, \gamma :(0,1)\rightarrow \mathbb{R} defined by \gamma (t):=t is a geodesic, but it is also clearly extendable. Indeed, it is extended by \tilde{\gamma}:\mathbb{R}\rightarrow \mathbb{R} defined by \tilde{\gamma}(t):=t. 1 Wald is a first rate relativist, and as such he is phrasing the concept of general covariance in terms of purely geometrical quantities, rather than resorting to the somewhat imprecise notion of coordinate transformations. In the discussion on pg. 57, he goes on to give an example of what it means to violate the principle of general covariance. In his ... 5 This is explained thoroughly in Thorne's book "The Science of Interstellar". There were two scientific papers based on the simulations: One in physics and one in computer rendering. The two circles are caused by gravitational lensing by a very rapidly spinning black hole. The radius of this black hole is 150 million kilometers with a mass of 100 million ... 13 First note that this is a fictional movie and the image is an artist's impression, not a detailed simulation. The public seems to think the movie is some sort of fictionalized documentary, which it never claimed to be. That said, the image is qualitatively conveying some of what happens near a black hole. The diagonal disk is the accretion disk -- this is ... 3 The horizontal circle is probably the accretion disc of the black hole. The vertical circle might depict the effect of gravitational lensing (although I am not sure this depiction is accurate). 1 First question Comparing a galaxy (or solar systems) to a merry-go-round is not a good comparison. Acceleration in a merry-go-round is proportional to the distance from the center of the merry-go-round, but acceleration in a central mass system is inversely proportional to the distance from the center. Single star solar systems are very close to central ... 1 This is really the same as a couple of the other answers, but I note that in the comments to those answers you are insistent that your experiment is a test of general relativity. However this is not the case. As long as spacetime is flat the experiment can be analysed using special relativity, and in this answer I shall explain why. It's commonly believed ... 2 Short answer: I'm afraid this is not a test of general relativity. I'll tell you why. I'll try to keep simple. You may use special relativity when your frame of reference is inertial. Let's say you see a Ultra-centrifuge spinning. You are experiencing no gravity at all (Earth's gravity is negligible for time dilation effects). You are experiencing no ... 0 A good start here would be to compute the time dilation effect expected from a centrifuge operating with a million Gs. With \frac{v^2}{r} = 10^6 g I would assume something like a radius of 10 cm, in which case v\approx 1000 m/s. We know gamma is \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, which means that gamma-1 is about 6\cdot 10^{-12}. Using ... 0 You recently commented I think the special relativity influence is absolutely negligible. It's exactly the opposite. It's the influence from general relativity that's absolutely negligible. Time dilation is a prediction of both general and special relativity. In general relativity, it's caused by an object being near a massive body. In special ... 0 It is a nice very idea for an experiment, but I don't think that radioactive decay would be an accurate enough 'clock' to use because generally with these types of measurements very small differences in time are detected - generally atomic clocks are used to measure the time in these experiments. With the radioactive decay process it is random decay and ... 0 There are a couple of reasons why your question isn't answerable. Firstly if you're sitting outside the horizon trying to measure the colour of the particle then you'll never see it reach the horizon, let alone cross it, so you would just measure a white particle as usual. If you were sitting alongside the particle as it fell then there would be no horizon ... 3 You say: I am NOT talking about Quantum Gravity but the change in strength of the strong and electroweak forces with energy is a quantum effect. It's due to a change in the coupling constant, which is known as running. There is no such effect in classical general relativity. The analogous effect in GR would be for Newton's constant G to be a ... 1 At the Event Horizon of Black Holes, the acceleration an object experiences is infinite (Here, I am NOT talking about freely falling objects; Freely Falling objects are in inertial motion in General Relativity). As for you, you can't feel that on the Event Horizon of Black Hole (Stephen Hawking was wrong in A Brief History of Time) because you won't survive ... 1 Say a planet has the same orbital period as Earth (365 days in a year, 24 hours in a day, etc.) but the planet has double the mass. For an even more direct comparison let's say that the latter planet has double the (average) density of planet Earth, so the surfaces of both have (as good as) equal circumference; and let's say that the durations of ... 0 For the first part you are correct but it is a bit of an illusion. If say a person fell into a black hole, then there are two perspectives: From the perspective of the person, they fall straight in (ignoring whether they survive to observe it or not). From an outside observer the person appears to slow down as they approach the event horizon of the black ... 1 All curves can be parameterised by an affine parameter (commonly written as \lambda in the GR books I have). However only for timelike curves does the parameter \lambda have a physical meaning i.e. it's the elapsed time \tau shown on a clock by the observer following the curve. So there's no mathematical difference between parameterising a timelike ... 1 Yes, the proper time along a timelike curve in relativity is very much analogous to the length of a curve. Just as the length of a curve is invariant under rotations, the proper time along a curve is invariant under Lorentz transformations. One difference with conventional length is that although a straight line is the shortest length between two points, a ... 2 General covariance basically means you can change your coordinate system arbitrarily and express the laws of physics in the new coordinates. Because of this freedom, the relationship between coordinate distances, angles, etc. and physical distances, angles, etc. is variable and is expressed by the metric. So the quoted statement is basically saying that ... 0 I will do this explicitly. In AdS_3, you can see we must have z=0 (x,y,z are symmetric so just choose z). For general cases, we must have u,v,x,y \neq 0, so we must have: \sin t \neq 0, \cos t \neq 0, \cosh \rho \neq 0, \sinh \rho \neq0, \sin \theta \neq 0, \cos \theta \neq 0, \cos \phi \neq 0. Thus we have \sin \phi =0, i.e, \phi =0 (n ... 1 There are two types of time dilation: dilation caused by being near a large body, and dilation caused by traveling very fast relative to another observer. Relativistic time dilation plays a bigger role for astronauts aboard a space station similar to the ISS. Even though velocity and gravity produce opposite time dilation, in this scenario, time dilation due ... 1 Have a look at my answer to Time Dilation Effects from simply being on a spinning planet orbiting a star in a rotating galaxy in an expanding universe.. Compared to an observer far from the Earth, time at the Earth's surface runs more slowly by a factor of 0.9999999993. Over a 70 year lifespan this makes a difference of about 1.5 seconds. So the man in ... 0 The event horizon is a surface from which light cannot move outwards from, all light here is either 'stuck' from the point of view of an observer at infinity, or moving inwards. Light emmited just outside this region can move outward very slowly, and moves outward faster the further it is away from the horizon. So yes it is related to redshift, the amount ... 3 The answer is it depends on which observer we are talking about - an observer "with" the collapsing mass sees it and them crushed to a singularity; an external observer "sees" (though see below) the mass frozen just at the event horizon. In GR and a standard black hole, there is only one future for a mass that finds itself at or inside the event horizon, ... 0 In the Friedman equation \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} what exactly is a? a=a(t) is called the "scale factor." It's part of the FLRW metric which describes the expanding universe. The scale factor determines spacial distances between points in space, and how those distances change ... 0 I'm not a GR expert (and there are some on this site), but I'll try to answer this. How does an object on the surface of earth stay fixed? Since there is no concept of "attraction" and only space time why cannot an object keep moving around on the surface of earth. The answer here is that there generally aren't any forces moving the object to the side. ... 4 is it possible to consider also the other fundamental forces [...] to be fictitious forces like gravity in the framework of general relativity? No, because the equivalence principle only holds for gravity. If we want a final unification of all fundamental forces, hasn't this feature of gravity to become a feature of the other forces as well? The ... 8 The classical theory of electrodynamics can indeed be written as a geometrical theory in a similar way to general relativity. As it happens there is a question and answer addressing just this, but it's in the Maths SE: Electrodynamics in general spacetime. Classical electrodynamics is an example of a class of theories called classical Yang-Mills gauge ... 8 On the quantum level, force is not acceleration. The concept of "fictitious force" makes no sense on a QFT level, because forces are interactions between quantum states, not the classical forces you might imagine. Quantum forces are not vector fields in space. The notion of "fictitious force" would mean that, e.g., the strong force is something influencing ... 1 If acceleration is equivalent to gravitation, it follows that the predictions of Special Relativity must also be valid for very strong gravitational fields. The curvature of spacetime by matter therefore not only stretches or shrinks distances, depending on their direction with respect to the gravitational field, but also appears to slow down the flow of ... 0 I lost one term, which is the one containing \nabla^t(\partial_t)^r=g^{tt}\Gamma^r_{tt}=-M/r^2, so this term is$$ -\frac{1}{8\pi}\int r^2\sin\theta \nabla^t(\partial_t)^r d\theta d\phi=\frac{M}{8\pi}\int \sin\theta d\theta d\phi=M/2  Add this term to previous one and correct!

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One can treat the spacetime coordinates as complex and therefore turn spacetime integrals into contour integrals in the complex plane. The value of the integral of a function that is holomorphic except at certain (singular) points is now determined by its singularity structure: the Cauchy's residue theorem tells us that the integral is given solely in terms ...

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You can find an explanation of scalar fields and associated quantum effects in the Schwarzschild background in chapter four of these lecture notes. The article also contains references which might be of use to you.

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I think the answer lies in the fact that the variational principle says is if there are two fixed points in space-time then there exists an open subset of space-time containing these two points such that amongst all curves contained in this open subset, a geodesic will be the curve of longest length between these two points. So, the variational principle ...

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c will always remain a constant with respect to space itself. We can't define a speed relative to space itself, only relative to an observer. In GR, we can only define speed relative to a nearby observer. To a nearby observer, the speed of light is always $c$, because GR is locally the same as SR. the only reason I understand for a change in the ...

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could something else be relative (other then time)? The speed of light is relative. In 1905 it was easy for Einstein to assume that the speed of light (relative to the observer) is independent of the speed of the light source - the assumption was false but sounded plausible since it was a tenet of the universally accepted ether theory. However this ...

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time slows in the presence of strong gravitational fields It's not the gravitational field that determines time dilation, it's the gravitational potential. The Newtonian approximation really isn't correct here, but let's use it anyway for insight: The potential falls off like $1/r$ with distance $r$. The field falls off like $1/r^2$. Tidal effects go like ...

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