New answers tagged

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The collapsing of the black hole is impossible. It is like trying to burn ashes. Ashes are already burnt. Similarly, the black hole is already collapsed. The black hole can evaporate and get bigger, but not collapse.


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I'm not quite sure how he gets a value of $36.36$ for that constant, because it depends a bit on exactly how he does his approximations and what values he assumes for different constants. But basically it's this: For a spherically symmetric mass distribution like the one considered here, the gravitational acceleration is simply, from Newton's law for the ...


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Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ...


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I am answering this because, even though you do not say in your profile, this is a proposal that would have some logic from somebody very young who is just learning. The idea I had was that whilst electrons usually orbit within the Bohr radius of atoms there is a chance of them appearing significantly further away than that, This is a quantum ...


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No, it cannot be so, for many reasons. One thing is, the amplitude of electron wave function in atom drops exponentially, which would make this effects vanishingly small, and would make any gravitation forces drop exponentially with distance.


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If you are using words like instantly and think it even means something in general relativity, then you need to learn more about general relativity. In special relativity you already learn that simultaneity depends on frame. Black holes form from collapsing matter. And the event horizon is a one way surface. Not because it is magic, but because it ...


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The "buoyant force" does not arise from a principle, but is what remains of the gravity force when you subtract from it an average hydrostatic pressure. Let's assume that you have some reference density $\rho_0$. Then if $\rho=\rho_0$ everywhere and the fluid is at rest, $p = \rho_0 g z$. Now let's call this baseline pressure $p_0(z)$, without loss of ...


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The "theory" describing the black hole interior (in the classical approximation) is the same theory that implies the existence of the black holes, namely the general theory of relativity. As the OP correctly said, the singularity at the event horizon is a coordinate singularity – one that is an artifact of a bad choice of coordinates. When a coordinate ...


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Yes, the acceleration of a freely falling particle anywhere else would be slightly altered. If the person jumping would be right next to the freely falling object, and one could neglect the maximum speed of propagation $c$, the falling object would be slightly pulled towards the jumping person. And thus change its falling acceleration. If this happens at a ...


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We won't see inside the horizon, but we will see up to close to the horizon. We will be able to see with much higher sensitivity when we deploy in a few years the eLISA gravitational observatory in space, with 3 satellites separated by 1 million Kms in a triad configuration. See http://www.livingreviews.org/lrr-2013-7 The article also describes the many ...


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Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ...


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Yes. You're exactly right, deviations from no-hair do occur for example after BH mergers --- and hints of the "quasi-normal" mode ("ringdown") were observed in the LIGO detection. The no-hair theorem is constructed for a static, stationary BH (i.e. fully settled). In general, deviations from no-hair (magnetic fields, asymmetry, etc) will be radiated away ...


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Any object with mass that accelerates (is it linear or angular acceleration) produces gravitational waves, though in most occasions those will be much too small to be detected. As @CuriousOne pointed out, same happens with electromagnetic waves and accelerating charges. The gravitational waves that can be detected usually come from very massive objects (such ...


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The asymmetry you observe will be the asymmetry of the configuration with the infalling matter - in this case, with the additional infalling black hole. It becomes more symmetric with time - a very short time, but not instantly.


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Not sure about the etiquette of this but I think I can now answer my own question. Please post if there is a better answer. The problem is that I misunderstood the meaning of the statement "the geodesic is the path the optimises the proper time". What this means is that given two endpoints, say $x_0^\alpha$ and $x_1^\alpha$ a geodesic is a path ...


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No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the ...


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Actually, only a gravitational wave and not a uniform gravitational field is composed of gravitons. I think that just like en electromagnetic wave, a gravitational wave is a sum of sinusodial waves with gravitons of different energy for different wavelengths in that sum. An isolated blackhole doesn't emit gravitons but a pair of orbiting blackholes does. I ...


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Please read the Laws of atomic gravitational fluctuation to understand how gravity works in the atom. http://www.windsorrealestateinfo.com/Laws-of-Atomic-Gravitational-Fluctuation-Tim-G.-Meloche


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This is really a difficult problem, but possibly not for the reason you imagine. The following naif criterion seems at first highly appropriate: take any pair of stars, subtract the center-of-mass motion, compute the kinetic ($T$) and gravitational ($W$) energies, and check whether $T + W < 0$. If so, the pair is bound, otherwise it is not. How can ...


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If you see the expression of gravitational force: it has a coefficient, two mass terms and one distance term. Now, G (gravitational constant) ~ 6.6x10-11 SI Units m1 & m2 (suppose mass of an electron) ~ 9.1x10-31 kg r (distance between two electrons) ~ 10-12 m Gives us force in range of 10-47 - 10-46 N which is very very less. And there are other ...


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There is no fundamental difference between the two terms, except that in certain situations one or the other have come to be used more often. Gravity is more often used to describe the concept ("Newtonian Gravity"), the force (the "Force of Gravity"). Gravitation is more often used for phenomena resulting from gravity ("Gravitational Waves", "Gravitating ...


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Gravity is the physical phenomena by which bodies attract themselves. It is the effect we observe. Gravitation is a model, a theory to explain the observed phenomena. The Newtonian Gravitation explains this phenomena in terms of attractive forces generated by massive bodies. It dos not depend whether the bodies are terrestrial or celestial. General ...


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A very peculiar fact is that in a compact space THERE IS a preferred inertial system. Indeed even if locally there is no way to single out a preferred inertial system, globally you can do it. Is the topology that tells you that an observer doing a loop around a torus is topologically different from an observer moving around simply connected loops. So for ...


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Some misunderstandings. Black hole can be round (we call them spherically symmetric, or spherical), or sort of ellipsoidal, or axially symmetric, if they have angular momentum (i.e., if they rotate). That is true for stationary black holes, i.e., after they achieve a stationary state. In getting to that state they can be very dynamic and have deformed ...


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Black holes, as seen in the picture are actually spheres formed by event horizon. The matter is all concentrated on the singularities (except for the matter that is falling into either singularity at a given point in time). So, individual black hole would be spherically round. During merger of two black holes, the event horizon can become non-round ...


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Just an idea I'm playing with, There is no dark energy/dark matter. The universe is not expanding into “infinite nothing”, but instead reducing into the amount of space it occupied at inception. Matter is reducing in scale because there is nothing in the universe requiring it to maintain a static size at the subatomic level. The gravitational effect of ...


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The measurement you made is time. When you setup you test, you are using another measurement height. With whatever instrument, measurement has error. This will affect test result. In this case, it is $a_2$ you calculated using the test data. I believe each time you repeat a test, the value $a_2$ varies a bit. Different people doing test will give different ...


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The surface has aged more : events move faster/happen quicker on the surface than in the core. So there would be a blue-shift in the light reaching the core. Light coming out of the core would be red-shifted. But in the case of the Earth (and even the Sun) the effect is incredibly small. https://en.wikipedia.org/wiki/Blueshift#Gravitational_blueshift


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Note - gravity is not a pull in GR. Anyway, I think the answer is no. If that was true, wouldn't all masses turn into black holes. If gravity pulled itself (because it is energy), then it would create more energy by pulling on itself, thus more gravity and so on.. eventually turning into a black hole. Gravity pulls on itself only in the sense that the ...


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Part 1: no, the amplitude decays as 1/r. The power or energy of the wave decays like the square of that, like 1/r^2 (one over r squared). This is the same as in electromagnetic waves, the gravitational wave amplitude (in General Relativity) components obey the same equations as that for electromagnetic waves for the electric or magnetic field. The reason is ...


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Gravity, the physical phenomena, does affect itself. On the other hand the physical theories expected to explain gravity may of may not admit this self interaction. The General Theory of Relativity does admit while Newtonian Gravitation does not. Newtonian Gravitation satisfies the Superposition Principle: If you have two masses $M_1$ and $M_2$ attracting a ...


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In relativity, gravity is the word we use to describe curved space. Don't think of curved space as simply 'no longer flat', also consider that it includes a gravitational gradient that will cause matter to move across that gradient. As an example, Earth bends the space around where you are right now, with a vertical gravitational gradient, causing you and ...


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Hawking, I believe, is referring to a more metaphorical 'hovering'. As light, or anything, approaches the event horizon, it becomes more and more redshifted---it's motion appearing to go slower and slower and slower, approaching zero apparent velocity to an outside observer (approximately) infinitely far away. Anything falling into a BH, thus appears to ...


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The short answer is special relativity. classically According to classical mechanics, there is nothing to prevent an object from accelerating faster than the speed of light. By conservation of energy you can calculate the speed of a dropped object by looking at the change in gravitational potential. As you note, the equation $PE=mgh$ needs to be modified ...


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The problem is that radioactivity is a poor way to measure time. Suppose you are trying to measure the current level of radioactivity from some source. If you measure $N$ counts then the standard error will be of order $\sqrt{N}$. The time dilation effects we are interested in are of order one part in $10^9$, so the number of counts we need to measure is of ...


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The surface area is $A=4\cdot \pi \cdot r^2$, so the distance in terms of the area is $r = \sqrt{A/4/\pi }$ Because $g=G\cdot m/r^2$ it follows that $g=4 \cdot \pi \cdot G \cdot m/A$ which means that the mass must increase by the same factor as the area in you want the same gravitational force on the surface. For the second part of your question see ...


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When you sit in your F1 car and turn right with 1g of lateral acceleration, you feel a centrifugal force (acceleration) of 1g pulling you in the opposite direction (left, in the car's frame). That is, there is a fictitious force caused by your curving reference frame. This is a Galilean point of view--an accelerated reference frame causes fictitious forces. ...


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You sound disappointed that the earth hasn't yet collided with the sun. Applying random formulas to a physical situation is never a good idea. I suspect that the formula you are having issues with is for two bodies in space which are not in orbit around one another, as unfortunately the earth is with the sun. When one body orbits another, the motion of ...


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The answer you refer to is to a question where both bodies are initially motionless. That's not the case for the Earth/Sun system. The Earth orbits the Sun. The gravitational force the Sun exerts on the Earth provides the centripetal force needed to keep the Earth in a stable orbit. In a sense the Earth is constantly free-falling towards the Sun but ...


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As stated in the answer you link, that formula is for 2 bodies starting from rest. The Earth and Sun are not at rest relative to each other, they are in orbit at a relative tangential speed of nearly $30\,{\rm km}\,{\rm s}^{-1}$.


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First of all, the Universe isn't expanding according to "current theories". It is an observational fact. Second, there is no center of the Universe. Space was created, and started expanding. This expansion pulls everything away from each other. Galaxies lie approximately still in space, but space is expanding. This means that no matter where you are located ...


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First of all, there is NO centre in the universe. I know it's not a good analogy, but think of the universe as the surface of a balloon. Forget the interior, we're only looking at 2 dimensions, whereas the real universe has 3 of them. Put some ink dots on the balloon, which represent galaxies (note: NOT planets). Now inflate that balloon. You'll see that ...


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Physics does not answer "why" questions at the level of basic laws and postulates, i.e. "why" this postulate. Newtons gravitational law : Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses but also inversely proportional to ...


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You say: My research into gravity indicates that warped spacetime, with time as the major influence, is gravity then ask: why would a mass seek a location where time runs slower? But the problem is that few of us will recognise the qualifier with time as the major influence, and few of us would describe the motion in a curved spacetime as mass ...


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Physics is full of theories that to the outsider seem to pose a chicken and egg problem, but there is no issue: both "mass tells space how to curve" and "space tells mass how to more" are axioms of a conceptual structure whose validation is that it's predictions agree with reality. There is no reason to expect one to come "before" the other because they are ...


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In general relativity 'gravity' is caused by following geodesics through curved space, so it is a pseudo force similar to the Coriolis or centrifugal forces. In quantum gravity, gravitons mediate the interaction between masses the same way gluon's mediate the strong force, etc. So in quantum theories, gravity is similar to other fundamental forces.


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Let's look at F = ma. This is not true for each force individually, for sure - each particle experiences many forces but only one acceleration. It's really Σ(all forces)F = ma. Even if we say just the force of gravity, we're looking at Σ(other objects)Fg + Σ(other forces)F = ma. So let's think about applying MOND. We have a particle that's ...


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As far as we know the classical (i.e. non-quantum) laws of gravity apply at all length scales. There are theoretical reasons to suppose that the classical description fails at scales approaching a Planck length, but this is far, far smaller than the size of a neutron. So inside a neutron we would expect the classical laws of gravity to apply, and in ...


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If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$ \mathbf{F}=m\cdot \mathbf{a} \tag{01} $$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...



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