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This is a curious question. Many black holes are detected and identified due to light emitted from material falling into them. When black holes accrete matter, conservation of angular momentum would usually lead to the formation of an accretion disk. The release of gravitational potential energy as material falls means that this disk can be hot, and it is ...

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Does the gravitational force of a massive, spherical body draw only towards its center of gravity, and to no other area? Yes. This is exactly what Newton's shell theorem says, and also Gauss' law of gravitation. To help illustrate: if you stand beside Mount Everest, there should be a pull towards the mountain even if its too slight to notice. Not ...

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The shell theorem ensures that the gravitational force of any spherically symmetric mass distribution acts towards the centre of mass. However, if spherical symmetry is broken, as in the example you quote, then indeed there can be other components to the gravitational force. For instance you can use Wolfram Alpha (for example) to tell you at what angle to ...

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The reason has to do with time dilation, and specifically, with the resulting red shift. A black hole forms from a collapsing star, which is of course made of brightly glowing matter. The event horizon forms in the centre and moves outwards while the star-matter falls towards it. Because of gravitational time dilation, the infalling matter never crosses the ...

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Weight is a subtle concept, because we are so used to it we don't even notice it anymore. You'll find surprising how badly some students grasp the concept of weight, despite the fact they are firmly sitting on a chair in the very same moment. Anyway, weight is the force an object experiences when inside a gravitational field. That means that me, you, ...

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While both are forces, weight is generally specific to any sum of forces you feel reciprocated as a normal force (or tension). I can feel heavy in a centrifuge because of the centrifugal force. When I find myself sitting in my chair (safely away from centrifuges) the astronauts on the International Space Station and I are both are subject to the force of ...

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After a bit of research, the key term here is "the secular dynamics of Mercury". With that, you can easily find course notes that cover the whole calculation: https://farside.ph.utexas.edu/teaching/celestial/Celestial/node118.html It's frowned upon to give a link-only answer, but this is a big reference so I think it's appropriate.

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I will admit that my understanding of General Relativity at the moment is limited, but this is my understanding: When dealing with curved coordinate systems, like space-time curved by a massive object, you can calculate what is known as a "geodesic" which is a straight line in curved coordinates. If you are dealing with a spherical space, all geodesics will ...

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As you described, if you have a metric, then you also have the manifold itself and all the points in it, and can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming no cosmological constant). But the reverse direction is very different. For instance, if you started ...

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In short: No. In addition to gravity pulling things together, the electroweak force pushes things apart. The more you push things together, the more the atoms the things consist of will resist the pushing together, meaning that unless the gravitational pressure is sufficient, gravitating objects are stable (asterisk). Essentially, your initial premise is ...

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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 ...

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Time is affected noticeably by gravity only in strong gravitational fields, i.e. in the vicinity of compact objects, so time doesn't run differently in voids from average regions. Dark matter (DM) and energy (DE), on the other hand, can to some extend be unveiled by studying voids. The morphology of the voids is affected by the nature of the DE, so ...

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OK,maybe I get where my mistake is. It's very important that $T^{(1)}_{ab}=G^{(1)}_{ab}=0$ but $T_{ab}\not=0$. From http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec14.pdf ,we can learn $$T^{03}=-\frac{1}{16\pi}[(\frac{\partial h_{11}}{\partial t})^2+(\frac{\partial h_{12}}{\partial t})^2]$$ (c=1 and G=1.)Because ... 0 Assuming the outer ring is stationary (an arbitrary assumption just to establish a range of reference), the inner magnet will either move up or down until the bottom of the inner magnet (or top if it moved down, but we'll assumed it moved up) is aligned with the top of the outer magnet. In an ideal experiment, the inner magnet would need to experience ... 0 In a conservative field (no dissipation of energy) whatever way follows the small sphere, i.e. rolling along another sphere, or other, the total energy (kinetic + potential + rotational) is conserved. 1 No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other ... 0 There are different measures of the strength of the gravitational field. One is the curvature, which is a nice local geometric invariant. However, the curvature only becomes "large" in very extreme situations, like near the singularity of a black hole. The curvature is actually not large near the event horizon (ie, you can't do local experiments to tell you ... 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. 0 The speed of light is invariant, so it is not possible to accelerate it above c. From Newton's law we know that the escape velocity v from any given mass is v_{esc}=\sqrt{\frac{2 G M}{r}} so if the mass M is high enough and the distance r to the mass small enough v>c. Since nothing can have more than c even light cannot escape from there. ... 5 Scientists asked the question "How does a body of arbitrary mass affect spacetime around it?" To answer this question, they took Einstein's General Relativity and applied it to the description of a spherically symmetric spacetime (meaning you can rotate any way you like and it looks the same) centered on a body of arbitrary mass, M. I'll spare you the ... 0 When the escape velocity of a body of mass is faster than the speed of light, you have a black hole. This is because gravity affects light the same way it affects matter. 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 By definition, work isW[\gamma] = \int_\gamma\mathbf F\cdot\mathbf v\text dt$$i.e. force times displacement. If an object is not moving, even if subjected to a force, then there is no work done by said force. Observe that, since the object is not moving, the sum of all the forces applied to a body must be zero. For example, a body on rest on a table is ... 1 Work= force x displacement force is weight and displacement is 0 so there is no work (in physics) 1 First, I highly suggest reading up on the concept of Locality. The issue is where you're measuring the speed from... Remember that it isn't so much that light can't escape due to the escape velocity, as it is that space itself is being dragged into the black hole (and anything residing in it), which happens to be falling in at the speed of light where you ... -3 Light will no longer be light as it gets absorbed in black hole . So there is nothing left to talk about its speed. 0 Fn is the normal reaction force and Fay is the y-component of the force Fa, since the block is not moving in the y-direction (+ve or -ve) and since we know that mg is acting downwards, there has to be some force(s) balancing out mg, and apart from the normal reaction, Fay is also acting upwards, so yea it's Fn + Fay = mg because both Fn and Fay are acting in ... 1 That equation expresses the fact that the forces on the block in the vertical direction add up to zero. F_n and F_{ay} are both forces pulling the block upwards, while the force of gravity on the block is mg downwards. We don't want the block to accelerate up or down, and so the net forces in the vertical direction must cancel out. 0 There is huge tension between the viewpoints. The current 'Firewall' debate - that QM predicts a huge amount of physics happening at the horizon where GR says that nothing at all should be happening is an example of the problems being faced in this area. 0 The gravitational force acting on the particle as function of radius r is [Gauss's law]$$F(r) = -\frac{GmM(r)}{r^2}$$where M(r) = \int_0^r4\pi \rho(r')r'^2dr' is the mass contained within a radius r. Note that for r>R we have \rho(r) = 0 and M(r) = M(R) \equiv M the total mass of the whole sphere. The potential energy is the work needed ... 1 I think it helps to look at why things get stuck at all. If you have two things that could together get to a lower energy level than they are at now, and they can give that excess energy up to something (like sending photons into deep space) that isn't going to give it back (at least for a long time), then they can get stuck (at least for a while). So why ... -2 I can think of a light beam as a pulsating stream of water from a hose traveling at the speed of light. If there's a hole in the side of my space ship and the hose of streaming water is pointed directly, perpendicular to my ship's direction of travel, at the hole, then only a portion of the pulsated water will enter the hole. Now concerning the portion of ... 3 The paradox you describe is even worse than the effect of gravity alone: Electrostatics works the same, attracting most matter at everyday distances very strongly by comparison. That is obvious for opposite charges, but even neutral matter attracts as electric interaction induces dipoles. Even where you have equal charges, which do repel each other, you do ... -1 Since relativity explains more what gravity does, but not necessarily what gravity is, your question is a valid one. Explaining gravity as some kind of residual EM force (Van der Waals, Casimir, and many other variations) actually has a long history, but they have all failed. They essentially give solutions with too large of a negative power term "n" and ... 6 Ignoring the quantum effects that make such a situation both improbable and overtaken by stronger but shorter acting forces - if we take Newton's gravitational equation with the inverse square law:$$ F = \frac{G \cdot M_1 \cdot M_2}{R^2}  and if, in theory, you get 2 objects to occupy the same exact space - (ignoring quantum and other difficulties there ...

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There is no established quantum theory of gravity. Hence, at the microscopic level of particle, we don't know what is going on gravitationally between particle, but it isn't going to be the "inverse square law" we know, just like electromagnetism between two charged particles is, on quantum scales, not just an "inverse square law", but a rich variety of ...

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First of all, infinitesimal small distances are not allowed in the quantum world simply due to the Heisenberg uncertainty principle - the Newtonian force law doesn't hold at these distances. Apart from that, new forces arise that repel at short distances, a simple $H_2$ molecule is an example for that. When you manage to collide particles instead, the whole ...

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Is there a way to quantify how the bending of the rod and the bending of the space at the same position are related? There is a way to quantify how the deviation of spacetime from being "flat" (in the region containing the given rod) and the deviation of the (eventual, static) "shape" of the rod constituents with respect to each other, which is ...

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Without GR, there is no reason for Mercury to not be tidally locked with the Sun, always showing the same face to the Sun (much as the Moon always shows the same face to us). There is also no reason for clocks at higher elevations to desynch with clocks at sea level. If you take these as strong evidence for general relativity, then all that the authors' ...

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I'll give this a try. Feel free to butt-in! Note that the rod is embedded within space. The bending of the rod is due to space itself being bent, hence the rod experiences no mechanical stress. As such, the degree of the rod bending has nothing to do with its elastic properties. Bending of space does not induce mechanical stress. I'd venture to say that, ...

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Forces cause acceleration, or changes in speed. If your speed is constant, the total force on you must be 0. Gravity is one force acting on you. The force of gravity is equal to your weight. It pulls you down. To make the total force add up to 0, you need another force equal to your weight in the upward direction. Often this upward force is provided by ...

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Take a simple example of an everyday situation where an object is kept from gaining or losing amplitude: Put the object on a shelf and let it sit there. Now think about this situation. What is the force applied by the shelf to the object to keep it from gaining or losing amplitude?

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Q1: I do not know why Sachs and Wu are treating the electromagnetic contribution separately from the other matters sources, but this is indeed non-standard. I think the physics convention is more natural, since the Einstein field equations are $G=T$, and the words associated with this equation is that the presence of matter ($T$) warps spacetime ($G$). Q2: ...

2

The answer is yes but only up to a point. If I jump in sunlight and under a sunshade, all other factors being equal, will I jump higher in the shade? Yes, you will jump higher. This is due to the radiation pressure that pushes you down when you are exposed to downwards-pointing sunlight. However, this effect is not at all likely to be measurable. ...

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There is no tension between the two viewpoints. In fact, the situation isn't much different from electromagnetism. The analogy is photon $\leftrightarrow$ graviton and electromagnetic field $\leftrightarrow$ metric tensor. Both of these theories, EM or GR, are field theories, and exhibit the phenomena of radiation. Gravity waves have not been detected yet ...

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Small point to add to this, but all matter has a Schwarzschild radius - the earth does (it's about 1/3rd of an inch in diameter), black holes do - it's their event horizon, and Neutron Stars do - but it's smaller than the star. Any non-black hole object has a Schwarzschild radius that is smaller than it is. Neutron Stars are dense enough that their ...

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There is not dark matter. There is matter with (non/ negative) divergence. Dark matter (neutron Cooper pairs) is isolated system (There are not fluctuation in dark matter area) and impact neutrino are aprox. 2-3 kelvin degree. There is a scalar field because "dark matter" particle attenuated neutrinino flux.

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The laser goes through the bagel hole each time until it hits the ground (assuming the mirrors are set up nicely orthogonal to your uniform gravitational field). To see why, the equivalence principle is all we need. You can imagine thrusters powering an elevator without gravitational fields, set up your mirrors and do the experiment, and that's what you ...

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… photons undergo twice the deflection from gravitational fields as do physical objects. I don't think this is a correct assessment of the situation. Rather, there is a Newtonian way to predict the deflection of light due to gravity: assume that the light is made of corpuscles with effective mass $m=E/c^2$ which enter the gravitational well with speed ...

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