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Milgrom compared MOND predictions to Galaxy-Galaxy Lensing (GGL) data and found reasonable M/L for the lensing galaxies gave good agreement. "Testing the MOND Paradigm Of Modified Dynamics with Galaxy-Galaxy Gravitational Lensing" (July 21, 2013) http://arxiv.org/pdf/1305.3516v2.pdf Notice that this does not explain the Bullet Cluster where the mass ...


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No. The Einstein field equations are the equation of motion for the metric (i.e. gravity) in the Einstein-Hilbert action. If you add other dynamical fields to the action, you not only change the stress-energy tensor appearing in the EFE, but you also have to vary the action with respect to the new fields to obtain e.o.m. for them.


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Correct, the standard Schwarzschild metric is asymptotically flat and indeed the time co-ordinate $t$ is the local time of an observer infinitely removed from the black hole and sitting in this flat space and so there is no pair of points outside the black hole's event horizon which ultimately cannot causally reach or signal each other. The de ...


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The relation between the metric and gravitational potential (and between Christoffel symbols and acceleration) is evident in the Newtonian limit of General Relativity. The basic assumptions of this approximation are: Weak gravitational field: the metric $g_{\mu\nu}$ differs from the Minkowski metric $\eta_{\mu\nu}$ only a small amount $$ g_{\mu\nu} = ...


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Although this is an old question, and has been answered satisfyingly, let me add a couple of things that are only tangentially related to the question. This whole thing about the metric being a gravitational "potential" and the Christoffel symbol being the "field strength" is just an analogy. The source of the analogy is exactly what you stated, the ...


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I'll assume that you're aware that all of this is described by differential equations. For instance, here's the equations for the inverted pendulum. What you seem to be interested in are the cases where the angle $\theta$ remains a constant. In the case of the needle on a flat surface, no magnetism, there are two steady states. A) The needle stays ...


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This, to me, depends on how uniform a magnetic field your magnet creates in reality. I am no expert, but if it's not, I can imagine you are doubling your chances of the needle falling, both from the unevenness of the surface and the possible unevenness of the magnetic field. Sorry for the short answer, but I hope you get a better one.


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Then, shouldn't this sphere be detectable via the way it lenses light coming from galaxies that lie behind it (relative to earth). Similarly, a mass of dark matter within our own galaxy, should be detectable via the way it lenses light from stars that lie behind it ( again relative to earth). Has anything like this ever been observed? Yes, it should do ...


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If you and then moon were together the earth would pull you both. It would pull the moon harder but since the moon is more massive this would (in the absence of other forces) produce the same acceleration as you undergo. If you factor in the moon pulling the earth up to it then since you and the moon are together the earth is pulled up to both of you. If ...


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The gravitational force is indeed bigger if the second object has a bigger mass, but the acceleration towards the main body (for example, the moon) remains the same for test bodies of all masses on its surface (assuming the bodies are small, not too far away from the surface, no air resistance etc.). You may recall this from the simple version of Newton's ...


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Neutrinos are produced in the atmosphere all the time as a consequence of cosmic ray interactions, and they mostly then fly right through the planet. So there is no problem with a source. In fact the "flying right through" bit is one of the problems: even the full diameter of a planet simply doesn't intercept a large enough fraction of the beam to make ...


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There will be spoilers if you keep reading Firstly, he is shown surviving inside black holes. From where did he got oxygen? Perhaps from oxygen bottles. But, in an intense gravitational pull, how he survives? He would have got torn apart! am I right? The popular press says the word black hole and it is a bit vague what they mean because there are some ...


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It depends on the scale. Some scales (called balance scales) balance your weight against the weight of a known mass. Your weight is a force, calculated as mass x acceleration. The weight of the known mass is calculated the same way. Generally, there's some kind of adjustable leverage for the known mass. The scale is calculated by measured weight of known ...


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Rod Vance's answer explains why your proposed explanation seems unlikely to many. I'd like to explain what the observations you allude to really show. The correlation between black hole mass and stellar velocities is known as the M-sigma relation. First, note that it does not involve the outer stars in the galaxy. While it is true that the outer stars (and ...


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The mass which fills 'empty' space is beginning to be referred to as the 'dark mass' in order to distinguish it from the baggage associated with dark matter. 'Dark Energy/Dark Mass: The Slient Truth' https://tienzengong.wordpress.com/2015/04/22/dark-energydark-mass-the-silent-truth/ "That is, all that we are certain about [is] the dark mass, not dark ...


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Yes. Gravity bends spacetime and anything occupying it. Edit: https://en.wikipedia.org/wiki/Gravitational_lens


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Limitations of $g \propto \frac{1}{r^2}$: The relationship $g(r) = G \frac{M}{r^2} \rightarrow g \propto \frac{1}{r^2}$ (where $g$ is the acceleration due to gravity, $G$ is the universal gravitational constant, and $r$ is the distance between the massive object and the accelerating object) is just fine in Newtonian physics—it doesn't need to be fixed. ...


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If you have a spherical body of radius $R$ with mass $M$, the gravitational field at any point at a radial distance $r$ is given by: $$\phi=\frac{GM(r)}{r^2}$$ where $M(r)$ is the mass enclosed inside a spherical shell of radius $r$. This is the only mass that matters in this case. (because of the Shell Theorem: https://en.wikipedia.org/wiki/Shell_theorem) ...


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There is no requirement for a central black hole in a dynamical sense. Many galaxies are not known to have one, or if they do, its mass is relatively small. The gravitational influence of the SMBH can be quite negligible at distances that are only a tiny fraction of the size of a Galaxy. What I mean by this is that say the BH at the centre of the Milky Way ...


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You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day. A tide is effectively ...


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Like most proposals, it is possible of course; in physics we must ultimately test proposals experimentally. In the meantime (i.e. in this case whilst waiting for experimental observation and study of dark matter here on Earth), one must resort to assessing plausibility in the light of what we already know. There are two ways your proposal, if true, could ...


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The stars in the galaxy don't really orbit the black hole in the center of the galaxy. They all orbit a common center of gravity. Obviously, a lot of the mass is in the black hole, and the center of gravity could very likely lie inside the black hole's event horizon, but it's not required. Look here for a cool animation of Pluto and Charon orbiting a center ...


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In Electrostatics you can write $F=qQ/(4\pi\epsilon_0 r^2)$ in 3D, or as $\vec\nabla \cdot \vec E = \rho/\epsilon_0$ and $\vec F=q\vec E.$ And the later generalize to 2d as a $1/r$ force. In Newtonian Gravity you can write $F=mMG/r^2$ in 3D, or as $\vec\nabla \cdot \vec C = \rho 4\pi $ and $\vec F=m\vec C.$ And the later generalize to 2d as a $1/r$ force. ...


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You are right: gravity acts on individual bits of mass, and is stronger towards the source of the gravitational field. The center of mass and center of gravity correspond if you assume constant gravitational field (and rigid bodies I would say).


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Imagine that you had a bunch of atoms all equally spaced on the x axis from $x=0 \to x=L$. The force you feel would be the sum of the forces from each one. And the potential would be the sum of the potentials from each one. That sum is what you approximate by the integral. (This is where early physics classes and early math classes can be opposite, in an ...


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The original Unruh & Wald (1983) paper addresses precisely this question. (You can find it here: http://www2.kau.se/tp/marcus/physics/lectures/unruhwald.pdf.) They discuss measurement of the thermal radiation by the accelerated observer via a two-state system, where the system jumps to a higher energy level on absorbing a quantum of the thermal ...


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Analogous to the tides of Earths oceans, do the Moon and Sun cause our atmosphere to bulge in what could be described as a low and high tide? The answer is yes, if you generalize beyond gravitation. Sunlight heats the atmosphere, and this causes atmospheric tides. The two dominant effects are absorption of visible and near infrared sunlight by water ...


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The differential force of gravity on the atmosphere works the same as it does for the rest of the earth (the oceans etc). However, moving the equipotential surface by a few m will be almost undetectable on the atmosphere, since the density of the atmosphere decreases so gradually – over many km. Contrast this with the surface of the ocean, which ...


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Yes. The short answer is you have one action you extremize to get Einstein's Field Equation $G_{\alpha\beta}=kT_{\alpha\beta}.$ Which you can think of as equations of motion for the gravitational metric $g_{\alpha\beta}.$ (They determine the second derivatives of the metric in terms of the matter fields and metric and the first derivatives of the metric.) ...


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Tyson goes onto explain that the bottom of the slinky simply does not know to fall until the arrival of the top which carries information of the gravitational force - it is essentially oblivious to the gravity exerted on it (I may be butchering his more eloquent explanation a bit). That isn't true. It's nothing to do with "information". The bottom of ...


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Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...


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Let's say you are holding the slinky and it isn't falling. Why isn't it falling? The same reason the sky isn't falling. The atmosphere has layers of air and the layer below it has more pressure than the layer above it so there is a net force on the air pushing it up that is just enough so it stays there. Same with the slinky. Let's talk a out the slinky ...


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You probably confused what he was saying a bit. The information that the bottom of the slinky is waiting for is the information that the top of the slinky is no longer supported. This travels down the slinky as the higher bits fall down and no longer exert a force (tension) supporting the lower bits. Incidentally in other circumstances the speed of gravity ...


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If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.


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SR is based upon two postulates: 1)The laws of physics are the same in all inertial frames of reference. 2)Speed of light $c$ is constant for all inertial frames of reference. To arrive at some results in relativity, one only needs to assume one of the postulates, other results are logical conclusions that require assuming the two postulates together. ...


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So Special Relativity states that for all non-accelerating objects of matter the laws of physics are the same. I think the point is just that the constants and the time and space derivatives that appear in a law of physics should not have to change the form of the equation if you measure the time and the space in two frames that move relative to each ...


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Due to gravitational time dilation, for an observer of the planet, the frequency of electromagnetic radiation would be slower. Visible light emitted from the planet would appear as infrared or micro-waves. The amplitude of the radiation would not change. Since frequency decreases while amplitude remains constant, the radiometer would receive less ...


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You say that you "can actually feel a substantial difference when I lift her, until her feet are not in contact with the ground anymore". That makes perfect sense i.e. she is also participating in the lifting process by pushing off the ground except when she's being difficult. Once she's off the ground, as long as you're holding her in the same way and she ...


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From Making yourself heavier Yes, it is possible to make yourself harder to lift. By shifting your center of gravity , which is usually referred to as your hips, down you can make it much harder to be lifted. It depends on which kind of lift, but this usually does the trick. Shifting your center of gravity does not make you heavier. It shifts the point ...


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Lets start with the definition of M theory: M-theory brought all of the string theories together. It did this by asserting that strings are really one-dimensional slices of a two-dimensional membrane vibrating in 11-dimensional space. So the universe which is at the microscopic level quantum mechanical, is composed of these ...


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The simple answer to both questions is, yes, you exert a force on Pluto and on the universe. By simple, I mean that the Pluto question is treated as a "two body" problem (and nothing else). This answer is obtained using the formula given by dotancohen. For the universe question, it should be easy to see that the same process can be applied to you and any ...


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To simplify notation let's denote $\gamma = \frac{1}{\sqrt{(1-0.99^2}}$, and let $m_{ball}, m_{Earth}$ be the rest (invariant) masses of the ball and the Earth. The way you write the gravitational force on the ball in the two FORs is: In the train: $F_{train} = g \frac{m_{ball}(m_{Earth}\gamma)}{d^2} = m_{ball} a_{train}$, hence $a_{train} = \gamma\;9.8$ ...


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When you talk about the speed of gravity, you are talking about the speed of gravity waves. If something jerks to one side or changes shape in a way that the change in gravity could be detected at a distance, the progress of that change through space is nothing other than a gravity wave. People have been trying unsuccessfully to directly detect gravity ...


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I shall first appeal to one of the key points of relativity: If you are 22 years old, like me, I can only have affected things up to 22 light years away from me. Things beyond that are untouched by my existence in every sense. If gravity were faster than light (by say a factor of 2) I could then have affected things up to 44 light years away! This is ...


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Antonio -- sure: your mass affects Pluto. Here's a line of thought that might get your head around this amazing concept: You probably agree that the Earth as a whole affects Pluto .. right? Simply, consider the earth chopped up in to little pieces each about the size and mass of yourself .. say 100kg. (Funnily enough .. there's actually NOT THAT MANY of ...


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So far, there is no experimental evidence that tells against the weak equivalence principle. The consequences of any real violation would be deep indeed: general relativity would instantly be falsified because GTR encodes the principle by saying for any path through spacetime, there is a Lorentz local frame (the "momentarily comoving inertial frame (MCIF)") ...


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For this answer, I'll assume that the center mass (absent any influence from the other two masses) is a perfect sphere, and that the other two masses are "above" and "below" the center mass. The middle mass will actually be pulled in both directions at once, because the top edge of the middle mass is closer to the mass on the top than to the mass on the ...


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Let's make this question a bit more operationally meaningful by asking if you can change the state of Pluto by choosing to do something here. As mentioned in the other answers, Pluto would feel the same force due to your mass even if you didn't exist, because the matter you consist of would be present on Earth anyway. However, you can still chose to move in ...


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Many things stop you. Firstly, it turns out forces do not cause accelerations. Forces increase momentum, and for slow speeds when you double the momentum you almost double your speed, and it is so close to doubling that for hundreds of years we thought it was doubling. But now that we've learned how to make things go fast and how to measure their momentum ...


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The mass that makes up your body and everything inside your body is attracting Pluto. If you did not exist, then that mass would take other forms, and it would still attract Pluto by the same amount. I don't want to offend, but you know where all your mass came from, and where it's going. It would likely be in one of those forms had you not been conceived, ...



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