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Upshot: The following two facts can be used to argue that the scalar correlator simplifies in the special case described above. When the scalar action is Weyl invariant, then the scalar equation of motion is covariant and we can use a Weyl transformation to simplify the equation. There is a Weyl transformation that maps AdS to flat space. The correlator ...


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Suppose that we align a perfectly cylindrical pencil with the z-axis. If the initial conditions are rotationally symmetric about that axis, then because the laws of physics are rotationally symmetric, the final state must also be symmetric under rotations about the z-axis. This means that the wavefunction of the universe will have to evolve into a ...


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I'll look at the question from multiple aspects. Classical mechanics, exact measurements, no thermodynamics, no perturbing forces In this fictitious universe it is possible to stand our perfectly balanced pencil exactly vertically and perfectly stationary. This is an unstable equilibrium position. With no perturbing forces, no thermodynamics, no quantum ...


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There are a number of issues with this question - it rests on false assumptions. The short answer is: No, we do not expect more forces to appear in any way. Yet, to say "the four forces used to be one force in the early universe" is overstating the knowledge we have. It is the basic idea of Grand Unified Theories to merge the three forces excluding gravity ...


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In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity. But there are forces afoot inside the pencil. Unless you also chill the pencil to absolute zero and thus stop all molecular activity, the trillions of atoms in the pencil are vibrating in random directions. It won't take long for the ...


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What you've fabricated is of course unrealistic in the physical world as CuriousOne stated, but not so in the virtual world of simulation. All the conditions you ask for can be arranged in a simulated universe. If perfectly balanced as its initial condition, the virtual pencil will not fall in this virtual world. It is unstable, but it will not fall until a ...


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Which way will the pencil fall, after you let go? Facetious answer: it would fall in the direction to which it was leaning when you let go. Okay, now to justify that: you cannot balance the pencil before letting go. For an object resting on a surface to be balanced, its centre of mass (a single point) must be directly above a point within the area of ...


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If you had a modular system of rods and connectors (like for making molecular models) then a certain number could be used to make a spherical shell of surface area $4\pi R^2$ that an observer far away would say has a mass M. You could then rearrange the parts to make a thicker shell with a smaller surface area. If you kept the parts at the same temperature ...


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I think if we consider a much bigger mass then the force becomes relativistically: $$ F=-GMm/r^2+4G^2M^2m/r^3/c^2 $$


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As per @lemon mentioned : You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring. So I just did like that only. Taking sphere as point mass then point mass will experiences one gravitational force of attraction that is: $$df=cos\theta$$ As vertical ...


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Actually you can formulate Gauss's law for the gravitational field as well: $$\oint_S \vec g \cdot d\vec A=-4\pi G M, $$ where on the left you have the gravitational flux through a closed surface and $M$ is the mass inside the volume. $G$ is the gravitational constant. When you call this quantity on the left $\Phi_G$ and write the mass as an integral of ...


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In analogy with Gauss's law for the electric field $\nabla\cdot \vec{E}=\rho/\epsilon_0$, the flux of the gravitational field through a closed surface is proportional to the mass contained inside the surface. There is an approximation to General Relativity called Gravitoelectromagnetism (see Wikipedia page of this name. It's relationship with Newton's law ...


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When the car takes the curve, think at the circle it goes on as a polygon with an infinite number of sides. Instead of continuing to go straight, the ground acts on the object with a force that increases the normal force and creates the centripetal force needed for the object to go circularly.


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The acceleration of gravity is that itself. The bigger the mass the more gravity pulls on the object. Especially with a force. This could be simplified into an equation we learned in Intro-Physics, F=ma or f=mg


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Many physicists are researching dark energy and the particle science behind it. There are many theories about what happens beyond the horizon of a black hole. Some people believe in something similar to your question. Its a theory about how things passing through a black hole travel into the multiverse and possibly creating more universes. This theory ...


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It sounds like you need something modelled on the controversial claims of Eugene Podkletnov involving a rotating superconducting disk and high voltage spark discharges resuting in anomalous pseudo-gravitational forces


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the simplest models for expanding universe (see frw metric) don't have any curvature term for 3 space, as constructed from experimental data. however using black holes to model systems do result in inserting curvature terms for 3 space, and as a result we can't use them to model expanding universe.


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An object becoming a black hole does not affect its gravitation strength. If the sun became a black hole, orbits would be the same.


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When we assume hydrostatic equilibrium, the pressure gradient is taken. Maybe the pressure should be 0 when $\xi>\xi_0$, so that constraint should be explicitly applied. This is generally understood as an implicit restriction/constraint of the hydrostatic equilibrium (HSE) condition for starting the Lane-Emden equation: HSE is for the star and not ...


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The assumption $$ P=K\rho^{1+1/n} $$ is a very simplistic mathematical model and should be applied with care. When the model predicts matter has negative density at some point, it is a sign that it is too simple to get it right there. The model neglects lots of things - radiation, magnetic field, solar wind, etc.


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I would go with the hydrostatic equilibrium condition being the source of the issue here. When there is no polytropic gas when $\xi \gt \xi_0$, there is no need for any hydrostatic equilibrium to be in place - the gravitational potential is free to take on any value, determined (together with boundary conditions) by Laplace's equation: $$\nabla^2\Phi = 0$$ ...


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If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.


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Air pressure exists because if we place something in a gas, then the molecules/atoms flying around will keep banging into it, and in this way produce a net constant force per unit area. As explained by @Chris2807 in the neat formula $P=n k_{B} T$, this is proportional to how many particles there are (since this is proportional to the amount of "banging" in ...


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Even though you were wrong, you nailed the critical question at the heart of the necessary understanding. Bold is my emphasis: Modelling this planet at my head then anywhere that effective gravity increases in strength is effectively lower potential energy "down" from anywhere that it doesn't. I would expect this to cause the planet to deform and reshape ...


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In some sense yes. Let me explain a little. If we were to take a sealed container of gas and put it into free space far away from other bodies so that the gravitational force on the box is negligible would you agree that there would still be some pressure in the container? If we assume we have an ideal gas then the pressure is simply given by $$P=nk_{B}T$$ ...


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In general, air pressure in the Earth's atmosphere is hydrostatic pressure, caused by the Earth's gravitational field. If there was no gravity then there wouldn't be any centripetal force and all the air molecules would just float away into space. This is why there is no atmosphere on the moon - because it doesn't have enough gravity to sustain one.


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Your intuition is correct. At that scale, the planet behaves essentially like a liquid, and it's shape will equalize into an ellipsoid where: The local gravity vector is everywhere perpendicular to the surface, and The downward force felt by a person standing on the surface is the same everywhere. But I don't think you could get an eccentricity as ...


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A partially reflective Dyson sphere is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The global effects, depend on the structure of a star and ...


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An analysis would have to look at the effects over time. With my limited understanding of physics and intuition, I see the outer layers of the star reabsorbing the rays. Where the outer portions of the star until now have experienced large amounts of energy flowing in one direction, now has a net outward energy flow of maybe 1/2 to 1/10 of what it used to. ...


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hahaha Actually, EMF is always applicable, even in a blackhole. All stars have perfect symmetry, to a degree, gravity is always trying to crush the star and thru fusion SNF / the EMF electrons repel this force outward causing this balance. Now you can have GRB's and the formation of blackholes if the star is large enough or if not Neutron stars or even ...


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Energy is only expended in movement. In a stationary configuration (holding a brick), energy is not expended. The brick is held back by a static force, which requires no input to maintain itself and expends no resources in doing so.


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You can't shield against gravity. If you could, you could just shield gravity from one side of a wheel and make a perpetual motion machine. Space elevators have been considered, but you still have to fight against gravity and friction. It just means that you can just climb a rope instead of using a rocket.


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f(friction) directly proportional to N(normal force acting, perpendicular to the plane of fridge in outward direction here) is the force that opposes motion, now friction will in this case be provided by the atomic attraction, only atomic attraction is taken into account here, because repulsion would not even let it attach to refrigerator, after the force is ...


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As you may know, the friction is proportional to the normal force of an object or in this case the force of attraction between magnet and refrigerator. If your force is strong enough then the friction will be sufficient and the magnet will not slip (on earth the force of friction must exceed the mass of your magnet multiplied by 9.81 m/s). If we assume ...


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It's probably easier to measure gravitational masses for anti-hydrogen or anti-ions, because those particles can be cooled and trapped using only electromagnetic fields. The problem is that antiparticles are generally "hot" when they are created. The processes that create antiparticles have an energy threshold beneath which they don't work at all (this is ...


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I am not sure what you mean by if something has more energy it should "slow time" more than something with low energy But even if I just neglect that for a moment. I'll use a description of waves for visual aid and simplicity. Imagine in "not slowed down time" a wave has a period (and frequency) of 1 in this time. So there is one wave crest and one ...


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It's probably best to not think of single photons as sources of gravitational energy. For one thing, most bulk electromagnetic fields are not eigenstates of the photon number operator. For another thing, the thing that couples to the gravitational field is the energy density of the field. This density is proportional to the intensity of the field, ...


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sound needs to bounce off of air and gravity. so if theirs no gravity sound can not bouse off of sound


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What if you had a proton travelling at .99999c towards a heavy object? Would it have to keep accelerating or would the acceleration of the proton slow down to zero and only it's mass would increase? It would keep accelerating, but there's a problem. See what Einstein said in the second paragraph about the speed of light being spatially variable: The ...


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Good question. Although the speed of gravity is approx. 3x10^8 meters per second, the acceleration at the Earth's surface is as you describe ~9.8m/s^2. If you choose the Earth as the heavy mass your proton in question is approaching, and forgo magnetic influence, then you would realize that the increase in acceleration is negligible as compared to the ...


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In order for that proton to reach the speed of light, it would need an infinite amount of energy. Perhaps if the proton reached the center of a black hole, converging with the singularity, it would cease to accelerate, but it would not travel neither faster or at the speed of light.


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You should look at it like an asymptote. Yes the proton would accelerate but it would probably accelerate to .999991c or more likely less due to the massive energy required to accelerate something so fast already. Therefore you could always keep accelerating your particle but it would never cross the Light-Barrier.


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I have to admire the fact that in the face of an infestation like this you wonder about the physics instead of setting the bed on fire to kill the termites... Looking at the video it seems to me that the pellets are not "dropped" but "ejected". If that is the case, then if the termites exit hole is slightly angled, they will eject their pellet not straight ...


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Having read John Rennie's answer above, I'm going to give an answer that's hopefully of the same sense, which hopefully makes sense, but which hopefully brings out an issue. 1. Is the free fall acceleration the same as the coordinate acceleration for a hypothetical observer at rest on the star surface? Yes and no. Yes because the falling body falls ...


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Termite pellets are the feces of a termite, the left overs of its eating wood and in no way of the size shown by the size of the bed. The pellets are from a colony of insects. Often, the only obvious signs of infestation are little mounds of fecal pellets building up underneath the infested wood or the appearance of “kick-out” holes in the surface of ...


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I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates): $$ ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1} $$ with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass: $$ ...


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Water forms close to perfect spheres in zero gravity due to it's surface tension. There's a variety of videos of water in the space station. Ice, assuming you start with one of those balls of water, you have to ask first, would it freeze outside in (say, the temperature of the station is dropped below 0 C), or would it freeze inside-out, say you stick a ...


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Sound will behave just like on earth provided it has a medium to travel through (Astronauts on the ISS can communicate normally; watch some videos)


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The quantum size of elementary particles is in conflict with the classical notion of "generating a gravitational field". The correct statement is that these bosons, as well as all other particles (even massless ones), do interact with the hypothesized massless spin two particle we call the graviton, through three or four point vertices.



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