# Tag Info

21

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...

14

The equation you are mentioning is the gravitation force derived by Newton. This force doesn't apply to particles such as photons for two reasons: Photons are too small, and you can't use Newtonian physics to describe their properties. Photons travel too fast (their velocity is the speed of light) and at such a velocity Newtonian mechanics cannot be ...

8

The annihilation produces gamma photons, whose total energy sums up to the total energy $E_0=\sqrt{p^2 \,c^2 + m^2\,c^4}$ formerly contained in the matter / antimatter kinetic energy (the $p\,c$ term) and that "frozen" in rest mass (the $m\,c^2$ term). So energy is conserved. As for gravity, the Einstein field equations "can't tell the difference" between ...

6

The idea that gravity is sourced just by mass only hold in the nonrelativistic limit. In a fully relativistic treatment, which is necessary to understand any process involving particle annihilation, you must use the full stress energy tensor. This tensor includes mass, but it also includes, for instance, the energy of any photons created in the annihilation. ...

5

According to general relativity the source for spacetime curvature/gravity is the stress-energy tensor. Mass contributes rest energy $mc^2$ but it is not the only form of energy that influences gravity. Therefore it's not exactly true to say that mass causes gravity, but energy does. The photons produced during annihilation will carry energy and so they will ...

5

Your logic is ultimately wrong because that equation doesn't reveal the true nature of gravity. According to general relativity, objects themselves bend space-time. Imagine space like a rubber sheet. If you stretch it and place a mass in the middle and roll a ping-pong ball past the mass, it will curve towards the mass. Similarly when space-time is being ...

4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...

3

Gravitational curvature is not fully described by the Einstein or stress-energy tensors. It is only fully described by the Riemann tensor. Consider, by analogy, a simple example from electromagnetism: $\nabla \cdot E =\rho/\epsilon_0$, Gauss's law for the electric field. The function $\rho$ describes the distribution of charge density. The LHS is zero ...

3

Your thought experiment of dropping an iron ball and a feather need not be in water; in fact, it is more commonly considered in air, but the pertinent facts are the same. All objects, regardless of their mass or composition, are accelerated identically by gravity. But within a particular medium, the acceleration of particular objects might be impeded by ...

3

To quote the Wikipedia article on gravitational shielding, "gravitational shielding is considered to be a violation of the equivalence principle and therefore inconsistent with both Newtonian theory and general relativity." You can also think of this as being due to the fact that mass only comes in one flavor, positive.

3

4.Without air resistance, how much energy would be required to get a given mass to this altitude? Energy Needed ($E$) = Potential Energy at L1 ($V_{L1}$) - Potential Energy at Earth's Surface ($V_e$) $$V_e = -Gm(\frac{M_e}{r_e} +\frac{M_l}{LD - r_e})$$ $$V_{L1} = -Gm(\frac{M_e}{d_{L1}} +\frac{M_l}{LD - d_{L1}})$$ where $m$ is the transported mass, ...

3

Here is an example for the Sun. The figure below plots a (reliable) estimate for the interior density profile of the Sun, $\rho(r)$. So for a given radius $a$, the mass interior to that radius is given by $$M(a) = \int^{a}_{0} 4\pi r^2 \rho(r)\ dr$$ And of course the gravitational field strength assuming spherical symmetry will be $$g(a) = - G ... 3 Ah, all that talk about curved space-time. Well, there is a simpler argument. The fundamental axiom of general theory of relativity, "principle of equivalence", says: The effect of a homogeneous gravitational field is equivalent to that of a reference frame in uniform acceleration in the direction opposite to that of the gravitational field. All that ... 3 In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by:$$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$If you do the same for two solar mass neutron stars ... 2 A Penrose diagram of a metric g_{ab} is used to represent the conformal structure of g_{ab}. Generally light rays move at \frac{\pi}{4} from the upward vertical and the spacetime considered is spherical symmetric. The metric, \overline{g_{ab}}, on the Penrose diagram satisfies: \overline{g_{ab}}=\Omega^{2} g_{ab}. This implies that timelike ... 2 The two situations are different. In the first case the star is radiating energy in the form of electromagnetic waves. Actually it isn't a good approximation to think of the star radiating photons because the energy it radiates is delocalised. If you insist on treating it as photons you would have to describe it as a superposition of all the possible photon ... 2 A physical theory is just a mathematical model that predicts the evolution of the physical system it is describing. The more successful theories tend to be motivated by experiment, but this doesn't have to be the case e.g. string theory. Any theory can be proposed, but at the end of the day the test is whether the theory makes predictions that match ... 2 The Torricelli Formula you used is certainly a good way to start if the level of the water source is stationary. In addition there will be losses which depend on geometry of your flow restriction and on Reynolds Number. Loss coefficients \zeta_{loss} for free jet discharge, valves, etc. can be found in literatute, for example here: ... 2 I think it has more to do with surface tension. Added: When an object is floating on the surface of water, it bends the surface to support it. This bending extends outward from the object decreasing with distance (much like gravity). When 2 objects are near each other, they feel the bent surface and move down the slope (they attract each other). 2 In 3+1 dimensions, the Ricci tensor vanishes when the stress-energy tensor vanishes indeed. Which means that, whenever there's a vacuum, the Ricci tensor vanishes as well. But it also happens that, in that number of dimension, the metric is not entirely determined by the Ricci tensor. The full curvature tensor (the Riemann tensor) is a mixture of the Ricci ... 2 Entropy is a natural tendency of the Universe to fall apart into disorder. In a reversible process, an increase in the Entropy of the system will be exactly equal to the Entropy decrease of the surroundings. Thus, the net change in the Entropy of the system and its surrounding will be zero. But in an irreversible process in an isolated system (for ... 2 As far as I know, the expansion of the universe contributes into creating more and more microstates. This is almost equivalent to saying that entropy increases (because S = k_Bln(\Omega)). We cannot be sure that the law of entropy applies to the whole universe (There is debate if the universe is a closed system or not, if its infinite or finite, etc..) ... 2 This is more of an extended comment than an answer, since I haven't had time to look up all the papers referenced in Felber's article. Hopefully this discussion will be a useful clarification if nothing else. It should be well known to any GR enthusiast (whether a formal student or not) that as viewed by a distant observer it takes an infinite time for any ... 2 One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric:$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ...

2

As you said, if feather has a stronger decelaration, is due to air friction (not because iron is more dense than feather or something). The same should hold in absence of gravity: if you give a boost to these balls, then the feather one will stop first. From this last consideration, I'd say that this medium does not affect gravitational force, but just it ...

2

The answer is yes, and in fact I've described how this works in my answer to another question of yours: If you shoot a light beam behind the event horizon of a black hole, what happens to the light?. I won't repeat the working from that question here, but it might be worth a comment on exactly how the idea works. When you solve the equations of GR you get ...

2

Found this interesting read on this website: "In a classical point of view, this question is based on an incorrect picture of gravity. Gravity is just the manifestation of spacetime curvature, and a black hole is just a certain very steep puckering that captures anything that comes too closely. Ripples in the curvature travel along in small undulatory packs ...

2

In Newtons theory of gravity photons are not affected by gravity (created by masses). So your conclusion is correct. But in General Relativity the curves of free objects like test particles or photons (geodesics) are determined by the space-time geometry. The geometry is described by the metric which is given by the energy and mass distribution of the ...

2

Newton's formula is an approximation of how "gravity really works". We actually still don't know how gravity really works, but we have vastly refined our understanding of it thanks to Einstein's general theory of relativity. Gravity is simply a measure of curvature of a 4-dimensional manifold we earthlings call space-time. Local concentrations of mass or ...

2

The formula $F=G \frac{m_1 \cdot m_2}{r^2}$ is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ...

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