# Tag Info

2

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 ...

0

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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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 ...

2

The math in that article is based in Cartesian space. Note specifically figure 4, where a portion of a Cartesian plane is pinched in at one side to show the supposed warping due to gravity. Using the shown transformation, she concludes that space is compressed near a black hole rather than stretched. The diagrams after that along with the process ...

2

Gravitational time dilation is not caused by acceleration. In most situations it is related to how deep the gravitational potential well is. So if two stars are orbiting you at a distance $d$ you will experience more gravitational time dilation than if you were a distance $d$ away from just one of them. So the time dilation is cumulative. Thus there will ...

1

yep, think of $\xi$ as a unit vector and replace all instances of it with $\epsilon \xi$ where $\epsilon$ is some small number. Then you will see that those two terms are second order in $\epsilon$.

-2

Charged versus rotating black holes as different kinds of wormholes There are no wormholes Timaeus. And no time travel either. There is no magic. Sorry. I've heard that a maximally extended charged black hole can be a traversable wormhole to the same universe And I've heard that when you die you go to heaven. Only I know that my fable isn't ...

1

To add to Ernie's Answer: Actually, tides on Earth are tantamount to an observation of a Moon-like object. Bernard Schutz in his book "A First Course in General Relativity" imagines an instance of your scenario whereby Earthlings lived under skies permanently shrouded completely and impenetrably by clouds: "The true measure of gravity on the Earth are ...

3

What's an observation? I think your question delves into the nature of "consciousness", a term which has never to my knowledge been satisfactorily defined. The seas observe the moon, and therefore there are tides. Whether or not a "conscious" being observes the seas is of no consequence in physics.

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Since another answer claims that a massive magic device would form in finite time I have to disagree. You have to wait forever, but only because your device is magic. The simplest problems are the spherically symmetric ones. And if you can get things close to an event horizon and magically bring them away as long as they stay outside then it is possible to ...

0

There are many different forces which can affect various types of particles: electromagnetism, gravitation, weak force and strong force. These forces act in different ways to change the state of other particles which are sensitive to that particular force and transfer information from one part of the system to another via these interactions. ...

1

Gravitational waves are not dark energy. Dark energy is closer to a fluid that is created when space expands and is destroyed when space is destroyed. That a fluid can do that requires a particular balance between energy and pressure one that is not normally achievable but if there were a fluid like that it would just keep filling everything. Classical ...

3

In case your ball is infinitely lighter than the black hole, the answer is infinity. You can never be sure it is not coming back. But in reality your ball has a finite mass which can not be neglected. Its mass is to be added to the black hole's mass M, therefore increasing its size. An outside observer will see that his ball got sucked into the black hole ...

3

If I understand correctly, you are just asking about the relation between energy and distances in both radiation and matter (and cosmological constant) dominated eras of the expansion of the universe. Consider the Einstein equation $$G_{\mu\nu} = 8\pi G T_{\mu\nu} \ ,$$ where $G$ is Newton's constant. In a FLRW unverse $G_{\mu\nu}$ is diagonal and using ...

2

Yes and no. A vacuum solution could be stationary, could be static, could be neither. For the orbiting black holes you end up with gravitational waves and gravitational radiation. You either have some going out, and the bodies in spiraling or you have some coming in and driving the system or both. And the point is that you have to specify the state of the ...

1

It just means the Schwarzschild coordinate system is faulty at the horizon because it assigns the same coordinates $(t,r,\theta, \phi)$=$(\infty,2m,\theta,\phi)$ to multiple events that are actually distinct events. If you look at the Kruskal-Szekeres coordinate system you can pick an event on the horizon and then draw the past light cone and those are ...

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How can black holes be observed to grow? Like a hailstone. See Kevin Brown's formation and growth of black holes. He says things like this: "Incidentally, we should perhaps qualify our dismissal of the "frozen star" interpretation, because it does (arguably) give a servicable account of phenomena outside the event horizon, at least for an eternal ...

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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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The equivalence principle only holds for extremely small regions of spacetime. Which means they hold for short time intervals as well as small spatial regions. Consider an event near the event horizon. If the event is outside the horizon there might be a frame moving away from it that cover a very small region that is also completely outside the horizon and ...

1

An object inside a black hole horizon cannot send signals outside of the horizon, but something falling into the black hole can fall through the signal. Take the example of the camera attached to the astronaut's foot. When the astronaut is halfway through the horizon, the camera is about a meter inside the horizon and the transmitter is about a meter ...

1

No, instead of the metric, the Vierbein enters, pulling back the gamma matrix, defined in the usual way in the tangent space, to the spacetime manifold. $$\bar \psi \, \gamma^\mu D_\mu \psi = \bar \psi \, \gamma^\alpha {e^\mu}_\alpha D_\mu \psi$$ This can be considered as the insertion of "half a metric", if one wishes. The Vierbein captures the ...

0

Is there anything else than spacetime? There's space and energy and fields and waves, but there is no actual spacetime. It isn't what space is, instead it's an abstract thing. See Ben Crowell's answer here. Objects don't move through spacetime. Objects move through space. The Earth is surrounded by space, not spacetime. Light waves move through space, ...

2

In a certain narrow sense, I think the answer to your question is believed by most physicists to be an emphatic no. That is, if by curvature you mean the fourth rank, valence $\left(\begin{array}{c}1\\3\end{array}\right)$ tensor $\mathbf{Riemann}$ that appears (through its contracted rank 2 covariant version the Ricci tensor) in the Einstein field equations. ...

1

Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning. The long one: What kind of systems does this apply to? This is not answered by referring to systems with ...

2

You are quite correct. For a system of two bodies the metric will be time dependant, and calculating a geodesic has to take into account the time dependance. With a few exceptions, time dependant metrics are even harder to calculate than time independant metrics. For example the merger of two black holes has to be studied by numerical calculations and isn't ...

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I'm going to answer this as a non- conformal field theorist, but I have been thinking about this stuff a bit lately. But I do believe I can answer the simplest of your questions: I'm sure an expert will put the following straight if there are mistakes. "So then a cordinate transformation which acts on the metric as a Weyl transformation is a conformal ...

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The acceleration they experience locally is different due to the different masses, whilst the acceleration they experience one to the other is of course equal (or at least synchronizing during any approach), otherwise A would hit B (when the planets crash together) later or sooner than B would hit A.

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Your misconception has nothing to do with gravity - you're just getting a little mixed up about acceleration vs. relative acceleration. Let's dispense with gravity, since it's a red herring here. Say there are two cars. Car A accelerates at $+3 ~\rm m/s/s$ (to the right). Car B accelerates at $-5 ~\rm m/s/s$ (that is, to the left). So far, so good, right? ...

2

It doesn't have to be modified, it's fine as it is. There's no paradox at all. The force that attracts both to each other is indeed $F=G\frac{M_A M_B}{r^2}$. But the acceleration they experience is not the same (at least provided $M_A \neq M_B$) because their inertias (masses) are not the same. For one $F=M_A a_A$, for the other $F=M_B a_B$. There's no ...

1

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.) ...

1

Consider the Nordström spacetime: $$ds^{2} = - f dt^{2} + \frac{1}{f}dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}$$ where $f = 1 - 2M/r + Q^{2}/r^{2}$, in units where $G = c =1$ and Gaussian units are chosen for the electric charge. While it's unclear what you mean by "curvature" since, generically, the curvature of this metric is a tensor ...

0

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. ...

1

GR can be recast into an equivalent but conceptually quite different form, using teleparallel gravity. This approach introduces the Weitzenboeck connection, which has no curvature, but has torsion. The presence of torsion indicates that gravity is not geometrized. Recall that in GR, we can always choose a locally inertial coordinate system such that the ...

2

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 ...

0

I wrote this answer assuming you don't know much GR The analogue of the Electromagnetic field in General Relativity is the metric tensor $g_{\mu\nu}$. The value of the metric is determined by the mass energy distribution via Einsteins equation, which is analogous to Maxwell's Equations. In the equation below $G_{\mu\nu}$ is a tensor which depends on ...

3

The Gauge Theory of Gravity (GTG) by Lasenby, Doran and Gull has a background spacetime with fields on it. It is basically derived from the same physical principles but as a background theory. It ends up not being the same theory, for instance it doesn't have the same isotropic solutions, and I think it does not allow time travel and such (unlike General ...

-4

Could we be on the inside of a concave hollow universe? Nope. Nor are we on the inside of a hollow Earth. Recently I was discussing this theory again (a little drunk, I admit) and then tried to find answers, but couldn't find anything satisfying. That's because it's bunk. There is a theory (or several theories) that we could be living on the ...

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For any transformed shape of the Universe, one may always easily define the theory in such a way that its results will be absolutely indistinguishable from the original theory. For example, one may describe the Earth and its vicinity by the polar coordinates $$R, \theta, \phi$$ so that $R\gt R_E$, the Earth's radius, corresponds to the space outside the ...

0

Sorry to be late to the party, I did an answer yesterday and then had some internet glitch and couldn't post it. I shall dust it off: Can we say that gravity (indirectly) is responsible for motion of electrons around nucleus? No, I'm afraid not. Note that an electron doesn't actually go round a proton (the simplest nucleus) like a planet round a Sun. ...

2

Only the symmetric stress tensor is physical, since thermodynamics demands a symmetric stress tensor. Since the symmetric stress tensor is unique, your option 2 is the correct one. (Canonical versions may be simpler but need not be physical; cf. the canonical momentum, which is often different from the physical momenetum.) This is completely unrelated to ...

1

Hawking radiation is indeed usually calculated as a semiclassical effect (classical gravitational field and quantum matter), which is invalid at high energy ($\langle T_{\mu\nu}\rangle >> 1$) and high fluctuations ($\langle T_{\mu\nu} T^{\mu\nu} \rangle - \langle T_{\mu\nu} \rangle \langle T^{\mu\nu}\rangle >>1$). Still, it is a result that is ...

1

So, can an area of space a few feet across (externally) be a light-year wide (internally) (TARDIS sort of thing)? Yes. It can and that's normal. If you made a region with less space inside than the surface area indicated that would require exotic matter. And with exotic matter you can make time machines. So ... Ironically you need something smaller on ...

3

Gravity is very, very weak compared to the electromagnetic force. Whatever curvature of space-time is induced inside an atom would be far too weak to hold mass-energy together at that scale. The electromagnetic force is 1/137 the strength of the strong nuclear force, while Gravity is 6 * 10^-39 the strength of the strong nuclear force. Here is a link that ...

8

While photons can in principle form a black hole, the black hole will not be massless. The mass of the black hole will be related to the energy of the photons that went into it by Einstein's famous equation $E = mc^2$. The black hole will be a regular black hole, and classically it has an infinite lifetime. Once you include Hawking radiation the black hole ...

4

To someone outside it looks the same as a regular, massive black hole. Classically the lifetime is the lifetime of the universe. It might merge with another black hole. It might last forever. It might meet a singularity in a big crunch if the whole universe contracts to a singularity. If you are worried that it can't decay by Hawking radiation because an ...

1

The mathematics of general relativity is clear and unambiguous. The trouble comes when you try and describe what is going on it non-mathematical terms, because there is no precise way to do it. Kip Thorne is attempting to talk about black holes in non-mathematical terms, and he is adopting a different perspective from (probably) most of us. That doesn't mean ...

3

A point to emphasize here is that one cannot separate the rest mass of the nucleons and electrons from that of field: rest masses aren't additive in this way and one can only state the rest mass of a system as a whole. The mass-energy of the fields is already built into the rest mass of the system. The rest mass of an electron includes the energy of the ...

1

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 ...

2

The easiest trick is to write (say, in three dimensions. For higher dimensions, add indices to the Levi-Civita symbol, and factors of g): $$g = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}g_{ax}g_{by}g_{cz}$$ Then, the variation is easy. I'll leave it as an excersise to work out the variation, and how to translate the result into factors of $g_{ab}$ and $g$

0

A brief intuitive answer. Black holes are enormously difficult to work with, the warped space and tidal forces are enormous, but I think you want to ignore the difficulties. A black hole with 100 solar masses would have a radius of about 300 KM, if you put it next to the Earth, it would be a bit smaller than Texas. Lets say your friend's ship orbits at ...

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