# Tag Info

6

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 ...

4

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 ...

4

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.

3

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 ...

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.

2

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 $\epsilon$.

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No, we can't say that. The vacuum of space cannot carry momentum. "Pushing against space itself" is just a roundabout way to say "pushing against nothing at all", or in other words "violating conservation of momentum".

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There are lots of things an EmDrive could 'push against' - some as yet unknown or ill understood field for instance (dark matter, dark energy, something else). It could also couple to the gravity field - in which case it would be pushing on space itself. The coupling to gravity would be miraculously strong though, so that's not likely. The main problem ...

1

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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