New answers tagged

2

It loses organization, e.g. matter changes into pure energy or some such thing. It's not entirely clear what form there is (some suggest there is no form at all, but it's obvious that it doesn't follow Pauli's exclusion principle). This is nothing special, it happens all the time - when you burn carbon, for example, you get a bit of disorganised energy (heat)...


2

I think I have a misunderstanding for the KK theory. In the KK theory, we are living in, say, a 5-dimensional spacetime with one dimension compactified. What's different from the brane-world theory is that, in brane-world theory, we are living on a 4-dimensional brane which is embedded in the 5-dimensional spacetime. So in the post, I can not assume that the ...


3

It is rather the other way around: our understanding of physics has enabled us to build GPS systems in the first place. It is correct however that GPS was direct confirmation of not only the theory of special relativity, but also of general relativity. Neglecting the time dilation from the difference in gravitational fields between the satellite's position ...


2

No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge. The area of each horizon ...


7

To expand on @dmckee's answer, if we have a speacetime that has the matter concentrated in a central area, we can definte an overall conserved energy-momentum vector called the ADM energy. It can be further shown that the ADM energy does not change when the matter falls into the black hole.


19

Energy (in any form) falling into a black hole contributes to the mass of the hole, and mass is one of the many forms that energy can take, using the usual conversion factor: $E = mc^2$.


0

Classically speaking, since total BH evaporation takes finite time, yet being at the event horizon stops all time (in your IRF), the black hole will evaporate before you can get to the event horizon.


3

Is there any other scientific theory that proves that big bang is the origin of time ? Here is a gross misunderstanding of what a scientific theory is. A scientific theory can never be proven. It is successful if it fits data and observations, then one says it is validated, and if its predictions are always validated. An invalid prediction requires drastic ...


0

Studying big bang itself gives a lot of evidences to believe in it.however redshift ,given by Edwin Hubble , is one of the major theory that supports it by giving evidence of expanding universe.cosmic wave background is also a big point to support it.


-1

The change in the geometry is not entirely certain. One can appeal to exact solutions. With the Schwarzschild solution for a non-rotating black hole the conformal diagram illustrates region I that is the exterior universe and region III for the black hole interior. The line separating them is the event horizon. There is another region II that is another ...


1

DISCLAIMER: I have never seen the below conlcusion made explicitly, but it seems to me to be a straight-forward consequence of established theory. I cannot find any error in my thinking, whence I will post it and await your judgement. The equivalence theorem, proved e.g. in Cartan's Leçons sur la géométrie des espaces de Riemann, states that in a rigid ...


2

So the answer is no. An electric field has energy and energy generates a gravitational field, just like any mass. See the charged black hole solution is the Wikipedia article https://en.m.wikipedia.org/wiki/Charged_black_hole The charge of a black hole, if nonzero, changes the metric and solution to account for the charge and electric field. That ...


1

The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes. Consider the standard twin 'paradox': twin a hangs around in free-fall; twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns. Well, ...


0

Why not negative energy? Energy is a scalar quantity that is conserved. Negative energy means that there exists a potential well which for bound quantum mechanical states has negative energy, and for free states positive. A matter of definition. Look at the hydrogen atom states Note the - sign for the energy levels In the case of the black hole the ...


1

The quote you give from Carroll about the covariant derivative is right: it quantifies the rate of change of a tensor field relative to parallel transport. The covariant derivative of a tensor at a point doesn't make sense. However, the commutator of covariant derivatives acting on a point does. The situation is analogous to the vector field commutator. ...


1

Consider a region in the manifold where a tensor field is everywhere well defined. Consider a point $x$ and a neighbouring point $x+dx$. The tensor field, say $V^\mu$ is given at both points as $V^\mu(x)$ and $V^\mu(x+dx)$. We can parallel transport the tensor(vector in this case) from $x$ to $x+dx$. This parallel transported vector $V_P^\mu(x+dx)$ is in ...


0

The attractivity effect depends on the time during the two masses are at proximity. You can experiment it in sending a ball through a water fall or a water flow at different speeds and look the direction the ball takes after crossing the water. The representation of space time curvature by a deformated plane around a planet is a farce. Try to represent a ...


8

Start by considering the ordinary Newtonian gravity. This tells us that the acceleration of a test mass due to our planet of mass $M$ is: $$ a = \frac{GM}{r^2} $$ The acceleration is the rate of change of velocity with time. A fast moving object spends less time near the planet than a slow moving object so its velocity changes less. That means fast moving ...


7

It is not true that there is a unique geodesic through every point. To understand it, imagine a point on a sphere (where geodesics are just great circles) or even on a plane (here geodesics are straight lines). Through that point you can draw infinitely many geodesics, e.g. infinitely many straight lines passing through this point. However, if you restrict ...


1

Object can't go with the speed of light on the Special Relativity, on which the General Relativity is based upon. The gravitational waves are predicted and calculated by the General Relativity, i.e. if you are talking about them, you are using the terminology of a theory which closes out the things going with the speed of the light. But we can talk about an ...


1

Toward position y. Gravity couples to itself (Einstein's equations are nonlinear), so yes, gravity can cause itself to curve. Another way to see it is that photons and gravitons are both massless and so they both travel along identical null geodesics, so wherever you see light coming from, you'll feel the gravity from the light source pulling you in the ...


1

This depends on how you do the analysis -- If you do it in the Kruskal spacetime, which is the more physical way, It's hard to talk about what "$r = 0$" is, and the question becomes somewhat ill-posed. If you naïvely do it in Schwarzschild coordinates, you'll find that $R$ has terms involving constants multiplied $\nabla^{2}\frac{1}{r}$, which an analysis ...


1

If we want to be formal, at $r=0$ there is a manifold singularity. So no tensor/scalar quantities are well defined there (i.e. our mathematical formalism does not work there). On the other hand, as physicists, we can look for different examples that behave in a similar manner. Also the (classical) electromagnetic field generated by a point-like charged ...


-1

When you solve the Einstein equations to find the black hole the energy-momentum tensor is vanishing everywhere, even at $r = 0$. At no moment you are imposing that it is a point source $T_{\mu\nu} \sim \delta(r)$, this would change the solution. The interpretation of the solution as a black hole with a central singularity comes by inspecting the properties ...


0

In your question you have proposed the statement: space geometry is changed if and only if there is a kinetic energy difference between them. I need to tidy this up just a bit, so let's consider this related statement: the geometry of space-time is changed if and only if one observer is moving (or even better yet, accelerating) relative to the ...


1

Suppose there is a flash event that we can represent as a light cone as the flash expands. There are three shutters with detectors around this flash event. The shutters open and close only once and this is almost instantaneous. One shutter opens in spacetime outside the light cone. One Shutter opens in spacetime inside the lightcone and the third opens ...


0

You are over complicating what appears to be your interest. You are asking in essence if there is an energy change in a particle as it moves in a gravitational field, i.e., a given spacetime metric. The answer is simple: except for singularities (i.e., what happens there, which is unknown, that's where any particle path ends in general relativity), a ...


0

Earth's gravitational field causes Earth to retain a gaseous atmosphere, which both absorbs light itself and refracts light towards the surface. Estimating the altitude of the optically thick part of the atmosphere as somewhere between 6 km and 60 km, this atmosphere effectively increases the cross-sectional area of the Earth for interacting with sunlight ...


0

This answer has been edited as I came to realise I had answered at first without understanding the question (the "no wait, I'm a moron!" experience). Therefore I will start with some unnecessary background. Let's agree to call a co-vector $p_i$ perpendicular (and leave the term normal for metrically normal vectors; we shall soon see that they are the same) ...


1

I think the resolution to the op is quiet simple and the entire paradox is because the solenoid in the third figure, should be moving to the left and not to the right! (If you are in a car moving to the right, then according to you the trees are moving in the opposite direction, ie to the left!) In other words, the system: in the reference frame of the ...


2

One talks about the size of the universe in the context of a model where spacetime is foliated by three-dimensional spacelike leaves. The "size of the universe" means the size of one of those leaves, not of all spacetime. For example: If you imagine spacetime to be filled with galaxies, the worldlines of those galaxies give a preferred global time ...


0

If you were falling into a black hole and you performed measurements of the speed of light in your spacecraft you would find it to always be the same $c~=~299,997km/s$, as I recall the more exact figure. If your spacecraft is small enough so there is no significant curvature across it then spacetime looks flat even as you cross the event horizon. This black ...


1

The components of $\text{Ric}$ transform during coordinate change $x^\mu\mapsto \tilde{x}^\mu$ as $\tilde{R}_{\mu\nu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}R_{\sigma\rho}$. This is just the usual transformation rule for coordinate-components of tensors. Contracting over the two indices gives $$ \tilde{...


3

It is special relativity that tells us space and time are different aspects of the same thing--spacetime. And indeed quantum mechanics does not respect the covariance principle of special relativity. That's exactly one of the reasons of why we need to invent quantum field theory(QFT) when we already have QM at hand. To make the space and time on the same ...


2

As is wellknown, the EFE is a PDE for the metric $g_{\mu\nu}$ / vielbein $e^a{}_{\mu}$, which plays the role of the dynamical field of GR. OP is apparently pondering whether the EFE's source term (i.e. the matter SEM tensor $T^{\mu\nu}$) is independent of the unknown fields (i.e. the metric $g_{\mu\nu}$ / vielbein $e^a{}_{\mu}$) and derivatives thereof? ...


1

In book "from eternity to here" by sean carroll he described distortion of spacetime near a black hole. also you can change the visualization to realize milder distortions. Before you go forward you should know about light cones in GR. it is explained in book but I assume you know it. He says: [But a real black hole, ..... It is a true region of no return—...


3

ANSWER WITHDRAWN I am withdrawing my answer because I am persuaded by Henning and others that I am mistaken about the impossibility of catching up with someone who has crossed the Event Horizon. I have also withdrawn my Vote-to-Close. Original Answer What do you mean that your friend has "fallen into a black hole"? If you mean that he has crossed the ...


0

Let us consider four dimensions with coordinates $(t,x,y,z)$. If the metric tensor is diagonal, then its inverse is also diagonal. Let the inverse be $g^{\mu\nu}=\{A,B,C,D\}$, where I have listed only the diagonal elements. Now, we know that $R_{\mu \nu \theta \phi}$ is antisymmetric in the first two indices. That is $R_{\mu \nu \theta \phi}=-R_{ \nu \mu \...


17

Assuming that the black hole is large enough that one can cross the event horizon without being spaghettified by tidal forces, and the that when the accident happened, the both of you were hovering in place above the black hole with your jetpacks, rather than orbiting it: You can still see your friend (no matter how long you dally, the part of his worldline ...


2

The answer by @peterh is accurate on the factual information about the Einstein Field Equations and that it describes how the matter distribution affects spacetime. There is more that may be added that hopefully will help understand more of it. First, just to be totally clear, gravity as described by GR (general relativity, through Einsteins Field ...


0

The charged accelerometer will register a non vanishing acceleration. The reason why the setup you are proposing (interaction via electric charge) gives a physically distinct result from the setup in the answer you linked (interaction via gravity), even though they can be described by the same mathematical force, is because the Equivalence Principle applies ...


2

In initially writing the Einstein-Hilbert action on a spacetime M: $$S=\intop_{M}(kR+l_{m})\sqrt{-g}d^{4}x$$ Where $R$ is the Ricci scalar, and $l_{m}$ is the matter lagrangian, we have k as an unknown (constant) quantity. Taking the variation of the action, with respect to $g^{\mu\nu}$ and dropping the integral, one obtains: $$k(R_{\mu\nu}-\frac{1}{2}...


7

We have Newton's law in the form $$ F = \frac{Gm_1m_2}{r^2}$$ which is the same as the field equation for the potential $$ \nabla^2 \phi = 4\pi G \rho $$ where $\rho$ is the mass density. The $4\pi$ here does indeed come from the solid angle of an entire sphere, but by redefining $G$ we could put it in Newton's law instead. This is connected to GR by ...


1

Pi crops up in mathematics and physics in all sorts of places that have absolutely nothing to do with circles and basic geometry, such as $e^{i\pi}=-1$, which is purely a statement about numbers. Don't think of it as the ratio of the circumference of a circle to its diameter. Think of it as a fundamentally important mathematical constant, and one of its ...


2

No, it depends on the metric of the Universe described by the Friedmann model. It is a general relativistic theory. In GR, gravity is not a force. Instead, there is actually two equations: The Einstein Field Equations, describing how the distribution of matter affects the geometry of the spacetime, There is also equations showing how matter moves in the ...


6

There is indeed a nett force on the body owing to the electrostatic attraction / repulsion. Therefore, there is nonzero four acceleration, and the body will have a different orbit from the ones defined by the spacetime geodesics for the metric describing the massive body's neighborhood. From the standpoint of an observer stationary with respect to the ...


3

from your profile you seem to be an amateur (gifted?, teenager?, still in school?) self-studying GR. Great! So drop the Old Man's Book (the Meaning of Relativity) and get yourself a modern intro to GR or - my suggestion - the wonderful MTW's Gravitation. I did the same when I was 16. It was published in 1973 and Kip Thorne himself told me two weeks ago that ...


2

I'd like to add to Anna V's complete answer, and CuriousOne's gem of a comment: Did general relativity mean the end of Newtonian mechanics? Of course not. The least useful of all theories will be the theory of everything. It will explain everything, but calculating even the most trivial problem will be such a hard thing to do that nobody even among those ...


0

I am not sure if GR explicitly says that gravity is not a force. It explains it in terms of curving of space and I assume, the curving of space provides the necessary force. But if GR explicitly says that "gravity is not a force", then, obviously forces can not be unified with something that is not a force itself. Therefore in order to have a unified ...


1

The history of physics shows that there is not really an end of physical theories, but it is a matter of regions of validity of the models in the space and time and energy momentum phase space. This is because physics theories are not just mathematical theories, but have extra postulates/laws which connect physical observables to the mathematical functions. ...



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