New answers tagged

2

at the moment after the big bang, how much gravity would a person 'standing', say a metre away experience The Big Bang didn’t happen at a point, so you can’t be one meter away from it. The Big Bang model is one of a homogeneous and isotropic universe in which spacetime curvature invariants are the same at every spatial location. The curvature gets weaker ...


0

Carroll is merely matching the Schwarzschild solution to the linearized weak field solution, treated as a consistent truncated Laurent series in $c^{-1}$, cf. this Phys.SE post. The main point is that the spatial components of the metric are subleading in an $c^{-1}$ expansion and may receive non-trivial contributions in order to maintain EFE.


2

However, using Newtonian mechanics, I calculated the G-field near an infinitesimally thin cylinder with an infinite radius to be $g=2πGσ$, where $σ$ is the surface density of the cylinder. I am really doubtful that GR predicts something far away from the Newtonian mechanics for this problem. There are some pathologies with static gravitating infinite plate ...


0

Addendum: Just another level of technicality The wave equation for the gravitational waves is, $$\left(\frac{1}{c^2} \frac{\partial^2}{\partial^2 t} - \nabla^2\right) \gamma_{\mu\nu} = 2\kappa T$$ just something to look at. $$\left(\frac{1}{c^2} \frac{\partial^2}{\partial^2 t} - \nabla^2\right) \vec{E} = 0$$ Any type of wave equation like this describes a ...


1

In General Relativity, which is our current description of gravity, changes in the gravity field is mediated by gravitational waves. 1. What are gravitational waves: If you have any medium, and you add some energy to it, it starts to oscillate or wiggle. In other words, there is sort of a disturbance in the medium. This is true for air, water and even ...


0

would eventually collapse into a black hole because the mass tends to infinity as well as the plate's spatial dimensions. These seem not good reasons to form a black hole, IMHO. A very large object with low density will not form a black hole. In order to form one you must put a large enough mass into a given distance, in practice you need the Schwarzchild ...


2

It is a fundamental supposition in physics: if you keep a detector satellite up there to detect the light, and if it odes detect light coming out from the hole, then that will be true for all observers; they will all see that the detector has detected something. I don't know the name of the principle, but you may call it something like 'universally agreed ...


0

A better mental picture would be that, after the star's radius shrinks below the Schwarzscild radius, the matter behind the horizon undergoes a complete gravitational collapse, forming the singularity in the center. There is no force "strong enough" to stabilize the star and have it be hidden behind the event horizon. As far as we know, the ...


4

An inertial frame (i.e. a Lorentz frame) is, by definition, one in free fall. In an inertial frame, an object which is not subjected to active forces stays at rest, or moves uniformly in a straight line according to Newton's first law. In a laboratory frame at rest on Earth, we are subject to active forces from the floor, pushing us upwards. So you are ...


0

In the The auto parallel equation, $\Gamma^\alpha_{\beta\gamma}$ is dependent on $x(t)$ alone, Along with it, there's also a quadratic equivalent of $\dot x(t)$, which are not dependent on x(t). And in the Newton's 1st law, the term seeming to be corresponding to the gamma terms in the auto parallel equation depends only on $x(t)$ and does not contain a ...


1

From a particle physics standpoint, the question is usually answered in the following way: Energy curves spacetime the same as mass, because they are the same thing, expressed in different units. This means that as we force more and more energy into a bunch of fundamental particles, there comes a point where their gravitational interactions become as strong ...


1

This is not a rigorous answer but I think when the energy of interaction $\sum_{\langle1,2\rangle}\frac{Gm_1m_2}{r_{12}}$ is comparable to the other scales in the problem, I would consider it relevant. For example, in atoms this is negligible compared the electric potential/kinetic energies, but for the sun and earth, it is not. Then again, energies are only ...


3

Say we drop an object of mass $m$ and it falls owing to its attraction to an object of mass $M$. The force on either object is $$ f = G M m / r^2 $$ where $r$ is the distance between the centre of one object and the centre of the other. Therefore the object whose mass is $m$ gets an acceleration $$ a_m = G M / r^2 . $$ Meanwhile the object of mass $M$ is ...


1

The following answer can be viewed as a more detailed explanation of the comments given by others. The main goal is if one performs the calculations in the original Schwarzschild coordinate system, as Sorey did, how to understand the apparent contradiction. I read through the question and found myself quite confused. I mostly repeated Sorey's calculations ...


0

In case further explanation is needed, here is how to solve such problems: Consider a cubical region submerged in water (or any other fluid): We will denote: H as height of the container $h_1$ as height from top of the container to the upper surface of cube $h_2$ as height from the top container to the bottom surface of cube $\rho_{fluid}$ as density of ...


1

If you're making the assumption $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$ then its very easy Landau, Lifshitz already gave us the ready to substitute formulae ($101.6$ in Vol2) here Because of the above assumption $$\Gamma^{\alpha}_{\mu\nu}=\frac{1}{2}\eta^{\alpha\rho}(\partial_{\mu}h_{\nu\rho}+\partial_{\nu}h_{\mu\rho}-\partial_{\rho}h_{\mu\nu})$$ where ...


0

Yes of course. In the atmosphere, pressure and density fall with altitude. So does temperature. It's all to do with the trade off between thermal and potential energy. Hence lapse rate $=\frac{\text{gravity}}{ \text{specific heat capacity}}$.


1

If I float [...] in outer space, I will measure spacetime to be approximately flat. [...] Strictly speaking, you floating all alone by yourself would not quite be capable of determining the curvature (values of curvature invariants) of the region that contains you, along your trajectory. Instead, you'd need the assistance of at least another four distinct ...


3

The rubber sheet analogy works only as an analogy of what happens. Remember that the universe is 3-dimensional while the trampoline has only 2 dimensions. The dimple stretches the trampoline into the 3rd dimension. In the real 3D universe, the dimple would be in the 4th dimension. Also, you may recall that the earth also makes a small dimple in the sheet. ...


1

The physics of the pseudo- tensor is quite independent of the weak field decomposition of the metric ($g_{\mu \nu}= \eta_{\mu \nu} +h_{\mu \nu}$). So, for the time being, lets forget about the decomposition and focus of the pseudo-tensor itself. The key facts about the quantity $$\tilde{t}^{\mu \nu}=\frac{1}{16\pi} \partial_\alpha \partial_{\beta} \left[(-g)...


3

You are saying "If I float motionless in outer space, I will measure spacetime to be approximately flat.". Let's say this is correct in the voids of intergalactic space. Now when you say "By the equivalence principle, I will get flat spacetime if I am free falling near earth.", this is not correct. Near earth, you are inside Earth's ...


2

The equivalence principle only applies if you use a region of spacetime which is too small to measure any curvature. @Dale I am slightly doubtful about the application of this condition. This statement is misused many times. I think if the observer, as well as the space station within which he floats, freely falls towards the earth, he always measures the ...


8

If I float motionless in outer space, I will measure spacetime to be approximately flat. By equivalence principle, I will get flat spacetime if I am free falling near earth. Is this right? otherwise I can distinguish between the two scenarios. Notice that the curvature of spacetime is a tensor, so it is a covariant quantity. In particular, if it is zero in ...


0

The post by Dilithium Matrix is rather helpful and fairly informative if what you truly care about here is the epistemological (experimental) value of general relativity. No physicist should be too worried about the experimental falsification of general relativity given the many numerous experiments that have been conducted in which it passed with flying ...


2

Science does not allow us to be sure of what things are, but rather of what will happen: scientific knowledge, and "truth", is about consequences, implications, and relations more than it is about "what things really are". The purpose of building theories is to try and describe patterns of cause and consequence, so that we may extrapolate ...


3

I found this question a very apt one to probe what it is we are saying about gravity in general relativity. My answer will agree with one already posted by Charles Francis, but perhaps I can help by spelling out the analysis in more detail. I would begin by constructing or imagining a spacetime diagram. Draw on the diagram the worldline of yourself and of ...


4

The equivalence principle applies only locally. An inertial reference reference frame near the Earth does not extend as far as centre of the Earth (which determines the position of the surface of the Earth). Spacetime is not flat over regions where tidal forces (differences in gravitational motions) can be detected. Compare this to observing that the surface ...


3

The structure you are describing sounds similar to a Dyson sphere, which has been proposed as a method of extracting energy from a star by an extremely advanced civilisation. For distances much larger than the Schwarzschild radius, a black hole has the same gravitational field as a star of the same mass, so the same physics applies*. It is theoretically ...


2

One must distinguish between the common meaning of curvature (called extrinsic curvature) from the mathematical meaning of curvature used in general relativity (intrinsic curvature). Intrinsic curvature can sometimes be represented as extrinsic curvature, but generally speaking this is unhelpful, and it is unfortunate that popular accounts tend to focus on ...


2

Will they both be the same age? No, they will not. The traveling twin will be substantially younger. To calculate the age of each twin simply integrate the metric over their worldline: $$\tau = \int d\tau = \int \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$$ (in units where c=1). This procedure is general, it works for any ...


1

Gravitational time dilation on the surface of the Earth vs. life at infinity is minuscule and has no impact on the Twin Paradox. Time dilation due to acceleration does not play a role in the age difference for the traveling twin: all that matters is that he changes direction after traveling at high speed. Including linear acceleration only confuses the ...


-1

Would it be true to say a person is less massive on the surface of the Earth than they are in space ...? Yes in the frame of the Earth (or in the frame of a distant observer); no in the frame of the person. Would the Earth also be reduced in mass slightly? Yes in the frame of the person (or in the frame of a distant observer); no in the frame of the Earth....


3

Have a look here : Physicists have observed quantized states of matter under the influence of gravity for the first time. Valery Nesvizhevsky of the Institute Laue-Langevin and colleagues found that cold neutrons moving in a gravitational field do not move smoothly but jump from one height to another, as predicted by quantum theory. The finding could be ...


1

Steve Carlip has answered in a more intuitive way the question of where the EM field and where the gravitational field "points" when the source is moving. He a short version on John Baez's physica FAQ http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html and a more technical paper https://arxiv.org/ftp/arxiv/papers/1912/1912.05439....


1

You have essentially discovered the need for field theories. Instead of knowing arbitrarily long sections of the particle's past world lines, the alternative is to specify the field variables on a time-slice.


11

The unique thing about gravity is that it is always attractive. Your intuition would be correct if the gas cloud was made purely of electrons, for example, as then the electromagnetic repulsion would be much greater than the gravitational attraction. But because there are both positive and negative charges in the gas cloud, the electromagnetic forces cancel ...


0

One can assume that as a photon goes up in a gravitational field, it gains potential energy and looses kinetic energy (and frequency). Then a distant observer would say that the system which emitted the photon was vibrating slower than it would in his position high in the field. There will also be a corresponding change in the wavelength. This approach ...


1

Let's call the falling direction $z$ ($-z$ to be exact). The rotation of the ball causes a (small) force perpenticular to the moving direction (so some combination of the $x$ and $y$ direction), due to the difference speed of the air flow to the sides of the ball Magnus effect. As soon as the ball gets velocity components in $x$ or $y$ direction, the force ...


2

No, the mass of the object doesn't change with its location in a gravitational field. Suppose for example a small body with a very elongated eliptical orbit around the sun. If we neglect the perturbation of the planets and other orbital bodies, its orbit can be determined knowing its position and velocity in a given instant. And it is independent of its mass ...


2

The short answer is that in introductory textbooks in GR, the only perturbations considered are those for which $h_{\mu \nu}$ and its derivatives are "small". In other words, not only do we have $|h_{\mu \nu}| \ll 1$ but also $|h_{\mu \nu,\rho}| \ll 1$, $|h_{\mu \nu,\rho \sigma}| \ll 1$, etc. This is not always stated explicitly. The ...


5

You are correct, you cannot assume that $h_{\mu\nu,\gamma} \ll 1$ based only on the fact that $h_{\mu\nu} \ll 1$. The fact that $h_{\mu\nu,\gamma} \ll 1/L$, and that $h_{\mu\nu,\gamma\delta} \ll 1/L^2$, where $L$ is a physical length of interest, are standalone assumptions that together yield the usual weak-field limit. There is another way the weak-field ...


4

It is definitely true that they are two different conditions as a counter example already mentioned in the comments proves. However, the approximation remains valid within a proper physical context. When one deals with weak field approximation, one usually also assumes that velocities of bodies involved in the problem are very small compared to that of the ...


1

Mass is proportional to internal energy in the rest frame by $E=mc^2$. For a massive body in an external potential the internal energy thus its mass does not change. For a bound system of two bodies, such as Earth and Moon, their mutual potential energy, as well their kinetic energy, do contribute to the internal energy of the total system hence to the mass ...


3

I assume that when you say that mass increase in regions of higher gravitational potential, you think that mass increases due to energy. This is the famous result from Special Relativity: $$m = \frac{E}{c^2}$$ But, there is a catch: you cannot use this in this situation. Here's why: Imagine Einstein throws an apple to Newton? What is the energy of the apple? ...


7

A few general comments first. The proper mass of any object is a property characterizing the object, and is invariant of its spatial location with respect to other objects. You might be confused between the weight of an object and its mass, where the weight is proportional to the gravitational force acting on the object. The person in your question has the ...


2

In Newtonian theory mass of an object is an intrinsic property so it does not depend on anything let alone potentials. What you might be alluding to is the mass energy equivalence of special relativity which being only a special case of a more general model, doesn't include gravity. Also potentials in classical physics are arbitrary to an extent because the ...


1

The gravitational force occurs due to curve in spacetime Curvature of spacetime is related to tidal forces. Because gravitational fields are non uniform, tidal forces are always present => spacetime curvature. . A curve in spacetime occurs due to energy yes, but more energy doesn't mean more curvature. Our Moon has more spacetime curvature at its ...


0

Gravitational forces don't depend on the motion of objects (except in extreme situations described in general relativity. If there was any such thing as a perfectly uniform gravitational field then uniform (parallel) relative motions would be maintained. This is Einstein's principle of equivalence, but it only applies in a locality. Here is a figure from my ...


2

Suppose you have two stationary objects and no forces. They don't move. Suppose you run by the same two objects. Now they move in parallel lines. Now add gravity. The stationary objects attract each other. They move toward each other and collide. If you run by them, they move in parabolic curves instead of straight lines.


-1

When a gyroscope is outside of a gravity field, the spinning disc has no weight. To my mind this means that no centrafugal force or any other force is produced so there is no effect. If you had a gyroscope on earth with no flywheel it would be the same thing, would that work !!


Top 50 recent answers are included