Tag Info

0

The distance between successive lines is 1,3,5... making their absolute position 1,4 (1+3), 9 (1+3+5)... This follows from double integration of the equation of motion with a constant force: $$v = \int \frac{F}{m} dt = a\; t\\ x = \int v\; dt = \int a\; t\; dt = \frac12 a t^2$$ So the position increases quadratically with time for constant acceleration ...

2

In the original $f(R)$ gravity, the action is $$S = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) \right] \, d^4x.$$ Here, $\psi$ represents the collection of matter fields present. The Euler-Lagrange equation for the metric resulting from this action is $$f'(R) R_{ab} - \frac{1}{2} f(R) g_{ab} + \left[ g_{ab} ... 1 The short answer is no. This link might help. Gravitational Radiation is to gravity what light is to electromagnetism. It is produced when massive bodies accelerate. You can accelerate any body so as to produce such radiation, but due to the feeble strength of gravity, it is entirely undetectable except when produced by intense astrophysical sources ... 2 No. But be careful about how you define the word vacuum. For example the Reissner-Nordström black hole has a non-zero stress-energy tensor because the SET includes a contribution from the electrostatic field. So although the vacuum contains no matter it isn't really a vacuum. 0 You feel the absence of it when in elevator or on a swing. Maybe they meant you are so used to it you don't consciously feel it. You are built to this particular gravity. Same way a car don't think about its brand or a more famous example a fish doesn't think of water. -1 Forces move objects in spacetime, and due to inertia you feel this. Gravity does not move objects in spacetime (hence it's not a force), instead gravity warps spacetime and you will warp with it provided nothing gets in your way, when something gets in your way (the ground at your feet), then you feel it. The ground at your feet is accelerating upward due ... 1 As said in the comments, there are only 4 known fundamental forces, gravity, strong and weak interaction, and electromagnetism. Since we are restricted to low energy classical mechanics, we can ignore two of them, and we are left only with gravity and electromagnetic forces (which includes, among others, magnetostatic and electrostatic forces). Moreover, ... 3 The stress-energy tensor is a local property. Its value at a point is determined by the properties at that point and not by the surroundings of the point. Consider a hypothetical object with a sharp edge i.e. the density changes discontinuously from a finite value to zero at the edge. In that case the stress-energy tensor will also change discontinuously. ... -2 Is it a continuous change or does it jump to zero? It jumps from zero. Think about a photon. It has an non-zero "active gravitational mass", and it's just an E=hc/λ wave in space. It doesn't have an outside, there is no discontinuity. But put this photon through pair production to create an electron and a positron. and each exists as a standing wave. We can ... 0 Fausto, I'm sorry, but there's a lot wrong here. You need to backtrack, and take one thing at once. For example, gravity is not a force in the Newtonian sense. No work is done on the descending photon. Throw a 511 keV photon into a black hole, and the black hole mass increase is 511keV/c². Conservation of energy applies. The descending photon doesn't gain ... 2 No. Take any 3 bodies, the sun, earth, and moon, for instance. Let's say that the sun is positive and the earth is negative. Then the sun and earth attract. But that means that, since the moon and earth also attract each other, the moon must be positive and be repelled by the sun. This does not happen - in fact, the gravitational attraction between the sun ... 0 I think the statement in Witten's paper, "Quantum Field Theory and the Jones Polynomial" saying the term is topological in the sense it is indeed independent of metric. However, he also mentioned, in order to make sense of this integration, i.e. make it to be a number, you need to choose a trivialization of tangent bundle, i.e. choose a framing. The tricky ... 2 When we use the terms "bending" or "warping" with respect to spacetime and gravity, you have to keep in mind that these words are not being used in a literal way. Since the majority of concepts in General Relativity are far beyond what our experiences allow us to comprehend, we have come up with a few ways of picturing these concepts in our minds, none of ... 5 I'll go into a bit more detail below but, basically, Einstein discovered that space and time can't be separated. They are part of the same thing, and gravity is actually produced be bending and warping this "spacetime," like a heavy ball sitting on a sheet of elastic. Classical (or Newtonian) Physics Time can be considered as a dimension, just like ... 0 I'm not quite sure of your level, so apologies if any of the following is telling you how to suck eggs. When people say informally that there is a vacuum inside a vessel, they most often mean that the pressure exerted by the gas on the inside of the vessel is less than that exerted by the surrounding gas on the outside. The latter is most often the Earth's ... 0 Assume the pontoon is a partially submerged cylinder. If we look at an individual section through the pontoon, we find a partially filled circle: The area of the submerged part (the blue piece) can be calculated as the area of the circle, minus the white "slice of pie", and minus the pink triangle:$$A = \frac{2\theta}{2\pi}\pi r^2 - r\sin\theta \cdot ...

0

I'm having difficulties understanding why a gravitational acceleration can be guaranteed to be locally equivalent to an accelerating frame. Actually, it can't. See section 20 of Relativity: the Special and General Theory where Einstein said this: “We might also think that, regardless of the kind of gravitational field which may be present, we could always ...

3

Electricity does not flow. Charge carriers (electrons) do if subject to an electromagnetic field. When all the charge carriers have flown from $A$ to $B$ such that they produce an electromagnetic field that evens out the initial one, then the system is in equilibrium, which is equivalent to saying that the two points of the conductor have reached the same ...

0

Firstly, when astronauts are in orbit around the Earth, they are effectively in free fall and feel weightlessness because of this, rather than feeling weightless due to being far from the Earth. Quite simply our sense of up and down is indeed solely determined by gravity. The world is round and what is up to me, is down to someone on the other side of the ...

0

If the floor of the elevator is exerting a force on me (due to some external force accelerating it) then this would be very different from a gravitational acceleration that would accelerate each part of my body equally. No, it wouldn't. The two situations are experimentally indistinguishable. That's one of the points of this thought experiment. An even ...

0

The answer is very simple and you may forget everything about lifts, elevators and so on and so forth. The gravitational force (which does exist, I wonder the comments above) is the only interaction in the universe where the dynamics does not depend on the mass of the particle (the so called statement that inertial mass is equivalent to gravitational mass), ...

1

It appears that all you have to do is integrate (6) to obtain: $$\alpha = \int dx \frac{2GM}{c^2}\frac{y}{(x^2+y^2)^\frac{3}{2}} = \frac{2GM}{c^2}\frac{xy}{y^2\sqrt{x^2+y^2}}=\frac{2GM}{c^2}\frac{x}{y\sqrt{x^2+y^2}} \equiv \frac{4GM}{c^2R}$$ So we obtain: $$R=\frac{2y\sqrt{x^2+y^2}}{x}$$

1

I agree with everything John Rennie said but let me just take a slightly different direction. Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show ...

-2

You can see horizontal rain in a tornado. I saw some this week while I was parked in a parking lot, probably an F1 tornado. It ripped trees apart mid-trunk--3-4 foot circumference trees. The roads were littered with trees, it took 1 hour to find a way home because all of the roads were blocked by felled trees--a 5 minute drive usually. When we came out ...

0

Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way. The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular ...

0

Let’s assume that by “swimming in space”, it is meant that there is no gravity whatsoever. A (basic) microscopic view of the pressure as the force exerted by the particles (e.g., molecules) of a fluid on a surface implies that still there will be a pressure on a macroscopic body in the fluid, regardless of external forces, like gravity — although in a static ...

0

Pressure of a gas comes from the average, collective change of momentum as its particles bounce against each other and the walls of anything encountering it. What gravity does is provide a counter-pressure - the weight of the gas itself - which stops it from expanding freely into space. The weight of the air above Earth's surface averages about 101,325 ...

2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...

1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...

11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...

-2

You are standing on surface of a planet of mass $M$ and radius $R$. With which velocity $v$ you need to throw away from the planet an object of mass $m$ that it will not return? The gravitational force is $F=G\frac{mM}{r^2}$. The work to be done to move the object in the gravitational field of the planet from distance R to infinity is $A = ... -4 There is an amusing "trick" (so is not meant as an answer to your question) in Newtonian gravity (plus a bit of special relativity) to derive that the Schwarzschild radius should be "special" Since self-gravitational energy of massive bodies is always negative (at least in Newtonian gravity) $$- \frac{G M^2}{R}$$ And massive bodies have positive rest ... 5 The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences. Start with a 2d plane, where we identify positions of points by their$(x, y)$coordinates. Suppose you move a distance$dx$then a distance$dy$, then the distance from your starting point,$ds$, is given by ... 1 Objects moving in an expanding universe experience an apparent drag force called the Hubble drag and given by: $$D = 2\frac{\dot{a}}{a}\mathbf{u}$$ where$\mathbf{u}$is the comoving velocity and$a$is the scale factor. This isn't a real force, it's the result of the universe expanding away from the moving object, but the end result is that in an ... 2 Is it possible that universe is not expanding but instead being dragged into singularity? Yes. That possibility is called the Big Rip. 'dragged into a singularity' can happen even while 'expanding', so they are not contradictory. It would occur if the rate of cosmic acceleration is exponential. In the Big Rip scenario, there is a finite time in the future ... 1 The scientific paper linked in the article you read is saying that it seems that there's a clustering of mass beyond the reach of current all-sky galaxy redshift surveys that is pulling the local volume more to one side. Such clusterings are possible in the framework of the standard$\Lambda$CDM cosmology, but a clustering large enough to explain the data is ... 0 First let me say that wormholes, event horizons and many other possible effects associated with collapsed stars are currently purely theoretical. We think these effects and other associated events may result from the collapse of a star with a mass in excess of the Chandrasekhar Limit We are fairly sure we are on the right track as to what what those effects ... 0 Your questions are indeed related. First let's describe spacetime as like a stack of photos, the photos being different times. Real things are in some region of one of the photos and in a possibly different region of a different photo but don't jump discontinuously from one region to another in between two individual photos. Note this is an analogy and lots ... -3 The answer is that it decreases your net density. The person has a density of 998 Kg/m3. The parachute has a mass of 7Kg so that increases the Kg by 7 and the m3 by 0.2*0.3*0.3 = 0.05 cubed New density =1006Kg per 1.07 m3 = 958.09 Kg per meters cubed. Next we add the air in the parachute at 1.225 Kg/meters cubed The volume =1/2 x 4/3 pi r cubed approximately ... 2 Is it possible that universe is not expanding but instead being dragged into singularity? No. We see galactic redshift every where we look, the galaxies are moving apart like the raisin-cake analogy. There is no overall gravitational field in the universe. You may have heard about "the big crunch" but I'm afraid it's popscience. The universe didn't contract ... 1 Hydrostatic pressure inside a pool on earth is given by:$p=ρgh +p_{atm}$, (g: gravity, h: depth, ρ: fluid density,$p_{atm}$: atmospheric pressure) Assuming 0 gravity and no atmosphere, there would be no pressure. You would feel no pressure at all whether you are on the surface or in the center of the sphere. Also, you wouldn't be able to float to the ... 11 Best answer is given from xkcd's what-if: https://what-if.xkcd.com/124/. It's not space, but it describes the fluid flow in lower gravity - such as how you could jump out of the pool just by performing aquadynamic maneuvers, or walk on the water. It is really a cool read. As mentioned in the xkcd article - diving and floating, being primarily about ... 0 Mainly because of surface tension, but ofcourse also because of Newtons laws. Water has a Surface tension of 72.8 mN/m, The unit, N/m can also be written kg/s^2 or J/m2 In this case the J/m2 is the most practical. The Smaller the hitting angle of the stone makes this area bigger providing more Energy to be returned on elastic collision. There is also ... 0 It doesn't work by using the internal energy of the water - it works by using the surface tension of the water. If we take a water droplet of radius$r$the pressure inside the droplet is: $$P = \frac{2\gamma}{r}$$ where$\gamma$is the surface tension of the water (about 0.073 N/m at room temp). Similarly if we take a column of water of height$h$the ... 0 When people talk about curvature of space, they are usually talking about two related ideas: 1) the curvature of world-lines in 3+1-space-time or 2) the curvature of light paths in 3-space. These ideas are related as light behaves as the high-speed, low-mass limit. 1) Consider a 2D surface embedded in a 3D flat space. At any point on the surface, there is a ... 2 The field line mind picture in General Relativity is probably not useful, at least not for me, aside from in very special cases, because The dimension of the pictures you're trying to see at is too high for our everyday spatial intuition to help us much - you're trying to visualize a rank 2,$4\times 4$tensor with 10 independent components (the metric), ... 2 The correct way is to define the reparametrization-invariant action $$S[X] = \int d\tau \sqrt{g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau} }.$$ Note that the choice of$\tau$is arbitrary. The system has a large group of gauge symmetries - those are reparametrizations of the worldline (different choices of$\tau\$). One way ...

5

When space bends, what are the lines that are being bent? The straight ones. For example: say you have two parallel laser beams traveling through space. Photons travel in a "straight line" more or less by definition, i.e. the path of a laser beam is the straightest line humans can produce, so it is a benchmark of sorts. Mathematically two parallel ...

2

If you plot grid-lines on a graph what are the grid-lines? Well, on a typical 2D plot with x increasing to the right and y increasing upward, the horizontal grid-lines are the loci of constant y and the vertical grid-lines are the loci of constant x. In 3D the grid lines are the loci of constant x and y for the ones that run parallel to the z-axis and of ...

1

Fortunately, you don't need any special relativity or general relativity. All you need to describe orbital motions (to a high enough degree for NASA to use) is classical mechanics. The key in classical mechanics is that rotating reference frames are not inertial. The laws of physics as you know them only hold in an inertial frame. Over time scales of a ...

Top 50 recent answers are included