# Tag Info

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The whole thing is much simpler. Communicating at a distance (tele) with gravitational waves would require astronomical or astrophysical bodies to generate those waves. The gravitational force is much weaker than electromagnetic, so it requires much more energy to generate and much bigger 'antennas' to detect them. The grav to EM force ratio is about $10^{-... 0 Although you didn't mention, I suspect under "it would be possible" you understand that there are no engineering advantages/disadvantages. These influence the actual engineering investitions often more as we would think. So, we have only a single difference between them, which is its largest advantage and also largest disadvantage, too: Unshieldability. ... 4 Susskind's original argument doesn't work. Alice just needs Bob to send a message to her saying "I'm still alive!" She doesn't have to illuminate Bob. Of course, it's hard to get a message out from the near-horizon region of a black hole because of the redshift, but there's no theoretical reason that this shouldn't work. Suppose you don't have a ... 2 Although your question isn't posed very clearly I worked on the problem of non-spherical objects sliding/tumbling down a slope and don't mind sharing the following insight with you. Left, a cuboid and right, a sphere of comparable dimensions and mass. The inclination has been chosen deliberately high. Let's look at the forces and torques acting on both ... 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 ... 0 As a star’s fusion fuel runs out the force of gravity takes over causing the star to collapse on itself. The star is rotating prior to the collapse and as it collapses this rotation at the core increases expediently. As the speed of the matter increases and gets closer to the speed of light, it's mass increases as well further fueling the collapse and in ... 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 I am not versed in general relativity but I see a problem with your scheme. Let us take the ensemble of distant stars as our inertial frame. Whenever a body has acceleration$\textbf{a}$w.r.t. inertial frame it experiences an inertial force equal to$\textbf{F}_{inertial}=-m\textbf{a}$, where$m$is body's mass. Now suppose that the large mass$M$in your ... 1 I think your right. Due to equivalence principle, to the free-falling body it will seem like it has no acceleration, as opposed to a body standing on ground which is equivalent to the ground pushing the body so that it will accelerate upward, and hence experience the normal force from the ground. Because the acceleration due to gravity will depend only on ... 3 Primary question: This is very similar to the old question of what would happen if you fell into a black hole. You are correct that you wouldn't feel the acceleration due to gravity per se, but you'd still need to worry about tidal forces. These have complicated geometric dependence - they're negligible near the center of a planar mass$M$, and for a ... 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 ... 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 ... 5 These notes put some numbers on @ACuriousMind 's answer: one needs to be looking at length scales of 100 Mpc and greater for the FLRW metric to be a realistic description of reality. That's a staggering distance, and equivalent to timescales amounting to the whole Mesozoic era, comprising the rise and fall of the Dinosaurs! So one cannot expect the ... 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. ... 2 You can only make a fall-directional cut if the trunk is vertical, because then the top part of the the tree is (roughly speaking) in an unstable gravitational equilibrium when cut and it can topple to either site. Here details such as the order of cuts makes a difference. In your case, however, the individual pieces of the cut are far from a gravitational ... 0 This is a complicated problem, it is not easily determined what forces are on the roots, and the scale is unclear. I would make the fall directional cut (or scarf) on the underside of the trunk, and the break cut on the topside, slightly higher (as the tree grew) up the trunk. I think this is the answer because the roots are likely bound in place by ... 3 The given answer is bad, and your version is correct. Newtons gravitational formula is very clear, and this is a trivial substitution in it, there isn't too many place of a mistake here. -1 They are, as Einstein pointed out, equivalent. So why distinguish between the two? Well the only real difference is that they are measured differently. To measure inertial mass, we exert a given force to something with an unknown mass. To measure gravitational mass we compare the force of gravity from an object with an unknown mass to the gravitational force ... 0 Gravity causes acceleration, but acceleration can happen from a lot of other things as well, for example, on electromagnetic effects. In most cases, the acceleration depends on some charge-like quantity. For example, a body with a mass of 1kg and with a charge of 1C will accelerate faster in the same electric field, as a body with 2kg of mass and with the ... 0 Applying Gauss at a distance$r$from the centre of the sphere you get$g(r)\;4\pi r^2 = - 4 \pi G M_{\text{enclosed}} \Rightarrow g = - \dfrac{GM_{\text{enclosed}}}{r^2}$where$g$is the gravitational field. So all you need to do is to find$M_{\text{enclosed}}$for each of the regions. What this equation tells you is that you can treat the enclosed mass$...

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This is a surprisingly complicated question, and I'm not sure there is a universally accepted answer. To see why this is turn off your magnetic field and give the electron enough velocity to keep it in orbit around the Earth. Now in the Earth frame the electron has a centripetal acceleration of $r\omega^2$ and therefore it should be emitting radiation. ...

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Yes, I also never liked the visualization with the 2D-plane and the ball. It is not even partially true. I think there is no possible way to visualize the mathematical and physical effects, because its mathematical formulation is so complicated that you wont ever have a 100% true visualization. But maybe this picture of a parralel transport of a vector on ...

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Visualization is a very personal thing and you must choose what works for you. Analogies can be good, bad but never wrong and science has always used analogies heavily to take its first steps into any field. In summary you need to ask: Is a visualization helpful or useful? and, in GTR, I'm strongly of the opinion that all everyday visualizations like balls ...

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Spacetime is four dimensional (three spatial dimensions and time) and hence so is gravity (as obtained from the metric tensor of spacetime) and we just can't visualize 4D spaces (much less spacetime!) so the best you can do is either 3 spatial dimensions (or with a timesliced video so you can view how gravity changes as a function of time) or 2 spatial ...

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I've included a couple of pictures that are a three-dimensional warping of spacetime. Obviously, these are artist's and mathematician's depictions, but perhaps they'll give you a better idea. Image 1 This image shows a ball (representing a massive object) warping spacetime around it. In your question, you mentioned seeing a massive object warping a two ...

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Have a read through the answers to How can you accelerate without moving? and If $F=ma$, how do can we experience both gravity and a normal force even though we are not accelerating as these explain in some details exactly what is meant by acceleration in relativity. If you are falling freely then by definition you are weightless i.e. your proper ...

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In Newtons three laws of motion, its the second law that introduces the concept of mass, here its the linking term between force applied on an object and the motion or acceleration that results. The more 'stuff' there in the object, that is the more mass it has, the harder it is to accelerate it. That is, it has more inertia; so we call this concept of mass, ...

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First of all note that human muscles require energy even when doing no work, so be a bit cautious about analysing any situation involving humans manipulating objects. See Why does holding something up cost energy while no work is being done? for more details. If we replace the human by some form of mechanical cantilever then you are quite correct that the ...

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Einstein had his "happiest thought" about his equivalence principle one day as he imagined himself in freefall towards the earth. A typical mind wouldve thought "okay, so i feel as though im floating" but Einstein and his fabulous mind thought about it a different way. Einstein connected his freefall with being in motion outside of a gravitational source ...

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A curious result of the physics involved is that the dropped mass oscillating up and down through the hole (say from North Pole to South Pole and back) would be matched exactly by a mass in a polar circular orbit at ground level. Note that the max velocity in the answer above is the same as the circular orbital velocity. If the object were dropped at ...

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The gravitational mass, $m_g$, gives you the strength of the gravitational interaction while the inertial mass, $m_i$, represents the inertia of the body. The first one is the mass appearing in the Universal Gravitation Law while the second one is the mass appearing in the Newton's second law. The equality between these masses is an empirical fact noticed ...

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This is a surprisingly simple thing to calculate. It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count." The acceleration of ...

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As the question already beyond exact science, I will take a stab at it. There are three components here - Kinetic Energy (KE), Energy of gravitational waves (GW), and lost mass in merger. KE and GW - As the GW are ripples in space (time), they have to be generated by motion and/or its disruption (i.e stoppage at moment of merger). Therefore, the energy ...

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Let me try to explain this by making an analogy with a simpler system i.e. a hydrogen atom. If you measure the mass of a hydrogen atom you find it is less than the mass of an electron plus the mass of a proton. In fact it is 13.6eV less. This happens because if you let a separated electron and proton fall together under their mutual electrostatic attraction ...

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You've forgotten an important player in the system: the gravitational field. Here's a pretty argument that gravitational fields are physically meaningful objects that carry energy: imagine two masses accelerating towards each other from rest, from a great distance away. The rest energy of the system is $E_\text{rest} = (m_1+m_2)c^2$; the kinetic energy is ...

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Image object $m$ is at some point $a$ and you were to supply a force opposite to the gravitational force caused by $M$. This force is equal to $F_{stop}=-F_g$ so that $m$ hovers completely still at point $a$. M ------------------- a ↑ m Obviously this force can't do any work because the object doesn't move and ...

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Saying that work done is +ve or -ve is a mathematical convention used for calculating energy transfers. It is +ve when it is done by us on the system, and -ve when it is done by the system on us. Positive work is done in pushing against a force to reach a configuration - eg pushing a car uphill. Negative work is done if a force pulls in the direction we ...

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The object would experience a net negative force, and be moved a negative displacement. Potential is defined in terms of the work done by an external force. The object has a negative force acting on it due to the gravitational attraction so the external force acting on the object must be in the positive direction to have a net zero force on the object. It ...

1

The work done by an external agency in bringing the body to a point is needed. Note that the process must be quasistatic, otherwise kinetic energy terms will be needed. Now gravity will tend to attract a mass. Thus to keep the process quasistatic, one must oppose this gravitational attraction. In other words, F is directed oposite to the gravitational ...

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No. Acceleration and velocity have different units, so their magnitudes cannot be compared. Whether one is larger than the other numerically depends on what units are used. Acceleration and velocity can have the same direction, but this is not necessary - eg a ball thrown upwards has upward velocity but downward acceleration, until it reaches maximum ...

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Gravity acceleration is... acceleration, measured in $\mathrm{m/s^2}$. It is the rate of change of the velocity.. Velocity is measured in $\rm m/s$. It is the rate of change of the position. The vertical velocity and the acceleration due to gravity of a body are collinear but they can have different magnitudes as well different orientations. Think about a ...

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It is an interesting idea and in principle you could indeed measure mass through the time dilation caused by the curvature of space-time. Experiments to measure the time dilation caused by large masses (the Earth) have also already been performed. But the focus of such a measurement is not the determination of the mass, but simply the confirmation of the ...

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The time dilation caused by the entire mass of the Earth was immeasurable until the invention of the atomic clock. The time dilation caused by a $1$ kg mass remains immeasurable today, and the time dilation caused by an atom of $^{12}$C weighing roughly $2 \times 10^{-26}$ kg is never going to be measurable. That's why we don't measure mass in time dilation ...

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I believe I understand what you are asking and the answer is that the observer does not see an accurate representation of what happens near the event horizon. In the first graph the line would curve up to infinity at the end because as the object fell into the hole the light would be slowed to a stop at the event horizon. As the object begins it's approach ...

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Actually, the critical mass is not affected by external gravitational fields. Gravity does not effect nuclear reactions. And, for the most part, gravity does not effect even chemical reactions either. Intermolecular forces (electromagnetic in nature) are vastly stronger than gravitational forces. Rather, the critical mass has to do with the effective ...

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