# Tag Info

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Remember that a curved path in spacetime is not necessarily a curved path in space. An object dropped to Earth from height $y_0$ has a height at time $t$ that's calculated as $y(t) = y_0 - 9.8 t^2$. An object dropped on the Moon from the same height follows the path $y'(t) = y_0 - 1.67 t^2$. If you plot the graphs of these two functions you'll see that they'...

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The commonly used image of a "bowling ball in a trampoline" model of curved spacetime is very deceptive because it implies a curved fall path, as you have described. I prefer to look at gravity as operating exactly like a low pressure system in the weather,. You can see below a typical weather low pressure system, and an image of the Earth as a ...

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Let's consider a gamma ray emitted from the surface of a $1.4$ solar mass neutron star, assuming the radius is 10 km (on the low end of what is consistent with observations from LIGO and NICER). Let's also assume the spin is small in units of the mass of the object, (a) for simplicity and (b) consistent with observations of known neutron stars. The redshift ...

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An object whose path is a straight line in the three spatial dimensions of spacetime can still be following a curve in the time dimension if it is accelerating. This is what happens when an object falls “straight down” under gravity. If an object followed a straight line in space and time then it would be moving in a straight line in space at a constant ...

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I think you are taking the term curvature too literal here. While general relativity says objects on which no forces (besides gravity) act, follow geodesics, which can be curved or more complicated paths, nothing prevents a geodesic from being a straight line in some special cases. This also does NOT mean that the geodesic is "extremely bent" (you ...

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In both cases, you break the force into two components, one in one direction and one at 90 degrees to it. However, in some situations, you can easily prove that one of those components is zero. In those cases, we skip a few steps and say "it just has one component." In the case of the object sliding on an incline, you have quietly made the choice ...

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It's only a trick for the "car on an incline" problem. One makes the incline problem simpler by rotating the coordinate system so that the incline is flat. This has the consequence of changing "down" from being just simply pointing in the negative y direction to pointing at an angle. But, gravity still points downward, our new ...

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Nothing on the other side of our cosmological event horizon can affect us gravitationally (or in any other way) by the definition of event horizon. The idea that virtual particles somehow change this is widespread, but wrong. There is no violation of light-cone causality in quantum field theory, nor in quantum gravity as far as anyone knows. Any faster-than-...

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A large mass A close to (but outside) the observable universe (OV) horizon can certainly influence the motion of an object B just inside the OV. Photons and gravitons from B can certainly influence Earth (if Earth's survival lasts long enough). Therefore, it seems obvious that A can also influence Earth (if Earth's survival lasts long enough).

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At present gravity has not been definitively quantized, and on cosmic scales it is the theory of Genera relativity that holds, and gravitational waves travel with the velocity of light at maximum. So if it is outside the cosmic event horizon which is moving faster than light , the masses out there cannot be detected the same way that the light from outside ...

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The equivalence principle dictates that objects in freefall in a gravitational field are not considered to be accelerating. To the extent that the gravitational field is uniform, two clocks, one above the other, will measure time at the same rate. This rate will be slower than a reference clock far above and outside the gravitational field. In practice, ...

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Virtual gravitons (inside the condensate of gravitons that constitutes the gravitational field) can travel faster than the speed of light. So you should be able to feel the presence of the big mass on Earth, even if it's behind the horizon. Real photons though can't travel faster than the speed of light, so the mass will not be visible. Gravitational waves (...

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Time dilation does not depend upon acceleration; this is has been experimentally verified in particle accelerators. So whether the clocks are undergoing acceleration or not does not affect their rates relative to distant clocks. The answer to your question therefore is that clocks closer to the center of the Earth undergo time dilation relative to clocks ...

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If you intend to place the clocks very close to each other, it does not make sense to speak about the gravitational time dilation. However, if you place the freely falling clocks a meter or a kilometer away from each other, the time dilation becomes tangible. Since the gravitational time dilation depends on the gravitational potential of the point at which ...

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According to mainstream cosmology even case $A$ can be an accelerating universe. If there is exponential growth of the scale factor $a$ as in case $A$, $a=a_0e^{Ht}$ where $H$ is the Hubble parameter, assumed constant here. the decelaration parameter $q$ https://en.wikipedia.org/wiki/Deceleration_parameter works out negative even for a constant $H$. Negative ...

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Negative mass is not a thing. Also, time begins at the singularity of a white hole so there is no sense in talking about "creating a white hole" (since that would require discussion of something occuring "before" a white hole).

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We'll define up as positive. So if a numerical value is supposed to be upwards/downwards, and we actually drew the arrow as upwards/downwards it will be positive. If the actual direction of the numerical value is the opposite of the direction we have made our arrow it will be negative. Normally, we would define our arrows for all known quantities in the ...

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I'm going to give you one quick answer that will help you to understand. Gravity bends spacetime so that EVERYTHING falls, from apples to light itself. You have probably heard that if you shoot a bullet horizontally and drop a bullet, they will both hit the ground at the same time. So from this, you can understand that if you had a really really big, ...

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The difference between photons and gravitons is in their probabilities to be emitted. Emmitting gravitons is highly improbable. But if one speaks of the attraction force, it is always here, (Newton law $F\propto G\cdot m_1 m_2/r^2$). Nevertheless, in atomic problems this force is still too small, and it never is taken into account.

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In quantum physics, gravitational force is exerted by two massive particles emitting gravitons and receiving each other's gravitons. The gravitons exist only while in transit between the two masses, and travel at (or very close to) the speed of light. The graviton is one of several force-carrying particles, collectively known as bosons. The others include ...

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As a non-physicist, I imagine that if we have an atom of lead and an atom of copper floating near to each other in deep space with no initial relative velocities, they will exert a gravitational pull on each other and accelerate towards each other. Yes this is right. Am I right? If so, where are the gravitons that cause this? Are there some in each atom? ...

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Let's say we "drop" two electrons in an uniform electric field. The distance between electrons does not change as the electrons "fall". For many people that have studied relativity this is just very weird, and impossible to understand. This is the Bell's spaceship paradox. Let's say we drop two masses in an uniform gravity field. The ...

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A falling rod length-contracts without any stress. When the coordinate speed of the rod is 0.86 * coordinate speed of a light pulse next to the rod, then the length of the rod is half of its rest-length. Accelerating observer observing a rod, observes the rod length-contracting without any stress. When the coordinate speed of the rod is 0.86 * coordinate ...

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To take an extreme example, keep increasing the speed of the block until it reaches orbital velocity. At that point the block is in orbit (even though it's right next to the surface, even touching the surface -- remember, you postulated no friction and no air resistance). For a body in orbit the normal force is 0. So the two extremes are normal force equal ...

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It's not about what causes the gravitational force to be greater. It's that the normal force will be less (these two statements are the same mathematically, but colloquially they change the focus). Let's look at a simpler example. Take a box and put it on the floor. We can say the normal force acting on the box is equal to its weight. Now let's apply an ...

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You treated the problem as if spacetime is Newtonian (non-curved). In reality, spacetime isn't Newtonian. If you approach the problem in a general relativistic way, you'll find that the orbits will be perfectly circular. The general relativistic calculation is complicated but you can intuitively state that the extra speed gained by the retardation (in the ...

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There's no contradiction here. You would observe a tangential acceleration, which would mean that the radius of the circle would increase, and the configuration would fly apart. If nothing else, using the retarded time makes the particles slightly farther apart than they would be using instantaneous time, which would make the gravitational force slightly ...

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When the photons are contained in a massless perfectly reflecting sphere, then the two masses (one of which is now contained inside the sphere in the form of photons) will still be accelerating towards each other. If the spare is small enough the acceleration will be the same as for the two earlier masses. If there is no sphere to contain the photons, the ...

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$E=mc^2$ is a formula coming from the mathematical constraints of special relativity. M1 accelerates towards M2 Is an expression valid in Newtonian mechanics. M1 converts totally into energy (which I believe would be photons) since you are talking about photons, they belong to quantum mechanics frame , where nothing spontaneously disappears without ...

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The theory of relativistic elasticity is one approach to extended bodies in GR. Carter & Quintana (1972) list two motivating applications: (a) the interaction of gravitational radiation with planetary type bodies such as the Earth, and (b) the vibrations and deformations of neutron star crusts. For rigid bodies, even the simplest examples are very ...

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You are describing LIGO: a beam of photons is sent through an empty region and allowed to interfere with a reference beam. When gravitational waves arrive they cause a shift of the interference fringes. You may want the wave to be fully inside the test chamber rather than affecting it as a whole, but note that the effect of the wave - periodic changes in the ...

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In your notation you are mixing up the integration of v. For the state space representation one would write: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$ This expresses the equations of motion of a ...

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The problem is not the particles, it is that one cannot control and focus a gravitational wave. The gravitational waves detected by LIGO, the first evidence of gravitational waves, were coming from the merging of two black holes . There is no way to detect or predict such occurrences, and the LIGO "screen" is continents long. Gravitons are another ...

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I don't think you can produce gravitational waves at will. At least, strong enough to be measurable. Also, the setup used to measure gravitational waves is extremely sensitive and precise, because they are very weak. Noticing if a gravitational wave has any impact on a beam of particles is basically impossible at the moment, considering on top of everything ...

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Whenever you have an attractive, conservative force between objects, the potential energy is smaller when the objects are closer together. Potential energy is energy and energy is mass, so the gravitational mass of the system is reduced when the objects are moved closer together. This is true regardless of the details of the force (and the Casimir force is ...

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The fact that a projectile lacks sufficient energy to orbit the earth, and will crash back into the earth, does not change the fact that the trajectory is actually an ellipse (assuming ideal conditions, such as no air resistance etc.). It is just an ellipse that intersects with the earth's surface. However, when the trajectory is small relative to the size ...

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Let me first go back to the theory of electromagnetism. I will first discuss the context of electromagnetism, and then, using that as starting point I will move to the context of general relativity. Maxwell's theory has been superseded by quantum theory, but as we know: Quantum theory should be thought of as a superset of Maxwell theory. Maxwell theory has ...

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Does the Earth really go around the Sun, or is this just a mathematical convenience? For some purposes we use a geocentric model and say "the sun rises in the East"; but creating a geocentric model which accurately reproduces all observations would be hideously difficult, and rather pointless, since the heliocentric model exists, gives accurate ...

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Assume you measure the length of each ruler by how long it takes a photon to travel from one end to the other. Due to time dilation near the black hole, the photon (as measured far away from the black hole), would take longer to traverse the ruler. This can be interpreted as the ruler near the black hole being longer (space expanding) or equivalently as time ...

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Mass and energy are equivalent, and affect gravity in the same way, so yes, you could produce gravity by putting enough energy on your spaceship. But then you're not producing "artificial" gravity, it would be real gravity, and that energy would have inertia just the same as the equivalent mass would. In fact matter is probably the most compact ...

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Is there any way to determine which spacetime relates to ours(2D) No: all these spaces are (globally) homeomorphic. They are not isotopic, but that is a property of embedded manifolds, not of all of spacetime, and general relativity depends only on the intrinsic geometry of spacetime. Can I find any physical observable in QFTs on such spacetime that is ...

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While it is true that the energy content of something strengthens its gravitation, this is just another way of saying that a more energetic object weighs slightly more than the same object in a lower energy condition. As such, these facts don't furnish us with any tools with which to create "artificial gravity".

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The action of $\Box$ on $x^\mu$ is defined by treating $x^\mu$ as a scalar (even though its not). Then, \begin{align} \Box x^\mu &= g^{\alpha\beta} \nabla_\alpha \nabla_\beta x^\mu \\ &= g^{\alpha\beta} [ \partial_\alpha \partial_\beta x^\mu - \Gamma^\lambda_{\alpha\beta} \partial_\lambda x^\mu ] \\ &= - g^{\alpha\beta} \Gamma^\lambda_{\alpha\...

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Integrating twice based on acceleration due to gravity. (Keplar could also be used, but the result is the same). m1 = mass of earth m2 = mass of object, so much smaller than mass of earth it can be ignored gravitational constant G = 6.6743 x 10^-11 m1+m2 = 5.9722 x 10^24 r is distance between the two objects v is combined closure rate of the two objects ...

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I think that a simpler example is two objects rotating in a circle, like the top and bottom stars in your diagram. This should be an allowable motion, but if we use a delayed gravity approach the motion cannot be maintained. That's why simply introducing a retarded gravity is not a solution to Newton's instantaneous action at a distance.

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QFTs are definitely sensitive to the topology of spacetime. As a matter of principle, you can of course measure this topology. You can even do it classically: get on a rocket and keep going; if the universe turns out to be a torus, you'll eventually come back home (and fly arbitrarily close to Henri Poincaré, infinitely many times). In practice, of course, ...

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When discussing things in general relativity, one generally strives to describe physical phenomena in a way independent of arbitrary coordinate choices, so that one can be sure they're extracting real physical information about the model rather than coordinate artifacts. Forces on point masses can be described in a relativistic setting in keeping with this ...

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If you want to figure this out "without calculus", there's a way you can get a good numerical approximation using a computer. If you take a "sufficiently small" interval of time $\tau$, then the acceleration and the velocity of the object will be approximately constant over that time period. This means that you can use the uniform ...

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Your argument is inconsistent because the concept of the center of mass, as used by you, implies instantaneous interaction. If you want to use it with retarded interaction, then the center of mass has to refer to the retarded position of the star at the top, not the instantaneous one (after all, that's what the stars interact with). So you have to move the ...

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I will only outline an answer for now. The object is on a suborbital trajectory. Classically this just means the objects or it is a 2 body gravitational orbit like earth and moon except the periapsis is lower altitude then the surface of the earth. Describing orbital characteristics is the same. Now there is usually no time parameterization of the orbit. One ...

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