New answers tagged

0

The lower case $\epsilon$, the way he defined it, is the dimensionless density parameter. The (first) Friedmann equation (neglecting the curvature term) is: $H^2 = \frac{8\pi G}{3}\rho$ Then denoting the Hubble parameter today as $H_0$ and the mean matter density today as $\rho_0$, we can see that $\epsilon=G\rho_0/H_0^2$ is a dimensionless parameter. Please ...


0

After a bit more than a year of studying the topic, I think I can answer my own question. The Post-Minkowskian approximation (PMA) starts with Minkowski spacetime at zeroth order and uses a weak gravity approximation in powers of $G$. This means that particles are alowed to move at relativistc speed and even have zero rest mass. However, since the zeroth ...


0

It sounds like you want to build a model of a solar system in a transparent box. There are devices which can levitate magnetic spheres, but generally these involve complex field configurations which offer only one location for a stable equilibrium at a small height. You might consider a simulation on a flat surface. My wife has a decorative piece which ...


0

Energy and mass are equivalent. If the electric field stores energy, then it stores mass. Weight is proportional to mass, other things being equal, so the electric field will contribute to the weight of the capacitor. Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal. It's really ...


1

It sounds like you want to build a simulated solar system using magnets instead of gravity. Your plan is to use magnetism as the attracting force, and then centrifugal force of planets spinning around the sun to offset and keep the planets in orbit at the appropriate distance. Is this what you want? If so, it would be extremely difficult, if not ...


0

This was tested experimentally by Kreuzer, Phys. Rev. 169 (1968) 1007. The "capacitors" in this experiment were actually atomic nuclei. The electric field of a heavy nucleus makes a fairly significant contribution to its own weight, which is easy to measure. Kreuzer used a Cavendish balance to test whether this was also true for the active ...


-3

IMO No. The electromagnetic energy from the source is used to coherently rearrange the existing charges in the circuit. It does not add more matter or energy in this open system. The electric potential energy of the source is transformed in kinetic energy of the moving charges and all of this energy is expelled out of the system as heat. After the capacitor ...


7

Ignoring your first paragraph because magic portals are governed by the whim of fantasy authors, not the laws of physics. Inertial (velocity-based, non-accelerating) time dilation goes as: $\Delta t' = \gamma\Delta t = \frac{\Delta t}{\sqrt{1-v^2/c^2}}$ where $\Delta t'$ is the elapsed time measured by a clock in the observed frame and $\Delta t$ is the ...


7

Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance. Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be $$ U = \frac12 C V^2 = \frac12\times10^6 \rm\,J $$ This is an awful lot of energy for a capacitor, ...


8

I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth. General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) ...


1

What you have to keep in mind is that General Relativity is built upon the Equivalence Principle. The strong version of this principle is that locally, whichever spacetime you're working with will be Minkowski spacetime, i.e. it will be locally flat. That's the key aspect of the (strong) Equivalence Principle, it is local. Therefore a local observer will not ...


1

How did even the ancients discover that the earth is spherical ? ? By using a mathematical model, that showed compared to measurements that the earth was not flat. In the same way, we are able to compare data with two models, the Newtonian gravity one and the General Relativity one , and see experimentally that the GR curvature exists.GPS is used by ...


2

So, how would an object or observer inside the distorted bit of space be able to tell that they had been stretched or squashed? It is important to understand that spacetime curves, which is not the same sort of deformation as stretching or compressing. Curved spacetime can lead to compressed or stretched matter as different parts of an extended object try ...


3

in the accelerated frame there is nothing that push us down. Actually this is not true! Newton's laws only hold in inertial (non-accelerating) reference frames. To use $F=ma$ in an accelerating frame, one needs to include so-called fictitious forces. The name is somewhat unfortunate; it means that there is no force from the point of view of an inertial ...


6

How can two things that are different be equal? That is essentially what all of physics is about. From $F=ma$ to $E=mc^2$, pretty much all of physics is about finding equalities between seemingly different things. The fact that two things are different in no way prevents them from being equal. It just means that there is a connection of some sort, possibly ...


2

Einstein thought of it in a different way: In a gravitational field (if we are not accelerating down), there is something pushing us up. In the accelerated frame, there is something that pushes us up. This image from researchgate.net shows the equivalence between the earth's gravitational field and a rocket accelerating at 'g'. The arrow on the ball on the ...


1

After accelerating, your spaceship has some velocity vector at some location (and it doesn't matter whether that is perigee, apogee, or any other point in the original orbit). By the very assumption of your problem, the spaceship is (still) on a closed, elliptical orbit. After one period, it therefore will be back at this location with that velocity. There ...


2

There are two conservation laws at work. By applying a tangential boost to the velocity you have applied a torque to the object and increased its angular momentum to $v_p r_p$ where $v_p$ is the new tangential velocity at perigee. This angular momentum must then be conserved such that at apogee $v_a r_a = v_p r_p$. Conservation of energy then means that when ...


3

It's a coincidence. There is no particular relationship between a planet's surface gravity (which depends on the planet's mass and radius) and the length of its year (which depends on the Sun's mass and the orbital radius). The agreement is not quite as good as you say: in fact $g\cdot 1\text{ yr} \approx 1.03c$. Also, this is not really the speed you would ...


1

The Moon has the same length of year as the Earth, as indeed do the various bits of space rubble in the L4 and L5 Lagrange points, and all these objects have different values for the gravitational acceleration at their surface. So there is nothing special about multiplying the surface gravity by the orbital period. The result you have obtained is just a ...


1

For an initial circular orbit, force balance shows that: $\frac{v^{2}}{r} = \frac{GM}{r^{2}}$ You basically ask, what if $\frac{v^{2}}{r} < \frac{GM}{r^{2}}$ for a particle with distance $r$ from the center of the Earth of mass M and tangential velocity $v$. This is a similar qualitative description as if the velocity suddenly slowed. The object would be ...


0

No, not in air. Helium is good. Hydrogen is better. Warm hydrogen has an even lower density and is thus even better. Vacuum is as low as it gets, but you need a massive hull to avoid crushing. All that matters for buoyancy is the difference in density between the inside of your balloon and the outside. So, take an air balloon and put it in water, and it'll ...


2

While it is true that sometimes weight and mass are referred to both with the same units (kg, grams etc.) this is technically incorrect (there really is no "technical versus non-technical definitions", but you are right in that the terms are frequently misused). Mass and weight are two distinct quantities. Mass is a measure of the total amount of ...


2

Newton isn't wrong at all. It is as you write: Einstein's GR simplifies to Newton's law of gravity in the respective non-relativistic limit. But no, we still can't come up with an algebraic expression for $G$ because that constant also appears in Einstein's equations and simply carries through to the Newtonian limit.


1

I interpret the question as: "Can you orbit an accelerating black hole?" The answer to this question is: That depends on the acceleration. There is a well known analytic solution to the vacuum Einstein equation, known as the "C-metric". In this metric the acceleration is power not by a laser (as in the OP's question), but by a conic ...


1

Spaghettification what will happen to an astronaut free falling into a black hole, is the vertical stretching and horizontal compression of objects into long thin shapes (rather like spaghetti) in a very strong non-homogeneous gravitational field; it is caused by extreme tidal forces. In the most extreme cases, near black holes, the stretching is so ...


1

Yes, we can move a spaceship by moving a black hole, but the best approach is not the one you described. The idea of using a black hole to power a spacecraft has been seriously proposed and studied in some detail: https://arxiv.org/abs/0908.1803 From a practical standpoint the size of the black hole is very small and the size of the spacecraft is relatively ...


1

Pushing the black hole means pushing both the ship and the black hole, because the ship is part of the system. That uses more energy than just pushing the ship itself with lasers (or whatever). You've just added extra mass to push by adding the black hole.


0

A simple and reasonably accurate approach: Consider a photon rising in a gravitational field. It gains gravitational energy (with m = hf/$c^2$) and loses kinetic energy (hΔf). A distant observer would say that the emitting atom was vibrating atom was vibrating at a lower rate.


0

"I picture a box that contains a particle that travels back and forth at the speed of light." Very good! "[...] We place this box on the surface of a large planet. The particles run upwards and down, perpendicular to the surface." As far as the relations between constituents of a box being held rigidly wrt. a mass can be approximated ...


2

I've read that if an object in space is moving in a relatively straight line, it will begin to follow a curved path around a nearby planet This is true in Newtonian Mechanics also, as seen by the orbits of the planets, and no need to involve "curvature of space", where time is just a parameter. In General Relativity the Newtonian orbits can be ...


1

You are thinking only in 3 dimensions. Remember that physicists all say that we are dealing in spacetime. So you have to start thinking in 4 dimensions by adding time to the equation. While your object is not moving in space, it is moving in time. In fact we travel through time at the speed of light. There are lots of science videos that show this. By ...


0

When the elevator of mass $m$ is pulled by the 'string' then after some time it will acquire some velocity, say $v$. At that point it will have a kinetic energy $K$: $$K=\frac12 mv^2$$ Now you cut the power. Inertia demands that the elevator will continue to travel upwards for 'a bit'. During this period, kinetic energy $K$ is converted to potential energy $...


0

Time dilation is widely misconceived, even in special relativity. You shouldn't think of it as time slowing down- instead, it is a consequence of the geometry of spacetime, in which different paths between two events can have different durations. If you and I take two different paths between two events, such that my path is twelve minutes long, say, and ...


0

The first kind of time dilation Einstein inferred was in special relativity, and due to relative motion. His light clock thought experiment shows frame shifts relate up-down and up-down-but-also-sideways motions: since $c$ is invariant, time dilation is equivalent to the difference in the paths' lengths. We can explain gravitational time dilation with a ...


3

If the speed of light is constant and it is predicted that the rate of time in a gravitational field will decrease, how is this accomplished? Speed is constant, length of time is increased, so that only leaves an increase in particle travel length. Has the box been elongated? Not exactly. The box isn’t simply elongated, the spacetime is curved. It is common ...


1

According to the equivalence principle, for a small box compared with the planet, the situation is like being inside an uniformed accelerated rocket. The time for the EM radiation "goes up" is longer than for "going down". The light speed in the vacuum is a constant everywhere in a Minkowski spacetime.


7

The Hawking effect is an effect associated to black hole spacetimes. Since every stationary black hole solution to the Einstein–Maxwell equations is essentially a rotating, charged black hole due to the no-hair theorems and the metric you provided is not that of a rotating, charged black hole, it does not characterize a black hole within the usual notion of ...


2

The aglet, on its lace, is performing forced oscillations. The periodic movement of your body as you walk causes a periodic force on the top of the lace, making the aglet swing like a pendulum. The oscillations are usually noticeably large only when the frequency of the periodic force from you is quite close to the natural frequency of the aglet and lace. We ...


2

When you walk normally at a steady pace, the motion of your body is fairly periodic. The motion of your arms and upper body for example, closely resembles harmonic motion, and so any object hanging from your body as in this case or a necklace, will move like a pendulum in rhythm with your body's motions.


1

Welcome to the Stack Exchange. I hope you can learn something here. What you have asked is pretty close to what is accepted, just backwards. Mass slows down the passage of time rather than speeding it up. As you know, it is normally considered that the warping of spacetime is what creates gravity. Near a small planet like Earth, 99.999% of that warping is ...


2

No, this doesn't work. For starters, we have measured the change in "speed of time" as you call it, as well as the change of space, and even both together, in a long list of experiments confirming the theories of special and general relativity. And what we find, in perfect agreement with what is predicted by general relativity, is that clocks that ...


1

If we presume a circular orbit for the moon, then the acceleration has a constant magnitude, from your question, of 0.0027 m/s^2, although the direction of the acceleration would be changing. If the moon were stopped relative to the earth, then released, the initial magnitude of the acceleration would be the same as it depends on the distance from the earth $...


0

The moon orbits Earth because of Centrifugal force balancing the gravitational force by Earth. And centrifugal force depends on the tangential velocity of the moon $$F_c = m\omega^2r = m\frac{v^2}{r} $$ where $m$ : mass of the moon $\omega$ : angular velocity $v$ : tangential velocity of moon $r$ : radius of orbit of moon (assuming that its orbit is circular ...


2

There are only two small problems: this approach has nothing to do with gravity, and it is not at all quantum :) Firt, your action doesn't describe gravity; it describes Yang-Mills theory with the group $GL(n) = U(1) \times SL(n)$. Not gravity. There's a formulation of gravity in the gauge theory language but it uses a different action: $$ S[e, A] = \int d^4 ...


2

To understand this consider the analogy of two cars driving north from the equator as discussed in the question: Why does the speed of an object affect its path if gravity is warped spacetime? The relevant diagram from that question is: Due to the curvature of the Earth's surface the two cars converge even though they started out parallel, and this is what ...


0

Here is a (perhaps oversimplified) way of looking at this. For any description of gravity which can be well-approximated with a 1/r^2 dependency, the presence of extra spatial dimensions can be inferred by "leakage" of gravity into the extra dimensions which manifests as a deviation from strict 1/r^2 dependence in the (3+1) dimensions we inhabit. ...


3

Unsurprisingly, GR recovers Newton. With $1$ time dimension and $n$ space dimensions, the Schwarzschild metric is $ds^2=-fdt^2+dr^2/f+r^2d\Omega_{n-1}^2$ with $f(\infty)=1,\,f^\prime\propto m/r^{n-1}$. The geodesic deviation equation $\ddot{x}^a=-\Gamma^a_{bc}\dot{x}^b\dot{x}^c$ includes the nonrelativistic special case$$\ddot{x}^r\approx -\Gamma^r_{tt}=\...


1

I find it helpful to think of gravity like a low pressure system in the weather. You don't have to consider the difference between the rate of time at altitude and at the earth surface. All you need to know is that the "pressure" of time immediately below the object is slightly lower than the pressure of time immediately above the object. So the ...


1

The resulting gravitational potential (GPE per unit mass) is also "small", compared with $c^2$... but in SI units, it may still be quite large. It's like asking how a speed can be measurably large when it's expected to be small compared with $c$. does this imply that gravity in Einstein's relativity requires time to pass to operate? It does in ...


Top 50 recent answers are included