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PeterDonis
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"Non-inertial" = "feels weight".valenumr said:Considering an object at rest on a massive body is non-inertial is my struggle.
Does that help?
"Non-inertial" = "feels weight".valenumr said:Considering an object at rest on a massive body is non-inertial is my struggle.
No problem with that, but I'm use to the idea that physics get broken in non-inertial frames. And I suppose that goes to my lack of understanding of coordinate choices, perhaps.PeterDonis said:"Non-inertial" = "feels weight".
Does that help?
Why would physics be "broken" in non-inertial frames? Where are you getting that idea from?valenumr said:I'm use to the idea that physics get broken in non-inertial frames.
I mean that from my understanding you can't count on conservation of energy. I can give an (perhaps misguided example).PeterDonis said:Why would physics be "broken" in non-inertial frames? Where are you getting that idea from?
And sorry, I didn't mean broken. Obviously that isn't the case. Not my understanding is broken.PeterDonis said:Why would physics be "broken" in non-inertial frames? Where are you getting that idea from?
Please do give an example, yes.valenumr said:I mean that from my understanding you can't count on conservation of energy. I can give an (perhaps misguided example).
Energy conservation works in non-inertial frames, as long as we include the work done by the so-called fictitious forces. The math is exactly that used to demonstrate conservation of energy on the surface of the earth: throw an object upwards and it gains potential energy as it loses kinetic energy, and the total (KE+PE) is conserved.valenumr said:I mean that from my understanding you can't count on conservation of energy. I can give an (perhaps misguided example).
That's going to take a minute.PeterDonis said:Please do give an example, yes.
There is no acceptable theory of quantum gravity yet. GR is the best we have; and in GR, whether you like it or not, gravity is spacetime curvature.f todd baker said:Although I have not read every posting on this thread, I find it surprising that so many state that gravity IS curvature of spacetime. ... In QFT, the gravitational field is just another force field, like the EM, strong and weak fields, albeit with a greater complexity that is reflected in its higher spin value of 2."
I think we need carefully define which is the 'system' involved. Potential energy for an object alone (i.e. without any internal structure) actually makes no sense: it applies only to systems made of parts since it is related to their configuration within the 'system' itself.Nugatory said:Energy conservation works in non-inertial frames, as long as we include the work done by the so-called fictitious forces. The math is exactly that used to demonstrate conservation of energy on the surface of the earth: throw an object upwards and it gains potential energy as it loses kinetic energy, and the total (KE+PE) is conserved.
Of course I know there is no acceptable quantization. That does not have to mean that GR is not a field theory.PeroK said:There is no acceptable theory of quantum gravity yet. GR is the best we have; and in GR, whether you like it or not, gravity is spacetime curvature.
Of course we do, but that doesn't change the point about energy conservation: the non-inertial frame is no impediment to conservation of energy if we do it right. A curved spacetime may be.cianfa72 said:I think we need carefully define which is the 'system' involved.
If we consider GR as an "effective field theory", i.e., as a low energy approximation to some underlying quantum theory, then yes, we can consider gravity as an "interaction" on that view. That is the view that Weinberg, for example, is advocating in the article you linked to. But that view is off topic in this forum; discussion of it belongs either in the quantum physics forum, or more likely in the Beyond the Standard Model forum since that's where discussion of quantum gravity in general belongs. See my post #3 for further discussion of this point.f todd baker said:What I have learned is that general relativity is a field theory and therefore may be thought of as a force.
Yes, definitely.Nugatory said:Of course we do, but that doesn't change the point about energy conservation: the non-inertial frame is no impediment to conservation of energy if we do it right.
Which is the problem curved spacetime bring in ?Nugatory said:A curved spacetime may be.
The difficulty in applying the integral form of the conservation law, second paragraph of #42 above.cianfa72 said:Which is the problem curved spacetime bring in ?
Do you mean an infinitesimal volume of spacetime or just a infinitesimal volume of a "space" slice (i.e. a spacelike hypersurface) ?Nugatory said:We need to be more careful with energy conservation in curved spacetimes, but even there the differential form of the conservation law (total amount of energy within an infinitesimal volume changes only when energy enters or leaves that volume) works.
Sorry, I guess I don't see why a comment about force in GR is "off topic" for a question about whether gravity is a force. It sure isn't worth splitting hairs over, though.PeterDonis said:If we consider GR as an "effective field theory", i.e., as a low energy approximation to some underlying quantum theory, then yes, we can consider gravity as an "interaction" on that view. That is the view that Weinberg, for example, is advocating in the article you linked to. But that view is off topic in this forum; discussion of it belongs either in the quantum physics forum, or more likely in the Beyond the Standard Model forum since that's where discussion of quantum gravity in general belongs. See my post #3 for further discussion of this point.
There are four fundamental forces in nature: strong and weak nuclear, electromagnetic and gravity. They all share one thing in common and that is at a lower energy level time goes slower. If a large cloud condenses into a star,even though the atoms in the star are the same as the ones in the original cloud, the mass of star is reduced because the atoms go through time slower. Think of a butterfly on a scale that flies off and on but is only on the scale half the time, the scale weight of the butterfly would then be cut in half. It is from the missing mass that the energy comes from to create the gravitational field. The gravitational field would be the same as it was outside the boundary of the original cloud but much greater between that boundary and the surface of the star. Visualize it this way, the side of the moon facing the Earth is traveling slower than the side away from earth. Now I am sure that someone will tell me I am all wrong but this is a simple way for me to logically understand gravity. Hope it helps.Schnellmann said:Summary:: Conflicting opinions on videos I’ve watched
I’ve watched a few videos recently that explained that gravity is not a force rather it is caused by time dilation because clocks tick slower closer to mass. Objects will follow a geodesic through spacetime and require a force to move them away from a geodesic - so the surface of the Earth is accelerating everything on the surface. If you fall into a hole you don’t experience a gravitational force pulling you down rather you feel the removal of the force that was pushing you away from your geodesic. I then watched a video on quantum gravity that said gravity was a force (although almost infinitely weaker than the other known forces) and the theoretical particle is the graviton.
So which is true?
On the subject of gravity not being a force - if we are being constantly accelerated by standing on the surface of the Earth doesn’t that require some energy? If yes where does that energy come from?
The point is that in classical GR, considered as an exact classical theory, without regard to whether it could be a low energy "effective theory" approximation to something else, gravity is simply not a force. Any view that considers gravity a force has to rely on some claim about GR being an "effective field theory" approximation to something else. Such claims are off topic in this particular forum because this particular forum is not about GR as a possible "effective field theory" approximation to something else; it's about classical GR, considered as an exact classical theory.f todd baker said:I guess I don't see why a comment about force in GR is "off topic" for a question about whether gravity is a force
In classical GR, as has already been remarked several times now in this thread, gravity is not a force. It is not like the other "fundamental forces" you describe.Robert Stenton said:There are four fundamental forces in nature: strong and weak nuclear, electromagnetic and gravity.
This is not correct. There is no analogue of gravitational time dilation for the strong, weak, or electromagnetic interactions.Robert Stenton said:They all share one thing in common and that is at a lower energy level time goes slower.
Indeed, yes, you are.Robert Stenton said:I am sure that someone will tell me I am all wrong
Personal theories and personal speculations are not allowed here at PF. You would be much better advised to learn a proper understanding from a GR textbook.Robert Stenton said:but this is a simple way for me to logically understand gravity.
phinds said:This, unfortunately, is utter nonsense, since it is backwards. Gravitational time dilation is CAUSED by gravity, it is not a cause OF gravity.
It's really disheartening to see someone talking about one particular solution of the Einstein Field Equation, described in one particular coordinate chart, as though it were "GR" and "gravity" without qualification.H_A_Landman said:It's really disheartening to see people still getting this exactly wrong and being loudly insistent that they are right. And other people upvoting them.
See the bolded addition I made above. It makes a huge difference. The Schwarzschild metric is just one solution of the Einstein Field Equation. There are many others. Most of them do not even have a simple concept of "gravity" as arising from a Newtonian potential.H_A_Landman said:In the Newtonian (weak-field, low-speed) limit of the Schwarzschild metric of GR
Again, see the bolded addition I made above. It makes a huge difference.H_A_Landman said:the reduced metric in Schwarzschild coordinates
Depends on how you define "gravity". By at least one fairly common definition (the one that underlies the equivalence principle), there is "gravity" in the non-inertial rest frame of a rocket accelerating in a straight line in flat spacetime.H_A_Landman said:consists of flat Minkowski spacetime (which causes no gravity by definition)
The ##- 2M / r## term in the ##g_{tt}## metric coefficient in Schwarzschild coordinates, which is what you are referring to here, is not a "time curvature" term. Curvature is expressed by the Riemann curvature tensor, not the metric. This term can be thought of, in this particular solution, in these particular coordinates, in this particular approximation, as arising from the Newtonian gravitational potential. But that just means that spacetime curvature, in its manifestation in this particular solution, in these particular coordinates, in this particular approximation as Newtonian gravitational potential, causes both time dilation for stationary observers (note that qualifier, it also makes a big difference--"time dilation" is observer-dependent) and "gravity" (more precisely, the fact that radial geodesics converge towards the central mass). It does not mean that time dilation causes gravity.H_A_Landman said:plus a time-curvature term, proportional to the Newtonian gravitational potential, that represents the time dilation field.
This claim, with appropriate caveats (just as for ##g_{tt}## above, the appearance of a term proportional to ##2M / r## in the spatial terms in the metric is not "space curvature", since that, as noted, is expressed by the Riemann curvature tensor, not the metric), is also only true for this particular solution, in these particular coordinates, in this particular approximation (weak field).H_A_Landman said:If you back out of taking the low-speed limit (so it's only weak-field), then there is a space curvature term that ranges from zero (at zero speed) to equal in size to the time-curvature term (at the speed of light).
Well, sure, it's easiest to see with the Schwarzschild metric, but that's a one-body metric. It also works with multiple bodies. You just add up the Newtonian potentials for each body and plug that in. You can do a whole galaxy (minus the strong-field parts) this way.PeterDonis said:See the bolded addition I made above. It makes a huge difference. The Schwarzschild metric is just one solution of the Einstein Field Equation. There are many others.
Which is what's on topic here; more precisely, mainstream physics regarding classical GR.H_A_Landman said:although what you say is certainly true for consensus mainstream physics
If you want to discuss these speculative theories, please do so in a new thread in the Beyond the Standard Model forum. They are off topic here.H_A_Landman said:there are multiple physicists who think otherwise.
There are no exact solutions for multiple bodies, so I'm not sure what you are basing this on.H_A_Landman said:It also works with multiple bodies.
In the weak field regime for an isolated system, you can get away with this because the non-linearities in the EFE can be assumed to be negligible. But there are still limitations with this approach, and you're still doing numerical solutions since no exact solutions are known.H_A_Landman said:You just add up the Newtonian potentials for each body and plug that in.
Please give a reference.H_A_Landman said:You can do a whole galaxy (minus the strong-field parts) this way.
Neither one is correct; note that I didn't say "time-only term". I said ##g_{tt}##, and I explicitly noted that it is for a particular system of coordinates.H_A_Landman said:If you're more comfortable with the phrase "time-only term" than "time-curvature term", I don't have any great objection.
It implies no such thing. There are "force-like effects" in an accelerating rocket in flat spacetime, or inside a rotating chamber in flat spacetime, for that matter.H_A_Landman said:"flat" implies no force-like effects
It implies no such thing. Spacetime curvature is geodesic deviation, which can be shown entirely from the behavior of freely falling objects that feel zero force.H_A_Landman said:"curved" implies force-like effects
All this is personal theory and you are getting very close to a warning at this point.H_A_Landman said:this clearly has force-like effects (curved geodesics). The metric is not flat, but the Minkowski part is flat. So all the non-flatness comes from the time dilation.
An infinitesimal volume of spacetime, since that is what is relevant for the differential conservation law.cianfa72 said:Do you mean an infinitesimal volume of spacetime or just a infinitesimal volume of a "space" slice (i.e. a spacelike hypersurface) ?
This is 100% completely wrong, even in the Newtonian limit. Causes always preceed effects. Time dilation and gravity occur at the same time, so they do not have a cause-effect relationship.H_A_Landman said:In the Newtonian (weak-field, low-speed) limit of GR, gravity is 100% caused by time dilation.
f todd baker said:Of course I know there is no acceptable quantization. That does not have to mean that GR is not a field theory.
Although this field theoretic approach, which has been advocated repeatedly by a num-
ber of authors, starts with a spin-2 theory on Minkowski spacetime,
it turns out in the end that the flat metric is actually unobservable,
and that the physical metric is curved and dynamical
Also, as I've noted, one can't explain spacetimes like FRW, which do not have the same conformal structure at infinity as Minkowski spacetime (i.e., are not asymptotically flat), with this approach.pervect said:it's probably not true that one could explain general strong-field space-times, for instance the space-time of a black hole, via this approach.
Sorry for the delayed response, just a bit busy the past few days. And apologies for some of the liberties below, please correct any misguided assumptions.PeterDonis said:Please do give an example, yes.
Of course not; the equivalence principle, which is what you are referring to, is local. It doesn't say everything everywhere is the same in both cases. It just says things inside the elevator, i.e., inside a small enough patch of spacetime that the effects of spacetime curvature are negligible, is the same in both cases. We do not expect that things will look the same once we look outside the elevator; obviously they will generally not be.valenumr said:that is fine *inside* the elevator, but not so much when taking into account everything *outside*.
Yes, both are inertial, but they have different 4-velocities.valenumr said:consider two (I think) inertial frames: one in circular orbit around a massive body, and one falling directly on a normal vector towards the same massive body. The former is inertial, not accelerating, not changing energy, etc. But the latter is also inertial (right?
"Inertial" means zero proper acceleration, yes.valenumr said:Does that imply not accelerating?)
Depends on what concept of "energy" you use. Both of them are on geodesic trajectories, meaning both have constant energy at infinity.valenumr said:but changing energy.
Using that, every object in the solar system has huge kinetic energy since the solar system barycenter is moving at about 600 km/s relative to the CMB rest frame, which is a far larger speed than any internal speed in the solar system.valenumr said:In both cases, the best I can say is energy is "with respect to the rest of the universe", which to be fair is ill defined, but say, take the CMB background as a standard rest frame.
Yes.valenumr said:considering an observer "at rest" on the same massive body. This is non-inertial?
Yes.valenumr said:So accelerating
Again, it depends on what concept of "energy" you use. "Energy" is not a single unique absolute quantity; there are different concepts of energy you could use. For example, relative to a free-falling observer, the object at rest on the Earth's surface is changing energy.valenumr said:but also not changing energy
There is a non-inertial frame in which this elevator is at rest, yes.valenumr said:So essentially the elevator "at rest" in a gravitational frame.
Why? "At rest" is frame-dependent; so is "energy". But nonzero proper acceleration is not; it's an invariant. So proper acceleration is a much better way to judge which observers are "more like" each other.valenumr said:It feels like this is essentially more like our orbiting observer than our free falling observer
So far nothing you have pointed out is any kind of problem. So I'm not sure what problem you think you are perceiving. The only issue I can see with anything you've said is that you need to be more careful to distinguish frame-dependent concepts (like "energy" and "at rest") from invariants (like proper acceleration). All of the actual physics is contained in the latter.valenumr said:all these various cases just are hard to reconcile in my personal head
This seems to me something of a mental block. Many of the ideas that you consider hard to reconcile appear in non-relativistic physics (e.g. that energy is not absolute but frame dependent; and, if we consider non-inertial frames, then even the change in energy is frame dependent). Take a look at this homework problem from elementary (Newtonian) kinematics and you'll see how many of the ideas you find difficult arise quite naturally in non-relativistic physics:valenumr said:So all these various cases just are hard to reconcile in my personal head. I'm not saying there is anything wrong, proposing alternative physics, or any such thing, or even that any of my assumptions are correct. I am just saying that these concepts are pretty hard to wrap my brain around.
Not exactly a mental block, but I suppose you might feel like your trying to explain that the Earth is a sphere to someone who never heard of such "craziness", having experienced their entire life perceiving it as flat.PeroK said:This seems to me something of a mental block. Many of the ideas that you consider hard to reconcile appear in non-relativistic physics (e.g. that energy is not absolute but frame dependent; and, if we consider non-inertial frames, then even the change in energy is frame dependent). Take a look at this homework problem from elementary (Newtonian) kinematics and you'll see how many of the ideas you find difficult arise quite naturally in non-relativistic physics:
https://www.physicsforums.com/threads/mechanics-calculating-launch-angle-of-projectile.1008645/
The "clever" solution is post #24. And, see my comment in post #25.
Gravity is a natural phenomenon by which all objects with mass are brought towards each other. It is one of the four fundamental forces in the universe and is responsible for keeping planets in orbit around the sun and objects on Earth from floating away into space.
Gravity can be thought of as both a force and an acceleration. As a force, it is the attraction between two objects with mass. As an acceleration, it is the rate at which an object falls towards the Earth due to the force of gravity.
Gravity works by the principle of mass attracting mass. The more massive an object is, the stronger its gravitational pull. This is why larger objects, such as planets, have a stronger gravitational force than smaller objects, such as rocks.
Currently, there is no known way to manipulate or control gravity. However, scientists are constantly researching and experimenting with different theories and technologies to better understand and potentially harness the power of gravity.
Gravity affects the motion of objects by causing them to accelerate towards each other. This acceleration is proportional to the mass of the objects and the distance between them. The greater the mass and the closer the distance, the stronger the gravitational force and the greater the acceleration.