- #51

Nugatory

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The difficulty in applying the integral form of the conservation law, second paragraph of #42 above.Which is the problem curved spacetime bring in ?

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- #51

Nugatory

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The difficulty in applying the integral form of the conservation law, second paragraph of #42 above.Which is the problem curved spacetime bring in ?

- #52

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Do you mean an infinitesimal volume of spacetime or just a infinitesimal volume of a "space" slice (i.e. a spacelike hypersurface) ?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.

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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.If we consider GR as an "effective field theory", i.e., as a low energy approximation to some underlying quantum theory,thenyes, 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.

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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.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?

Robert Stenton

- #55

PeterDonis

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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.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

- #56

PeterDonis

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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.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.They all share one thing in common and that is at a lower energy level time goes slower.

Indeed, yes, you are.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.but this is a simple way for me to logically understand gravity.

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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 people still getting this exactly wrong and being loudly insistent that they are right. And other people upvoting them.

In the Newtonian (weak-field, low-speed) limit of GR, gravity is 100% caused by time dilation. That is, the reduced metric consists of flat Minkowski spacetime (which causes no gravity by definition) plus a time-curvature term, proportional to the Newtonian gravitational potential, that represents the time dilation field. And the geodesics of that metric are the arcs of Newtonian gravity. So there isn't any "gravity" left over to cause anything, let alone the time dilation which is already baked into the metric.

So, from the GR perspective, Newtonian gravity is a theory of curved time only (in flat space). The curved time is what bends paths. This isn't news. It's been known for over half a century. Even Feynman spent some time explaining it (Lectures II 42-8).

Around a perfectly spherical non-rotating Earth, this would be accurate to within 1 part per billion. Around real Earth, it's more like 1 part per million.

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). This is why the bending of light is double the Newtonian prediction. But the time curvature still causes at least 50% in all cases, even for light.

Please note that I am not making any claims about what happens in the strong-field regime, e.g. close to a black hole or neutron star. That's above my pay grade.

- #58

PeterDonis

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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.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 justIn the Newtonian (weak-field, low-speed) limit ofthe Schwarzschild metric ofGR

Again, see the bolded addition I made above. It makes a huge difference.the reduced metricin 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.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, isplus 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).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).

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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.See the bolded addition I made above. It makes a huge difference. The Schwarzschild metric is justonesolution of the Einstein Field Equation. There are many others.

If you're more comfortable with the phrase "time-only term" than "time-curvature term", I don't have any great objection. But "flat" implies no force-like effects, and "curved" implies force-like effects, and this clearly has force-like effects (curved geodesics). The metric is not flat, but the Minkowski part

- #60

PeterDonis

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Which is what's on topic here; more precisely, mainstream physics regarding classical GR.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.there are multiple physicists who think otherwise.

- #61

PeterDonis

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There are no exact solutions for multiple bodies, so I'm not sure what you are basing this on.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.You just add up the Newtonian potentials for each body and plug that in.

However, this is still limited to the weak field regime and an isolated system. You can't do it in the strong field regime, e.g., for a neutron star. And you can't do it for, e.g., FRW spacetime.

Please give a reference.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.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."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."curved" implies force-like effects

All this is personal theory and you are getting very close to a warning at this point.this clearly has force-like effects (curved geodesics). The metric is not flat, but the Minkowski partisflat. So all the non-flatness comes from the time dilation.

- #62

PeterDonis

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An infinitesimal volume of spacetime, since that is what is relevant for the differential conservation law.Do you mean an infinitesimal volume of spacetime or just a infinitesimal volume of a "space" slice (i.e. a spacelike hypersurface) ?

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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.In the Newtonian (weak-field, low-speed) limit of GR, gravity is 100% caused by time dilation.

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Of course I know there is no acceptable quantization. That does not have to mean that GR is not a field theory.

It's true that GR is a field theory. For instance, the Lagrangian approach derives GR from a Lagrangian, though the Lagrangian involves the curvature scalar R of space-time.

While it is true that GR is a field theory, it is rather questionable if one can regard it as a field theory in the Minkowskii space-time of special relativity. If one ignores the global topological aspects, approaches such as Straumann in "Reflections on Gravity", https://arxiv.org/abs/astro-ph/0006423, can work, and they may feel more comfortable to those who aren't familiar with differential geometry. The "fields" in Straumann's approach modify the lengths of rulers, and the ticking of clocks. Straumann does not address the global topological issues, unfortunately.

Some quotes from Straumann's abstract.

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

So in the weak-field case, one can reproduce GR with Straumann's "fields" on a Minkowskii space-time, but 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.

Furthermore, Straumann himself notes that the underlying Minkowskii space-time in his approach is not observable, one is inevitably led to curvature as soon as one settles on physical measurements of distance or time, such as the SI meter or the SI second.

In the end, one probably needs to learn differential geometry at some point to deal with GR. And that is certainly the "standard" approach to teaching the theory.

- #65

PeterDonis

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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.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.

- #66

PeterDonis

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https://www.physicsforums.com/threads/gauge-theory-of-gravity.1008720/

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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.Please do give an example, yes.

Anyhow, I think of the elevator analogy, i.e. experiments conducted in a frame "at rest" on a massive body (in an accelerated gravitational field?) should be indistinguishable from those conducted in an properly accelerated (equivalent) frame. So that is fine *inside* the elevator, but not so much when taking in to account everything *outside*. The former can pop the hatch of the elevator and the measured net kinetic energy of the universe is essentially the same, but this won't be the case for the latter (they would likely have a bad dose of gamma radiation). At least I think that's the case; either way, their respective measurements will be vastly different after a significant amount of time.

But from another point of view, 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? Does that imply not accelerating?), but changing energy. 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.

Finally, considering an observer "at rest" on the same massive body. This is non-inertial? So accelerating, but also not changing energy. So essentially the elevator "at rest" in a gravitational frame. It feels like this is essentially more like our orbiting observer than our free falling observer, but I'm pretty sure that is not right.

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.

- #68

PeterDonis

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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.that is fine *inside* the elevator, but not so much when taking in to account everything *outside*.

Yes, both are inertial, but they have different 4-velocities.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.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.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.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.considering an observer "at rest" on the same massive body. This is non-inertial?

Yes.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.but also not changing energy

There is a non-inertial frame in which this elevator is at rest, yes.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.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.all these various cases just are hard to reconcile in my personal head

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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: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.

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.

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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.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.

Anyhow, I am working on learning the actual math, albeit slowly. I am still learning about covectors, tensor inner product, tensors as metrics and such things. Any other suggested topics would be welcome.

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