Spacetime curvature due to acceleration causing gravity?

[Buckethead]

In reviewing one of Einstein's thought experiments, the accelerating elevator in space, and the resulting bending of light passing through the elevator, Einstein's predicted that light will bend in gravity. Now Einstein's original prediction was off by a factor of 2 because he hadn't yet realized that spacetime is curved in a gravitational field. When he realized this he made the correct prediction. However, this means his original thought about how much light bends in an accelerating elevator was wrong since (I assume) he did not (yet?) predict an accelerating elevator would curve spacetime (if indeed it does).

Due to the Equivalence Principle we know that the degree to which light bends in a gravitational field is the same as the degree it must bend in a moving elevator which leads me to the thought that an accelerating elevator in space must cause a warping of spacetime so that the degree of bending of light in both situations (gravity or acceleration) will be the same.

Is this correct? If there is an actual curving of spacetime around an accelerating mass, wouldn't this curving produce gravity and couldn't it be this gravity then that is the cause of what we view as inertial mass (resistance to acceleration)?

 

[Dale]

However, this means his original thought about how much light bends in an accelerating elevator was wrong since (I assume) he did not (yet?) predict an accelerating elevator would curve spacetime (if indeed it does).

The elevator calculations are correct because an accelerating elevator does not bend spacetime.

 

[Buckethead]

I'm not sure I understand. If light bends a certain amount in a gravitational field with a certain acceleration and this differs by a factor of 2 when compared to the same acceleration of an elevator in space, why doesn't this contradict the equivalence principle?

 

[Dale]

I'm not sure I understand. If light bends a certain amount in a gravitational field with a certain acceleration and this differs by a factor of 2 when compared to the same acceleration of an elevator in space, why doesn't this contradict the equivalence principle?

Because the equivalence principle only applies locally. Meaning over regions small enough that curvature can be ignored

 

[kent davidge]

an accelerating elevator does not bend spacetime

I thought anything having a mass would bend space-time, including an elevator, accelerating or not.

 

[Buckethead]

Because the equivalence principle only applies locally. Meaning over regions small enough that curvature can be ignored

So If you have a 10 meter wide box on Earth and a 10 meter wide box in space, and light passes through both, assuming both have the same acceleration, light will bend less in the elevator in space within that 10 meter wide box?

 

[Dale]

So If you have a 10 meter wide box on Earth and a 10 meter wide box in space, and light passes through both, assuming both have the same acceleration, light will bend less in the elevator in space within that 10 meter wide box?

No, that will be the same. 10 meters is small enough that curvature is negligible.

The difference is that if you place a string of 10 meter boxes together accelerating in flat spacetime the boxes are all aligned together, but a string of 10 meter boxes in gravity will not all align with each other.

Across each 10 m box the behavior is the same. But across the string of boxes it differs

 

[PeterDonis]

I thought anything having a mass would bend space-time, including an elevator, accelerating or not.

In order to apply the equivalence principle at all, spacetime has to be flat (to a good enough approximation) within the elevator. That already assumes that the elevator does not bend spacetime, whether it is accelerating or not. More precisely, it assumes that the elevator does not bend spacetime enough to be detectable. Since it takes a huge amount of mass (much more than an elevator) to bend spacetime detectably, that is a good assumption.

 

[Buckethead]

I see, so the idea of tidal forces is part of this. So the light going a long distance past the spherical Earth as a whole is going to be different because the Earth is in fact a sphere and not an endless length mass. Is this correct?

I remember something about imagining an infinite flat mass is space (so there would be no tidal forces). In such a scenario is spacetime above the surface of this plane flat? Wouldn't this mean such an object would exert no gravity? I would think it would, but this would mean the spacetime would be curved.

 

[PeterDonis]

Einstein's original prediction was off by a factor of 2 because he hadn't yet realized that spacetime is curved in a gravitational field.

No. It was off by a factor of 2 because he hadn't yet realized that spacetime is curved if tidal gravity is present. (A modern relativity physicist would say that spacetime curvature is tidal gravity.) Tidal gravity will be present if a massive body like a planet or star is present. But the whole point of the EP is that, if you're inside the elevator and don't look outside or make any measurements of outside, you can't tell whether tidal gravity is present or not--so you don't know whether the elevator is accelerating through flat spacetime far from any gravitating bodies, or accelerating in order to maintain a constant altitude above a planet or star.

so the idea of tidal forces is part of this

Yes, see above.

I remember something about imagining an infinite flat mass is space (so there would be no tidal forces).

"I remember something about..." isn't a good reference. But there have been papers published investigating such models. The upshot of those papers is basically that, unlike in Newtonian gravity, there is no solution in GR that corresponds to "an infinite flat mass with a uniform gravitational field but no tidal forces above it". We've had previous threads about that here on PF, though it was a while ago.

 

[Buckethead]

"I remember something about..." isn't a good reference.

Haha, yes, sorry, I was in fact referring to an old discussion here on PF. I can't remember where that discussion led although I do recall the importance of tidal forces.

A modern relativity physicist would say that spacetime curvature is tidal gravity.)

So just to be clear, if no tidal forces are present then (at least as far as a modern physicist is concerned) there is no gravity?

 

[PeterDonis]

if no tidal forces are present then (at least as far as a modern physicist is concerned) there is no gravity?

No, if no tidal forces are present, there is no tidal gravity. "Gravity" in general is a vague term and has different meanings, one of which is what you observe if you drop a rock in an accelerating elevator, even if you don't know whether the elevator is accelerating way out in deep space or accelerating to maintain constant altitude above a planet or star. So that kind of "gravity" can be present when there are no tidal forces.

 

[Buckethead]

No, if no tidal forces are present, there is no tidal gravity. "Gravity" in general is a vague term and has different meanings, one of which is what you observe if you drop a rock in an accelerating elevator, even if you don't know whether the elevator is accelerating way out in deep space or accelerating to maintain constant altitude above a planet or star. So that kind of "gravity" can be present when there are no tidal forces.

That's a good point. So I'll be more specific. Is gravity present above a vast flat mass (with an unvarying thickness) that extends to infinity in the x and y directions and is not accelerating?

 

[PeterDonis]

Is gravity present above a vast flat mass (with an unvarying thickness) that extends to infinity in the x and y directions and is not accelerating?

What do you mean by "gravity"?

 

[Buckethead]

What do you mean by "gravity"?

If I'm standing on this mass and drop a rock, will it accelerate toward the mass?

 

[PeterDonis]

If I'm standing on this mass and drop a rock, will it accelerate toward the mass?

Yes. Note that there is, IIRC, more than one solution to the EFE that can be described as "an infinite flat mass", but all of them share this property.

 

[Buckethead]

Yes. Note that there is, IIRC, more than one solution to the EFE that can be described as "an infinite flat mass", but all of them share this property.

This is a surprising answer. In thread #10 you said:

But there have been papers published investigating such models. The upshot of those papers is basically that, unlike in Newtonian gravity, there is no solution in GR that corresponds to "an infinite flat mass with a uniform gravitational field but no tidal forces above it". We've had previous threads about that here on PF, though it was a while ago.

If I'm reading that correctly, you are saying there are solutions to GR regarding such a mass that do indicate tidal forces will be present above this mass? And the rock will be attracted because of this?

 

[PeterDonis]

If I'm reading that correctly, you are saying there are solutions to GR regarding such a mass that do indicate tidal forces will be present above this mass?

Yes.

And the rock will be attracted because of this?

No. You keep on confusing tidal gravity with "acceleration due to gravity" (the thing that makes the rock fall). They're not the same. The rock falls when you drop it for the same reason it does on the surface of the Earth: because you, standing there, are accelerating.

 

[PeterDonis]

This previous thread contains discussion of an "infinite flat slab" in GR, with a link to a relevant paper:

https://www.physicsforums.com/threads/einsteins-elevator-gravity-without-curvature.919092/

 

[Buckethead]

No. You keep on confusing tidal gravity with "acceleration due to gravity" (the thing that makes the rock fall). They're not the same. The rock falls when you drop it for the same reason it does on the surface of the Earth: because you, standing there, are accelerating.

But the acceleration due to something like the Earth is caused by the curvature of spacetime. If there is no curvature there is no acceleration near the Earth. In post 10 you say "A modern relativity physicist would say that spacetime curvature is tidal gravity." So that led me to think that it must be tidal gravity, causing curvature of spacetime causing the rock on the infinite mass to fall.

 

[Buckethead]

This previous thread contains discussion of an "infinite flat slab" in GR, with a link to a relevant paper:

https://www.physicsforums.com/threads/einsteins-elevator-gravity-without-curvature.919092/

Thanks for the link, I'll have a go at re-reading it.

 

[PeterDonis]

the acceleration due to something like the Earth is caused by the curvature of spacetime. If there is no curvature there is no acceleration near the Earth.

You're confusing "acceleration due to gravity" with tidal gravity. Tidal gravity is the curvature of spacetime.

You're also confusing proper acceleration with coordinate acceleration. A rock falling near the Earth has zero proper acceleration; it's weightless, in free fall. So there is no "acceleration" of the rock to be due to the curvature of spacetime or anything else.

 

[Dale]

I see, so the idea of tidal forces is part of this.

Yes. Spacetime curvature essentially is tidal gravity

 

[Buckethead]

You're also confusing proper acceleration with coordinate acceleration. A rock falling near the Earth has zero proper acceleration; it's weightless, in free fall. So there is no "acceleration" of the rock to be due to the curvature of spacetime or anything else.

My apologies. I know this, I just wasn't careful about my wording.

But where I am still confused is this: Is it true that curved spacetime (such as caused by the Earth) is the cause that an object standing on the Earth feels a force otherwise known as proper acceleration? If so, doesn't the lack of curved spacetime indicate that there will not be a proper acceleration felt by that object? If this is true and if curved spacetime is Tidal gravity, then Tidal gravity is the cause of proper acceleration to this same object. It then follows that if the infinite mass mentioned earlier does not have Tidal gravity, then it cannot have curvature, and therefore an object standing on the mass would not feel proper acceleration. Which means a rock would not fall.

Although you said earlier that there is a solution to GR that indicates that such a mass does have tidal gravity, so yes, the rock would fall.

 

[stevendaryl]

But the acceleration due to something like the Earth is caused by the curvature of spacetime. If there is no curvature there is no acceleration near the Earth. In post 10 you say "A modern relativity physicist would say that spacetime curvature is tidal gravity." So that led me to think that it must be tidal gravity, causing curvature of spacetime causing the rock on the infinite mass to fall.

There are a bunch of effects that we would say were "due to gravity":

1. You drop an object. It falls down.
2. Clocks that are higher up seem to run faster.
3. Clocks that are lower down seem to run slower.
4. Light beams are bent downward.

However, all of these effects are ALSO true inside an accelerating rocket (or elevator, as Einstein said). That means that they are not ACTUALLY effects of spacetime curvature. These are all effects that the mere presence of a "gravitational field" causes.

Spacetime curvature describes how the apparent "gravitational field" changes from place to place. The apparent gravitational field of an accelerating rocket points in one direction: toward the floor of the rocket. But if you look at the planet Earth, you'll see that the gravitational field points in different directions at different points. It always points toward the center of the Earth, but what direction that is depends on where you are. That's a manifestation of spacetime curvature--when the gravitational field changes from point to point.

 

[stevendaryl]

My apologies. I know this, I just wasn't careful about my wording.

But where I am still confused is this: Is it true that curved spacetime (such as caused by the Earth) is the cause that an object standing on the Earth feels a force otherwise known as proper acceleration?

No, it is not true. Try jump out of a tree. During the second it takes to fall to the ground, you will feel no gravity. But of course, the "spacetime curvature" is the same whether you are falling or whether you are standing on the ground.

What you "feel" when you feel gravity is actually the feeling of the floor pressing up against your feet. That feeling is exactly the same aboard an accelerating rocket: what you feel is the floor of the rocket pressing up against your feet.

 

[Buckethead]

However, all of these effects are ALSO true inside an accelerating rocket (or elevator, as Einstein said). That means that they are not ACTUALLY effects of spacetime curvature. These are all effects that the mere presence of a "gravitational field" causes.

Point taken, but in the case of the flat infinite plane, in order to generate a "gravitational field", it would either have to be accelerating (which I specify it is not) or else it must generate a curved spacetime because AFAIK these are the only two things that can generate what you are calling a gravitational field.

No, it is not true. Try jump out of a tree. During the second it takes to fall to the ground, you will feel no gravity. But of course, the "spacetime curvature" is the same whether you are falling or whether you are standing on the ground.

Yes, I understand this, but I didn't mention anything about gravity. What I asked precisely was this:

"Is it true that curved spacetime (such as caused by the Earth) is the cause that an object standing on the Earth feels a force otherwise known as proper acceleration?"

 

[stevendaryl]

"Is it true that curved spacetime (such as caused by the Earth) is the cause that an object standing on the Earth feels a force otherwise known as proper acceleration?"

And I answered: No, it is not true.

 

[Buckethead]

And I answered: No, it is not true.

This confuses me. Isn't this exactly what Einstein said? That curved spacetime is what causes things to fall to the Earth? If you are out in Minkowski space where there is no mass, you don't feel proper acceleration (unless of course you are being accelerated by an outside force).

 

[stevendaryl]

This confuses me. Isn't this exactly what Einstein said? That curved spacetime is what causes things to fall to the Earth?

No, he didn't say that. Spacetime curvature is not a force. The apparent force that you feel when you stand on the floor is not due to spacetime curvature.

If you are out in Minkowski space where there is no mass, you don't feel proper acceleration (unless of course you are being accelerated by an outside force).

Whether you're in curved spacetime or in flat spacetime, you don't feel proper acceleration unless something is pushing up on you. On the Earth, you only feel proper acceleration because the floor is pushing up on you. The same is true in outer space: You feel proper acceleration because the floor of the rocket is pushing up on you. In both cases, if the floor gives way, then you will no longer feel proper acceleration. So absolutely no: spacetime curvature is not what causes you to feel proper acceleration.

The more accurate way to think about it is this: There is a certain type of motion called "inertial motion", which is the motion of an object that is not acted on by any forces. You only feel proper acceleration when something forces you to travel in a noninertial way.

The difference between curved spacetime and flat spacetime is that in flat spacetime, inertial motion has a very nice property: If you start off putting two objects a small distance apart and initially traveling in the same direction at the same speed, then under inertial motion, they will continue traveling in the same direction, and will remain the same distance apart. In contrast, curved spacetime has an effect called "geodesic deviation". If you start two objects off traveling in the same direction at the same speed a small distance apart, they will not continue to travel in the same direction and will not continue to stay the same distance apart. The most stark example is: You drop two objects from rest far above the surface of the Earth. Initially, they have the same velocity, namely 0. But as time goes on, they will converge toward the center of the Earth. The distance between them will get smaller, and their velocities will point in different directions.

That's a key indicator of spacetime curvature, is that geodesics (the paths of objects traveling inertially) do not remain parallel when they start out parallel. This fact has nothing directly to do with "feeling proper acceleration". You feel proper acceleration when you are forced to move noninertially.