I Local meaning of the equivalence principle

[craigthone]

[Moderator's note: Spun off from another thread due to topic change.]

Not quite. It says that accelerating in flat spacetime is locally equivalent to being at rest in a uniform gravitational field.

Can I say that "being at rest in a uniform gravitational field is locally equivalent to accelerating in flat spacetime, free falling in a uniform gravitational field is locally equivalent to nonacceleration in flat spacetime ?"

 

[craigthone]

So am accelerating electron (in an elevator) will not produce any radiation that can be detected by a co-accelerating detector. By the equivalence principle the same is what we expect for the charge and detector at rest in gravity.

I think you give the answer to the original question. I have a further question, from an old post: https://www.physicsforums.com/threa...ction-of-unruh-radiation.883227/#post-5994170
Some firend, vanadiam 50 , said that
"The equivalence principle does not say you can swap gravity for acceleration. It takes 16 numbers to describe gravity and 3 to do acceleration. It's really more the reverse - it tells how to tell apart gravity from acceleration."
But for me locally it seems that we can change the gravity and acceleration. Is there any exception that I can not change them? I am puzzled.

 

[craigthone]

Yes.

[Edit: removed final statement, not correct.]

Thanks for your clarification! Does this mean that we can swap gravity for acceleration locally?

 

[Dale]

But for me locally it seems that we can change the gravity and acceleration. Is there any exception that I can not change them? I am puzzled.

I think that you would need to ask him for an explanation. For me the equivalence principle says that for local experiments (no tidal effects) there is no difference between a gravitational field and a uniformly accelerating reference frame. The force of gravity is equivalent to the fictitious force in every way.

 

[craigthone]

I think that you would need to ask him for an explanation. For me the equivalence principle says that for local experiments (no tidal effects) there is no difference between a gravitational field and a uniformly accelerating reference frame. The force of gravity is equivalent to the fictitious force in every way.

Thanks for your explanation.

 

[PeterDonis]

Does this mean that we can swap gravity for acceleration locally?

I don't know what you mean by "swap gravity for acceleration". I stated what the equivalence principle says.

 

[PeterDonis]

for me locally it seems that we can change the gravity and acceleration

What do you mean by "locally change the gravity and acceleration"? I asked you this in post #23. Until you answer it I have no way of answering your question about the EP.

 

[craigthone]

What do you mean by "locally change the gravity and acceleration"? I asked you this in post #23. Until you answer it I have no way of answering your question about the EP.

My understanding is that locally gravity and acceleration have no difference. There should be correspondence in acceleration phenomena and gravity phenomena locally, for example, Unruh's effect and Hawking's effect. I am not sure whether this is the motivation of Unruh's discovery.

 

[atyy]

My understanding is that locally gravity and acceleration have no difference. There should be correspondence in acceleration phenomena and gravity phenomena locally, for example, Unruh's effect and Hawking's effect. I am not sure whether this is the motivation of Unruh's discovery.

It depends on how "local" you go. There is interesting discussion by Blandford and Thorne in http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1225.2.K.pdf.

Section 25.3
"It is instructive to compare Eq. (25.9a) for the metric in the local Lorentz frame of a freely falling observer in curved spacetime with Eq. (24.60b) for the metric in the proper reference frame of an accelerated observer in flat spacetime. Whereas the spacetime curvature in (25.9a) produces corrections to... of second order in distance from the world line, the acceleration and spatial rotation of the reference frame in (24.60b) produces corrections of first order. This remains true when one studies accelerated observers in curved spacetime (e.g. Sec. 26.3.2)."

Section 25.7
"Thus, when the local, nongravitational laws of physics are known in frame-independent form, one need not distinguish between whether they are special relativistic or general relativistic.
In this conclusion the word local is crucial: The equivalence principle is strictly valid only at the spatial origin of a local Lorentz frame; and, correspondingly, it is in danger of failure for any law of physics that cannot be formulated solely in terms of quantities which reside at the spatial origin—i.e., along a timelike geodesic"

The Unruh effect was indeed inspired by the Hawking radiation and the equivalence principle. There is a very interesting discussion by Rovelli and Smerlak about what is going on in the Unruh effect in https://arxiv.org/abs/1108.0320.

 

[PeterDonis]

My understanding is that locally gravity and acceleration have no difference.

This doesn't answer my question. You still haven't explained what you mean by "locally change gravity and acceleration". What specific experiment or scenario are you talking about?

 

[pervect]

My view of the matter goes something like this. Suppose we have a stationary space-time geometry. Then we can think of objects "at rest" in the geometry. There's a formal definition of what I mean by "stationary" and "at rest" that I can give, but I'm not sure if it'd be helpful. Loosely speaking, though, the stationary geoemtry is "not changing with time". In this stationary geometry, we can find the weight of test particles, the force on a non-spinning test particle required to keep it at rest.

With certain modest assumptions about smoothness, the weight of sufficiently nearby particles will all be very close, there won't be any sudden jumps. And the weight will be proportional to the mass of the test particle. We can then say that the ratio of the weight of the test particle (the force required to hold it in place) to the mass of the test particle represents "gravity", and that with this definition of "gravity", an accelerating elevator has "gravity", and so does being near a large mass. And we can equate the two.

If we get more sophisticated though, we might ask how the weight of a test particle varies with its position, or with its spin. Or we might inquire about the force on a moving test particle, rather than a stationary one. When we ask these more sophisticated questions, we need a more complex model of gravity.

These more sophisticated notions of gravity usually involve space-time curvature and the Riemann curvature tensor, and we take the viewpoint that in flat space-time the Riemann curvature tensor is zero regardless of whether an object is accelerating or not. The Riemann curvature tensor formulation allows us to answer the questions I asked above - how much does the force on a test particle change when we change it's position slightly? How much does it change when the particle is moving vs when it is still. What is the effect of spin on the weight of the particle?

At the more sophisticated level, we realize that the answers to these extended questions depend on whether or not we are in an accelerating elevator, or near a large mass. So we can, for instance, by using a gravity gradiometer such as the "Forward Mass Detector", we can distinguish the cases of standing on the Earth, and being in an accelerating spaceship.

 

[sweet springs]

Can I say that "being at rest in a uniform gravitational field is locally equivalent to accelerating in flat spacetime, free falling in a uniform gravitational field is locally equivalent to nonacceleration in flat spacetime ?"

I do not think so. Think of two men: one is on the flat of accelerating rocket, the other is on the wall of rotating disk. Even if they feel same amount of force and reaction force, the physics are different, for an example, the former observe Lorentz contraction of other bodies moving any direction, however, the latter may observe inflation of bodies moving in a certain direction. It annoys me to which acceleration system I should replace gravitation field.

 

[craigthone]

I do not think so. Think of two men: one is on the flat of accelerating rocket, the other is on the wall of rotating disk. Even if they feel same amount of force and reaction force, the physics are different, for an example, the former observe Lorentz contraction of other bodies moving any direction, however, the latter may observe inflation of bodies moving in a certain direction. It annoys me to which acceleration system I should replace gravitation field.

The effects you mentioned do depend on velocity rather than acceleration.

 

[MikeGomez]

If we get more sophisticated though, we might ask how the weight of a test particle varies with its position, or with its spin. Or we might inquire about the force on a moving test particle, rather than a stationary one. When we ask these more sophisticated questions, we need a more complex model of gravity.

These more sophisticated notions of gravity usually involve space-time curvature and the Riemann curvature tensor, and we take the viewpoint that in flat space-time the Riemann curvature tensor is zero regardless of whether an object is accelerating or not. The Riemann curvature tensor formulation allows us to answer the questions I asked above - how much does the force on a test particle change when we change it's position slightly? How much does it change when the particle is moving vs when it is still. What is the effect of spin on the weight of the particle?

At the more sophisticated level, we realize that the answers to these extended questions depend on whether or not we are in an accelerating elevator, or near a large mass. So we can, for instance, by using a gravity gradiometer such as the "Forward Mass Detector", we can distinguish the cases of standing on the Earth, and being in an accelerating spaceship.

@pervect

Yes, you can distinguish between the cases of standing on the Earth and being in an accelerating spaceship due to the non-vanishing of the Riemann curvature tensor in the first case and zero curvature tensor in the later. However, as you move the test particle from one location to another in sufficiently small regions of spacetime, doesn’t the metric tensor reduce to the Minkowski metric? That means that each point in the curved spacetime in a sufficiently small region has an equivalence to a flat spacetime acceleration. It seems to me that you are implying that since globally each equivalence case is different, that gravity is different in curved spacetime than in flat spacetime, or to put another way, you are saying that gravity which can be transformed away is different from gravity which can not.

That is the mistake that Reichenbacher and Kottler made when they denied the validity of the equivalence principle. I know that you are not invalidating the EP, but it seems that you have the same argument as they did regarding gravity which can be transformed away. Here is Einstein’s response.

http://einsteinpapers.press.princeton.edu/vol7-trans/220

“I now turn to the objections against the relativistic theory of the gravitational field. Here, Herr Reichenbacher first of all forgets the decisive argument, namely, that the numerical equality of inertial and gravitational mass must be traced to an equality of essence. It is well known that the principle of equivalence accomplishes just that. He (like Herr Kottler) raises the objection against the principle of equivalence that gravitational fields for finite space-time domains in general cannot be transformed away. He fails to see that this is of no importance whatsoever. What is important is only that one is justified at any instant and at will (depending upon the choice of a system of reference) to explain the mechanical behavior of a material point either by gravitation or by inertia. More is not needed; to achieve the essential equivalence of inertia and gravitation it is not necessary that the mechanical behavior of two or more masses must be explainable as a mere effect of inertia by the same choice of coordinates. After all, nobody denies, for example, that the theory of special relativity does justice to the nature of uniform motion, even though it cannot transform all acceleration-free bodies together to a state of rest by one and the same choice of coordinates.” - Albert Einstein

 

[PeterDonis]

as you move the test particle from one location to another in sufficiently small regions of spacetime, doesn’t the metric tensor reduce to the Minkowski metric?

Only within a single local inertial frame. But within a single local inertial frame, you can't measure any spacetime curvature.

That means that each point in the curved spacetime in a sufficiently small region has an equivalence to a flat spacetime acceleration.

No, it means that a particular accelerating object, within a single local inertial frame, can't tell whether it is accelerating in a flat or curved spacetime. Points in spacetime themselves don't accelerate.

gravity is different in curved spacetime than in flat spacetime

In the sense that curved spacetime has a different geometry from flat spacetime, of course this is true. But the term "gravity" has multiple possible meanings. You can make the quoted sentence true or false by picking different meanings for "gravity"; and you can easily confuse yourself by interpreting what someone else has said using a different meaning than they were using. That is what you are doing here with what @pervect said. He is not saying that "gravity" in the sense of the acceleration you feel and the phenomena you can measure within a single local inertial frame is different in curved spacetime than in flat spacetime; of course it isn't, that's the whole point of the equivalence principle. He is only saying that "gravity" in the sense of spacetime curvature is (obviously) different in curved spacetime than in flat spacetime.

 

[MikeGomez]

No, it means that a particular accelerating object, within a single local inertial frame, can't tell whether it is accelerating in a flat or curved spacetime. Points in spacetime themselves don't accelerate.

Yes, thank you, I could have worded that better.

In the sense that curved spacetime has a different geometry from flat spacetime, of course this is true. But the term "gravity" has multiple possible meanings. You can make the quoted sentence true or false by picking different meanings for "gravity"; and you can easily confuse yourself by interpreting what someone else has said using a different meaning than they were using. That is what you are doing here with what @pervect said. He is not saying that "gravity" in the sense of the acceleration you feel and the phenomena you can measure within a single local inertial frame is different in curved spacetime than in flat spacetime; of course it isn't, that's the whole point of the equivalence principle. He is only saying that "gravity" in the sense of spacetime curvature is (obviously) different in curved spacetime than in flat spacetime.

Only pervect can say what his exact meaning is.

Globally, yes we agree that there is an obvious difference between curved spacetime and flat spacetime. But labeling this difference as "different types of gravity" seems a poor semantic choice and not a good description of the physics, because locally (small regions where EP applies) what we call gravity (whether it is the gravity of curved spacetime which can not be transformed away, or gravity in flat spacetime that we call inertia) is identical in all cases due to the EP. As per Einstein, gravity and inertia are phenomenon identical in nature.

 

[PeterDonis]

labeling this difference as "different types of gravity" seems a poor semantic choice

I agree, but the multiple uses of the word "gravity" are not my idea; that's been the case in the literature for decades.

locally (small regions where EP applies) what we call gravity (whether it is the gravity of curved spacetime which can not be transformed away, or gravity in flat spacetime that we call inertia) is identical in all cases due to the EP.

No, the gravity that is curved spacetime is not detectable locally (so you can't say it's "identical in all cases"), within a single local inertial frame; again, that's the whole point of the EP. The gravity that is inertial is identical in all cases, yes; that's one way of stating the EP.

 

[MikeGomez]

...The gravity that is inertial is identical in all cases, yes; that's one way of stating the EP.

Let’s go with the gravity that is inertial, and run a more sophisticated global experiment here on Earth as pervect suggests, which can’t be performed in flat spacetime. At the start of the experiment we are in a small enough region that things behave exactly as in an equivalent accelerating rocket.

Now we move the experiment to the top of a mountain. The experiment might have different results at this new location because, because this new location is equivalent to an accelerating rocket that is accelerating slightly less than the previous position, but we the gravity at this location is again gravity that is inertia.

Of course the experiment might be sensitive to being moved from one location to another, and these effects (tidal) due to continuously changing gravitation might not be able to be reproduced in flat spacetime, but at each instant along the way there exists an equivalent flat spacetime scenario and therefore we only have gravity that is inertia.

So I can see that the global effects of varying inertia can produce results that are not found in flat spacetime, but I only see gravity that is inertia. I don’t see any other kind of gravity that is not inertia, and that is why I keep repeating what Einstein said, that gravity and inertia are phenomenon identical in nature, and that is what I think Einstein is telling Reichenbacher when he says that it makes no difference whatsoever that gravity can not be transformed away in extended regions.

 

[PeterDonis]

I don’t see any other kind of gravity that is not inertia

Tidal gravity is not inertia; it's spacetime curvature.

 

[PeterDonis]

that is what I think Einstein is telling Reichenbacher when he says that it makes no difference whatsoever that gravity can not be transformed away in extended regions

But if gravity is just inertia, then it makes no sense to even talk about it not being transformed away in extended regions. That whole concept only applies to gravity as spacetime curvature.

 

[MikeGomez]

But if gravity is just inertia, then it makes no sense to even talk about it not being transformed away in extended regions. That whole concept only applies to gravity as spacetime curvature.

Why should it make no sense? That whole concept applies to gravity as spacetime curvature, the inertia that has not been transformed away.

 

[PeterDonis]

gravity as spacetime curvature, the inertia that has not been transformed away

Spacetime curvature is not inertia.

 

[MikeGomez]

Spacetime curvature is not inertia.

You have said that before. I am asking for an explanation of how it is not, but I think I am afraid I might be irritating you at this point and I don't want to get in trouble, so let's call it quits for now.

I really do appreciate all your help.

 

[PeterDonis]

I am asking for an explanation of how it is not

How spacetime curvature is not inertia? Isn't that obvious? Inertia is a property of individual objects. Spacetime curvature is a property of the spacetime geometry. They're different things.

 

[PeterDonis]

I am afraid I might be irritating you at this point and I don't want to get in trouble

You're not irritating me and you won't get in trouble; the topic of discussion is fine and you aren't speculating, you're just apparently using language in a very different way from me so I'm trying to understand what you mean by the terms you're using.

For example: what actual experimental observations do you associate with the terms "inertia" and "spacetime curvature"?

 

[MikeGomez]

For example: what actual experimental observations do you associate with the terms "inertia" and "spacetime curvature"?

I think of inertia as per Newton’s 1st, the tendency of a body to remain in state of rest or uniform motion in a straight line in Newtonian mechanics, or in GR as the tendency of a body to follow a geodesic unless acted upon by an external force.

I think of inertial forces (fictitious forces) as forces due to the inertia of bodies as just described

I thought that GR unifies inertial forces and gravity.

I think of spacetime curvature in 2 ways. The first has to do with the tidal forces the same as for Newtonian mechanics, which I think of as the inverse square law, and I think of it as forces due to geometry.

The second aspect of spacetime curvature that I think of is also geometric, but non-Euclidian geometry. This is the effect for example, of the circumference of the Earth not being equal to 2π times the radius.

How spacetime curvature is not inertia? Isn't that obvious? Inertia is a property of individual objects. Spacetime curvature is a property of the spacetime geometry. They're different things.

I thought that GR geometrises gravity.

 

[PeterDonis]

I think of inertia as per Newton’s 1st, the tendency of a body to remain in state of rest or uniform motion in a straight line in Newtonian mechanics, or in GR as the tendency of a body to follow a geodesic unless acted upon by an external force.

Ok, good. That's basically how I would put it as well (except I wouldn't use the word "tendency", since it's an exact law).

I think of inertial forces (fictitious forces) as forces due to the inertia of bodies as just described

I thought that GR unifies inertial forces and gravity.

It does in the sense that it says neither of them are forces. A "force" in GR is something that produces nonzero proper acceleration, i.e., something that makes the trajectory of an object not a geodesic. "Fictitious forces" (including gravity) are "forces" that arise because of a particular choice of coordinates and how those coordinates describe geodesic motion, i.e., motion with zero proper acceleration and therefore zero actual force (in the GR sense) applied.

I think of spacetime curvature in 2 ways. The first has to do with the tidal forces

Ok so far.

the same as for Newtonian mechanics, which I think of as the inverse square law

The law of tidal gravity, even in Newtonian mechanics, is not an inverse square law. That's the "fictitious" gravity force.

I think of it as forces due to geometry

I would not recommend thinking of it this way since tidal gravity manifests itself in the behavior of geodesics, which, as above, are the trajectories of objects with zero force acting on them.

There is a common confusion here which is worth going into. When we speak of "tidal forces", or for that matter when we speak of a "fictitious" force such as centrifugal force or the force of gravity, we are really being sloppy. Consider the following three cases:

(1) A person standing at rest on the surface of the Earth.

(2) A person pressed against the wall of a rotating cylindrical chamber (like those amusement park rides where you stand against the wall, the chamber starts rotating, and then the floor drops out from under your feet but you stay pressed to the wall and don't fall).

(3) An extended object free-falling radially towards Earth and being stretched by tidal gravity, setting up stresses (i.e., internal forces) inside the object.

In all three of these cases, there are forces present, but they are not, according to GR, properly described as "the force of gravity", "centrifugal force", or "tidal force". They are, respectively:

(1) The force of the Earth's surface pushing up on the person's feet, keeping them from following a geodesic path (which would be free fall towards the center of the Earth).

(2) The force of the chamber wall pushing on the person's back, keeping them from following a geodesic path (which, if we imagine the chamber far out in deep space, so there is no large gravitating mass present, would be to fly off in a straight line tangent to the chamber wall).

(3) The force of the internal bonds in the material that makes up the object, keeping its parts from following geodesic paths (which would be for the parts of the object to diverge from each other).

The second aspect of spacetime curvature that I think of is also geometric, but non-Euclidian geometry. This is the effect for example, of the circumference of the Earth not being equal to 2π times the radius.

No, this isn't spacetime curvature. It's spatial curvature, but in a particular set of coordinates (in this case, Schwarzschild coordinates). You can make this spatial curvature go away by choosing different coordinates (for example, in Painleve coordinates space around a spherically symmetric gravitating body is ordinary Euclidean 3-space).

 

[MikeGomez]

No, this isn't spacetime curvature. It's spatial curvature, but in a particular set of coordinates (in this case, Schwarzschild coordinates). You can make this spatial curvature go away by choosing different coordinates (for example, in Painleve coordinates space around a spherically symmetric gravitating body is ordinary Euclidean 3-space).

Thank you. Will research that more.

There is a common confusion here which is worth going into. When we speak of "tidal forces", or for that matter when we speak of a "fictitious" force such as centrifugal force or the force of gravity, we are really being sloppy. Consider the following three cases:

(1) A person standing at rest on the surface of the Earth.

(2) A person pressed against the wall of a rotating cylindrical chamber (like those amusement park rides where you stand against the wall, the chamber starts rotating, and then the floor drops out from under your feet but you stay pressed to the wall and don't fall).

(3) An extended object free-falling radially towards Earth and being stretched by tidal gravity, setting up stresses (i.e., internal forces) inside the object.

In all three of these cases, there are forces present, but they are not, according to GR, properly described as "the force of gravity", "centrifugal force", or "tidal force". They are, respectively:

(1) The force of the Earth's surface pushing up on the person's feet, keeping them from following a geodesic path (which would be free fall towards the center of the Earth).

(2) The force of the chamber wall pushing on the person's back, keeping them from following a geodesic path (which, if we imagine the chamber far out in deep space, so there is no large gravitating mass present, would be to fly off in a straight line tangent to the chamber wall).

(3) The force of the internal bonds in the material that makes up the object, keeping its parts from following geodesic paths (which would be for the parts of the object to diverge from each other).

I like your explanation here a lot. The commonality is that all three cases involve bodies (or particles) being pushed off their geodesics. All three cases are different manifestations of the same underlying phenomenon. Can I say it like that?

Or is it more accurate to say different views of the same phenomenon, like the three blind men describing an elephant, one describes the tail, another describes the trunk, and the third describes the feet.

 

[PeterDonis]

The commonality is that all three cases involve bodies (or particles) being pushed off their geodesics. All three cases are different manifestations of the same underlying phenomenon. Can I say it like that?

Yes, that would be a valid description.

is it more accurate to say different views of the same phenomenon

That could be appropriate too, depending on the context.

 

[MikeGomez]

My view of the matter goes something like this. Suppose we have a stationary space-time geometry. Then we can think of objects "at rest" in the geometry. There's a formal definition of what I mean by "stationary" and "at rest" that I can give, but I'm not sure if it'd be helpful. Loosely speaking, though, the stationary geoemtry is "not changing with time". In this stationary geometry, we can find the weight of test particles, the force on a non-spinning test particle required to keep it at rest.

With certain modest assumptions about smoothness, the weight of sufficiently nearby particles will all be very close, there won't be any sudden jumps. And the weight will be proportional to the mass of the test particle. We can then say that the ratio of the weight of the test particle (the force required to hold it in place) to the mass of the test particle represents "gravity", and that with this definition of "gravity", an accelerating elevator has "gravity", and so does being near a large mass. And we can equate the two.

If we get more sophisticated though, we might ask how the weight of a test particle varies with its position, or with its spin. Or we might inquire about the force on a moving test particle, rather than a stationary one. When we ask these more sophisticated questions, we need a more complex model of gravity.

These more sophisticated notions of gravity usually involve space-time curvature and the Riemann curvature tensor, and we take the viewpoint that in flat space-time the Riemann curvature tensor is zero regardless of whether an object is accelerating or not. The Riemann curvature tensor formulation allows us to answer the questions I asked above - how much does the force on a test particle change when we change it's position slightly? How much does it change when the particle is moving vs when it is still. What is the effect of spin on the weight of the particle?

At the more sophisticated level, we realize that the answers to these extended questions depend on whether or not we are in an accelerating elevator, or near a large mass. So we can, for instance, by using a gravity gradiometer such as the "Forward Mass Detector", we can distinguish the cases of standing on the Earth, and being in an accelerating spaceship.

After reading this again, I think that pervect is not making the Reichenbacher/Kottler error, and I apologize for implying that in my post # 14. Pervect's contributions here are extremely high quality and make up a significant portion of my personal notes that I take from this site. Thank you for all your contributions.