I Time in gravitational reference frame?

[tim9000]

Hi,
so Newton said that gravity was like or equivalent to a force?

When I learned about SR we were talking about the passing of time as defined by a photon bouncing between two parallel mirrors. So when we're sitting our two parallel mirrors in a gravitational field, even though I suppose mathematically the experience of sitting in the field is like accelerating, the mirrors aren't actually moving. So why does the photon bouncing around in them see time change? If the mirrors aren't actually moving?

Thanks

 

[Markus Hanke]

Always provided that your reference frame is small enough, locally there will be no change whatsoever in the way time is measured and experienced; your light clock will continue to operate just as it always did, i.e. it will continue to measure exactly "one second per second". Time dilation becomes apparent only if you step outside your small local frame, and compare the light clock to some other reference clock which is in a different frame - you will then notice differences in the clock readings.

Time dilation is not something that involves local "changes" of any kind, but rather it is a relationship between at least two reference frames in space-time. You can never experience time dilation by using a single, isolated clock. In essence, time dilation is a measure of how events are related in space-time, and that relationship is affected by the presence of sources of energy-momentum - that is an aspect of what we call gravity.

 

[A.T.]

Hi,
so Newton said that gravity was like or equivalent to a force?

When I learned about SR we were talking about the passing of time as defined by a photon bouncing between two parallel mirrors. So when we're sitting our two parallel mirrors in a gravitational field, even though I suppose mathematically the experience of sitting in the field is like accelerating, the mirrors aren't actually moving. So why does the photon bouncing around in them see time change? If the mirrors aren't actually moving?

Thanks

In non-inertial frames time passes at different rates depending on position, not just movement.

 

[Ibix]

so Newton said that gravity was like or equivalent to a force?

Newton described gravity as a force. It's not at all clear how light and gravity interact in a Newtonian picture. Einstein describes gravity in terms of curvature of spacetime.

When I learned about SR we were talking about the passing of time as defined by a photon bouncing between two parallel mirrors. So when we're sitting our two parallel mirrors in a gravitational field, even though I suppose mathematically the experience of sitting in the field is like accelerating, the mirrors aren't actually moving. So why does the photon bouncing around in them see time change? If the mirrors aren't actually moving?

What do you mean by the mirrors "not actually moving"? According to a free-falling observer the mirrors (and the ground) are accelerating upwards. This is a particular case of a more general point - "not actually moving" is not a well defined concept, not even in Newtonian physics.

You will not see any difference in operation of your light clock, whether you place it at the bottom of a mountain or the top. However, if you use a telescope to observe some other guy's clock at the top of the mountain from the bottom, you'll notice his clock ticking fast. You can interpret this as being because you are in an accelerating rocket, and are "catching up" with the other guy's clock (more precisely, compared to some inertial frame you were going faster at the second tick than the first) or because you are deeper in a gravitational field and are subject to time dilation due to the curvature of spacetime. But neither interpretation involves the photons in the light clock "seeing time change".

 

[tim9000]

Thanks for the replies!

Ok so I'm floating in space (minding my own business as a non-interial frame of reference), then a planet with a photon clock on it falls towards me. As the planet is falling towards me the mirror face that is furthest from the planet's surface is moving away from the planet, towards me, so it will take longer for the photon to reach that top mirror? Thus, I see the planet's clock tick slower than the photonic-Fob watch chained to my pocket?
So is that one ANALOGY for how it works, but the actual explanation is that the more distorted spacetime is from a large mass, the 'time' part of spacetime gets accelerated from the perspective of a non-enertial frame of reference?

[Mentor's note: some personal theorizing in violation of the Physics Forums rules has been removed from this post]

 

[Ibix]

Ok so I'm floating in space (minding my own business as a non-interial frame of reference),

If you're floating in space then you're in an inertial frame. The simple test is: if I stand on a spring balance, does it say I weigh anything? If the answer is no, I'm free falling and in an inertial frame.

then a planet with a photon clock on it falls towards me. As the planet is falling towards me the mirror face that is furthest from the planet's surface is moving away from the planet, towards me, so it will take longer for the photon to reach that top mirror? Thus, I see the planet's clock tick slower than the photonic-Fob watch chained to my pocket?

That's a slightly different scenario from two clocks in a tower. But assuming you are only moving radially with respect to the Earth, yes. If you're moving tangentially as well, I think it depends on how deep in the field you and the other clock are (I think).

So is that one ANALOGY for how it works, but the actual explanation is that the more distorted spacetime is from a large mass, the 'time' part of spacetime gets accelerated from the perspective of a non-enertial frame of reference?

I don't think analogy is the right word. "I'm in a rocket" and "I'm on the surface of a planet" are two competing hypotheses. Locally, both are equivalent. There's no way to tell the difference, whereas an analogy is an approximate description that's evocative rather than precise. If you start to look over a larger region you will notice that the apparent g-field is uniform or not, which will tell you whether you are in curved spacetime or not, and the "in a rocket" explanation is either falsified or not.

I don't think that "the 'time' part of spacetime gets accelerated" makes any sense. How could you describe time being accelerated? What would you differentiate, and with respect to what?

[Mentor's note: quotes of deleted text from a previous post have been removed]

 

[PeterDonis]

So is that one ANALOGY for how it works, but the actual explanation is that the more distorted spacetime is from a large mass

No, the analogy is not correct. Your analogy is with a light clock in flat spacetime, but your actual scenario is in curved spacetime (because of the presence of the planet). Flat spacetime and curved spacetime are not the same.

You are also confusing inertial motion with accelerated motion. In the usual light clock scenario in flat spacetime, the light clock is moving inertially. In the usual equivalence principle scenario, the observer is accelerated--he feels weight. Locally he can't tell whether the weight he feels is because he is inside a rocket whose engine is accelerating it, or inside a room at rest on the surface of a planet, with the planet's surface pushing up on him. But neither of those cases is the same as moving inertially. Similar remarks would apply to a light clock sitting on the surface of a planet--you would have to analyze it as an accelerated light clock, not an inertial light clock.

 

[peety]

I'd just like a little clarity about the right terminology. Alfred is at rest in a train at a station. On his left Henry is in a train that suddenly accelerates forward with the result that he spills his coffee. Henry concludes that he is at rest in a gravitational field that results in the acceleration backwards of Alfred, along with his train and the platform. None of the passengers in Alfred's train spill their coffee because... they are in free fall, they are in an inertial frame?

 

[PeterDonis]

Alfred is at rest in a train at a station.

Just to be clear, we are treating the station as floating somewhere out in space, so there is no Earth gravity involved. (Most thought experiments gloss over this, but for this one we need to make it explicit.) The station is in free fall, at rest in an inertial frame.

On his left Henry is in a train that suddenly accelerates forward with the result that he spills his coffee. Henry concludes that he is at rest in a gravitational field

Yes. And being at rest in a gravitational field, he feels acceleration, and can observe acceleration-related effects, such as spilling his coffee.

that results in the acceleration backwards of Alfred, along with his train and the platform.

That results in the coordinate acceleration backwards of Alfred, etc. But coordinate acceleration is not the same as proper acceleration, which is what Henry is experiencing. The term "acceleration" is often used sloppily, in a way that ignores this crucial distinction.

None of the passengers in Alfred's train spill their coffee because... they are in free fall, they are in an inertial frame?

Yes. And they can tell because they don't feel acceleration, whereas Henry does. The felt acceleration, or lack thereof, is an invariant, and doesn't change if we change frames, or if we introduce "gravitational fields" because we've chosen a non-inertial frame (such as Henry's frame when he spills his coffee).

 

[peety]

Just to be clear, we are treating the station as floating somewhere out in space, so there is no Earth gravity involved. (Most thought experiments gloss over this, but for this one we need to make it explicit.) The station is in free fall, at rest in an inertial frame.
Yes. And being at rest in a gravitational field, he feels acceleration, and can observe acceleration-related effects, such as spilling his coffee.
That results in the coordinate acceleration backwards of Alfred, etc. But coordinate acceleration is not the same as proper acceleration, which is what Henry is experiencing. The term "acceleration" is often used sloppily, in a way that ignores this crucial distinction.
Yes. And they can tell because they don't feel acceleration, whereas Henry does. The felt acceleration, or lack thereof, is an invariant, and doesn't change if we change frames, or if we introduce "gravitational fields" because we've chosen a non-inertial frame (such as Henry's frame when he spills his coffee).

Thanks - that's very clear and helpful. I always thought setting these thought experiments on the Earth confused things, but I like familiar territory.

 

[tim9000]

Its been so long that I'm looking over what I wrote almost as an impartial observer.

I don't think that "the 'time' part of spacetime gets accelerated" makes any sense. How could you describe time being accelerated? What would you differentiate, and with respect to what?

[Mentor's note: quotes of deleted text from a previous post have been removed]

Yeah I agree with you. I think what I was saying was: it's like time is a 'thing that happens' and when spacetime is warped, the time thing just does it's thing, but faster. However I agree, that time is only 'a thing' when relative to more than one reference frame, so yeah you're right.

Also, For the life of me I can't remember what I wrote, do you remember what my 'personal theorising' was?

So if I'm in a rocket accelerating at 9.8m/s/s, does that locally warp spacetime?

No, the analogy is not correct. Your analogy is with a light clock in flat spacetime, but your actual scenario is in curved spacetime (because of the presence of the planet). Flat spacetime and curved spacetime are not the same.

You are also confusing inertial motion with accelerated motion. In the usual light clock scenario in flat spacetime, the light clock is moving inertially. In the usual equivalence principle scenario, the observer is accelerated--he feels weight. Locally he can't tell whether the weight he feels is because he is inside a rocket whose engine is accelerating it, or inside a room at rest on the surface of a planet, with the planet's surface pushing up on him. But neither of those cases is the same as moving inertially. Similar remarks would apply to a light clock sitting on the surface of a planet--you would have to analyze it as an accelerated light clock, not an inertial light clock.

could you please explain what inertial motion is? Is it just moving at a constant speed?
and the difference between coordinate acceleration and proper acceleration?

Thanks

 

[Ibix]

Also, For the life of me I can't remember what I wrote, do you remember what my 'personal theorising' was

It'd just get deleted again even if I could remember.

So if I'm in a rocket accelerating at 9.8m/s/s, does that locally warp spacetime?

No. In both cases you feel a "force of gravity" because the floor is pushing you out of your natural free-fall path. The curvature of spacetime is why you are falling into the floor when you're on a planet. In the rocket, you are falling into the floor because there's a rocket motor pushing the floor upwards. No spacetime curvature needed.

could you please explain what inertial motion is? Is it just moving at a constant speed?

Motion with no forces acting on you. So rockets off, no electromagnetic fields, not moving through a medium (remember gravity is not a force in relativity).

and the difference between coordinate acceleration and proper acceleration?

Proper acceleration is something you measure with an accelerometer. Stand on a weighing scale. Do you weigh anything? If yes, you're undergoing proper acceleration.

Coordinate acceleration is when the rate of change of your coordinates isn't constant. If I pick a coordinate system where x=0 here, x=1 is 1m away, x=2 is 1.5m away and x=3 is 1.75m away, then you walk along at constant speed (no proper acceleration - hold a weighing scale vertically in front of you and ask if it shows anything) then you get from x=1 to x=2 in less time than from x=0 to x=1. But this is just an artifact of the coordinate system, not anything real.

That was obviously a contrived example. But non-trivial coordinate systems do arise in curved spacetime and it's important to distinguish between "my coordinate system isn't the best for what I want to measure" and "I am actually accelerating".

 

[PeterDonis]

The curvature of spacetime is why you are falling into the floor when you're on a planet.

More precisely, the curvature of spacetime is why people all over the planet can be "falling into the floor", even though "into the floor" is a different direction in different parts of the planet. There's no way to reproduce that global effect in an accelerating rocket, no matter how large you make it.

Proper acceleration is something you measure with an accelerometer.

And just to add a clarification, this measurement is how we can tell, physically, whether or not something is moving inertially. If an accelerometer attached to an object reads zero, that object is moving inertially; if not, not.

 

[MikeGomez]

The curvature of spacetime is why you are falling into the floor when you're on a planet

What if our planet were an (almost) infinite plane? Would the curvature of spacetime still be the reason why we are falling into the floor?

 

[MikeGomez]

More precisely, the curvature of spacetime is why people all over the planet can be "falling into the floor", even though "into the floor" is a different direction in different parts of the planet. There's no way to reproduce that global effect in an accelerating rocket, no matter how large you make it.

Could you reproduce that global effect with millions of accelerating rockets, one for each person?

 

[Ibix]

What if our planet were an (almost) infinite plane? Would the curvature of spacetime still be the reason why we are falling into the floor?

I'm told you can't make a uniform gravitational field. I think that this is because there are limits to material strength in relativity that you can't ignore and a disc bigger than a certain diameter will simply collapse under its own weight. So there's always curvature - even at the symmetry axis of such a disc there are non-zero second derivatives.

Might be worth waiting for (e.g.) @PeterDonis to confirm that...

 

[Nugatory]

I'm told you can't make a uniform gravitational field. I think that this is because there are limits to material strength in relativity

It's more than that - a uniform gravitational field is not a vacuum solution of the Einstein field equations, you can't arrange matter in a way that produces a uniform field around it.

 

[Nugatory]

Could you reproduce that global effect with millions of accelerating rockets, one for each person?

You could reproduce the behavior in which dropped objects fall towards the floor of the ship they're in, no matter where on the sphere the ship is. However, objects at rest on the floors of different spaceships will move apart from one another, and this doesn't happen in the gravitational case - so still no global equivalence.

 

[Ibix]

It's more than that - a uniform gravitational field is not a vacuum solution of the Einstein field equations, you can't arrange matter in a way that produces a uniform field around it.

I was trying to work out why not. Symmetry suggests the good old infinite thin plane ought to produce a similarly symmetric field, but it would also have to have infinite extent in time too. I presume there's some reason why that doesn't work.

 

[PeterDonis]

Could you reproduce that global effect with millions of accelerating rockets, one for each person?

No, because the rockets would be accelerating in different directions and so they would move apart, whereas people all standing on the surface of the Earth are at rest relative to each other--they don't move apart.

 

[phinds]

What if our planet were an (almost) infinite plane? Would the curvature of spacetime still be the reason why we are falling into the floor?

Yes. What else do you think it could be?

EDIT: that is, in General Relativity (which describes the real world more correctly than Newtonian mechanics) the curvature of spacetime IS gravity.

 

[PeterDonis]

Symmetry suggests the good old infinite thin plane ought to produce a similarly symmetric field

It does in Newtonian gravity, but IIRC there isn't a corresponding consistent solution in GR.

 

[MikeGomez]

I'm told you can't make a uniform gravitational field.

From what I have researched, I am convinced that what you were told is true. There is no uniform gravitational field, and (as I have seen here at PF and elsewhere) gravity is curvature. These two statements, that there is no uniform gravitational field, and that gravity is curvature, are consistent with the statement made by PeterDonis in another post, which is that all curvature is tidal. What is not consistent (or at least is a point of confusion for me) is that people seem to want to qualify certain types of gravity as due to curvature, and other types as not due to curvature, as you have done in post #12, when you say that no spacetime curvature is needed in the case of the rocket.

I could be wrong, but it seems to me that you are saying (and PeterDonis when he points out that globally the two situations are different) that the non-uniformity in the horizontal direction (tangential to the planet) is the cause of gravity for the vertical direction (radial from the center of the planet). That is why I brought up the example of the (almost) infinite plane. I said “almost” because it could not be perfectly infinite, and it would need some depth, so we have an extremely large planet which is not spherical. It’s true that the field is not perfectly uniform in the vertical direction, but in the horizontal direction it is, and it seems to me that the tidal gravity of the kind that you are referring to (that which is due to the spherical nature of the planet) is completely removed. In this scenario, is there a difference, even globally, between the case of the rocket or at the surface of the planet? Objects dropped to the floor in either case will not approach each other.

 

[MikeGomez]

Yes. What else do you think it could be?

You misunderstand. I agree that spacetime curvature is the reason for gravity in the case of the infinite plane. See my post #23 for the reason I brought it up.

 

[PeterDonis]

gravity is curvature

More precisely, tidal gravity is spacetime curvature. But the term "gravity" can have meanings other than "tidal gravity". For example, in the case of the accelerating rocket, the "gravity" observed is "acceleration due to gravity" (or more precisely coordinate acceleration due to gravity), which can, as that case illustrates, be present even in flat spacetime. So the issue you are seeing is one of terminology, not physics; the term "gravity" can have different meanings in different contexts, so you have to be clear about which meaning is being used.

the non-uniformity in the horizontal direction (tangential to the planet) is the cause of gravity for the vertical direction (radial from the center of the planet)

No; both of them are manifestations of tidal gravity, i.e., spacetime curvature. Neither one causes the other.

It’s true that the field is not perfectly uniform in the vertical direction, but in the horizontal direction it is

As I and others have commented, AFAIK there is no valid solution to the Einstein Field Equation that has this property. However, we can consider the Newtonian version, at least as some sort of approximate case.

it seems to me that the tidal gravity of the kind that you are referring to (that which is due to the spherical nature of the planet) is completely removed

Not necessarily; you would have to look at how the field varies in the vertical direction. Only one particular kind of vertical variation is consistent with zero tidal gravity, i.e, zero spacetime curvature.

 

[MikeGomez]

No, because the rockets would be accelerating in different directions and so they would move apart, whereas people all standing on the surface of the Earth are at rest relative to each other--they don't move apart.

The equivalence principle holds good locally, for each rocket. You object that the million rockets scenario is not globally equivalent to the spacetime curvature of the planet because the rockets move apart from each other, but the change in distance between any two rockets is a coordinate transformation. Is that why spacetime curvature is attributed to “falling to the floor” on the surface of the planet (but not falling to the floor in the spaceship) as Ibix says in post #12, and you in post#13?

 

[PeterDonis]

The equivalence principle holds good locally, for each rocket.

Yes, but "locally" means local in spacetime, not just space. See below.

the change in distance between any two rockets is a coordinate transformation

No, it isn't, it's a direct observable. The rockets can exchange light signals and observe that the round-trip travel times of the signals are changing--that's how they can tell they are accelerating rockets in flat spacetime, not rockets holding station above a spherical mass in curved spacetime. This would be true even for a neighboring pair of rockets--i.e., a pair separated by a small enough distance that, spatially speaking, they would be considered "local" to each other.

However, spotting the change in the round-trip travel time of the signals itself takes time; and the closer together the pair of rockets is, the longer it will take for them to be able to measure a change if there is one. So if we restrict ourselves to a small enough interval of time for a neighboring pair of rockets, they won't be able to distinguish the two cases and the EP holds.

 

[MikeGomez]

However, spotting the change in the round-trip travel time of the signals itself takes time; and the closer together the pair of rockets is, the longer it will take for them to be able to measure a change if there is one. So if we restrict ourselves to a small enough interval of time for a neighboring pair of rockets, they won't be able to distinguish the two cases and the EP holds.

What makes a person fall to the floor locally on any rocket, does not depend on the global configuration with respect to the other rockets. I am not seeing anything that would be different at the surface of the planet.

More precisely, the curvature of spacetime is why people all over the planet can be "falling into the floor", even though "into the floor" is a different direction in different parts of the planet. There's no way to reproduce that global effect in an accelerating rocket, no matter how large you make it.

At the planet, a global effect which causes people at different locations to fall into the floor in different directions, does not mean that the same global effect has anything to do with why they fall into the floor locally. In the case of the planet which approaches an infinite plane in shape, people in different locations fall into the floor in the same direction, so the “different directions in different parts of the planet” argument doesn’t apply here. In all cases you can say that spacetime curvature causes the person to fall into the floor (or that the floor pushes them up),

 

[PeterDonis]

What makes a person fall to the floor locally on any rocket, does not depend on the global configuration with respect to the other rockets. I am not seeing anything that would be different at the surface of the planet.

That's correct, provided "locally" is interpreted as meaning local in both space and time, as I said before.

At the planet, a global effect which causes people at different locations to fall into the floor in different directions, does not mean that the same global effect has anything to do with why they fall into the floor locally.

But it does have something to do with why all of those local fallings-into-the-floor fit together globally the way they do.

In the case of the planet which approaches an infinite plane in shape, people in different locations fall into the floor in the same direction, so the “different directions in different parts of the planet” argument doesn’t apply here.

That's correct, and it means that spacetime curvature, at least of the form that is present around an ordinary spherical planet, disappears in this limit. But that's irrelevant to why things fall to the floor locally, by your own argument. See below.

In all cases you can say that spacetime curvature causes the person to fall into the floor (or that the floor pushes them up),

No, you can't. You said that what makes a person fall to the floor locally does not depend on the global configuration, and I agreed (see above). But spacetime curvature is the global configuration. So spacetime curvature can't be what makes a person fall to the floor locally, by your own logic.

 

[PeterDonis]

the curvature of spacetime IS gravity.

No, it isn't. It's tidal gravity. But tidal gravity is not what makes things fall into the floor locally on the surface of a planet (or in an accelerating rocket).

 

[phinds]

No, it isn't. It's tidal gravity. But tidal gravity is not what makes things fall into the floor locally on the surface of a planet (or in an accelerating rocket).

Thank you Peter. I KEEP getting that wrong.

 

[MikeGomez]

I don’t have a problem with the statement that gravity is spacetime curvature, because I don’t feel the need to qualify that with “tidal” gravity. We had the discussion in earlier posts that no uniform gravitational field exists and I assumed that was true because nobody challenged it, and that indicates to me that all gravity is tidal.

I guess there is a terminology issue that I have been missing regarding spacetime curvature. If spacetime curvature is present globally then it seems to me it must exist locally. By my own logic, spacetime curvature would be a broad term with many aspects, some of which have a global effect on separated spacetime events, and some of which have an effect locally in a area where the separation is small enough that the EP holds. When you say “… spacetime curvature, at least of the form that is present around an ordinary spherical planet, disappears in this limit” I thought that the one form of spacetime curvature which disappears in the limit is due to the spherical shape of the planet, and not not fundamentally different from any other form of spacetime curvature (or gravity) which remains. Anyway, I guess that’s wrong. Could you please discuss a little about this non-tidal gravity which makes things fall to the floor locally.

Thanks

 

[PeterDonis]

I don’t have a problem with the statement that gravity is spacetime curvature, because I don’t feel the need to qualify that with “tidal” gravity.

I'm not sure this is a good policy on your part. See below.

We had the discussion in earlier posts that no uniform gravitational field exists and I assumed that was true because nobody challenged it, and that indicates to me that all gravity is tidal.

The "gravity" that makes a rock fall locally on the surface of a planet is not tidal gravity, as I have already said. But the term "gravity" is very commonly used to refer to this phenomenon, so the statement "all gravity is tidal" is, at the very least, liable to cause confusion.

It is true, AFAIK, that "no uniform gravitational field exists" in the sense that there is no known solution to the Einstein Field Equation that can correctly be described as a "uniform gravitational field". But going from that to "all gravity is tidal" is a logic error; the only correct logical deduction you can make from it is "all gravitational fields are non-uniform". (We're also assuming here that the term "gravitational field" has an unambiguous meaning, which is not really true; that term also has multiple meanings in the literature.) As I noted just now, there are phenomena taking place in non-uniform gravitational fields which are referred to as "gravity" but are not "tidal gravity".

If spacetime curvature is present globally then it seems to me it must exist locally.

The term "locally" is yet another term with multiple meanings. In one sense of the term, spacetime curvature is indeed local, since it is described by a tensor, the Riemann tensor, and tensors are "local" objects--they are defined in the tangent space at each point of spacetime.

However, in another sense of "local", spacetime curvature is not local, because it involves second derivatives of the metric, and in a small enough patch of spacetime around a given point, we can always adopt coordinates (local inertial coordinates) in which the effects of second derivatives of the metric are negligible. But phenomena that are referred to as "gravity" are still detectable in that small patch of spacetime--for example, bricks fall when dropped in an accelerating rocket, and this can be described, within a local inertial frame, as the rocket accelerating upward while the brick remains at rest, just as a brick falling on Earth can be described, in a local inertial frame, as the surface of the Earth accelerating upward while the brick remains at rest. All of this holds even though the effects of second derivatives of the metric are negligible within a local inertial frame, i.e., spacetime curvature is not "local" in this sense.

some of which have an effect locally in a area where the separation is small enough that the EP holds.

No effects of spacetime curvature are detectable in a patch of spacetime small enough that the EP holds. That is the definition of what "small enough" means in terms of the EP--if any effects of spacetime curvature are detectable, it means you aren't focusing on a small enough patch of spacetime for the EP to apply.

spacetime curvature, at least of the form that is present around an ordinary spherical planet, disappears in this limit”

We should probably drop that particular example, since I've already said several times that AFAIK it doesn't correspond to any valid solution of the EFE. Saying that "well, you just take a spherical planet and take the limit as its radius goes to infinity" doesn't mean that limit corresponds to anything valid.

Could you please discuss a little about this non-tidal gravity which makes things fall to the floor locally.

See above.

One other note: you appear to me to be getting hung up on terminology instead of focusing on the physics. This is one reason why discussions of this sort are best done with math, not ordinary language. The math underlying everything I've said is completely unambiguous, and raises no awkward questions about what "gravity" means or what "local" means or what "spacetime curvature" means.

 

[MikeGomez]

The term "locally" is yet another term with multiple meanings. In one sense of the term, spacetime curvature is indeed local, since it is described by a tensor, the Riemann tensor, and tensors are "local" objects--they are defined in the tangent space at each point of spacetime.

However, in another sense of "local", spacetime curvature is not local, because it involves second derivatives of the metric, and in a small enough patch of spacetime around a given point, we can always adopt coordinates (local inertial coordinates) in which the effects of second derivatives of the metric are negligible.

Is this the vanishing of the Christoffel symbols (or making them small enough to be negligible)?

But phenomena that are referred to as "gravity" are still detectable in that small patch of spacetime--for example, bricks fall when dropped in an accelerating rocket, and this can be described, within a local inertial frame, as the rocket accelerating upward while the brick remains at rest, just as a brick falling on Earth can be described, in a local inertial frame, as the surface of the Earth accelerating upward while the brick remains at rest. All of this holds even though the effects of second derivatives of the metric are negligible within a local inertial frame, i.e., spacetime curvature is not "local" in this sense.

So what would be a good name for this type of gravity? Having to say “the gravity that makes you fall into the floor” is a mouthful, and “apple falling gravity” or “apple gravity” sounds dumb.

One other note: you appear to me to be getting hung up on terminology instead of focusing on the physics. This is one reason why discussions of this sort are best done with math, not ordinary language. The math underlying everything I've said is completely unambiguous, and raises no awkward questions about what "gravity" means or what "local" means or what "spacetime curvature" means.

Very well then. A few terminology issues remain (for me) but that can be postponed for another time. What I am still hung up on is Ibis’s comments in post #12, which seem to indicate that the physics (and not just terminology) are different for the two cases of falling to the floor on the planet and falling to the floor in the rocket. If that is what he is really saying, I don't think the math will support his argument.

 

[Nugatory]

Is this the vanishing of the Christoffel symbols (or making them small enough to be negligible)?

No. You can make the curvature tensors vanish even when the Christoffel symbols do not. Heuristically speaking, the Christoffel symbols capture the effects of both curvature of the manifold and the effects of our choice of coordinates; the Riemann tensor is built out of combinations of these symbols in which the coordinate effects cancel out leaving only the coordinate-independent effects of curvature. For example, in a Euclidean plane the Christoffel coefficients are non-zero in polar coordinates and zero in cartesian; but the plane is flat and the Ricci and Riemann tensors are zero everywhere.

So what would be a good name for this type of gravity?

"Gravity" works fine, and we can use "tidal gravity" for the non-local curvature-dependent effects. (Consider that in natural language no one has any trouble with the convention that "milk" comes from cows even though "soy milk" does not).

Very well then. A few terminology issues remain (for me) but that can be postponed for another time. What I am still hung up on is Ibis’s comments in post #12, which seem to indicate that the physics (and not just terminology) are different for the two cases of falling to the floor on the planet and falling to the floor in the rocket. If that is what he is really saying, I don't think the math will support his argument.

The two cases are different. If I drop two objects from my two outstretched hands, and measure the distance between where they hit the floor... On the accelerating spaceship that distance will be equal to the distance between my outstretched hands, and on the surface of a planet it will be slightly less. The equivalence principle just says that we can make that difference arbitrarily small.

 

[MikeGomez]

The two cases are different. If I drop two objects from my two outstretched hands, and measure the distance between where they hit the floor... On the accelerating spaceship that distance will be equal to the distance between my outstretched hands, and on the surface of a planet it will be slightly less.

I understand that. It is a tidal effect. It doesn’t mean that dropping an object from one hand is any different between the two scenarios.

"Gravity" works fine, and we can use "tidal gravity" for the non-local curvature-dependent effects. (Consider that in natural language no one has any trouble with the convention that "milk" comes from cows even though "soy milk" does not).

But milk from cows and milk from soy are pretty different. With gravity, it’s not like that. Perhaps more like the difference in the air between when it is blowing and when it is not, or the difference in the bath water when the water swirling or when it is not, or the difference between the electrons in AC or DC.

No. You can make the curvature tensors vanish even when the Christoffel symbols do not. Heuristically speaking, the Christoffel symbols capture the effects of both curvature of the manifold and the effects of our choice of coordinates; the Riemann tensor is built out of combinations of these symbols in which the coordinate effects cancel out leaving only the coordinate-independent effects of curvature. For example, in a Euclidean plane the Christoffel coefficients are non-zero in polar coordinates and zero in cartesian; but the plane is flat and the Ricci and Riemann tensors are zero everywhere.

Most excellent! Thank you.

 

[MikeGomez]

BTW, Einstein addressed this issue when Hans Reichenbach made this very same mistake. When he (Einstein) says that it is of no importance whatsoever that the gravitational fields in general cannot be transformed away, he is speaking about the difference between tidal gravity and non-tidal gravity. Is he not?

“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.”
-A. Einstein

 

[PeterDonis]

what would be a good name for this type of gravity?

The term "acceleration due to gravity" is sometimes used for it; also "the force of gravity". Both of these have issues as well, of course.

When he (Einstein) says that it is of no importance whatsoever that the gravitational fields in general cannot be transformed away, he is speaking about the difference between tidal gravity and non-tidal gravity. Is he not?

Yes. Tidal gravity is the part that can't be transformed away; but the EP only refers to the other part, the part that can.

 

[MikeGomez]

However, in another sense of "local", spacetime curvature is not local, because it involves second derivatives of the metric, and in a small enough patch of spacetime around a given point, we can always adopt coordinates (local inertial coordinates) in which the effects of second derivatives of the metric are negligible. But phenomena that are referred to as "gravity" are still detectable in that small patch of spacetime--for example, bricks fall when dropped in an accelerating rocket, and this can be described, within a local inertial frame, as the rocket accelerating upward while the brick remains at rest, just as a brick falling on Earth can be described, in a local inertial frame, as the surface of the Earth accelerating upward while the brick remains at rest. All of this holds even though the effects of second derivatives of the metric are negligible within a local inertial frame, i.e., spacetime curvature is not "local" in this sense.

So then speaking in the proper language, what makes an object fall to the floor is due to the first derivatives of the metric tensor, not spacetime curvature, which are the second derivatives of the metric tensor. It doesn’t matter if the planet is round, infinite plane, or unicorn shaped.

Yes. Tidal gravity is the part that can't be transformed away; but the EP only refers to the other part, the part that can.

And Einstein is pointing out that the part that can’t be transformed away is of no consequence to how the part that can be transformed away is equivalent for the planet and the rocket.

 

[PeterDonis]

So then speaking in the proper language, what makes an object fall to the floor is due to the first derivatives of the metric tensor

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve. And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

spacetime curvature, which are the second derivatives of the metric tensor

More precisely, spacetime curvature is described by the Riemann tensor, which is a particular combination of second derivatives (and first derivatives) of the metric tensor that transforms as a tensor itself.

Einstein is pointing out that the part that can’t be transformed away is of no consequence to how the part that can be transformed away is equivalent for the planet and the rocket.

Yes.

 

[MikeGomez]

And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

And is that because the first derivatives of the metric tensor are transformed away?

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve.

Perhaps that is how it must be viewed in order to use the mathematical tools of GR, but there must be a better way to describe it than that if we are free to choose whatever reference frame we like. We can say the object is falling to the floor, just as well as the floor is accelerating up. The object is deflecting the planet from its freefall path in addition to the planet deflecting the object from its freefall path.

 

[PeterDonis]

is that because the first derivatives of the metric tensor are transformed away?

Yes; the transformation from a generic coordinate chart to a local inertial coordinate chart makes all of the first derivatives of the metric vanish.

The object is deflecting the planet from its freefall path in addition to the planet deflecting the object from its freefall path.

The planet, or at least the small piece of it that we're talking about, is not on a free-fall path to begin with. It's on an accelerated path. And the object doesn't deflect the planet's surface, because it's not free to be deflected; it's being pushed on by the rest of the planet. So the situation is not symmetric between the planet's surface and the object.

 

[MikeGomez]

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve. And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

I’m certain this is correct (and extremely useful for doing physics) but I have doubts as to whether this is the definitive answer for what makes an object fall to the floor. It seems to be giving preferential treatment to a particular frame of reference in a relative situation. The planet is in freefall. Does choosing a piece of the surface of the planet as having an accelerated path have any more physical significance than selecting a frame of reference, or is there a deeper physical significance here that I am missing?

 

[PeterDonis]

It seems to be giving preferential treatment to a particular frame of reference in a relative situation.

That's because the description "fall to the floor" does the same thing: it gives preferential treatment to the frame in which the planet's surface is at rest. In the object's local inertial frame, the planet is rising up to hit it.

If you really want a "definitive" answer that doesn't make any use of particular frames, you would have to describe the worldlines of the object and the piece of the planet's surface and how they intersect at some particular event.

The planet is in freefall.

The center of mass of the planet is in free fall. But a particular piece of the surface of the planet is not. It has nonzero proper acceleration. Put another way, it feels a force--the force of the piece of the planet below it pushing up on it. The object that is falling, by contrast, feels no force; it is in free fall.

Does choosing a piece of the surface of the planet as having an accelerated path have any more physical significance than selecting a frame of reference, or is there a deeper physical significance here that I am missing?

There is a deeper physical significance: as above, the piece of the planet's surface feels a force, while the falling object does not. That is an invariant statement, independent of any choice of coordinates.