Understanding the General Relativity view of gravity on Earth - Comments

[harrylin]

That doesn't answer the question how what I wrote contradicts the Landau quote. And I'm not trying to avoid anything, but clarify by acknowledging conceptual differences.

Again you put words in my mouth that I did not say ("not according to" is not synonymous with "contradict"!). In the cited part they completely avoid the use of "inertial frame" which different people may interpret differently; rather they use terms that are understood the same by everyone.

 

[A.T.]

they completely avoid the use of "inertial frame"

If neither that quote nor the Newton quote use the term "inertial frame", then there can obviously be no contradiction in how they use it. That doesn't change the fact that the term is being used differently in classical Mechanics and GR.

 

[PAllen]

As far as I know, "inertial frame" was not part of the vocabulary at that time, and it is besides the point. A group of free falling bodies towards a planet could according to Newton's mechanics be used for local calculations as if they are in straight uniform motion, discounting the acceleration from the planet's gravitation. On that point there is no disagreement between Newton and Einstein.

PS compare with modern usage:

"in a gravitational field the particle moves so that its world point moves along an extremal or, as it is called, a geodesic [..]; however, since in the presence of the gravitational field space-time is not galilean, this line is not a "straight line", and the real spatial motion of the particle is neither uniform nor rectilinear. [..]by a suitable choice of the coordinate system one can always [turn] an arbitrary point of pace-time [into] a locally-inertial system of reference [which] means the elimination of the gravitational field in the given infinitesimal element of space-time"
- Landau & Lifchitz (Fields)

Not sure if L & L cover this later, but you can do much more than that. For an inertial particle (free fall, geodesic), you can introduce inertial coordinates that are spatially local but temporally global. That is, in mathematical terms, the metric remains diag(1,-1,-1,-1) and the connection components vanish, at the spatial origin, for all time. These are Fermi-Normal coordinates.

 

[harrylin]

If neither that quote nor the Newton quote use the term "inertial frame", then there can obviously be no contradiction in how they use it. That doesn't change the fact that the term is being used differently in classical Mechanics and GR.

Once more, that depends on the definitions. In their "mechanics" book, L&L describe "inertial frames" as "Galilean" reference systems (it's even the term they use in the older English version) and as you saw, they state in their "fields" book that the real spatial motion of the particle is neither uniform nor rectilinear while you state that it is actually considered inertial. In order to distinguish the concepts, they use "local inertial frames".

 

[A.T.]

In their "mechanics" book, L&L describe "inertial frames" as...

According to your interpretation of their definition, which frame is inertial:
- A frame at rest to the surface of a non-rotating planet?
- A frame free falling towards that planet?
- Both?

 

[PAllen]

As far as I know, "inertial frame" was not part of the vocabulary at that time, and it is besides the point. A group of free falling bodies towards a planet could according to Newton's mechanics be used for local calculations as if they are in straight uniform motion, discounting the acceleration from the planet's gravitation. On that point there is no disagreement between Newton and Einstein.

PS compare with modern usage:

"in a gravitational field the particle moves so that its world point moves along an extremal or, as it is called, a geodesic [..]; however, since in the presence of the gravitational field space-time is not galilean, this line is not a "straight line", and the real spatial motion of the particle is neither uniform nor rectilinear. [..]by a suitable choice of the coordinate system one can always [turn] an arbitrary point of pace-time [into] a locally-inertial system of reference [which] means the elimination of the gravitational field in the given infinitesimal element of space-time"
- Landau & Lifchitz (Fields)

I would have to say that parts of this wording are not modern, common usage. Especially e.g. "real spatial motion" is a concept with no plausible definition. Neither can "straight line" be defined in some way other than geodesic to make the statement that a geodesic is not straight. I would call modern books on GR as e.g. Wald, Carroll, Straumann.

 

[Dale]

Once more, what most matters for physics is the methods of calculation; and Landau gives a good example of vocabulary that is reasonably theory neutral.

I agree, that is what matters most. The disagreement is (at this level) a purely semantic one. The semantics are different, so I tried to capture that.

Note, that the quote from Newton is not limited to gravity. For example, a bunch of electrons interacting with each other in the presence of an external uniform E-field would also satisfy the description given by Newton and could use the same simplification mentioned by him. However, neither GR nor Newtonian physics would consider such a frame to be an inertial frame. So I do not think that the description is intended to be a definition of an inertial frame. I think it is intended to be a description of the utility of non-inertial frames.

What is the Landau vocabulary that you are talking about?

 

[PAllen]

To clarify terminology, I would like to propose a specific situation and what I think is essentially universal modern terminology. For any that disagree, this should hopefully focus discussion. Consider:

1) A lab sitting sitting on Earth under the fictitious assumption that Earth is isolated in an empty universe, and is not rotating.

2) A space lab in orbit around earth.

Newtonian terminology:

1) The Earth lab is an inertial frame, and as with any true inertial frame in Newtonian physics, it is global in extent - covering the whole universe.

2) The space lab is an accelerating frame, with the special feature that it locally only can be treated as if it were inertial. Unlike a true inertial frame, there is no way to give it global extent while preserving the characteristics of simulating an inertial frame.

Relativistic Terminology:

1) A lab sitting on Earth is an accelerated frame. It can be made part of some natural coordinate system that is asymptotically Minkowski at infinity because the spacetime is asymptotically flat. There are several such natural coordinates (e.g. standard exterior Schwarzschild, isotropic exterior Schwarzschild, etc.).

2) The space lab is an inertial frame. Like any frame in GR, it is local. There are no useful global coordinates based on the space lab. The quasi-local coordinates that correspond to this inertial frame are Fermi-Normal coordinates, which will be indistinguishable in properties from Minkowski coordinates over the space station, for all time (not just for some finite period). This latter feature is due to this being a true inertial frame in GR. (The Fermi-Normal coordinates are primarily useful within a reasonable size world tube encompassing the space lab's history).

 

[Dale]

That seems right to me. The only minor detail is that I wouldn't say that a physical object is a reference frame, but I realize that saying things like "the frame of the lab" or "the frame where the lab is at rest" makes the wording more cumbersome.

 

[PAllen]

That seems right to me. The only minor detail is that I wouldn't say that a physical object is a reference frame, but I realize that saying things like "the frame of the lab" or "the frame where the lab is at rest" makes the wording more cumbersome.

Agreed.

 

[PAllen]

...

Relativistic Terminology:

1) A lab sitting on Earth is an accelerated frame. It can be made part of some natural coordinate system that is asymptotically Minkowski at infinity because the spacetime is asymptotically flat. There are several such natural coordinates (e.g. standard exterior Schwarzschild, isotropic exterior Schwarzschild, etc.).

...

I should add that the Earth lab coordinates that best capture experience of 'inertial except for nearly uniform acceleration' are also Fermi-Normal coordinates in which the metric is almost Minkowski for the lab (exactly at e.g. lab center), for all time, but there are time independent, nearly constant, connection coefficients. These coordinates are NOT useful globally, and are NOT the same as any of the convenient global coordinates. This is important to note in reference to all frames being local in GR. Note also, such coordinates are close to Rindler coordinates in flat spacetime with the origin suitably translated - up to the order of tidal effects.

 

[PeterDonis]

Fermi-Normal coordinates in which the metric is almost Minkowski for the lab (exactly at e.g. lab center), for all time, but there are time independent, nearly constant, connection coefficients.

I assume that by "almost Minkowski" you mean "Minkowski except for the extra term in $g_{00}$", correct?

 

[vanhees71]

That seems right to me. The only minor detail is that I wouldn't say that a physical object is a reference frame, but I realize that saying things like "the frame of the lab" or "the frame where the lab is at rest" makes the wording more cumbersome.

I'd say, a reference frame is always determined by some physical object. How else should it be realized?

 

[PAllen]

I assume that by "almost Minkowski" you mean "Minkowski except for the extra term in $g_{00}$", correct?

No, g00 will take the approximate form 1 + 2a z, which is 1 for z=0. Note:

- It is a universal feature of FN coordinates that the metric is exactly Minkowski at the origin, for all time (but not so for connection coefficients, unless the origin world line - time axis - is an inertial world line).

- To see this from Rindler coordinates, translate the origin to z = 1/g (the world line whose proper acceleration is g). Get g00 = (1 + gz)², consistent with the general first order value for g00 in FN coordinates (In general, for FN coordinates, we have proper acceleration as a function of time, and and varying in direction over time. Then, the first order form of g00 is 1+2 a_j(t)x^j, with summation implied. )

[edit: I guess another aspect of almost Minkowski is that deviation from Minkowski is first order proportional to distance from the origin. The smaller the 'lab' the less deviation anywhere from Minkowski metric. Again, the connection coefficients encode the acceleration, which does not vanish locally. This is mathematically what I mean by 'almost Minkowski except for acceleration'].

 

[Dale]

I'd say, a reference frame is always determined by some physical object. How else should it be realized?

Consider an otherwise isolated system of two equal-mass classical charges. What would you consider to be the most natural reference frame? I would consider the inertial center of momentum frame most natural, not one of the non inertial frames attached to the charges.

 

[PeterDonis]

g00 will take the approximate form 1 + 2a z, which is 1 for z=0

Ah, right, got it.

 

[vanhees71]

Consider an otherwise isolated system of two equal-mass classical charges. What would you consider to be the most natural reference frame? I would consider the inertial center of momentum frame most natural, not one of the non inertial frames attached to the charges.

Sure! But to realize such a frame you need indeed some materialization of it, i.e., a measurement apparatus prepared such that it is at rest relative to the center-of-momentum frame, which then should be an inertial frame (which of course can be tested either as soon as you have materially realized this frame).

 

[A.T.]

Sure! But to realize such a frame you need indeed some materialization of it, i.e., a measurement apparatus prepared such that it is at rest relative to the center-of-momentum frame

Not sure what you mean by "realize a reference frame". But to define a reference frame, one certainly doesn't need any physical object to be at rest in that frame.

 

[Dale]

to realize such a frame you need indeed some materialization of it, i.e., a measurement apparatus prepared such that it is at rest relative to the center-of-momentum frame.

No, you don't. We know quite well the solar system's center of momentum frame despite not having any measurement apparatus at rest relative to the frame. Similarly, the GPS Earth centered inertial frame is extremely well realized without any at-rest measurement apparatus.

This started from an exceedingly minor nitpick about a well written post, and I don't want to make any more of it than we already have. Please do not read more into it than was there.

 

[vanhees71]

Sure, but the GPS satellites are a material realization of some reference frame (or many local reference frames), from which you can evaluate the coordinates in whatever other frame you want. Without a material realization of some reference frame you cannot measure anything. As theoretical physicists we often forget that the world does not consist of quadruples of numbers (coordinates) but that you have to realize reference frames in order to map the space-time manifold to subsets of $\mathbb{R}^4$.

 

[Dale]

As theoretical physicists we often forget that the world does not consist of quadruples of numbers (coordinates)

I don't think that anyone here is making that mistake. However, avoiding this mistake does not require making the alternate mistake of saying that a physical object "is" a related mathematical quantity.

from which you can evaluate the coordinates in whatever other frame you want

Precisely. This can be done because the frame is not the material object, it is a mathematical quantity.

 

[vanhees71]

Then explain to me, how to define a frame in practice without realizing it somehow as a material object.

 

[A.T.]

Then explain to me, how to define a frame in practice without realizing it somehow as a material object.

One can define a reference frame, without any physical objects at rest in that frame, simply by stating that the frame moves a velocity v relative to some physical object.

 

[Dale]

Then explain to me, how to define a frame in practice without realizing it somehow as a material object.

The way GPS does it is a good example. Do you not understand that there is no material object in the GPS system which "is" the ECI frame?

But in any case this misses the heart of the issue. Even when you do attach a reference frame to a material object, the frame and the object are not the same thing. The frame is a mathematical tool for the analysis, the object is a physical material thing.

Do you not understand the difference between a material object and a mathematical tool? Were you not the one who was complaining that people mistake the world for the coordinates? It seems like that comment was introspective, because I don't see anyone else here showing any indication of that confusion besides yourself.

 

[vanhees71]

The GPS satellites are not nothing but made of something. I'm not familiar with the technical details and how the reference frame(s) are defined, but for sure the set of satellites realize a reference frame.

To the contrary, I'm complaining about people that take the coordinates for the world, but I think it's only a semantical issue, and we shouldn't discuss it further here.

 

[Dale]

The GPS satellites are not nothing but made of something.

This is a complete strawman. Nobody is claiming this.

I'm not familiar with the technical details and how the reference frame(s) are defined, but for sure the set of satellites realize a reference frame.

No part of the GPS system is at rest in the ECI frame. And there is no sense in which the material of the GPS system "is" the ECI frame.

To the contrary, I'm complaining about people that take the coordinates for the world,

This is another strawman. Nobody is doing that here.

but I think it's only a semantical issue, and we shouldn't discuss it further here.

I certainly agree with that.

 

[harrylin]

I agree, that is what matters most. The disagreement is (at this level) a purely semantic one. The semantics are different, so I tried to capture that.

Yes, I simply gave a suggestion for less ambiguous phrasing.

[..] What is the Landau vocabulary that you are talking about?

As I said, Landau uses the term "locally inertial system of reference" (similarly others use "local inertial frame") for non-Galilean reference systems that locally can be used just like Galilean reference systems.

PS: and once more, that skillfully avoids the contradictory definitions that PAllen described in post #43

 

[A.T.]

Once more, Landau uses the term...

Once more, how would you call the following frames according to Landau's conventions?
- A frame at rest to the surface of a non-rotating planet
- A frame free falling towards that planet

 

[PAllen]

Yes, I simply gave a suggestion for less ambiguous phrasing.

As I said, Landau uses the term "locally inertial system of reference" (similarly others use "local inertial frame") for non-Galilean reference systems that locally can be used just like Galilean reference systems.

PS: and once more, that skillfully avoids the contradictory definitions that PAllen described in post #43

How does it avoid it? The Earth lab is an inertial frame per Newton and an accelerated frame per relativity.

 

[Dale]

Yes, I simply gave a suggestion for less ambiguous phrasing.

As I said, Landau uses the term "locally inertial system of reference" (similarly others use "local inertial frame") for non-Galilean reference systems that locally can be used just like Galilean reference systems.

PS: and once more, that skillfully avoids the contradictory definitions that PAllen described in post #43

I actually think that it makes the situation worse, not better. With the purported Landau definition of a local inertial frame you have that in Newtonian mechanics the apple frame and the ground frame are both local inertial frames. Since the two sets of frames accelerate wrt each other locally I think that is more confusing and contradictory than the usual terminology.

 

[PWiz]

I actually think that it makes the situation worse, not better. With the purported Landau definition of a local inertial frame you have that in Newtonian mechanics the apple frame and the ground frame are both local inertial frames. Since the two sets of frames accelerate wrt each other locally I think that is an untenable situation.

So in these kind of situations, GR says that if two or more observers which have 0 proper acceleration being acted upon them notice each other's paths to diverge/converge (i.e. notice coordinate acceleration of the other observer and can read each other's accelerometer's reading as being 0), they can conclude that spacetime around them must have non-zero intrinsic spacetime curvature, right?

 

[Dale]

So in these kind of situations, GR says that if two or more observers which have 0 proper acceleration being acted upon them notice each other's paths to diverge/converge (i.e. notice coordinate acceleration of the other observer and can read each other's accelerometer's reading as being 0), they can conclude that spacetime around them must have non-zero intrinsic spacetime curvature, right?

Yes, exactly.

 

[Ramanujan143]

I don't understand what it means that " 5° N line is constantly turning to the north and the 5° S line is constantly turning to the south".
Could someone explain?

If it is hard to see at first then consider the 89.9 degree latitude line. This is a tight little circle around the pole, so to stay on the latitude line you have to constantly turn towards the pole.

The same thing happens on the 5 degree latitude line, it just is not as tight of a turn.

sorry i still don't get it. I understand that the north latitude line is turning in a circular path, but it's not really turning towards the north pole just towards the axis of the north pole. and the same with the south latitude line, and arent the north and south pole on the same axis? so doesn't that mean both latitude lines are turning in the same direction. or am i thinking this because I'm visualizing this in three dimensions? thanks.

 

[A.T.]

or am i thinking this because I'm visualizing this in three dimensions?.

Yes, you seem to think about the 3D embedding space, which has no physical significance. The axis is not part of the 2D surface which represents curved space-time here. Only that 2D surface matters in this analogy. Try this applet, which shows the space-time geometry along a radial line:

http://www.adamtoons.de/physics/gravitation.swf

 

[Dale]

sorry i still don't get it. I understand that the north latitude line is turning in a circular path, but it's not really turning towards the north pole just towards the axis of the north pole. and the same with the south latitude line, and arent the north and south pole on the same axis? so doesn't that mean both latitude lines are turning in the same direction. or am i thinking this because I'm visualizing this in three dimensions? thanks.

For understanding Riemannian geometry on a sphere you have to consider only the 2D curved surface of the sphere, not the 3D flat space it is embedded in. The axis is not part of the surface, so in the geometry of the surface it is not something you can turn towards.

On a sphere the "straight lines" (aka geodesics) are great circles. All other paths must turn, including latitude lines other than the equator.