Insights Understanding the General Relativity view of gravity on Earth - Comments

[harrylin]

And what does "free motion" mean here? The employed model of gravity (Newtonian force vs. GR) determines which object is "force free".

Surely you can answer that question yourself: are systems that are affected by gravitational fields, generally in "constant straight line and uniform motion relative to each other"?

Anyway, for sure the elaborations here were many times more than what textbook authors assume to be sufficient; it won't be useful to comment or clarify more.

 

[Dale]

Free falling reference systems only mimic Galilean frames locally for the physics.

Galilean frames clearly mimic Galilean frames also.

Once more: the ECI frame is a free falling reference system of the Earth, but does not correspond to the free falling local reference system of a group of particles near the Earth.

Yes, that is the problem. Two different free falling reference frames do not correspond to each other and are not equivalent even though they both cover some of the same events.

 

[vanhees71]

But it's clear that an observer on the surface of the Earth is not inertial by definition, because he is not freely falling because of the electromagnetic interactions of the material (together with Pauli blocking for that matter) around with the observer. I think, it's very clear that the ECI is not a local inertial reference frame. Why some authors call such a frame Galilean is one of the great mysteries of the textbook writers, which I never understood. Galilei-Newton spacetime is very different from the general-relativistic (Einstein-Hilbert) spacetime. I'd prefer to call the reference frames that are realized by freely falling bodies (note again, that's a very real issue!) local inertial frames.

 

[Dale]

I think, it's very clear that the ECI is not a local inertial reference frame... I'd prefer to call the reference frames that are realized by freely falling bodies (note again, that's a very real issue!) local inertial frames.

On the contrary, it is very clear that the ECI is a local inertial reference frame (after all, the "I" in "ECI" is for "Inertial"). It is freely falling around the sun and therefore clearly qualifies as a "local inertial frame" per your usage and per the Landau usage.

In fact @harrylin is incorrect in claiming that the ECI is a "Galilean reference frame", but I share your distaste for the term. The ECI is a local inertial frame: it is free-falling around the sun, which is in turn free falling around the galaxy, ...

But it's clear that an observer on the surface of the Earth is not inertial by definition, because he is not freely falling because of the electromagnetic interactions of the material (together with Pauli blocking for that matter) around with the observer.

Sure, (neglecting rotation) such an observer is not inertial. However, in Newtonian mechanics they are at rest in an inertial frame, the ECI. They are acted on by two real forces, the contact force and gravity, which cancel each other out. So although the observer itself is not inertial, their rest frame (the ECI) is inertial.

 

[PeterDonis]

it is very clear that the ECI is a local inertial reference frame (after all, the "I" in "ECI" is for "Inertial")

It might be "inertial" in the Newtonian sense, yes, but not in the GR sense. Which, of course, just underscores the ambiguity in terminology that has driven much of this thread.

ECI stands for "earth centered inertial" frame, and as used with GR, it has a metric varying radially from the center, with connection coefficients becoming non-vanishing away from the center.

This means the ECI is not a local inertial frame in the standard GR sense; such a frame would have vanishing connection coefficients everywhere within its domain. (The fact that the connection coefficients must vanish, to the accuracy of measurement, is what restricts the domain of a local inertial frame to a small patch of spacetime.) What you're describing, in GR terms, are more like Fermi normal coordinates centered on a freely falling worldline; such coordinates are not a local inertial frame because they can cover an entire "world tube" centered on the worldline, not just a small patch centered on a particular event. and the connection coefficients can become non-vanishing off the centered worldline because of spacetime curvature.

Also, as I understand it, the ECI frame, from a GR point of view, takes the metric for Fermi normal coordinates centered on a freely falling worldline, and adds in the Earth's gravitational potential "by hand" in the appropriate metric coefficients. (See, for example, the treatment in section 3 of the Living Reviews article on relativity in the GPS http://relativity.livingreviews.org/Articles/lrr-2003-1/fulltext.html .) This means that, in GR terms, the ECI is not even an inertial frame in a small patch of spacetime; its metric is not Minkowski anywhere, because of the gravitational potential.

So the only sense in which the ECI could be said to be "inertial" is the Newtonian sense in which DaleSpam is using the term here. (The main intent of the "I" in ECI appears to be to signify that it is non-rotating, as opposed to the ECEF frame which rotates with the Earth. In GR terms, once again, this would mean Fermi normal coordinates, not local inertial coordinates--but then we still have the Earth's gravitational potential added in, as above.)

 

[Dale]

It might be "inertial" in the Newtonian sense, yes, but not in the GR sense. Which, of course, just underscores the ambiguity in terminology that has driven much of this thread.

Yes. The GPS system does not use GR in its computations, other than a correction for time dilation. Other than that, it treats spacetime as flat.

 

[PeterDonis]

The GPS system does not use GR in its computations, other than a correction for time dilation.

If you mean that the calculations of the satellite orbits (which are essential to the position data sent to receivers) don't require GR, that's true; GR effects are much too small to matter for anything other than time dilation.

Other than that, it treats spacetime as flat.

That's not quite true; if you look at the metric in the Living Reviews article I referenced, it has a correction term in the spatial part of the metric as well. But that correction term turns out to be small enough that it can be ignored (it's a factor of $c^2$ smaller than the correction to $g_{00}$). So in practical terms, yes, the GPS coordinates are assumed to be Euclidean in the spatial part.

 

[PAllen]

Well, to Peter's comment on my description, I make the following notes:

MTW does not define a proper reference frame ultralocal at all. It defines it geometrically, with the result being Fermi-Normal coordinates extended to allow for rotation of the tetrad relative to Fermi-Walker transport of a starting tetrad. In inertial frame is simply the special case where connection components vanish exactly at the origin world line (the metric is Minkowski exactly at the origin world line in all cases). Thus, ECI would a an inertial frame per this definition.

Note, from the Living Review article you reference, the author seems to agree:

"For the GPS it means that synchronization of the entire system of ground-based and orbiting atomic clocks is performed in the local inertial frame, or ECI coordinate system "

"because in the underlying earth-centered locally inertial (ECI) coordinate system"

Finally, the metric given, if naturally extended to the center (they don't bother with this since subterranean GPS is not a realized product), would have Minkowski metric an vanishing connection at the origin.

 

[PeterDonis]

MTW does not define a proper reference frame ultralocal at all.

But MTW also makes a sharp distinction between a "local inertial frame" (my term, I'd have to go back and look to see exactly what term(s) MTW uses for this), which only covers a small patch of spacetime, and what you are calling a "proper reference frame", which MTW calls Fermi normal coordinates and which can cover a "world tube" around any chosen worldline. The ECI frame is definitely not a local inertial frame by MTW's definition. Whether it qualifies as Fermi normal coordinates is more problematic, because of the extra terms in the metric due to the gravitational potential. See below.

from the Living Review article you reference, the author seems to agree

He agrees on an ambiguous use of terminology, yes. He is using "local inertial frame" in the Newtonian sense (or perhaps the "Fermi normal" sense--but see below), not the GR sense (i.e., the MTW sense I referred to above). Unfortunately this seems to be very common.

the metric given, if naturally extended to the center (they don't bother with this since subterranean GPS is not a realized product), would have Minkowski metric an vanishing connection at the origin.

It would if you rescaled the potential $V$ to be zero at the center of the Earth, yes (the standard ECI frame does not do this; the potential is effectively zero on the geoid, so it would be negative at the center of the Earth).

However, that isn't enough to make ECI coordinates the same as Fermi normal coordinates. At least as I read MTW's discussion of those, they assume that the object following the chosen worldline is a test object, and does not produce any spacetime curvature on its own; the nonzero connection coefficients as you move away from the chosen worldline can only be due to spacetime curvature from other sources (for example, the Sun). The Earth clearly does not meet this requirement.

 

[PAllen]

But MTW also makes a sharp distinction between a "local inertial frame" (my term, I'd have to go back and look to see exactly what term(s) MTW uses for this), which only covers a small patch of spacetime, and what you are calling a "proper reference frame", which MTW calls Fermi normal coordinates and which can cover a "world tube" around any chosen worldline. The ECI frame is definitely not a local inertial frame by MTW's definition. Whether it qualifies as Fermi normal coordinates is more problematic, because of the extra terms in the metric due to the gravitational potential. See below.

I don't recall an MTW definition of local inertial frame separate from the section on Proper Reference Frame (which has inertial frame as a special case). However, I can't check right now because I have no access to my books.

He agrees on an ambiguous use of terminology, yes. He is using "local inertial frame" in the Newtonian sense (or perhaps the "Fermi normal" sense--but see below), not the GR sense (i.e., the MTW sense I referred to above). Unfortunately this seems to be very common.

Of course all we can do is guess what Ashby meant. I don't see his usage as Newtonian.

It would if you rescaled the potential $V$ to be zero at the center of the Earth, yes (the standard ECI frame does not do this; the potential is effectively zero on the geoid, so it would be negative at the center of the Earth).

Yes, I agree on this. Almost as soon as I wrote my prior post I realized you would have to reset the zero point of the potential (which is arbitrary anyway).

However, that isn't enough to make ECI coordinates the same as Fermi normal coordinates. At least as I read MTW's discussion of those, they assume that the object following the chosen worldline is a test object, and does not produce any spacetime curvature on its own; the nonzero connection coefficients as you move away from the chosen worldline can only be due to spacetime curvature from other sources (for example, the Sun). The Earth clearly does not meet this requirement.

My read is different from yours. The frame is based on a world line not a body of any kind (test or otherwise). The center of the Earth is perfectly ok. Nothing in their derivation restricts the world line to being in vacuum (any mix of Weyl and Ricci curvature is accommodated by their construction).

 

[Dale]

Does anyone know if Newton Cartan spacetime is curved, or if it is only space that is curved since the space and time parts are metrically distinct? (If metrically distinct is even a valid term)

I don't know much about Newton Cartan gravity, but it might go a long way to harmonizing terminology like this.

 

[PeterDonis]

I don't recall an MTW definition of local inertial frame separate from the section on Proper Reference Frame

I'll check my copy when I get a chance.

Nothing in their derivation restricts the world line to being in vacuum (any mix of Weyl and Ricci curvature is accommodated by their construction).

I'll take a look, it's been a while since I reviewed that chapter and I may be misremembering.

 

[PAllen]

My read is different from yours. The frame is based on a world line not a body of any kind (test or otherwise). The center of the Earth is perfectly ok. Nothing in their derivation restricts the world line to being in vacuum (any mix of Weyl and Ricci curvature is accommodated by their construction).

Ok, so I guess a precise description of ECI is 'almost' what I wrote many posts ago, which was:

" the (inertial) frame of a non-spinning observer at the center of the earth" [you can think of this as a test body or small lab, though only an origin world line figures in the construction]

The correction is that ECI starts from the above, and rescales such that g00 is 1 at the geoid rather than the center, for more convenient usage at and outside the Earth's surface. To me, it is still functionally an inertial frame of a center of Earth observer, despite this coordinate transform, but this is a judgement.

 

[cianfa72]

A comment on this old insight.

@Dale said

Inertial frame: a coordinate system where inertial objects have no coordinate acceleration.

I believe no coordinate acceleration refers to 2nd derivative of spacelike coordinates w.r.t. timelike coordinate.

In inertial frames, the line formed by an inertial object’s coordinates is a straight line.

This should be just a necessary condition.To be an inertial reference frame (aka inertial coordinate chart/system) we must add the requirement of vanishing geodesic deviation for them (i.e. straight worldlines of inertial objects starting parallel remain parallel).

 

[Dale]

To be an inertial reference frame (aka inertial coordinate chart/system) we must add the requirement of vanishing geodesic deviation for them (i.e. straight worldlines of inertial objects starting parallel remain parallel).

Hmm, is that condition actually required? Is it possible to have a region of curved spacetime with geodesic deviation and a chart where all inertial objects have worldlines that are straight lines in the chart? I can’t think of an example, but certainly admit that there may be one that I haven’t considered.

 

[cianfa72]

Is it possible to have a region of curved spacetime with geodesic deviation and a chart where all inertial objects have worldlines that are straight lines in the chart? I can’t think of an example, but certainly admit that there may be one that I haven’t considered.

For sure you can map a region of curved spacetime with a chart in which a full congruence of timelike geodesics are "at rest" in it (i.e. they have constant spatial coordinates in that chart).

 

[Dale]

a full congruence of timelike geodesics are "at rest"

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

 

[cianfa72]

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

Ah ok, I see. I don't know if it is actually possible.

About the first statement in #104 do you actually refer to zero coordinate acceleration as vanishing 2nd coordinate derivative w.r.t. the timelike coordinate and/or other coordinates in the given coordinate chart ?

 

[Dale]

About the first statement in #104 do you actually refer to zero coordinate acceleration as vanishing 2nd coordinate derivative w.r.t. the timelike coordinate

Yes, this was one of the other “important concepts” bullet points.

Note, I didn’t intend for those to be rigorous definitions. I was covering too many to be rigorous. I just wanted to get the “kernel” of the idea to an audience with some calculus and physics background (e.g. two semesters undergraduate-level physics and calculus) but no GR background.

 

[cianfa72]

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

The point discussed in this thread was that in flat spacetime (i.e. Minkowski) there is only a family/congruence of timelike geodesics filling the entire manifold "at rest" in a given specific coordinate chart. If we further add the condition of zero shear/expansion and zero vorticity for those timelike geodesics we get the definition of coordinate chart as global inertial.

Now I believe the point to be investigated is whether those additional conditions are actually equivalent to require that all timelike geodesics are actually straight in that chart (even if not all "at rest" in it).

 

[cianfa72]

Any thought w.r.t. the previous post ?

 

[Dale]

I was considering not just accelerometers at rest. I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation. I cannot prove it, but the linked post is specifically dealing with a congruence of “at rest” objects, so the requirement there is not applicable.

 

[cianfa72]

I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation.

Reading again this post, I had the following doubt: consider an inertial object at rest in a given frame (coordinate chart). The above definition of inertial coordinate chart requires that for this 'at rest' inertial object the coordinate acceleration must vanish. In this special case the spatial coordinate acceleration (i.e. the derivative of spatial coordinates w.r.t. coordinate time) is of course null by definition of 'at rest in it'.

So, for this case, which is the meaning of zero coordinate acceleration ?

 

[Dale]

So, for this case, which is the meaning of zero coordinate acceleration ?

Zero coordinate acceleration means that the coordinate system consists of one timelike coordinate $t$ and three spacelike coordinates $\vec x$ and that $$\frac{d^2\vec x}{dt^2}=0$$

 

[DrGreg]

$$\frac{d\vec x}{dt}=0$$

I'm sure that's a typo and you meant to say $$\frac{d^2\vec x}{dt^2}=0$$

 

[Dale]

I'm sure that's a typo and you meant to say $$\frac{d^2\vec x}{dt^2}=0$$

Oops, yes, I corrected it

 

[PeterDonis]

I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation.

I think it does, since the standard FRW coordinates used in cosmology are coordinates in which inertial objects--comoving objects--have zero coordinate acceleration, but there is nonzero geodesic deviation.

The usual definition of an inertial chart in SR does include an (often unstated) assumption of zero geodesic deviation, since of course such a chart is intended for use in flat Minkowski spacetime.

 

[cianfa72]

Zero coordinate acceleration means that the coordinate system consists of one timelike coordinate $t$ and three spacelike coordinates $\vec x$ and that $$\frac{d^2\vec x}{dt^2}=0$$

So that actually means we cannot decide whether a coordinate chart is inertial or not by looking just at the coordinate acceleration of inertial objects at rest in it -- indeed by definition inertial objects at rest in any (1 timelike + 3 spacelike) chart have zero coordinate acceleration in it.

 

[cianfa72]

I think it does, since the standard FRW coordinates used in cosmology are coordinates in which inertial objects--comoving objects--have zero coordinate acceleration, but there is nonzero geodesic deviation

Yes but comoving objects are actually at rest in standard FRW coordinates. If we include the entire class of inertially moving objects (not just those at rest) then is the requirement about nonzero geodesic deviation still needed to declare inertial a coordinate chart ?

 

[Dale]

So that actually means we cannot decide whether a coordinate chart is inertial or not by looking just at the coordinate acceleration of inertial objects at rest in it

No, that is incorrect.

indeed by definition inertial objects at rest in any (1 timelike + 3 spacelike) chart have zero coordinate acceleration in it.

Yes, but it is not possible to find inertial objects at rest is all charts. For example, in a rotating coordinate system an object at rest is not inertial and inertial objects have non-zero coordinate acceleration. So the proposed definition correctly identifies the rotating chart as non-inertial