On the coordinate independence of general relativity

I've been having a bit of trouble with the idea of coordinate independence in general relativity. Let me start with a simple example that I think illustrates my question conceptually:

Say you have two objects, A and B. A is at rest and B accelerates directly away from A. From a mathematical point of view, there is absolutely nothing which distinguishes the two objects - A is accelerating away from B as much as B is from A. To distinguish the two, we need something extra - something empirical which tells us that B feels a force acting on it and A does not.

In this sense, there are infinitely many coordinate motions which might be the truly unaccelerated ones, and they can only be distinguished by the physical fact that they feel no forces. So, at some point, doesn't any physical law, even if it is written in such a way that it is satisfied in all coordinate systems, have to make reference to truly unaccelerated frames, or something similar? In other words, in order to connect an abstract mathematical description of motion, which in some sense can be transformed in any way, to physical observations, doesn't one have to anchor the mathematical reference somehow to physically meaningful concepts like unaccelerated frames?

Now, as I've been studying the Einstein field equations, I've wondered about the following: if you compare the equations for empty, flat space, in inertial coordinates versus some weird accelerating coordinates they'd be exactly the same - the space must be Ricci flat, or $R^{\alpha \beta} = 0$. Of course, there are many solutions of $R^{\alpha \beta} = 0$ (including non-flat ones, but ignore those for a second). Without some physical knowledge, wouldn't it be impossible to say which coordinates lead to $g_{\alpha \beta} = \eta_{\alpha \beta}$, and which coordinates lead to $g_{\alpha \beta}$ being some complicated (still Ricci flat) function of the coordinates? I imagine that this means that while the field equations must be satisfied, they alone are not enough to say what someone will observe - one must also know how his coordinates relate to locally inertial ones.

So I now ask: by the "coordinate independence of General Relativity", does one really just mean that the expression for curvature makes no reference to a coordinate system - that the expression for curvature is coordinate independent, and so the law relating curvature and mass-energy, the Einstein field equation, is valid in all coordinate systems? And yet, even though the law must be satisfied, in order to know what one will observe, it's necessary to also know which coordinate systems are (locally) inertial?

Finally, if you're feeling up to it, is this a real point I'm making in general? Don't the mathematical laws of physics all have to "give up" and some point and make reference to the solely empirically-defined concept of unaccelerated motion?

EDIT: I don't think I made myself totally clear - let's just say this. If I asked you to solve Einstein's field equations in flat space in the coordinates $x^\mu$, wouldn't you, in general, have to say that there's not enough information? How could you possibly know whether $x^\mu$ were inertial coordinates and $g_{\mu \nu} = \eta_{\mu \nu}$, or if $x^\mu$ were some weird accelerated coordinates, and $g_{\mu \nu}$ were the same flat metric but written in these accelerated coordinates? Both are solutions of the Einstein field equations. In this way, aren't the field equations underdetermined? Don't they have to be supplemented with information about how the coordinates relate to locally inertial coordinates?

• You assume B feels a force, but suppose A and B are in orbit at different heights. Both are in free fall so neither feel any force, yet there is a relative acceleration between them. So which one is accelerating? – John Rennie Jul 26 '14 at 6:19
• @JohnRennie - True, but the point I was making was not that one must feel a force, but that giving their motions in some set of coordinates without saying how the coordinates relate to inertial ones, there is no way at all to decide what they will feel. Right? – Melativity Jul 26 '14 at 14:42
• Flat space is <b>Riemann</b> flat, not Ricci flat. All Riemann flat spacetimes differ from the Minkowski metric by a coordinate change. It's actually true that Einstein's equation will have multiple solutions, IN GENERAL for a given stress-energy tensor. You also need to specify initial conditions and boundary conditions for the metric. – Jerry Schirmer Jul 28 '14 at 14:59
• In particular, there are several solutions to Einstein's equation in vacuum that are NOT just Minkowski space, because they contain gravitational radiation, for example (other, more exotic contents are possible) – Jerry Schirmer Jul 28 '14 at 16:47

I wonder if your question is a form of the hole argument. It's true that you can't always find the geodesics (i.e., the unaccelerated worldlines) in certain coordinates given only the Riemann curvature in those coordinates, but that's just a gauge redundancy, not a physical ambiguity. There are up-to-diffeomorphism uniqueness results for the GR initial-value problem analogous to the uniqueness theorem for classical electromagnetism. I don't know the details, but you could look at Theorem 1 of arXiv:1304.1960.

• Ah! This is exactly what I was talking about. I'm glad I now have a name I can give to this issue. I'll read up. Thanks. – Melativity Jul 28 '14 at 5:20
• That hole argument still confuses the bejeezus out of me – lurscher Jul 28 '14 at 14:54

In other words, in order to connect an abstract mathematical description of motion, which in some sense can be transformed in any way, to physical observations, doesn't one have to anchor the mathematical reference somehow to physically meaningful concepts like unaccelerated frames?

Take something simpler that General Relativity.

Consider cartesian coordinates $ds^2= dx^2+dy^2$, it is an inertial frame (relatively to coordinates $x,y$).

However polar coordinates $ds^2=dr^2 + r^2 d\theta^2$ corresponds to a non-inertial frame (relatively to coordinates $r,\theta$), with $g_{\theta\theta} = r^2$. There are non-zero Christoffel symbols , even if the Riemann/curvature tensor is zero here.

However, the laws of physics expressed in polar coordinates, are no more better or worse that the laws of physics expressed in cartesian coordinates. You have only a diffeomorphism between coordinates, which translate the laws of physics, from one reference frame to the other.

First off, the question focuses a lot on the issue of how to tell whether a particular set of coordinates is inertial. This is the wrong thing to focus on, and it wouldn't even normally occur to a relativist to ask whether a certain set of coordinates was inertial. Coordinate systems are global things, whereas an inertial frame of reference only exists locally in GR. Coordinates aren't even mandatory in relativity. You can do relativity without ever picking any coordinates.

What we really care about is whether a particular world-line is inertial. Here are a couple of equivalent ways of defining this:

(1) An inertial world-line is one that extremizes $\int ds$. This can be determined either from the metric or empirically using clocks.

(2) Take a test particle and shield it from all nongravitational forces. Whatever trajectory it takes, that's inertial.

Say you have two objects, A and B. A is at rest and B accelerates directly away from A. From a mathematical point of view, there is absolutely nothing which distinguishes the two objects - A is accelerating away from B as much as B is from A. To distinguish the two, we need something extra - something empirical which tells us that B feels a force acting on it and A does not.

It makes a difference here whether you're talking about a gravitational force or a nongravitational one. If B is subject to a nongravitational force and A is not subject to any nongravitational forces, then it's A that's inertial. The two equivalent definitions above provide two ways of determining that this is the case: (1) A's proper time is extremal while B's is not; (2) A has been shielded from nongravitational forces while B has not.

I don't think I made myself totally clear - let's just say this. If I asked you to solve Einstein's field equations in flat space in the coordinates $x^\mu$, wouldn't you, in general, have to say that there's not enough information? How could you possibly know whether $x^\mu$ were inertial coordinates and $g_{\mu \nu} = \eta_{\mu \nu}$, or if $x^\mu$ were some weird accelerated coordinates, and $g_{\mu \nu}$ were the same flat metric but written in these accelerated coordinates? Both are solutions of the Einstein field equations. In this way, aren't the field equations underdetermined?

I wouldn't say that both are solutions of the field equations. I would say that both are different descriptions of the same solution of the field equations. Both are correct.

Don't they have to be supplemented with information about how the coordinates relate to locally inertial coordinates?

That information is contained in the metric.

Say you have two objects, A and B. A is at rest and B accelerates directly away from A. From a mathematical point of view, there is absolutely nothing which distinguishes the two objects - A is accelerating away from B as much as B is from A. To distinguish the two, we need something extra - something empirical which tells us that B feels a force acting on it and A does not.

You should look up Mach's principle, which postulates a unique unaccelerated frame through the mass distribution of the whole universe (let's call it the Mach frame).

Machs principle defines an inertial frame, not by measuring motions or forces, but only by mass distribution. This is one possible empirical way of distinguishing A and B.

In your example, A is not accelerated w.r.t. the Mach frame, while B is. By this, you could say that the one that is a force exerted on is the one who is accelerated w.r.t. the Mach frame.

In Newtonian mechanics, this principle is obviously not implemented. In general relativity, i found the following in Mach's Principle: From Newton's Bucket to Quantum Gravity (Birkhäuser, 1995, page 10 line 4):

The proposal's most prominent sponsor was Albert Einstein. In the early years of his work on general relativity, he believed that his theory implemented the proposal, although he completely lost this believe in his later years.

So if we believe in the late Einstein, general relativity does not allow to distinguish whether A or B is accelerated without introducing an unaccelerated frame.

A word on transformations: Some people distinguish active and passive transformations Frewer 2008. Coordinate transformations are passive. They have nothing to do with physics, but are merely an identical conversion of the equations.

• Mach's principle is neat but I really don't think this comes anywhere close to answering the question. – Brandon Enright Jul 26 '14 at 16:17
• This is not even a "link-only" answer, because there is no link... – Bernhard Jul 27 '14 at 7:49

Hi Melativity and welcome to Physics StackExchange. I see that you are new to the StackExcahnge community, so let me suggest you fill your profile with some information about your scientific interests and knowledge. This will help people better fine-tune the level of detail/formalism in the answers they give you. You can see my profile (or many others) for an example of what I mean.

EFE (Einstein's field eqs.) is a system of 10 2nd order PDE (partial differential eqs.) , 6 of which are functionally independent. The functional dependence is given by the automatic conservation of the source as MTW's Gravitation puts it at page 408. What this means is that EFE include (as a bonus) the conservation of the stress-energy tensor $T$. This is a set of 4 PDE describing the zero divergence of $T$ Since the unknowns are the 10 independent components of the metric, it is clear that 4 degrees of freedom are left to represent the freedom in selecting a coordinates frame.

Something similar happens with Maxwell's eqs. where the charge conservation eqs. are automatically satisfied and this gives rise to gauge invariance.

Besides this peculiarity of EFE, we have to take into account that all (systems of) differential eqs. in any branch of science are not enough - by themselves - to uniquely determine a solution, unless we add some initial value or boundary value eqs. (IVE or BVE). This is a mathematical fact that holds true everywhere from Newton's eqs. to drum eqs. to ...Schrödinger's eq.

So, all considered , it is not surprising that the Initial Value Data problem is rather important in GR. You can read about it in the above mentioned MTW's Gravitation Chap. 21 and Wald's General Relativity Chap. 10.

I would like to address another issue. You wrote:

Say you have two objects, A and B. A is at rest and B accelerates directly away from A. From a mathematical point of view, there is absolutely nothing which distinguishes the two objects - A is accelerating away from B as much as B is from A. To distinguish the two, we need something extra - something empirical which tells us that B feels a force acting on it and A does not.

This is not correct. If you are given:

1- 4 symbols representing an otherwise unspecified coordinate system

2- the 10 algebraically independent components of the metric expressed as functions of these 4 symbols

3- the world lines of A and B expressed as coordinates (the 4 symbols of point 1) vs affine parameter/s

you can in principle decide which particle, if any, is in free-fall. You do this by using the geodesic eqs., the line element and the $\Gamma$s obtained by the metric.

Point 1 and 2 uniquely describe the physical background.

In general, General Relativity is a very well thought theory and if you can physically distinguish 2 situations, you can certainly distinguish them mathematically

Acceleration, in the sense of moving non-inertially, is an absolute concept , while velocity is a relative concept. I can match someone's velocity (so she is at rest, at least for a moment) by flying in an appropriate inertial system, but I cannot match her acceleration even for a moment by flying in any inertial system.

Your misguided perception is what is called the "twin paradox" which is not a paradox at all, once you understand the subtleties of GR.

• This doesn't answer the question. – Ben Crowell Jul 28 '14 at 16:23
• @BenCrowell Really??? That's curious because I am restating and expanding on the hole argument (a terminology not found in the standard textbooks mentioned in my answer) mentioned by benrg in his accepted answer and that the OP declared to be very happy with. It seems you either did not understand the question or benrg.s answer or mine or...neither of them. Indeed your own answer is merely a restatement of the second part of my answer. – magma Jul 28 '14 at 18:46