STRACT: Inertial Frames in Special and General Relativity

[Dale]

SR always applies locally, and the GR version of Newton's laws always applies.

 

[madness]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

 

[bcrowell]

I like that point of view too but unfortunately you can still tell whether or not you are in an inertial frame

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

Your first statement sounds like you're talking about using observations to tell whether you're in an inertial frame or not, where "in an inertial frame" presumably refers to the frame defined by the floor in your lab. But your second statement doesn't refer to observables, it refers to a choice of coordinates.

As DaleSpam points out, any laboratory, and any set of experiments in that lab, can be described in locally Minkowski coordinates -- or in coordinates that are not Minkowski.

 

[madness]

Think about the famous thought experiment with a man in a lift. If he's floating, and objects float around him then he is in an inertial frame. If he is stuck to the floor and objects around him fall to the floor, then he is in a non-intertial frame. The difference is that one frame has a vanishing connection at that point and the other doesn't. In the latter case case you can transform to a frame where the connection does vanish, but for the man in the lift he is still in a non-inertial frame.

 

[Fredrik]

If he's floating, and objects float around him then he is in an inertial frame.

You need to be more careful with the terminology. A person or an object never just "is in an inertial frame". What you're trying to say is probably that both his velocity and acceleration are 0 in an inertial frame.

 

[madness]

"You need to be more careful with the terminology. A person or an object never just "is in an inertial frame". What you're trying to say is probably that both his velocity and acceleration are 0 in an inertial frame"

What I mean is that his frame of reference is inertial. If he is stuck to the floor of the lift and objects fall to the floor, it is possible to make a change of coordinates such that at his space-time point the connection vanishes, but the new coordinates are not his frame of reference. In other words, if we switch to a frame which is accelerating downwards with respect to him, then events at his space-time point appear to obey the laws of SR, but that for him he is still in a non-inertial frame.

 

[bcrowell]

In other words, if we switch to a frame which is accelerating downwards with respect to him, then events at his space-time point appear to obey the laws of SR, but that for him he is still in a non-inertial frame.

A lot of this is just a matter of taste. However, it's important to realize that the distinction you're drawing between inertial and noninertial frames is very different from, and much weaker than, the one made in Newtonian mechanics.

In Newtonian mechanics, you can (1) determine whether a certain object's motion is inertial. If it is, you can (2) build a global set of coordinates tied to that object. If it's not, then (3) the laws of physics are false if expressed in coordinates tied to that object.

In general relativity, you can still do (1), but you can't do (2), and (3) is no longer an issue. In Newtonian mechanics, 2 and 3 are the reason you care so much about 1. In general relativity, neither one constitutes a good reason to care about 1.

The way Einstein liked to present it was that because 2 and 3 are no longer considerations, 1 loses its interest entirely. You don't have to agree with his emphasis, but he clearly has a strong point.

 

[madness]

I don't see how it's a matter of choice and not a matter of fact. If there were no such thing as an inertial frame then the theory would be fully Machian. Then for example the twin paradox really would be a paradox in GR.

 

[bcrowell]

I don't see how it's a matter of choice and not a matter of fact. If there were no such thing as an inertial frame then the theory would be fully Machian. Then for example the twin paradox really would be a paradox in GR.

All you really need in order to resolve the twin paradox is the idea of inertial motion, not the idea of an inertial frame. The traveling twin can tell that his motion is noninertial, because he feels the floor pressing against his feet. If you want to start talking about frames, then that's a matter of taste, but the frames will lack properties 2 and 3 from #37, which to my taste makes them fairly uninteresting.

 

[Dale]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

I can see that definition, it is not common but it makes sense. Usually when most people talk about SR they just mean flat spacetime; at least that is what I meant by SR always applies locally. But I do understand your point.

My point is just that the GR versions of Newton's laws always apply as do the GR versions of Maxwell's equations, which IMO would be the more direct analogue of the original Newtonian meaning of "inertial frame". This is why I am of the opinion that the concept of inertial frame is not very significant in GR. Using your definition of inertial frame you can certainly say that they do exist in GR, but since GR can handle frames with non-vanishing connections just fine such frames are not terribly important. Certainly not the central role that they play in Newtonian mechanics.

 

[atyy]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

I can see that definition, it is not common but it makes sense. Usually when most people talk about SR they just mean flat spacetime; at least that is what I meant by SR always applies locally. But I do understand your point.

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame". But the derivatives of the Christoffel symbols, ie. second derivatives of the metric, cannot be set to zero, and so the "local inertial frame" that exists anywhere in curved spacetime only pertains to measurements which do not measure curvature.

 

[pervect]

In coordinate independent language, one can simply talk about an observer following a geodesic. There's no need to talk about the specific coordinates this observer uses to define the notion of a "free-falling" or "geodesic" observer.

Note that there may be several coordinate systems through any given point that make the connection vanish. Even if you specify the basis vectors at that point, there are more than one set of coordinates that make the connection vanish.

For instance, the following quote from MTW's "Gravitation" discussing Riemann normal coordinates.

Is this the only coordinate system at P0 that is locally inertial at P0 ... and is tied to the basis vectors there? No. But all such coordinate systems, (called 'normal coordinates') will be the same to second order.

In the elipses of the above quote (...) MTW explains that a coordinate system that is locally inertial at a point has a vanishing connection.

 

[Altabeh]

Note that there may be several coordinate systems through any given point that make the connection vanish. Even if you specify the basis vectors at that point, there are more than one set of coordinates that make the connection vanish.

Would you mind giving us an example of what you just claimed above?! It sounds completely wrong to me as if you are ready, I'm going to explain it matheamtically!

AB

 

[Altabeh]

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame". But the derivatives of the Christoffel symbols, ie. second derivatives of the metric, cannot be set to zero, and so the "local inertial frame" that exists anywhere in curved spacetime only pertains to measurements which do not measure curvature.

Completely true! You don't need to even think of "local flatness" in terms of Christoffel symbols when in GR. One just needs to set the first derivatives of metric tensor equal to zero which fits within the definition of "geodesic coordinates" which is not actually an actual coordinate system (read the point 1 below)! But remember that there is a strongly local coordinates in the sense that just at a point you can make the values of metric equal those of Minkowski metric. Here two points must be made:

1- We don't have a "coordinate transformation" x^{\mu}\rightarrow {\bar{x}}^{\mu} by which at a point P, one has g^{\mu\nu}(P)={\eta}^{\mu\nu}(P) iff g^{\mu\nu}(P) is symmetric. If it is diagonal, the number of degrees of freedom of its matrix would be the same as the number of equations of its transformation so there is just one set of coordinates being in charge of the transformation g^{\mu\nu}(P)\rightarrow {\eta}^{\mu\nu}(P) and of course for this case one can show a coordinate system.

2- Geodesic coordinates just account for the first derivatives of metric being all equal to zero, so this way of leading to the local flatness at some point is another alternative.

3- Following 1, in the neighborhood of P the spacetime is "nearly" flat within a range that the equivalence principle issues. This is because we don't know what is meant by "neighborhood" in GR and this is only evaluated\estimated by EP. This might not be applicable for the case 2.

AB

 

[Dale]

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame".

Yes, they can be made to vanish anywhere through suitable choice of coordinate system, but there do exist coordinate systems where they do not vanish at any given point. Madness is defining such coordinate systems as locally non-inertial, which is not unreasonable IMO.

 

[madness]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.
I studied Ray d'Inverno's book before, which is where I got the idea about vanishing affine connections and locally inertial frames. He shows that at any point on a manifold, it is possible to make a change of coordinates such that the connection vanishes at that point. This can be generalised to a curve but not usually to the whole manifold. I interpreted this as transforming to a locally inertial frame.
Further, d'Inverno seemed to present the affine connection in the geodesic equation as describing non-inertial forces or gravitational forces, which lends to the idea that a vanishing affine connection means a locally inertial frame.

 

[Fredrik]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.

They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details).

 

[madness]

"They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details)."

Well for any LIF, there are infinitely many other LIF's at that point related by a Lorentz transform. I'm not sure if that's why you mean by unique.

 

[Altabeh]

"They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details)."

Well for any LIF, there are infinitely many other LIF's at that point related by a Lorentz transform. I'm not sure if that's why you mean by unique.

Madness, you can find one and only one "coordinates transformation" (be careful with this; read my last post here) that makes your connection (first derivatives of metric tensor) vanish at some point. We never ever can use Lorentz transformations at that point to make lots of frames since in general those coordinates transformations which would meet our demands here, are not explicitly written or shown in simple expressions.

AB

 

[bcrowell]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.

The "L" in LIF is one reason that LIFs are much less central than in Newtonian mechanics. The other reason they're less important is that the laws of GR (unlike Newton's laws) are still valid in coordinates that don't describe an LIF.

If you look at a presentation of relativity like Einstein's "The Foundation of the General Theory of Relativity," ( http://hem.bredband.net/b153434/Works/Einstein.htm ) that doesn't mention the concept of inertial frames (or mentions them only in order to pooh-pooh the concept), what's happening is that he makes the equivalence principle more central conceptually. One way of stating the equivalence principle is that LIFs exist.

 

[Altabeh]

I'd like to modify my last post here and sorry for the inconvenience:

Completely true! You don't need to even think of "local flatness" in terms of Christoffel symbols when in GR. One just needs to set the first derivatives of metric tensor equal to zero which fits within the definition of "geodesic coordinates" which is not actually an actual coordinate system (read the point 1 below)! But remember that there is a strongly local coordinates in the sense that just at a point you can make the values of metric equal those of Minkowski metric. Here two points must be made:

1- In general, we can hardly determine a "coordinate transformation" x^{\mu}\rightarrow {\bar{x}}^{\mu} by which at a point P, one has g^{\mu\nu}(P)={\eta}^{\mu\nu}(P). If the metric is diagonal, the number of equations of its transformation would be much less than the case when metric is considered to be symmetric. This is because, for instance, if we count the number of equations involved in a symmetric metric trans., that is,

n(n+1)/2,

then you must fit at least this number of arbitrarily-chosen contants within the coordinates transformation at any given point to form a set of equations with the same number of unknowns and equations. But in the diagonal case, this number reduces to n, so the system of equations gets much simpler to be solved.

2- Geodesic coordinates just account for the first derivatives of metric being all equal to zero, so this way of leading to the local flatness at some point is another alternative.

3- Following 1, in the neighborhood of P the spacetime is "nearly" flat within a range that the equivalence principle issues. This is because we don't know what is meant by "neighborhood" in GR and this is only evaluated\estimated by EP. This might not be applicable for the case 2.

AB

 

[DrGreg]

Note that there may be several coordinate systems through any given point that make the connection vanish. Even if you specify the basis vectors at that point, there are more than one set of coordinates that make the connection vanish.

Would you mind giving us an example of what you just claimed above?! It sounds completely wrong to me as if you are ready, I'm going to explain it matheamtically!

It's not clear to me whether an example has been given, so let me give one.

Let (t,x) be Minkowski coordinates in flat spacetime. (To save typing I'll ignore y and z but you can add them back if you want.) They define an everywhere-inertial frame, not just a locally inertial frame.

Now (with the convention c=1) define new coords T = t − x³/3, X = x in a region around the origin. The inverse transformation is t = T + X³/3, x = X.

The metric is

ds^2 = dt^2 \, - \, dx^2 = dT^2 \, + \, 2X^2dT\,dX \, - \, (1-X^4)dX^2​

At any event where X=0, but nowhere else, the metric in (T,X) coordinates takes the Minkowski form, and its first-order coordinate derivatives vanish (and hence the connection vanishes).

Thus (T,X) defines a locally inertial frame but doesn't define an everywhere-inertial frame.

So we have two different locally-inertial frames, which establishes the non-uniqueness claim.

 

[Altabeh]

It's not clear to me whether an example has been given, so let me give one.

Let (t,x) be Minkowski coordinates in flat spacetime. (To save typing I'll ignore y and z but you can add them back if you want.) They define an everywhere-inertial frame, not just a locally inertial frame.

Now (with the convention c=1) define new coords T = t − x³/3, X = x in a region around the origin. The inverse transformation is t = T + X³/3, x = X.

The metric is

ds^2 = dt^2 \, - \, dx^2 = dT^2 \, + \, 2X^2dT\,dX \, - \, (1-X^4)dX^2​

At any event where X=0, but nowhere else, the metric in (T,X) coordinates takes the Minkowski form, and its first-order coordinate derivatives vanish (and hence the connection vanishes).

Thus (T,X) defines a locally inertial frame but doesn't define an everywhere-inertial frame.

So we have two different locally-inertial frames, which establishes the non-uniqueness claim.

Sorry I don't know what you are even trying to do!

1- You translate Minkowski spacetime uniformally into a flat spacetime through some coordinate transformation; so, by looking at the "coordinates" you introduce, at the origin "local inertia" is always guaranteed as both metrics coincide there!

2- Nonetheless, though the connection in your spacetime is not globally zero, but the Riemann tensor vanishes and thus it is a special class of spacetimes in witch your example works. I think Pervect brings up his claim in any spacetime of GR with curved metrics such as Schwartzchild metric.

3- Furthermore, assuming that we are given

ds^2 = dT^2 \, + \, 2X^2dT\,dX \, - \, (1-X^4)dX^2,

how is it possible to transform this metric into Minkowski at a given point P without using X=x, T=t-x^3/3!? By "making a connection vanish at some point" one means that using a coordinate system (I'd like to say that this coordinate system is unique) the metric is able to be transformed into \eta_{\mu\nu}.

AB

 

[Altabeh]

I might be getting a little bit acute here, but I like to hear your ideas about DrGreg's example about the non-uniqueness of "coordinate transformation" that makes the connection vanish at some point. I think his example looks like this:

Imagine the Schwartzchild metric

ds^2 = (1-2m/r)dt^2- (1-2m/r)^{-1}dr^2-r^2(d{\theta}^2+\sin^2(\theta)d{\phi}^2).

When m=0, then obviously we have the minkowski spacetime. Also, if r\rightarrow \infty, then the metric does not give us the Minkowski metric. Now let's transform this metric into its isotropic form:

ds^2 = \frac{(1-m/2r')^2}{(1+m/2r')^2}dt^2- (1+m/2r')^{4}(dx^2+dy^2+dz^2),

where r'=f(r). So here if r'\rightarrow \infty, one would lead to the Minkowski metric. But isn't this the so-called "Begging the Question fallacy"? We know that for m=0 and r'\rightarrow \infty, the Minkowski metric reveals itself and thus claiming that "local inertia" must happen to exist under those conditions is vacuously true and of course is non-unique.

In my opinion, the problem of finding a "coordinate transformation" or "coordinates" which can transform a given metric into Minkowski locally requires neccessarily uniqueness unless it is said that a "nearly locally inertial" frame is seeked out in a small region of spacetime around a central point at which |g_{\mu\nu}-\eta_{\mu\nu}| is minimum.

AB

 

[atyy]

Is the question whether Riemann normal coordinates for a given point are unique?

 

[Altabeh]

Is the question whether Riemann normal coordinates for a given point are unique?

Yes!

AB

 

[atyy]

I looked up the Riemann normal coodinates construction in Eq (2.35) to Eq (2.35) at http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html .

Say we want to get RNC at point p. Suppose you start with some non-RNC coordinates x. Any new coordinates y can be specified by arbitrarily choosing the terms A, B, C, D, E,... in a Taylor expansion x=A+By+Cyy+... where A are the 16 first derivatives dx/dy evaluated at p, B are the 40 second derivatives dxdx/dydy evaluated at p, etc.

The new metric G can be Taylor expanded G=Gp+dGp.y+ddGp.yy+... For RNC at p we require Gp=diag(-1,1,1,1) and dGp=0 as constraints on our choice of A,B,C etc. The Gp=diag(-1,1,1,1) are 10 constraints on the 16 numbers A, so they are underspecified. Choosing all 16 numbers I think corresponds to what Pervect meant by "specifying the basis at that point". dGp=0 are 40 constraints on our choice of the 40 numbers in B, which are thus exactly specified. It looks to me like at this point we have constructed an RNC, while C, D, E,... are still unconstrained. So I guess RNC for a point are not unique.

 

[Fredrik]

I checked Spivak's definition of "Riemannian normal coordinates". He defines them in a way that makes them unique up to an O(3,1) transformation. (Actually, since we're only interested in coordinate systems with the t axis coinciding with the tangent of the world line at some specified point p, the only non-uniqueness that remains is an O(3) transformation, or SO(3) if we want to preserve the orientation).

Another source told me about something called "Fermi normal coordinates", which is something very similar, but instead of using all the geodesics through p to define the coordinate system, we look at an arbitrary point q on the tangent of the world line at p, and use the geodesics in the hypersurface through q that's orthogonal to that curve to construct a part of the coordinate system. To define the whole coordinate system, we have to do this for every q on the tangent. (I think we have to use the tangent of the world line rather than the world line itself, if we want something that resembles an inertial frame, but I don't remember if the definition says that we should).

So there appears to be many different types of "normal" coordinate systems that someone could claim is the "local inertial frame" we're looking for. The definition that atyy found seems to be describing only the property that they all have in common: the condition on the components of the metric and their derivatives. I think it would be appropriate to define the term "normal coordinates" by saying that any coordinate system that satisfies that condition is a normal coordinate system. Riemann and Fermi normal coordinate systems are then special cases of that.

So which one is the real "local inertial frame"? I suspect that this is a meaningless question.

 

[Altabeh]

I looked up the Riemann normal coodinates construction in Eq (2.35) to Eq (2.35) at http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html .

Say we want to get RNC at point p. Suppose you start with some non-RNC coordinates x. Any new coordinates y can be specified by arbitrarily choosing the terms A, B, C, D, E,... in a Taylor expansion x=A+By+Cyy+... where A are the 16 first derivatives dx/dy evaluated at p, B are the 40 second derivatives dxdx/dydy evaluated at p, etc.

The new metric G can be Taylor expanded G=Gp+dGp.y+ddGp.yy+... For RNC at p we require Gp=diag(-1,1,1,1) and dGp=0 as constraints on our choice of A,B,C etc. The Gp=diag(-1,1,1,1) are 10 constraints on the 16 numbers A, so they are underspecified. Choosing all 16 numbers I think corresponds to what Pervect meant by "specifying the basis at that point". dGp=0 are 40 constraints on our choice of the 40 numbers in B, which are thus exactly specified. It looks to me like at this point we have constructed an RNC, while C, D, E,... are still unconstrained. So I guess RNC for a point are not unique.

All right, let's shed more light on what we want from the discussion here. In my penult post, I said that to obtain at a given point P "locally flat" or "locally inertial" spacetime, one has to find a coordinate transformation by which the equality

{\bar{g}}_{\mu \nu}(P)={\eta}_{\mu \nu}(P), (1)

can be gained. As Schutz puts forward, there are only 6 independent values left in (1) which correspond to 6 degrees of freedom in the Lorentz transformations.

But in the course of Riemannian normal coordinates (RNC) one sees that there is a much deep concept which is known to be "geodesic coordinates"! I don't want to go into its details here and address readers to the Eisenheart's book "Riemannian Geometry" page 56 where he shows that a transformation of the form

x^{\alpha}={\bar{x}}^{\alpha}_P+{\bar{x}}^{\alpha}-\frac{1}{2}(\Gamma_{\mu \nu}^\alpha)_P{\bar{x}}^{\mu}{\bar{x}}^{\nu}, (2)

at some point P where the coordinates of P in the {\bar x}^{\alpha} vanish, leads to

({\bar{\Gamma}}_{\mu \nu}^\alpha)_P=0

and thus giving rise to

({\partial}_{\kappa}{\bar{g}}_{\mu \nu})_P=0. (3)

Now we have a locally inertial frame at a given point P which is deduced from the unique coordinates (2). RNC are the geodesic coordinates for which the connection vanishes for x^{\alpha}=0. Now the moral we can dig from the whole discussion about local flatness or locally inertia in GR at a given point would be briefly discussed in two lines:

1- The concept of "local inertia" in GR can only revive at a given point so the coordinates that make such a revivification would not be applicable and meaningful anywhere else. This has a simple reason: If not so, then the whole spacetime is globally inertial. Besides, all information would only be valid according to the coordinates of the point at which we adjust those free values (6 degrees of freedom of Lorentz group in RNC) to have (1) hold.

2- The "uniqueness" of existence of such a coordinates system is guaranteed if the coordinates system is only directed to make the connection vanish at some point. If it comes to (1), where it also yields (3), then the coordinates are not unique since 6 degrees of freedom exist in (1).

Thank you all for participating in this great discussion.

AB

 

[atyy]

On a non-geodesic worldline, Fermi normal coordinates make the metric Minkowski at a point on the worldline, but the first derivatives of the metric don't vanish at that point. Is this sufficient to count as locally inertial in some sense, eg. the local speed of light is c?