STRACT: Inertial Frames in Special and General Relativity

[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 teh 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?

 

[Altabeh]

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?

Fremi normal coordinates are just defined along geodesics (basically timelike geodesics)! The big difference between FNC, RNC and geodesic coordinates (GC) is that the two latter ones make the spacetime along a geodesic locally flat, i.e. at some point, but FNC makes the spacetime globally flat.

AB

 

[atyy]

Fremi normal coordinates are just defined along geodesics (basically timelike geodesics)! The big difference between FNC, RNC and geodesic coordinates (GC) is that the two latter ones make the spacetime along a geodesic locally flat, i.e. at some point, but FNC makes the spacetime globally flat.

I was thinking of Fermi normal coordinate on a timelike curve, not necessarily geodesic, like in section 3.2 of http://relativity.livingreviews.org/Articles/lrr-2004-6/ .

 

[Altabeh]

I was thinking of Fermi normal coordinate on a timelike curve, not necessarily geodesic, like in section 3.2 of http://relativity.livingreviews.org/Articles/lrr-2004-6/ .

Well, I've read his own book "A Relativist's toolkit" and there he doesn't bring up this stuff and so it is new to me and if I were able to find some time to spend on it, then I would have something to share!

AB