Is this right? [relativity]

[Surlethe]

Given a 4-manifold M equipped with the Minkowski metric, all of the orthogonal coordinate systems are precisely those which induce metric components

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[-1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ 0 0 0 1 ]

and, given any one orthogonal coordinate system, the orbit of the system under the Lorentz group is all of the coordinate systems.

[CaptainChewbacca]

Looks good to me. What's this for?

[Kuroneko]

What you've described are orthonormal, rather than orthogonal, coordinate systems. At a point, they define three spatial and one temporal axes, all mutually orthogonal, so that of course they are related by a Lorentz transformation, by definition. However, this statement is only locally true, even in the plain Minkowski case--e.g., in normalizing polar coordinates, there is no global Lorentz transformation between the resulting system and the Cartesian one, or the Rindler coordinates corresponding to a uniformly accelerated observer.

I'm not sure what you're doing with this, but perhaps one thing to keep in mind is that the Lorentz group is a bit more complicated than the "boost plus rotation" treatment given in many introductory texts. For example, a rotation about a null vector, e.g., one mapping the standard basis vectors to {e0 = [3;-1;2;1]/2, e1 = [1;1;2;0]/2, e2 = [1;-1;1;0], e3 = [0;0;0;1]}, is a Lorentz transformation that is not in that form.

[Surlethe]

What you've described are orthonormal, rather than orthogonal, coordinate systems. At a point, they define three spatial and one temporal axes, all mutually orthogonal, so that of course they are related by a Lorentz transformation, by definition. However, this statement is only locally true, even in the plain Minkowski case--e.g., in normalizing polar coordinates, there is no global Lorentz transformation between the resulting system and the Cartesian one, or the Rindler coordinates corresponding to a uniformly accelerated observer.

There is surely a way to generalize Lorentz transformations to all local coordinate patches. My hunch (I am about to try to prove this) is that given a nondegenerate bilinear form on a manifold (e.g., the Minkowski metric) the intersection of some arbitrary set of coordinate patches at a point induces a group of Lorentz transformations. In particular, given two overlapping coordinate patches, you have a lifting from the manifold to the Lorentz group which is not in general constant. Right?

I'm not sure what you're doing with this, but perhaps one thing to keep in mind is that the Lorentz group is a bit more complicated than the "boost plus rotation" treatment given in many introductory texts. For example, a rotation about a null vector, e.g., one mapping the standard basis vectors to {e0 = [3;-1;2;1]/2, e1 = [1;1;2;0]/2, e2 = [1;-1;1;0], e3 = [0;0;0;1]}, is a Lorentz transformation that is not in that form.

You mean a hyperbolic reflection? Can't you characterize the Lorentz group as all of the hyperbolic rotations and reflections that fix the light cone?

[Kuroneko]

I'm unclear as to your purpose. If you just want to reconstruct the Lorentz group, then a spacetime by definition is a certain type of Lorentzian manifold, meaning its tangent planes are Minkowski, and the Lorentz group is just the set of all linear isometries.

There is surely a way to generalize Lorentz transformations to all local coordinate patches.

It sounds like you're trying to rediscover the Cartan formalism. If you have one timelike and three spacelike linearly independent vector fields eiμ, then they form a tetrad or vierbein that represents an inner product preserving map between the tangent plane and Minkowksi spacetime:
[1] gμν = ημνeiμejμ,
where the eiμ components are just a matrix inverse of eiμ. The coordinates corresponding to the frame axes are meaningful only as a local inertial frame, unlike the background coordinates xμ, but this formalism is also able to represent global geometric objects, since the vector fields we started with are independent.

My hunch (I am about to try to prove this) is that given a nondegenerate bilinear form on a manifold (e.g., the Minkowski metric) the intersection of some arbitrary set of coordinate patches at a point induces a group of Lorentz transformations.

The frame metric doesn't have to be Minkowski, but the condition that a transformation of the frame axes preserves the frame metric is
[2] ηij' = ΛikΛjlηkl = ηij,
so for the Minkowski metric they are precisely the Lorentz transformations. This also means the tetrad frame axes are orthonormal. Some GTR books call this "orthonormal frame" and distinguish it from a "coordinate frame", which is rather confusing because all they mean by the latter is "the usual coordinates."

In particular, given two overlapping coordinate patches, you have a lifting from the manifold to the Lorentz group which is not in general constant. Right?

If the above is what you're looking for, yes, since Λ can vary from point to point.

Tensors work exactly in the same way as in coordinates, with the vierbein switching between them. For example, ∂i = eiμ∂μ would be the derivative with respect to proper time for an observer at rest and comoving with the frame, while the corresponding covariant derivative would be for one instantaneously at rest but inertial.

As an interesting example, consider the not uncommon description of a black hole as having "space falling into the singularity" or some-such, and this effect being superluminal inside the horizon. This I found extremely puzzling at first, because outside the event horizon, the Schwarzschild metric clearly static outside the horizon. But what it refers to is the Gullstrand-Painleve metric, which can be interpreted as a foliation of Schwarzschild spacetime by spatially flat hypersurfaces. The corresponding vierbein is the frame of an inertial observer falling from rest at infinity, and in it, the situation looks like space is flowing inward at Newtonian escape velocity. I can fill out some of the details if you're interested.

You mean a hyperbolic reflection? Can't you characterize the Lorentz group as all of the hyperbolic rotations and reflections that fix the light cone?

If you allow all possible compositions while including elliptic rotations (in space), yes. Hyperbolic rotations have two null eigenvectors, whereas the Lorentz group, neglecting reflections and time-reversals, includes transformations that have only one null eigenvector. But this probably has no relevance to what you're doing, so never mind.

[Surlethe]

While I work through the Cartan formalism (shame on you for presenting it in coordinate form!), here's a question to keep the thread going: is there any physical meaning at all to reflections across the light cone? (In other words, toss out temporal causality.)

[Kuroneko]

There may be an ambiguity in the above discussion of 'characterizing' the Lorentz group that I should have caught earlier, and possibly misinterpreted your statement. Because of Thomas-Wigner rotation, we can get rotations by the composition of three boosts, so in that sense your earlier statement was correct if you were thinking of arbitrary boosts along arbitrary axes. However, we cannot get the full Lorentz group starting with arbitrary boosts along (say) the usual Cartesian axes; arbitrarily composing them gives a three-dimensional subgroup instead of the full six-dimensional Lorentz group.

While I work through the Cartan formalism (shame on you for presenting it in coordinate form!), ...

Oh dear...

... is there any physical meaning at all to reflections across the light cone? (In other words, toss out temporal causality.)

Well, the direct meaning is just that Maxwell's equations are invariant under time-reversal, but the way this works out in practice is actually pretty interesting. If you have a source four-current J, then the electromagnetic field Fμν = ∂[μAν] has ∂μFμν = Jν, up to some constants (this is equivalent to Gauss and Ampère-Maxwell laws, as I've noted before). Fixing the Lorentz gauge (∂μAμ = 0), this reduces to ∂μ∂νAα = Jα. This is exactly the wave equation with a source term.

The solution to this inhomogeneous system of partial differential equations is the solution to the homogeneous part plus Int[ G(x-x')J(x') d4x' ], where G = G(t,r) is a Green functio of ∂μ∂ν. In this case we have two, proportional to δ(t+r)/r and δ(t-r)/r (if the field was scalar, that would just be it; since it's not, there's an extra term I'm ignoring), the advanced and retarded Green functions, respectively. There is no mathematical reason why a charge has to radiate only retarded waves, but for reasons of causality, standard electrodynamics disregards the backward-time advanced solutions completely. But there is an interesting reformulation of electrodynamics, the Wheeler-Feynman absorber theory, in which emission is always half-retarded and half-advanced, and the advanced portion of the absorber response is responsible for the other half.

Another interesting application of time-reversal symmetry is from quantum mechanics, where it can be used to prove that a system of half-integral spin and a Hamiltonian invariant under time reversal cannot have a nondegenerate energy eigenstate. If you're interested, look up Kramer's theorem.

There are some interesting connections between electromagnetism and Lorentz transformations, beyond their service regarding inertial reference frames, because their infinitesimal generators are Killing vector fields, and any such field corresponds to a solution of the source-free Maxwell's equations. In GTR, this has direct relevance to the electrovacua, but for reasons of time I won't go into that right now.

Elliptic Lorentz
The most straightforward example is a spatial rotation about the x-axis, which can be represented by the usual Euclidean rotation matrix, and preserves et and ez. Its derivative at zero rotation angle is:

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[ 1  0        0       0 ]     [ 0  0  0  0 ]
[ 0  +cos(p) -sin(p)  0 ] --> [ 0  0 -1  0 ]
[ 0  +sin(p) +cos(p)  0 ] --> [ 0  1  0  0 ]
[ 0  0        0       1 ]     [ 0  0  0  0 ]

The infinitesimal generator can be read straight off the right-hand matrix, giving K = x∂y - y∂x, and we can recover the original transformation through Λ = exp(pK). If we take A = (φ,A) = (B/2)K as the electromagnetic four-potential, then clearly the electrostatic potential vanishes and B = ∇×A = (B/2)(1 - -1)ez = Bez. Thus, elliptic rotations generate a magnetostatic field.

Hyperbolic Lorentz
A Lorentz boost along the x-axis has the form

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[ cosh(p) sinh(p)  0   0 ]     [ 0  1  0  0 ]
[ sinh(p) cosh(p)  0   0 ] --> [ 1  0  0  0 ]
[   0       0      1   0 ] --> [ 0  0  0  0 ] 
[   0       0      0   1 ]     [ 0  0  0  0 ]

Thus, its infinitesimal generator x∂t + t∂x. The potential A = (E/2)K has E = -∇φ - ∂A/∂t = Eex and B = ∇×A = 0. This is an electrostatic field in the direction of the boost.

Parabolic Lorentz
Lorentz transformations generally have two null eigenvectors of reciprocal eigenvalues. The exceptions are the purely spatial elliptic case, and the null rotations, which can be thought of as the composition of boosts that share exactly one eigenvector with reciprocal eigenvalues. The following transformation preserves et + ez:

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[ 1+p²/2  p  0   -p²/2 ]     [ 0  1  0  0 ]
[   p     1  0    -p   ] --> [ 1  0  0 -1 ]
[   0     0  1     0   ] --> [ 0  0  0  0 ]
[  p²/2   p  0  1-p²/2 ]     [ 0  1  0  0 ]

It has K = x(∂t+∂z) + (t-z)∂x.
Exercise: Verify that this produces a plane wave.

Spacetime Inversion
This is obviously not a Lorentz transformation, but it is part of the fifteen-dimensional conformal group, under which Maxwell's equations are invariant. Briefly, a scale deformation dx'μdx'μ = λdxμdxμ preserves the light for any smooth positive function λ = λ(x). The full conformal group is generated by dilations and inversions, along with the Lorentz transformations and translations that generate the Poincaré group.

In Euclidean space, r→R²/r inverts across the sphere r=R maps between generalized circles: it maps lines to either other lines or circles. In Minkowski space, the inversion of the spacetime interval s→R²/s inverts across the hyperboloid s=R (in your -+++ convention, so that real s means spacelike separation) preserves the light cone, mapping lines to either other lines or hyperbolas. From the boost along the x-axis above, it should be clear that if p is proportional to proper time along the worldline, thus being uniformly boosted in proper time, the resulting curve is a hyperbola in the (t,x)-plane. In other words, hyperbolic motions in spacetime are precisely the uniformly accelerated worldlines, and the transformation between such worldlines and the inertial ones preserves Maxwell's equations.

Exercise: According the (Abraham-)Lorentz-Dirac equation, a uniformly accelerated charge experiences no radiation reaction, which is perhaps anticipated by the above reasoning. But does it actually mean that a uniformly accelerated charge does not radiate?

[Surlethe]

Re. the Cartan formalism: So vierbiens are simply a different way of looking at (some) coordinate systems? Given a vierbien on a Lorentzian manifold, you should be able to go to a coordinate system on the open patch on which it is defined simply by picking an arbitrary point and identifying the flow of each vector field from that point with the coordinate axes under the system. You'll get that subset of the global coordinate systems which correspond exactly to physically realistic ones.

[Kuroneko]

So vierbiens are simply a different way of looking at (some) coordinate systems? Given a vierbien on a Lorentzian manifold, you should be able to go to a coordinate system on the open patch on which it is defined simply by picking an arbitrary point and identifying the flow of each vector field from that point with the coordinate axes under the system.

In a sense yes, although I suppose in practice this is analogous to the difference between active and passive transformations in other contexts. Transformations of the vierbein/tetrad frame genuinely distort the geometric objects on the manifold, whereas coordinate transformations do not.

You'll get that subset of the global coordinate systems which correspond exactly to physically realistic ones.

Right. If you want to know what a result means physically, you better have an orthonormal vierbein ready. It's a very natural axiomatization of the physical 'frame of reference'.

Mathematically, we can define a 'vector field' as a section of the tangent bundle, and a 'frame' as a section of the bundle of all ordered bases of the tangent spaces. Making it orthonormal can always be done if desired.

[Surlethe]

In a sense yes, although I suppose in practice this is analogous to the difference between active and passive transformations in other contexts. Transformations of the vierbein/tetrad frame genuinely distort the geometric objects on the manifold, whereas coordinate transformations do not.

I don't follow. Is the convention to require that for each vierbein frame the metric is always tr(1,-1,-1,-1)? (I.e., given a vierbein frame (d/dx0,d/dx1,d/dx2,d/dx3), pick g such that g(d/dxi,d/dxj) = 0, i/=j, g(d/dxi,d/dxi) = 1 if i=0, -1 else?) It seems to me that you can fix a metric, which fixes a certain class of orthonormal frames (e.g., in the Euclidean metric on an n-dim manifold, those invariant under O(n)), or you can fix a vierbien and declare it orthonormal, leading to some (unique) metric?

Mathematically, we can define a 'vector field' as a section of the tangent bundle, and a 'frame' as a section of the bundle of all ordered bases of the tangent spaces.

Ah, that follows very nicely. And then you make it orthonormal by magicking a metric into existence, as above?

[Kuroneko]

Note: As I implied before, but perhaps wasn't clear, there is an unfortunate lack of uniform terminology. For relativists, the common names are 'tetrad' and 'vierbein', which mean the same thing (some use the latter for the particular matrix of components while the former for the field itself, many don't bother with an explicit distinction), and some other texts just call 'orthonormal basis.' In Riemannian geometry, the usual term is a 'moving frame.' Historically, this method was introduced decades before ∇.

I don't follow. Is the convention to require that for each vierbein frame the metric is always tr(1,-1,-1,-1)?

It's a very common conventional restriction, but it's not necessary. Typically, the whole point of is to choose a tetrad/vierbein that has some particularly nice adaptation to the problem at hand, and if we're dealing with physics, it is very commonly orthonormal. Not always, though--for example, if we were studying radiation, it is frequently more convenient to use a null tetrad instead. An analogous situation exists in Riemannian geometry--e.g., we can have a moving frame on the entire torus or sphere, whereas coordinates necessarily fail to be global on those manifolds.

It seems to me that you can fix a metric, which fixes a certain class of orthonormal frames (e.g., in the Euclidean metric on an n-dim manifold, those invariant under O(n)), or you can fix a vierbien and declare it orthonormal, leading to some (unique) metric?

Sure, you can do it both ways. Since the frame fields form a basis at every tangent plane, one can define the metric as whatever makes some arbitrary frame orthonormal. That would involve deforming the manifold, though.

Ah, that follows very nicely. And then you make it orthonormal by magicking a metric into existence, as above?

Yes, but in practice, the vierbein encodes the metric automatically (computationally in the sense that the components eiμ do), so one isn't free to redefine it once that's written down.

[Surlethe]

It's a very common conventional restriction, but it's not necessary. Typically, the whole point of is to choose a tetrad/vierbein that has some particularly nice adaptation to the problem at hand, and if we're dealing with physics, it is very commonly orthonormal. Not always, though--for example, if we were studying radiation, it is frequently more convenient to use a null tetrad instead. An analogous situation exists in Riemannian geometry--e.g., we can have a moving frame on the entire torus or sphere, whereas coordinates necessarily fail to be global on those manifolds.

So, for instance, if one were to study a rotating black hole, one would pick a rotating frame; if one were to study a neutron star binary, one would pick rotating-rotating frames, etc.

Sure, you can do it both ways. Since the frame fields form a basis at every tangent plane, one can define the metric as whatever makes some arbitrary frame orthonormal. That would involve deforming the manifold, though.

Isn't this what you meant when you said that changing between frame fields changed the geometry?

Edit:

While I work through the Cartan formalism (shame on you for presenting it in coordinate form!), ...

Oh dear...

http://hphotos-snc3.fbcdn.net/hs088.snc ... 3049_n.jpg
http://hphotos-snc3.fbcdn.net/hs108.snc ... 8443_n.jpg

[Kuroneko]

Isn't this what you meant when you said that changing between frame fields changed the geometry?

No. Doing so does not change the geometry of the manifold, but it does change the frame field on that manifold--I only meant that the frame field is a geometrical construction itself. This is in contrast to coordinates, which do not change any vector fields, just their expression in the chosen coordinates (and indeed coordinate transformations do not change the tetrad-frame components of tensors, once they are expressed in that way).