Lorentz transformations: 1+1 spacetime only

[PeroK]

Yes. In a way, that is my point.

As I understand it, the claim is, starting with frame A:

1) Do a Lorentz boost along in an arbitrary direction - call this frame B.

2) Do another Lorentz boost in an arbitrary direction - call this frame C.

3) The transformation from B to C is not a Lorentz transformation?

3b) Alternatively, the motion of the origin of frame B, as measured in frame C, is not constant velocity (speed and direction)?

 

[robwilson]

As I understand it, the claim is, starting with frame A:

1) Do a Lorentz boost along in an arbitrary direction - call this frame B.

2) Do another Lorentz boost in an arbitrary direction - call this frame C.

3) The transformation from B to C is not a Lorentz transformation?

3b) Alternatively, the motion of the origin of frame B, as measured in frame C, is not constant velocity (speed and direction)?

More or less. I wouldn't say I'm necessarily claiming that, but I am asking the question, and I'm asking for the experimental evidence for whatever answer is provided. So far, I don't find the experimental evidence for the Lorentz group particularly convincing, when compared to the other groups that are available.

 

[vanhees71]

From what I can gather, the issues seem to be:

1) The Lorentz group is the wrong group for spacetime symmetries.

2) If you compose two Lorentz Transformations (in 2D or 3D) you get a non-inertial, rotating coordinate system.

3) The Lorentz Group is not a global spacetime symmetry group when gravity is involved.

4) An exact (mathematically consistent) global spacetime symmetry group is sought for the precise spacetime of the universe as it is.

1) there is not a single hint from observations that this might be wrong, despite it's tested in several ways
2) the full Poincare group transforms from one inertial frame to another. The composition of two rotation-free boosts in different directions is not a rotation-free Lorentz boost but one followed by a rotation ("Wigner rotation"). That doesn't mean that Lorentz transformations lead to a rotating coordinate system since the rotations are "static", i.e., the Lorentz transforms including rotations transform Minkowski-orthogonal tetrads to Minkowski-orthogonal tetrads but do not lead to a time-dependent rotation of the spatial coordinates.
3) is correct.
4) I don't understand the question. In GR the space-time geometry has no fixed geometry nor symmetry but it depends on the distribution of "matter", which may or may not be symmetric in some sense.

 

[stevendaryl]

I'm coming into this discussion pretty late, but for the record, it's not the case that a combination of two Lorentz transformations is another Lorentz transformation. There are 10 independent linear transformations such that any two inertial coordinate systems are related by some combination of the 10:

1. A translation in the x-direction (that is, the transformation $x' = x + \delta x$.
2. A translation in the y-direction
3. A translation in the z-direction
4. A translation in the t-direction ($t' = t + \delta t$)
5. A rotation about the x-direction (the transformation $y' = y cos(\theta) + z sin(\theta), z' = z cos(\theta) - y sin(\theta)$)
6. A rotation about the y-direction
7. A rotation about the z-direction
8. A boost in the x-direction ($x' = \gamma (x - v t), t' = \gamma (t - \frac{v}{c^2} x)$
9. A boost in the y-direction
10. A boost in the z-direction

If you combine boosts in two different directions, the result is not a boost, but a combination of a boost and a rotation.

Lorentz transformations themselves don't form a group (in more than one spatial dimension), but only the combination of Lorentz transformations + rotations.

 

[stevendaryl]

As I understand it, the claim is, starting with frame A:

1) Do a Lorentz boost along in an arbitrary direction - call this frame B.

2) Do another Lorentz boost in an arbitrary direction - call this frame C.

3) The transformation from B to C is not a Lorentz transformation?

3b) Alternatively, the motion of the origin of frame B, as measured in frame C, is not constant velocity (speed and direction)?

Are these rhetorical questions? If they are real questions, the answer is: The relationship between B and C is not a Lorentz transformation, but a combination of Lorentz transformation and rotation. It certainly is not the case that a combination of Lorentz transformations can produce a relationship that is not a constant-velocity relationship.

 

[PeroK]

It certainly is not the case that a combination of Lorentz transformations can produce a relationship that is not a constant-velocity relationship.

Yes, that's what we have been arguing. Or, to avoid the double negatives:

The product of two Lorentz boosts produces a constant velocity relationship (Lorentz boost plus fixed rotation).

 

[vanhees71]

I'm coming into this discussion pretty late, but for the record, it's not the case that a combination of two Lorentz transformations is another Lorentz transformation. There are 10 independent linear transformations such that any two inertial coordinate systems are related by some combination of the 10:

1. A translation in the x-direction (that is, the transformation $x' = x + \delta x$.
2. A translation in the y-direction
3. A translation in the z-direction
4. A translation in the t-direction ($t' = t + \delta t$)
5. A rotation about the x-direction (the transformation $y' = y cos(\theta) + z sin(\theta), z' = z cos(\theta) - y sin(\theta)$)
6. A rotation about the y-direction
7. A rotation about the z-direction
8. A boost in the x-direction ($x' = \gamma (x - v t), t' = \gamma (t - \frac{v}{c^2} x)$
9. A boost in the y-direction
10. A boost in the z-direction

If you combine boosts in two different directions, the result is not a boost, but a combination of a boost and a rotation.

Lorentz transformations themselves don't form a group (in more than one spatial dimension), but only the combination of Lorentz transformations + rotations.

Of course the Lorentz group is a group and thus the composition of two Lorentz transformations is again a Lorentz transformation. It's all linear transformations which leave the Minkowski product invariant. For four-vector components it reads
$$V^{\prime \mu} = {\Lambda^{\mu}}_{\nu} V^{\nu},$$
and the matrix must fulfill the pseudo-orthogonality relation
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}$$
with $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.
These matrices build the group $\mathrm{O}(1,3)$. Physically relevant for the space-time description is a priori only the subgroup connected continuously to the identity, and that's the proper orthchronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$, i.e., all Lorentz-trafo matrices with determinent 1 and ${\Lambda^0}_0 \geq 1$.

While the rotations form a subgroup the rotation-free boosts are not a subgroup; only those in one fixed direction form an Abelian subgroup. The composition of two rotation-free boosts in different directions is of course again a Lorentz transformation but not a rotation-free boost, but a rotation-free boost followed by a rotation (the Wigner rotation).

Minkowski space is the affine pseudo-Euclidean space with a fundamental form of signature (1,3), also called an affine Lorentzian space, and as such the full symmetry group is the Poincare group and is the corresponding semidirect product generated by Lorentz transformations and spatio-temporal translations. Again a priori physically relevant is the proper orthochronous Poincare group, and indeed Nature is described well with a space-time model obeying this symmetry group. The larger group built from the proper orthochronous Poincare group by including time, space, and space-time reflections is not a symmetry group of Nature. The weak interaction violates both time reversal as well as space reflections. Within relativistic local QFT you have, however, necessarily an additional symmetry, which is charge conjugation, i.e., where all particles in a reaction are exchanged by their antiparticles. The weak interaction also breaks the C-symmetry. Today, it has been independently observed that the weak interaction breaks all these discrete symmetries, i.e., T, P, and CP. Relativistic local QFT predicts however that necessarily the "grand reflection" CPT must be a symmetry, and so far this symmetry indeed has passed all tests.

Together with all the other tests of Lorentz symmetry, it is pretty safe to say that Poincare symmetry is obeyed by all phenomena with an amazing accuracy, as is the extension to GR and the corresponding gauge symmetry leading to it from SR and its global Poincare symmetry.

 

[Ibix]

Of course the Lorentz group is a group and thus the composition of two Lorentz transformations is again a Lorentz transformation.

Just to check: "Lorentz transform" as a technical term includes boosts, rotations, and translations?

 

[stevendaryl]

Of course the Lorentz group is a group and thus the composition of two Lorentz transformations is again a Lorentz transformation.

You're using "Lorentz transformation" in a way that includes boosts and rotations. Which is fine, but it certainly is not what is meant by "Lorentz transformation" when it is first introduced.

 

[stevendaryl]

Just to check: "Lorentz transform" as a technical term includes boosts, rotations, and translations?

I think that when people are first introduced to Lorentz transformations, they don't mean rotations. Rotations have to be included if you are going to consider combinations of boosts along different axes.

 

[vanhees71]

Well, I might have again a too narrow terminology. For me

Lorentz transformations are the transformations which keep the origin of a reference frame of Minkowski space fixed and keep the Minkowski product of all vectors unchanged. In the fundamental representation that's the matrix group O(1,3) (using the west-coast convention of the metric, i.e., $(\eta_{\mu \nu})=(1,-1,-1,-1)$. This group has of course several subgroups. One is SO(1,3), i.e., all Lorentz-transformation matrix with determinant +1 (any Lorentz-trafo matrix must have determinant +1 or -1 of course). The physically most important one is $\mathrm{SO}(1,3)^{\uparrow}$, which in addition has ${\Lambda^0}_0 \geq 1$, which keeps also the time-direction unchanged an is the subgroup of the Lorentz group that is continuously connected with the identity. Of course also the rotations form subgroups (O(3), which are those including space reflections and SO(3), which contains only the proper rotations). The Lorentz transformations build the symmetry group of Minkowski space in the sense of a vector space.

Boosts are proper orthochronous Lorentz transformations which describe a change from one inertial frame to another moving against it with a constant velocity without rotating the spatial axes. They do not form a subgroup (only if you restrict yourself to the boosts in one fixed direction they form a one-parameter subgroup).

The Poincare group then also includes temporal and spatial translations and together with the Lorentz transformations generate the symmetry group of Minkowski space in the sense of an affine Lorentzian manifold.

 

[PeroK]

I think that when people are first introduced to Lorentz transformations, they don't mean rotations. Rotations have to be included if you are going to consider combinations of boosts along different axes.

In Sean Carroll's Spacetime and Geometry, he defines a Lorentz Transformation as satisfying: $$\eta = \Lambda^T \eta \Lambda$$ So, that's not just boosts, but everything in the Lorentz Group.

In truth, the terminology probably does change between elementary texts (where Lorentz Transformations are 1D boosts) and more advanced texts where the Lorentz Group is used.

 

[stevendaryl]

In Sean Carroll's Spacetime and Geometry, he defines a Lorentz Transformation as satisfying: $$\eta = \Lambda^T \eta \Lambda$$ So, that's not just boosts, but everything in the Lorentz Group.

In truth, the terminology probably does change between elementary texts (where Lorentz Transformations are 1D boosts) and more advanced texts where the Lorentz Group is used.

Yes, but (maybe this is an unwarranted assumption on my part) if someone is asking questions about how to generalize the 1-D Lorentz transformations to 3-D, then they probably are not using Carroll's definition of a Lorentz transformation.

 

[PeroK]

Yes, but (maybe this is an unwarranted assumption on my part) if someone is asking questions about how to generalize the 1-D Lorentz transformations to 3-D, then they probably are not basing their question on Carroll's Spacetime and Geometry

You may be right. Although, perhaps you ought to read posts #66 and #68.

 

[DrGreg]

Looking back on this thread, I think there may have been a confusion between two different meanings of the word "rotation":

1. (kinematic) where the angle between two things is continuously changing as a function of time, we describe the changing orientation of one relative to the other as "rotation";
2. (static) where the angle between two things is constant, we may be able to describe the transformation from one to the other as "a rotation"

The "rotations" within the Lorentz group are of the static type.

 

[PeterDonis]

I am interested in a mathematically consistent globally correct answer. SR and GR taken together do not provide that.

Sure they do. The mathematically consistent globally correct answer for a general spacetime (i.e., one that doesn't have to be flat, so there can be gravity and non-negligible stress-energy present) just doesn't have Lorentz transformations, or even the more general group of Poincare transformations, as a symmetry group. So what?

 

[PeterDonis]

The question is, what group do you need if you want the speed of light to be finite and the same for all observers.

If by "speed of light" you mean the coordinate speed, there is no such group in a general curved spacetime. In flat spacetime, the Lorentz group (correctly understood to include spatial rotations so that you have group closure) is the correct group (or the Poincare group if you want to include translations).

 

[PeterDonis]

Lorentz transformations themselves don't form a group (in more than one spatial dimension), but only the combination of Lorentz transformations + rotations.

"Lorentz transformations" in the context of group theory, as I understand it, is standardly taken to include spatial rotations, so that you have group closure. The fact that this issue does not arise in 1+1 spacetime, where there are no spatial rotations, does not mean it is simply ignored, as the OP of this thread appears to believe.

To put it another way, the term "Lorentz group", as I have always understood it, is the subgroup of the Poincare group (i.e., the group of all boosts, spatial rotations, space and time translations, and parity and time reversals) that (a) keeps the origin fixed, and (b) is continuously connected to the identity. (I guess to be completely pedantically correct, this group should be called the "proper orthochronous Lorentz group".)

[Edit: I see @vanhees71 has already posted along the same lines.]

 

[PeterDonis]

Moderator's note: The subthread on "what is a rotating worldline" has been moved to a separate thread, here:

https://www.physicsforums.com/threads/what-is-a-rotating-worldline.1000875/

 

[PeterDonis]

if someone is asking questions about how to generalize the 1-D Lorentz transformations to 3-D, then they probably are not using Carroll's definition of a Lorentz transformation.

That seems like an issue with pedagogy, not physics. The obvious pedagogical solution is to teach them the proper definition, the one that has "Lorentz transformations" still forming a group when we go from 1 to 3 spatial dimensions. For bonus points, you could explain why it's so convenient to have the transformations form a group, and why you have to include the spatial rotations to do that.

 

[PeterDonis]

1+1 dimensions, where we are talking about 2 independent observers. I struggle in 2+1 dimensions, where we have three independent observers. In 3+1 dimensions, with four independent observers

This makes no sense to me. What does the number of spacetime dimensions have to do with the number of observers?

 

[PeterDonis]

from the one-dimensional Lorentz transformations it is impossible to infer what the two-dimensional group is. Therefore there is a physical assumption going into the process somewhere.

I would assume that the added assumption has to do with invariance under purely spatial rotations, which don't exist in 1+1 spacetime but do exist when there are two or more spatial dimensions. We know that spatial rotations form a group, so then we just have to figure out what larger group includes spatial rotations as a subgroup and also includes Lorentz boosts. The usual definition of "Lorentz transformations" includes spatial rotations (I have seen the group of such transformations referred to as "spacetime rotations") because that group (the proper orthochronous Lorentz group, as I referred to it in a previous post) is the smallest one that includes both spatial rotations and boosts (but, as already noted by @vanhees71, boosts only form a subgroup if we restrict to boosts in a single direction).

 

[robwilson]

The OP considers that this thread has outlived its usefulness. I thank those people who provided useful responses. Those from physicists and engineers trying to teach a group theorist how to do group theory were not useful, and were frankly insulting. Those from people unwilling to distinguish between physical reality and our current best fit mathematical models of reality merely muddied the waters. I notice that many of you have visited my blog, in which I explain the problem as I see it, and the wider context, and perhaps some of you have looked at my three recent papers on the arXiv, where I explain the group theory in detail. I do not claim to have any answers, but I do claim to be asking the right questions.

 

[PeterDonis]

my three recent papers on the arXiv, where I explain the group theory in detail. I do not claim to have any answers, but I do claim to be asking the right questions.

In any case, you are pursuing your research, and time will tell whether it is useful. Best of luck to you in your endeavors.

Thread closed.