Interpreting Lorentz Transformation Rotations

[David Lewis]

Are Lorentz transforms actual "rotations" in the commonly understood sense, or a non-intuitive formal mathematical operation?

 

[PeterDonis]

Are Lorentz transforms actual "rotations" in the commonly understood sense

It depends on what you think the "commonly understood sense" of "rotations" is.

Pure spatial rotations are a subset of Lorentz transformations. So if pure spatial rotations is what you think the "commonly understood sense" of "rotations" is, then some Lorentz transformations are rotations, but not others.

Pure Lorentz boosts are "rotations" in a sense--they are hyperbolic rotations in a 2-plane--but you might or might not think that falls within the "commonly understood sense" of "rotations".

 

[Dale]

Are Lorentz transforms actual "rotations" in the commonly understood sense, or a non-intuitive formal mathematical operation?

Can any geometric operation in spacetime be considered “the commonly understood sense”? I mean spacetime itself isn’t commonly understood so how could any operation involving spacetime be commonly understood? Even a purely spatial rotation gets weird since in four dimensions it is no longer a rotation about an axis.

 

[pervect]

Are Lorentz transforms actual "rotations" in the commonly understood sense, or a non-intuitive formal mathematical operation?

You might try "Parable of the Surveyor", from the introduction of Taylor & Wheeler's "Spacetime physics". Version 1 is available online, from http://www.eftaylor.com/pub/stp/STP1stEdThruP20.pdf. Their explanation is less abstract than the one I'll sketch out below, but quite a bit longer.

The way I would describe (which is a bit oversimple) is to consider the 2-space case. Let it have cartesian coordinates x and y. Then rotations are a one-parameter group that leaves dx^2 + dy^2, which is distance, unchanged or invariant.

We contrast this to the 1-space + 1-time case of the Lorentz transform. Let the coordiantes be x and t. Then the Lorentz boost is a 1 paramter group that leaves dx^2 - dt^2 constant.

The 3d and 4d cases are a bit more complex to describe, but basically the group of 3d spatial rotations still leaves distances unchanged, and the group of 4d Lorentz boost still leaves the Lorentz interval unchanged.

THe place where I've oversimplifed things is that both rotation and translation leave distances unchaged, and I haven't described the details of how one distinguishes rotations from translations. Similar remarks apply to the Lorentz group - the Poincare group , composed of both translations and boosts, leaves the Lorentz interval unchanged, and I haven't described how to separate out the translations from the boosts.

Note that 3d spatial rotations are a subgroup of the 4d Lorentz group, because 3d spatial rotations leave both distances and the Lorentz interval invariant.

 

[vanhees71]

There is some analogy between rotations and Lorentz transformations. Both are defined as transformations which leave a fundamental form on a real vector space invariant.

For rotations the fundamental form is a scalar product (i.e., a positive definite bilinear form). There you can define orthonormal bases (Cartesian bases) $\vec{e}_j$ which fulfill
$$\vec{e}_j \cdot \vec{e}_k=\delta_{jk}.$$
Using this basis the fundamental form can be written in terms of the components of vectors wrt. to this basis as
$$\vec{V} \cdot \vec{W}=V^{j} V^{k} \vec{e}_j \cdot \vec{e}_k=\delta_{jk} V^{j} V^{k}.$$
I use the usual Einstein summation convention.

Now a transformation between two such Cartesian bases,
$$\vec{e}_j={D^k}_j \vec{e}_k'$$
leads to the transformation of vector components
$$\vec{V}=V^j \vec{e}_j=V^j {D^k}_j \vec{e}_k'=V^{\prime k} \vec{e}_k' \; \Rightarrow \; V^{\prime k}={D^k}_j V^j.$$
And now for all vectors $\vec{V}$ and $\vec{W}$
$$\vec{V} \cdot \vec{W} = \delta_{jk} V^{\prime j} W^{\prime k} = \delta_{jk} {D^j}_{l} {D^k}_m V^l W^m =\delta_{lm} V^l W^m.$$
Since this has to hold true for all vectors, you have
$$\delta_{jk} {D^j}_l {D^k}_m=\delta_{lm},$$
which defines a orthogonal matrix. Writing this in matrix notation it means
$$\hat{D}^{\text{T}} \hat{D}=\hat{1} \; \Rightarrow \; \hat{D}^{\text{T}}=\hat{D}^{-1}.$$
These orthogonal matrices (in $d$ dimensions) build a group, the orthogonal group $\mathrm{O}(d)$.

The very same calculation goes through for the Minkowski space and Lorentz transformations. The only difference is that the fundamental form is no longer positive definite but has signature (1,3) (in the west-coast convention). A pseudo-orthogonal basis (a Galilean basis) is thus defined as
$$\vec{e}_j \cdot \vec{e}_k=\eta_{jk}, \quad (\eta_{jk})=\mathrm{diag}(1,-1,-1,-1).$$
The Lorentz transformation between two such Galilean bases must fulfill (in complete analogy to the rotations above!)
$$\eta_{jk} {\Lambda^j}_{l} {\Lambda^k}_m=\eta_{lm}.$$
In matrix-vector notation this reads
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta},$$
or, because $\hat{\eta}^2=\hat{1}$,
$$\hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda} = \hat{1},$$
i.e.,
$$\hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta} = \hat{\Lambda}^{-1}.$$
The Lorentz-transformation matrices also build a group, the pseudo-orthogonal group for the fundamental form of signature (1,3), the O(1,3).