Is every Lorentz transformation a pure boost plus some rotation?

Question

I'm self-studying Friedman and Susskind's Special Relativity and Classical Field Theory.
They define a general Lorentz transformation (which keeps the origin fixed) to be a Lorentz boost in $x$-direction composed with some spatial rotation(s). I've proved myself that such a general Lorentz transformation $L$ (leaving origin unchanged) corresponding to velocity $\mathbf v = (v_x, v_y, v_z)$ (using $c=1$) is given by

$$ L = \begin{pmatrix} \gamma & -\gamma v_x & -\gamma v_y & -\gamma v_z\\ -\gamma v_x & 1+(\gamma -1){v_x^2\over v^2} & (\gamma-1){v_xv_y\over v^2} & (\gamma-1){v_xv_z\over v^2}\\ -\gamma v_y & (\gamma-1){v_xv_y\over v^2} & 1+(\gamma-1){v_y^2\over v^2} & (\gamma-1){v_yv_z\over v^2}\\ -\gamma v_z & (\gamma-1){v_xv_x\over v^2} & (\gamma-1){v_yv_z\over v^2} & 1+(\gamma-1){v_z^2\over v^2}\\ \end{pmatrix}, $$

such that for any 4-vector $x$, the same vector as observed by the moving observer is given by $Lx$. ($\gamma = 1/\sqrt{1-\mathbf v^2}$.)

Now, many resources (like the accepted answer in this former SE post of mine) define a Lorentz transformation matrix (still origin fixed) to be any matrix $\Lambda$, satisfying $\Lambda^T\eta\Lambda = \eta$, for the Minkowski metric $\eta$. I've proved that this is a necessary and sufficient condition for leaving the inner products invariant.

But, corresponding to any such $\Lambda$, does there exist an $L$ of the form above?

Answer 1

No, the Lorentz group contains all rotations, boosts, and compositions of rotations and boosts. What you have written is the most general form for a boost. However, a pure rotation, like

\begin{equation} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & 0 & \sin \theta \\ 0 & 0 & 1 & 0 \\ 0 & -\sin \theta & 0 & \cos \theta \end{pmatrix} \end{equation} can not be written in that form.

Edit: I have never seen the most general form of a Lorentz transformation written. I would imagine it is probably a messy expression, depending on three Euler angles (for rotations) and three velocity components (for boosts). It would be heavily dependent on how you want to parameterize it, and for most uses I'm not sure such an expression would be useful (except maybe to satisfy some intellectual curiousity, and that does count for something).

I will say, though, that a good way to "parameterize" groups in general is by using the "generators" of the group, and exponentiating them (using the matrix exponential).

For example, the Lorentz group is "generated" by the six matrices $K_x, K_y, K_z, L_x, L_y, L_z$, corresponding to infinitessimal boosts and rotations, \begin{equation} K_x = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} L_x = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \end{equation}

\begin{equation} K_y = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} L_y = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \end{equation}

\begin{equation} K_z = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} L_z = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation}

For example, a rotation around the $y$ axis by a tiny angle $\theta$ could be written as \begin{equation} I + \theta L_y + \mathcal{O} (\theta^2) \end{equation} which is why we say that $L_y$ "generates" rotations around the y-axis. For a finite (not infinitessimal) angle, we could write the rotation in terms of the matrix exponential as

\begin{equation} e^{\theta L_y} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & 0 & \sin \theta \\ 0 & 0 & 1 & 0 \\ 0 & -\sin \theta & 0 & \cos \theta \end{pmatrix}. \end{equation}

Boosts can be written the same way using the rapidity of the boost, which I'll call $\phi$. For instance, a boost along the z direction can be written as \begin{equation} e^{\phi K_z} = \begin{pmatrix} \cosh \phi & 0 & 0 & \sinh \phi \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh \phi & 0 & 0 & \cosh \phi \end{pmatrix}. \end{equation}

Now, finally I will explain how to "parameterize" the Lorentz group. Every Lorentz transformation can written as

\begin{equation} \Lambda = e^{\theta (\hat n \cdot \vec{L}) + \phi (\hat m \cdot \vec{K})} \end{equation} where $\hat n$ and $\hat m$ are both unit vectors ($|\hat n| = |\hat m| = 1$) and $\vec L$ and $\vec K$ is just notation I am using for a "vector of matrices," so for instance $\vec L = (L_x, L_y, L_z)$ and $\hat n \cdot \vec L = n_x L_x + n_y L_y + n_z L_z$. This parameterization can be thought of as sort of "continuously" performing a rotation by $\theta$ while also "boosting" by the rapidity $\phi$ at the same time. It doesn't decompose nicely into a composition of a boost and a rotation though, because the rotation and boost generators do not commute. \begin{equation} e^{\theta (\hat n \cdot \vec{L}) + \phi (\hat m \cdot \vec{K})} \neq e^{\theta (\hat n \cdot \vec{L}) }e^{\phi (\hat m \cdot \vec{K})} \end{equation}

Answer 2

To add to the discussion what @user1379857 did not regarding "what does the most general proper Lorentz transformation look like written down explicitly", I will provide this as a derivation.

The three rotation matrices (note here that I prefer the Tait-Bryan angles instead of the more "physicist-y" Euler angles because I find them more intuitive as you can think of aiming a pointed craft) for an object in 3D space, initially pointing along the $z$-axis, where the $y$-axis is vertical, the $x$-axis is horizontal, and the $z$-axis points into the screen, look like

$$R_R(\theta_R) := \begin{bmatrix} \cos(\theta_R) && -\sin(\theta_R) && 0 \\ \sin(\theta_R) && \cos(\theta_R) && 0 \\ 0 && 0 && 1 \end{bmatrix}$$

for roll, and

$$R_P(\theta_P) := \begin{bmatrix} 1 && 0 && 0 \\ 0 && \cos(\theta_P) && -\sin(\theta_P) \\ 0 && \sin(\theta_P) && \cos(\theta_P) \end{bmatrix}$$

for pitch, and finally

$$R_Y(\theta_Y) := \begin{bmatrix} \cos(\theta_Y) && 0 && -\sin(\theta_Y) \\ 0 && 1 && 0 \\ \sin(\theta_Y) && 0 && \cos(\theta_Y) \end{bmatrix}$$

for yaw. The relevant order of application is to first roll (bank) your craft side to side, then to pitch it up or down, and finally to yaw it left or right:

$$R_{RPY}(\theta_R, \theta_P, \theta_Y) := R_Y(\theta_Y) R_P(\theta_P) R_R(\theta_R)$$

is the total rotation in 3D space. If you do those matrix multiplications, ideally with the help of a computer(!), you get this awful mess:

$$R_{RPY}(\theta_R, \theta_P, \theta_Y) = \begin{bmatrix} \cos(\theta_Y) \cos(\theta_R) - \sin(\theta_Y) \sin(\theta_P) \sin(\theta_R) && -\sin(\theta_Y) \sin(\theta_P) \cos(\theta_R) - \cos(\theta_Y) \sin(\theta_R) && -\sin(\theta_Y) \cos(\theta_P) \\ \cos(\theta_P) \sin(\theta_R) && \cos(\theta_P) \cos(\theta_R) && -\sin(\theta_P) \\ \sin(\theta_Y) \cos(\theta_R) + \cos(\theta_Y) \sin(\theta_P) \sin(\theta_R) && \cos(\theta_Y) \sin(\theta_P) \cos(\theta_R) - \sin(\theta_Y) \sin(\theta_R) && \cos(\theta_Y) \cos(\theta_P)\end{bmatrix}$$

Going to spacetime, we must add the temporal coordinate, viz.

$$^{(4)}R_{RPY}(\theta_R, \theta_P, \theta_Y) := \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && \cos(\theta_Y) \cos(\theta_R) - \sin(\theta_Y) \sin(\theta_P) \sin(\theta_R) && -\sin(\theta_Y) \sin(\theta_P) \cos(\theta_R) - \cos(\theta_Y) \sin(\theta_R) && -\sin(\theta_Y) \cos(\theta_P) \\ 0 && \cos(\theta_P) \sin(\theta_R) && \cos(\theta_P) \cos(\theta_R) && -\sin(\theta_P) \\ 0 && \sin(\theta_Y) \cos(\theta_R) + \cos(\theta_Y) \sin(\theta_P) \sin(\theta_R) && \cos(\theta_Y) \sin(\theta_P) \cos(\theta_R) - \sin(\theta_Y) \sin(\theta_R) && \cos(\theta_Y) \cos(\theta_P)\end{bmatrix}$$

But, as you noted, to get a general Lorentz transformation, we actually don't now need an arbitrary boost matrix - we just need one, and this is where the "craft" intuition from the Tait-Bryan angles becomes even more helpful. Think about it as flying a relativistic spaceship, which starts off initially "at rest" in some ground frame, pointing along its $z$-axis. If you want to go somewhere, you would first orient your ship according to some positioning $(\theta_R, \theta_P, \theta_Y)$, and then you would engage your thrusters, bringing your rapidity up to some $\phi$ for the cruise. From your point of view, your ship's nose is always facing down your $z$-axis, so we need only consider the $z$-boost:

$$^{(4)}B_z(\phi_z) := \begin{bmatrix} \cosh(\phi_z) && 0 && 0 && \sinh(\phi_z) \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ \sinh(\phi_z) && 0 && 0 && \cosh(\phi_z) \end{bmatrix}$$

Now since we're imagining this as happening from the POV of the pilot, but we produced the rotation matrices from our outside point of view, we should pull-back through the rotation matrix, so hir Lorentz frame shifts via:

$$^{(4)} L(\theta_R, \theta_P, \theta_Y, \phi) :=\ ^{(4)} B_z(\phi)\ ^{(4)} R^{-1}_{RPY}(\theta_R, \theta_P, \theta_Y)$$

corresponding to the craft orientation change followed up by the thruster engagement.

Now we could write that out too, but you can kind of get where this will be going given what I gave above for the $^{(4)}R_\mathrm{RPY}$ matrix and ... yeah.

But yes, then, it is a boost plus a rotation for the proper group - the remainder of the Lorentz group also will involve a further reflection in both space and in time, which gives 4 possible combinations (spatial orientation reversed/not reversed + temporal orientation reversed/not reversed) so 4 more "sheets" of transformations (c.f. Lie group).

---

ADD: As mentioned in the comments, you actually do need the two remaining boost matrices - if Susskind's book suggested you can get away with he $x$-axis (here $z$-axis) boost alone, it was wrong! To continue with the above illustration, imagine that after firing its engines, the ship is "shoved" from either the left or right or top or bottom by some force, thus hitting it with additional rapidities $\phi_x$ and $\phi_y$. We then need

$$^{(4)}B_x(\phi_x) := \begin{bmatrix} \cosh(\phi_x) && \sinh(\phi_x) && 0 && 0 \\ \sinh(\phi_x) && \cosh(\phi_x) && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$$

$$^{(4)}B_y(\phi_y) := \begin{bmatrix} \cosh(\phi_y) && 0 && \sinh(\phi_y) && 0 \\ 0 && 1 && 0 && 0 \\ \sinh(\phi_y) && 0 && \cosh(\phi_y) && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$$

and now we have the full Lorentz boost matrix (note that it will differ if you use a different sequence, but I imagine the shove from the left/right coming first followed by that from the top/bottom):

$$^{(4)} B_{xyz}(\phi_x, \phi_y, \phi_z) :=\ ^{(4)} B_y(\phi_y)\ ^{(4)} B_x(\phi_x)\ ^{(4)} B_z(\phi_z)$$

which, again using our computer, is

$$^{(4)} B_{xyz}(\phi_x, \phi_y, \phi_z) = \begin{bmatrix} \cosh(\phi_y) \cosh(\phi_x) \cosh(\phi_z) && \cosh(\phi_y) \sinh(\phi_x) && \sinh(\phi_y) && \cosh(\phi_y) \cosh(\phi_x) \sinh(\phi_z) \\ \sinh(\phi_x) \cosh(\phi_z) && \cosh(\phi_x) && 0 && \sinh(\phi_x) \sinh(\phi_z) \\ \sinh(\phi_y) \cosh(\phi_x) \cosh(\phi_z) && \sinh(\phi_y) \sinh(\phi_x) && \cosh(\phi_y) && \sinh(\phi_y) \cosh(\phi_x) \sinh(\phi_z)\\ \sinh(\phi_z) && 0 && 0 && \cosh(\phi_z)\end{bmatrix}$$

and then the true "full" Lorentz quarter-group is

$$^{(4)} L(\theta_R, \theta_P, \theta_Y, \phi_x, \phi_y, \phi_z) :=\ ^{(4)} B_{xyz}(\phi_x, \phi_y, \phi_z)\ ^{(4)} R^{-1}_{RPY}(\theta_R, \theta_P, \theta_Y)$$

where the pilot first orients, then hits the engines, gets a left kick and then a bottom kick, which is gonna be really awful to write out! But you can do it, and then you'll have almost (up to some reflections) the most general Lorentz group element.