Intuitive argument for symmetry of Lorentz boosts

Question

The Lorentz boosts are represented by symmetric $4\times4$ matrices. Though the most general Lorentz transformations has no obvious symmetry property, can the symmetry (under transpose) of the Lorentz boost matrices be understood intuitively? Like, from considerations of principle of relativity (that the inverse transformation can be obtained by the transformation $v\leftrightarrow-v$).

Answer 1

I don't think that may be there exists any intuitive argument for the symmetry of Lorentz boosts. But at least I try to think as follows.

First consider the 1+1-Lorentz boost $\mathbb{L}_{_{2\times2}}$ \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'\vphantom{\dfrac{a}{b}}\\ ct'\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} L_{11} & L_{14}\vphantom{\dfrac{a}{b}}\\ L_{41} & L_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x\vphantom{\dfrac{a}{b}}\\ ct\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \mathbb{L}_{_{2\times2}} \mathbf{X} \tag{01}\label{01} \end{equation} Since special relativity unifies space and time in an entity, an argument would be that the Lorentz boost \eqref{01} must be symmetric under exchange of $x$ and $ct$. So applying the matrix \begin{equation} \sigma_1\boldsymbol{=} \begin{bmatrix} \:\:0 & 1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1 & 0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}\,,\qquad \sigma^2_1\boldsymbol{=}\rm I \tag{02}\label{02} \end{equation} on equation \eqref{01} we have \begin{equation} \begin{bmatrix} ct'\vphantom{\dfrac{a}{b}}\\ x'\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:0 & 1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1 & 0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x'\vphantom{\dfrac{a}{b}}\\ ct'\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \overbrace{ \begin{bmatrix} \:\:0 & 1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1 & 0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} L_{11} & L_{14}\vphantom{\dfrac{a}{b}}\\ L_{41} & L_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \:\:0 & 1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1 & 0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}}^{\sigma_1 \mathbb{L}_{_{2\times2}}\sigma_1} \begin{bmatrix} ct\vphantom{\dfrac{a}{b}}\\ x\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03}\label{03} \end{equation} So we must have $\sigma_1 \mathbb{L}_{_{2\times2}}\sigma_1\boldsymbol{=}\mathbb{L}_{_{2\times2}}$ or \begin{equation} \sigma_1 \mathbb{L}_{_{2\times2}}\boldsymbol{=}\mathbb{L}_{_{2\times2}}\sigma_1 \tag{04}\label{04} \end{equation} The Lorentz boost matrix $\mathbb{L}_{_{2\times2}}$ must commute with the $\sigma_1 $ matrix (that the latter is a Pauli matrix is irrelevant here) \begin{equation} \begin{bmatrix} L_{41} & L_{44}\vphantom{\dfrac{a}{b}}\\ L_{11} & L_{14}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} L_{14} & L_{11}\vphantom{\dfrac{a}{b}}\\ L_{44} & L_{41}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{05}\label{05} \end{equation} From above equation \begin{equation} L_{14}\boldsymbol{=}L_{41} \quad \text{and} \quad L_{11}\boldsymbol{=}L_{44} \tag{06}\label{06} \end{equation} So the matrix $\mathbb{L}_{_{2\times2}}$ must be symmetric with equal elements on the diagonal. Setting \begin{equation} \!\!\!\!\!\!L_{11}\boldsymbol{=}L_{44}\boldsymbol{=}\xi\ge 1 \:\:\texttt{(orthochronus)} \:\text{and} \: L_{14}\boldsymbol{=}L_{41}\boldsymbol{=}\eta\stackrel{\det \mathbb{L}_{_{2\times2}}\boldsymbol{=+}1}{\boldsymbol{=\!=\!=\!=\!=\!=}}\boldsymbol{}\pm\sqrt{\xi^2-1} \tag{07}\label{07} \end{equation} we have \begin{equation} \mathbb{L}_{_{2\times2}}\boldsymbol{=} \begin{bmatrix} \:\:\xi & \eta\:\:\vphantom{\dfrac{a}{b}}\\ \:\:\eta & \xi\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}\,, \qquad \eta\boldsymbol{=}\pm\sqrt{\xi^2-1} \tag{08}\label{08} \end{equation} Given that $y'\boldsymbol{=}y,z'\boldsymbol{=}z $ the corresponding $4\times4$ matrix is \begin{equation} \mathbb{L}_{_{4\times4}}\boldsymbol{=} \begin{bmatrix} \:\:\xi & \:\:0\:\:& \:\:0\:\:& \eta\:\:\vphantom{\dfrac{a}{b}}\\ \:\:0 & \:\:1\:\:& \:\:0\:\:& 0\:\:\vphantom{\dfrac{a}{b}}\\ \:\:0 & \:\:0\:\:& \:\:1\:\:& 0\:\:\vphantom{\dfrac{a}{b}}\\ \:\:\eta & \:\:0\:\:& \:\:0\:\:& \xi\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}\,, \qquad \eta\boldsymbol{=}\pm\sqrt{\xi^2-1} \tag{09}\label{09} \end{equation} By a pure rotation in space we end up with a symmetric matrix for the Lorentz boost. To see how take a look in SECTION B of my answer as "user82794" here Two sets of coordinates each in frames O and O′ (Lorentz transformation).

Answer 2

Here is one way of understanding the group of Lorentz transformations: It is comprised of essentially two types of transformations, rotations in 3 dimensions and boosts. Hopefully rotations are familiar, so I will focus on understanding boosts. Specifically, I will talk about the boosts in the $x$-direction because a boost in any other direction can be constructed by first rotating the direction we want to boost along to point in the $x$-direction, apply a boost along the $x$-direction, and then rotate back.

With these things in mind, Lorentz boosts are the unique transformations $L(v)$ which satisfy the requirements to form a group, and two other conditions:

a) Boosting by zero does nothing: $L(0)=I$

b) Boosting by $-v$ is the same as the inverse boost: $L^{-1}(v)=L(-v)$

c) Associativity (follows automatically from dealing with matrices)

d) Closure: For any two velocities $u,v$ there exists some other velocity $w$ (we don't make any claims on what it should look like) such that $L(u)L(v)=L(w)$

e) Boosting by $v$ from stationary should make us move with velocity $v$: $L(v)\left(\begin{array}{c}t\\0\end{array}\right)=\left(\begin{array}{c}t^\prime\\ vt^\prime\end{array}\right)$ for some $t^\prime$ (again, no claims on what it should be).

f) The final condition is equivalent to the following: Suppose $P$ is the parity matrix, meaning it negates all the spatial directions. Then we should have $$ PL(v)\left(\begin{array}{c}t\\0\end{array}\right)=L(-v)P\left(\begin{array}{c}t\\0\end{array}\right)=L(-v)\left(\begin{array}{c}t\\0\end{array}\right). $$ In other words, applying a boost to a stationary frame and then applying parity should be the same as just applying a boost in the opposite direction.

With these conditions, the generic form of a boost along the $x$-direction is completely fixed except for a free constant, which is equivalent to the speed of light (and taking this constant to infinity gives back Galilean boosts too!).

The requirements are stated somewhat differently, but the steps to actually compute the Lorentz boost from the above data can be found here.

Answer 3

As Frobenius says in a comment, drop to the 1+1-D case. One dimension of time $w=ct$, one dimension of space $x$.

Then you want to preserve the velocity of light and this means $\hat w\pm \hat x$ are eigenvectors of the Lorentz transform: orthogonal (in the Euclidean sense) eigenvectors with real eigenvalues. So it’s Hermitian but it's real: so it's symmetric. Gives you the hint that it's not symmetric if you choose, say, $w=2ct$. But had you chosen $w=2ct$ the resulting matrix would still obey the inversion principle $L(v)L(-v)=I$ so that cannot be sufficient.

Now, that's kind of cheesy, you could say it needs to be “symmetrizable” in some sense, so let me give more substantial examples. First, the 2x2 rotation matrix by angle $\theta=\tan^{-1}(v/c)$ satisfies $R_\theta R_{-\theta}=I$, but $R_\theta$ is not symmetric. Second, the Galilean transformation matrix satisfies $G(v)G(-v)=I$ but it's not symmetric.

Looking at Richard Myers’ answer these appear to satisfy in addition all axioms (a)-(f) and he gives a hint that the second example corresponds to an infinite speed of light, so it seems plausible that the first example is instead something like an imaginary speed of light?

Answer 4

The transformation matrix of the four-vectors in RR is: $$ \Lambda^\mu_{\phantom{\mu}\nu} = \pmatrix{ \gamma & -{\gamma \over c} \bf{v} \cr -\gamma {\bf{v} \over c} & \bf{L} \cr } = \pmatrix{ \gamma & {\gamma\over c}v_1 & {\gamma\over c}v_2 & {\gamma\over c}v_3 \cr -{\gamma\over c}v_1 & 1+(\gamma-1){v_1^2\over v^2} & (\gamma-1){v_2v_1\over v^2} & (\gamma-1){v_3v_1\over v^2} \cr -{\gamma\over c}v_2 & (\gamma-1){v_1v_2\over v^2} & 1+(\gamma-1){v_2^2\over v^2} & (\gamma-1){v_3v_2\over v^2} \cr -{\gamma\over c}v_3 & (\gamma-1){v_1v_3\over v^2} & (\gamma-1){v_2v_3\over v^3} &1+(\gamma-1){v_3^2\over v^2} \cr } $$

$\bf{L}$ is a space operator having eigenvectors $\parallel\bf{v}$ belonging to the eigenvalue $\gamma(v)$ and eigenvectors $\perp\bf{v}$ belonging to the eigenvalue $1$. The symmetry of $\Lambda^\mu_{\phantom{\mu}\nu}$ depends on the symmetry $ (x\leftrightarrow ct)$ of the standard Lorentz transformations: $$ {R^\mu}' = \Lambda^\mu_{\phantom{\mu}\nu} \, R^\nu \qquad\qquad \pmatrix{ct' \cr x' \cr y' \cr z' \cr} = \pmatrix{ \gamma & -{\gamma\over c}v & 0 & 0 \cr -{\gamma\over c}v & \gamma & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr } \pmatrix{ct \cr x \cr y \cr z \cr} $$

To understand this symmetry it's appropriate to derive the above result with a method due (I believe) to Ignatowsky (1910). It's easy to see that, to safeguard the concept of inertial reference system, the coordinate transformation formulae must be pre-formatted (with one indeterminate parameter $\gamma$) as follows: $$ \eqalign{ & x' = \gamma (x - vt) \cr & y' = y \cr & z' = z \cr & t' = \gamma \left[t-\left(1 - 1/\gamma^2 \right){x\over v} \right] \cr } $$ The only relevant assumption needed to obtain this result is that the inverse transformations are gained by changing the sign of time, ie by changing the sign of the relative speed of the refernce systems $(\bf{v} \leftrightarrow \bf{-v})$. Consistently the velocity transformation formula results $$ u' = {u -v\over 1 - \left(1-{1\over \gamma^2} \right){u\over v}} $$

Such formulas are independent of any principle of relativity. For $\gamma = 1$ they provide Galileo's transformations.

Let us now ask ourselves if there can exist a velocity $c$ invariant, that is having the same value in all inertial reference systems: $\forall v: u' = u =c$. For $u'=u=c$ that implies: $$ 1-{1\over \gamma^2} = {v^2 \over c^2} $$ If $\gamma > 1$ this relationship gives us a positive real value for $c$. In this way we obtain the Lorentz transformations in standard configuration, which have the required symmetry $ (x \leftrightarrow ct)$.

Conclusion: if and only if we admit the existence of an invariant velocity the transformation matrices are symmetrical. Therefore the symmetry of the matrix $\Lambda^\mu_{\phantom{\mu}\nu} $ can be seen as an expression of the postulate of constancy of the speed of light.