# Intuitive argument for symmetry of Lorentz boosts

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