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Please can someone give the proof of all restricted Lorentz transformation matrix can uniquely decomposed into product of pure rotation and pure boost? Or at least give some hint how to proceed to prove it?

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If I understood your question, what you should do is write the Lorentz transformation $\Lambda$ as the exponential of a matrix $e^{\mathbb{L}}$, as is the form of the transformation of a Lie group and from the generic condition of all Lorentz transformation $\Lambda^T\mathbb{G}\Lambda=\mathbb{G}$ you should arrive to prove that $(\mathbb{G}\mathbb{L})^T=-\mathbb{G}\mathbb{L}$, where $\mathbb{G}\doteq(G_{\alpha\beta})$ is the metric tensor (while $\mathbb{L}$ matrix has components of the form ${{\mathbb{L}}^\alpha}_\beta$). You should conclude that the matrix has this form $$ \mathbb{L} = \begin{pmatrix} 0&{\mathbb{L}^0}_1&{\mathbb{L}^0}_2&{\mathbb{L}^0}_3 \\ {\mathbb{L}^0}_1&0&{\mathbb{L}^1}_2&{\mathbb{L}^1}_3 \\ {\mathbb{L}^0}_2&-{\mathbb{L}^1}_2&0&{\mathbb{L}^2}_3 \\ {\mathbb{L}^0}_3&-{\mathbb{L}^1}_3&-{\mathbb{L}^2}_3&0 \end{pmatrix} $$ and that's it. The matrices associated to ${\mathbb{L}^0}_1,{\mathbb{L}^0}_2,{\mathbb{L}^0}_3$ parameters are the generators of the boosts, the others represent the generators of rotations.

Edit: \begin{gather*} \boldsymbol{\beta} \doteq \left( {\mathbb{L}^0}_1, {\mathbb{L}^0}_2, {\mathbb{L}^0}_3 \right), \, \boldsymbol{\alpha} \doteq \left( -{\mathbb{L}^1}_2, {\mathbb{L}^1}_3, -{\mathbb{L}^2}_3 \right) \\ \mathbb{K}_1 \doteq \begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}, \, \mathbb{K}_2 \doteq \begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}, \, \mathbb{K}_3 \doteq \begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix} \\ \mathbb{J}_1 \doteq \begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}, \, \mathbb{J}_2 \doteq \begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}, \, \mathbb{J}_3 \doteq \begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix} %\end{drcases} \\ \mathbb{L} \equiv \beta^i\mathbb{K}_i +\alpha^i\mathbb{J}_i \end{gather*} As you can see $\boldsymbol{\alpha}$ are the parameters of rotation, $\boldsymbol{\beta}$ of boost. The following commutation relations hold \begin{gather*} \begin{cases} \left[ \mathbb{K}_i, \mathbb{K}_j \right] = -{\epsilon_{ij}}^k \mathbb{J}_k \\ \left[ \mathbb{J}_i, \mathbb{J}_j \right] = {\epsilon_{ij}}^k \mathbb{J}_k \\ \left[ \mathbb{J}_i, \mathbb{K}_j \right] = {\epsilon_{ij}}^k \mathbb{K}_k \end{cases} \end{gather*}

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  • $\begingroup$ Nice argument. I got (GL)^T= - GL. by expandindg" exp(L)" and '" exp(L^T) " in terms of "L" and " L^(T)", in generic Lorentz transformation condition, then comparing linear terms in "L" and " L^(T)" on both side of the equation.Then all other higher terms cancel with each other due to the obtained condition .Is this the correct way ? $\endgroup$
    – Rakesh
    Commented Oct 27, 2020 at 19:08
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    $\begingroup$ @ Rob Tan , But how your answer shows that general Lorentz matrix can uniquely decomposed into pure rotation and pure boost?Here it only shows that Lorentz transformation has 6 generators. $\endgroup$
    – Rakesh
    Commented Oct 27, 2020 at 19:25
  • $\begingroup$ You demonstrate it through the fact that the generic Lorentz transformation has the property $(\mathbb{G}^{\alpha\beta})\Lambda^T\mathbb{G}=\Lambda^{-1}$ and noticing that $\Lambda^T=e^{\mathbb{L}^T},\Lambda^{-1}=e^{-\mathbb{L}}$ $\endgroup$
    – Rob Tan
    Commented Oct 27, 2020 at 21:58
  • $\begingroup$ The transformation as you say has $6$ generators, $3$ for boosts in the $3$ spatial directions, $3$ for the rotations around the $3$ planes $x/y,x/z,y/z$. But (I admit I'm not absolutely sure) the generic Lorentz transformation cannot be separated in a pure rotation and a pure boost: this is because the commutation relations between boost generators give rotation generators, try it!! I think this is also the reason for the so-called Thomas precession $\endgroup$
    – Rob Tan
    Commented Oct 27, 2020 at 22:04
  • $\begingroup$ I edited the answer hoping to be more clear $\endgroup$
    – Rob Tan
    Commented Oct 27, 2020 at 22:14

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