# Deriving $\Lambda^i_{\,j}$ components of the Lorentz transformation matrix

I am trying to follow Weinberg's derivation (in the book Gravitation and Cosmology) of the Lorentz transformation or boost along arbitrary direction. I am having trouble deriving the $$\Lambda^i_{\,j}$$ components. Here's how I am trying, \begin{align} \eta_{0i}=0&=\eta_{\alpha\beta}\Lambda^\alpha_{\,0}\Lambda^\beta_{\,i}\nonumber\\ &=\eta_{00}\Lambda^0_{\,0}\Lambda^0_{\,i}+\eta_{jk}\Lambda^j_{\,0}\Lambda^k_{\,i}\nonumber\\ &=-c^2\gamma\left(-\frac{1}{c^2}\gamma v_i\right)+\eta_{jk}\left(-v^j\gamma\right)\Lambda^k_{\,i}\nonumber\\ &=\gamma^2 v_i-\gamma v_k\Lambda^k_{\,i} \end{align} Then we have, $$$$\begin{gathered} v_k\Lambda^k_{\,i}=\gamma v_i \end{gathered}$$$$ How do I go from the above equation to the solution below? $$\Lambda^i_{\,j}=\delta^i_{\,j}+\frac{v^i v_j}{\mathbf{v}^2}\left(\gamma-1\right)$$ I am a newbie in the subject and please show the in between steps.

I would like to see a general quick derivation that follows on exactly in the spirit of Weinberg.

Recall this involves considering a particle in a frame $$\mathcal{O}$$ in which the particle appears to have no velocity and also in a frame $$\mathcal{O}'$$ in which it appears to have velocity $$\mathbf{v} = (\frac{dx'^i}{dt'}),$$ and then using $$dt' = \Lambda^0_{\,\,\,0} dt,$$ $$dx'^i = \Lambda^i_{\,\,\,0} dt,$$ along with the $$00$$ component of the identity $$\Lambda^{\alpha}_{\,\,\,\gamma} \Lambda^{\beta}_{\,\,\,\delta} \eta_{\alpha \beta} = \eta_{\gamma \delta}$$ to show that a Lorentz transformation $$\Lambda^{\alpha}_{\,\,\,\beta}$$ between these frames must satisfy $$\Lambda^0_{\,\,\,0} = \gamma$$ $$\Lambda^i_{\,\,\,0} = \gamma v^i,$$ while the remaining $$\Lambda^{\alpha}_{\,\,\,\beta}$$ components are not uniquely fixed as above since for any rotation $$R^{\alpha}_{\,\,\,\beta}$$ we have that both $$\Lambda^{\alpha}_{\,\,\,\beta}$$ and $$\Lambda^{\alpha}_{\,\,\,\gamma} R^{\gamma}_{\,\,\,\beta}$$ will transform from the frame where the particle appears with zero velocity to the frame where it appears to have velocity $$\mathbf{v}$$.

One way, which is similar to this, is to first consider the special case (boo) of a frame in which $$v^1 = v, v^2 = 0, v^3 = 0$$ and then try to write the results so that it easily applies for any $$\mathbf{v}$$. Thus, given $$\Lambda^{\alpha}_{\,\,\,\beta} = \begin{bmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ 0 & \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ 0 & \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix}$$ use the fact that \begin{align} 1 &= \det(\Lambda^{\alpha}_{\,\,\,\beta} ) \\ &= \gamma \begin{bmatrix} \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} - \gamma v \begin{bmatrix} \gamma v & 0 & 0 \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} \\ &= \gamma \begin{bmatrix} \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} - \gamma^2 v^2 \begin{bmatrix} \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} \end{align} to motivate choosing the rotation $$R^{\alpha}_{\,\,\,\beta}$$ so that the obvious relations $$\Lambda^1_{\,\,\,1} = \gamma$$ $$\Lambda^2_{\,\,\,2} = \Lambda^3_{\,\,\,3} = 1$$ $$\Lambda^1_{\,\,\,2} = \Lambda^1_{\,\,\,3} = ... = 0$$ make the above determinant relation an identity. Thus we work with $$\Lambda^{\alpha}_{\,\,\,\beta} = \begin{bmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ Now, since the $$3 \times 3$$ spatial part of the matrix should reduce to $$I$$ when $$\mathbf{v} = (v,0,0)$$ is zero, we simply try to re-write it as in terms of the identity and a part that depends on $$\mathbf{v}$$ in a way that will easily generalize to arbitrary $$\mathbf{v}$$'s via \begin{align} \Lambda^{i}_{\,\,\,j} &= \begin{bmatrix} \gamma & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= I + (\begin{bmatrix} \gamma & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - I) \\ &= I + \begin{bmatrix} \gamma - 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &= I + (\gamma - 1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &= I + (\gamma - 1) (1,0,0) \otimes (1,0,0)^T \\ &= I + (\gamma - 1) \frac{1}{v^2} v(1,0,0) \otimes v (1,0,0)^T \\ &= I + (\gamma - 1) \frac{1}{\mathbf{v}^2} \mathbf{v} \otimes \mathbf{v}^T \\ &= \delta^i_{\,\,\,j} + (\gamma - 1) \frac{1}{\mathbf{v}^2} v^i v_j \end{align} This is Weinberg's (2.1.20), where I still have $$\mathbf{v} = (v,0,0)$$, but now the relation is a vector relation independent of the form of $$\mathbf{v}$$ so you can just set $$\mathbf{v} = (v^1,v^2,v^3)$$.

The choice $$\mathbf{v} = (v,0,0)$$ means $$\Lambda^i_{\,\,\,0} = \gamma v^i$$ reduces down to $$\Lambda^i_{\,\,\,0} = \gamma v^i = (\gamma v,0,0)$$ but it is easily generalized to a general $$\mathbf{v}$$ frame again by setting $$\Lambda^i_{\,\,\,0} = \gamma v^i$$ for general $$\mathbf{v}$$, which is (2.1.21).

• Thanks! I've accepted the answer. But I've only one remaining doubt. As you write "to motivate choosing the rotation $R^\alpha_{\,\,\beta}$ so that the obvious relations..." Do you mean I can choose a rotation matrix such that any general matrix $\Lambda^i_{\,\,j}$ becomes, $$\Lambda^1_{\,\,\,1} = \gamma$$ $$\Lambda^2_{\,\,\,2} = \Lambda^3_{\,\,\,3} = 1$$ $$\Lambda^1_{\,\,\,2} = \Lambda^1_{\,\,\,3} = ... = 0$$ Oct 24, 2020 at 7:33
• Yes. I've found a few sets of notes which cite Weinberg such as this (see equations 2.3 and 2.18) and this (see equations 10 and 56) and you see they always refer to the formula I derived above (they just state it or ask you to prove it) as a way to motivate it's general form so while still hopeful I'm doubtful one can do much better than this. Oct 24, 2020 at 18:21

As Weinberg says there in that section (page 29), only $$\Lambda^0_{\ 0} = 1$$ and $$\Lambda^{i}_{\ 0} = \gamma v_i$$ are uniquely determined - the other $$\Lambda^{\alpha}_{\ \beta}$$ are not uniquely determined (the reason for this being that if $$\Lambda^{\alpha}_{\ \beta}$$ carries a particle from rest to velocity $$\mathbf{v}$$, then so does $$\Lambda^{\alpha}_{\ \delta} R^{\delta}_{\ \beta}$$ where $$R$$ is an arbitrary rotation).

The convenient choice that Weinberg write down is $$\Lambda^{i}_{\ j} \ = \ \delta_{ij} + \frac{ v_{i} v_{j} }{ v^2 } (\gamma - 1)$$ is just a choice.

EDIT: To see that this choice is consistent with your equation, you can write the above as a $$3 \times 3$$ matrix: $$\tilde{\Lambda} = \mathbb{I} + \frac{\gamma - 1}{v^2} \mathbf{v}\mathbf{v}^{T}$$ where $$\mathbf{v}\mathbf{v}^{T}$$ is an outer product, and you can verify that the components of this matrix agree with the above. Notice that $$v_{k} \Lambda^{k}_{\ j} = \gamma v_{j}$$ can be written as $$\mathbf{v}^{T} \tilde{\Lambda} = \gamma \mathbf{v}^{T}$$ or because $$\Lambda$$ is symmetric, you can also write this as $$\tilde{\Lambda} \mathbf{v} = \gamma \mathbf{v} \ .$$ Plug in the above matrix to the LHS and you get: $$\text{LHS} = \left( \mathbb{I} + \tfrac{\gamma - 1}{v^2} \mathbf{v}\mathbf{v}^{T} \right) \mathbf{v} = \mathbf{v} + \tfrac{\gamma - 1}{v^2} \mathbf{v} \mathbf{v}^{T} \mathbf{v} = \gamma \mathbf{v}$$ which is $$=$$RHS, where the last equality uses $$\mathbf{v}^{T}\mathbf{v} = v^2$$.

EDIT 2: I am wondering if Weinberg has a typo, where he says that $$\tilde{\Lambda} R$$ also satisfies the equation. I think it should rather be $$\tilde{\Lambda}' = R^{T} \tilde{\Lambda} R$$, which solves the equation $$\tilde{\Lambda}' \mathbf{v} = \gamma \mathbf{v}$$ whenever $$\tilde{\Lambda}$$ does.

• Then please at least show, how this choice is consistent. For example, for $i=j$, I see that the choice is consistent. However, when $i\neq j$, I don't see how to show that the choice is consistent. Oct 22, 2020 at 18:04
• Moreover, if $\Lambda^\alpha_{\,\,\beta}$ carries a particles from rest to $\mathbf{v}$, then shouldn't the transformation $\Lambda^\alpha_{\,\,\gamma}R^\gamma_{\,\,\beta}$ carry the particle from rest to a velocity different in direction than $\mathbf{v}$? I am confused. Can you demonstrate this mathematically without making qualitative/cryptic statements as Weinberg does. As I mentioned, I am a newbie! Oct 22, 2020 at 18:16
• I've written out how Weinberg's choice is consistent. For your other question, it's not that the velocity is physically rotated, it's that your coordinate system you use to describe the $\mathbf{v}$ gets rotated (the particle always moves with the same velocity no matter what's the orientation of your coordinate system) Oct 22, 2020 at 19:06
• Question regarding Edit 3: "$\bar{\Lambda}^\prime=R^T\bar{\Lambda}R$ solves equation $\bar{\Lambda}^\prime\mathbf{v}=\gamma\mathbf{v}$ whenever $\bar{\Lambda}$ does". But I find this, $$\bar{\Lambda}\mathbf{v}=\gamma\mathbf{v}\implies R^T\bar{\Lambda}RR^T\mathbf{v}=\gamma R^T\mathbf{v}\implies \bar{\Lambda}^\prime\mathbf{v^\prime}=\gamma \mathbf{v^\prime},$$ with $\mathbf{v^\prime}=R^T\mathbf{v}$. Therefore, this equation is different from $\bar{\Lambda}^\prime\mathbf{v}=\gamma\mathbf{v}$. Please, elaborate this discrepancy. Oct 24, 2020 at 10:03

The components $$\Lambda^i_{\,\,\,j}$$ cannot be uniquely determined. The best way you can motivate the form of these components are given in bolbteppa's answer. Perhaps this is the best you can do. However, this can still feel like cheating, especially when you generalize the results from $$(v,0,0)$$ to the case of $$\mathbf{v}$$. So, this answer will complement the previously cited answer in this generalizing. As we now know our convenient form the components $$\Lambda^i_{\,\,\,j}$$, we can do the following algebra,

$$$$\begin{gathered} v_k\Lambda^k_{\,\,\,i}=\gamma v_i\\ v_k\Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)v_i+v_k\delta^k_{\,\,\,\,i}\\ v_k\Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)v_i\frac{v_kv^k}{\mathbf{v}^2}+v_k\delta^k_{\,\,\,\,i}\\ v_k\Lambda^k_{\,\,\,i}=v_k\left[\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i}\right]\\ v_k\left[\Lambda^k_{\,\,\,i}-\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i}\right]=0 \end{gathered}$$$$ Now, as $$v^k$$ is arbitrary, we must have, \begin{align} \Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i} \end{align}

$$$$\begin{gathered} \text{Assume }{\Lambda^k}_i = \delta^k _i +c^k _i \end{gathered}$$$$

$$$$\begin{gathered} v_k\Lambda^k_{\,\,\,i}=\gamma v_i \end{gathered}$$$$

$$$$\begin{gathered} v_k\Lambda^k_{\,\,\,i}-\gamma v_i =0 \end{gathered}$$$$

$$$$\begin{gathered} v^i v_k\Lambda^k_{\,\,\,i}-\gamma v^i v_i =0 \end{gathered}$$$$

$$$$\begin{gathered} v^i v_k(\delta^k _i + c^k _i) -\gamma v^2 =0 \end{gathered}$$$$

$$$$\begin{gathered} (v^2 + v^i v_k c^k _i) -\gamma v^2 =0 \end{gathered}$$$$

$$$$\begin{gathered} v^i v_k c^k _i =\gamma v^2 -v^2 =v^2 (\gamma -1) \end{gathered}$$$$

$$$$\begin{gathered} v_i v^i v_k c^k _i = v_i v^2 (\gamma -1) \end{gathered}$$$$

$$$$\begin{gathered} v_k c^k _i = v_i (\gamma -1) \end{gathered}$$$$

$$$$\begin{gathered} v^k v_k c^k _i = v^k v_i (\gamma -1) \end{gathered}$$$$

$$$$\begin{gathered} c^k _i = \frac{v^k v_i}{v^2} (\gamma -1) \end{gathered}$$$$

Consider two inertial systems $$Oy_1y_2y_3t$$ and $$O'y'_1y'_2y'_3t'$$ with common axis $$\left(y'_1,y_1\right)$$ and parallel axes the pairs $$\left(y'_2,y_2\right),\left(y'_3,y_3\right)$$. If the primed system $$O'y'_1y'_2y'_3$$ is moving with uniform velocity $$\mathbf v_0=v\,\mathbf e_1$$ along the common axis with respect to the system $$Oy_1y_2y_3$$ then the $$1+1-$$ Lorentz transformation is \begin{align} y'_1 & = \gamma\left(y_1-v\,t\right) \tag{01a}\label{01a}\\ y'_2 & = y_2 \tag{01b}\label{01b}\\ y'_3 & = y_3 \tag{01c}\label{01c}\\ t' & = \gamma\left(t-\dfrac{v}{c^2}\,y_1\right) \tag{01d}\label{01d}\\ \gamma & = \left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}} \tag{01e}\label{01e} \end{align} This boost along the $$y_1-$$axis is in matrix form $$$$\mathbf Y'= \begin{bmatrix} \vphantom{\dfrac{a}{b}}\\ \mathbf y'\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ c\,t'\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} y'_1\vphantom{\dfrac{a}{b}}\\ y'_2\vphantom{\dfrac{a}{b}}\\ y'_3\vphantom{\dfrac{a}{b}}\\ y'_4\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} \hphantom{-}\gamma & \:\:0\:\: & \:\:0\:\: & -\dfrac{\gamma v}{c}\vphantom{\dfrac{a}{b}}\:\:\\ \hphantom{-}0 & \:\:1\:\: & \:\:0\:\: & \hphantom{-} 0 \vphantom{\dfrac{a}{b}}\:\:\\ \hphantom{-}0 & \:\:0\:\: & \:\:1\:\: & \hphantom{-} 0 \vphantom{\dfrac{a}{b}}\:\:\\ -\dfrac{\gamma v}{c} & \:\:0\:\: & \:\:0\:\: & \hphantom{-} \gamma \vphantom{\dfrac{a}{b}}\:\: \end{bmatrix} \begin{bmatrix} y_1\vphantom{\dfrac{a}{b}}\\ y_2\vphantom{\dfrac{a}{b}}\\ y_3\vphantom{\dfrac{a}{b}}\\ y_4\vphantom{\dfrac{a}{b}} \end{bmatrix} =\mathcal B\,\mathbf Y \tag{02}\label{02}$$$$ where $$$$\mathcal B= \begin{bmatrix} \begin{array}{ccc|c} \hphantom{-}\gamma & \:\:0\:\: & \:\:0\:\: & -\dfrac{\gamma v}{c}\vphantom{\dfrac{a}{b}}\:\:\\ \hphantom{-}0 & \:\:1\:\: & \:\:0\:\: & \hphantom{-} 0 \vphantom{\dfrac{a}{b}}\:\:\\ \hphantom{-}0 & \:\:0\:\: & \:\:1\:\: & \hphantom{-} 0 \vphantom{\dfrac{a}{b}}\:\:\\ \hline -\dfrac{\gamma v}{c} & \:\:0\:\: & \:\:0\:\: & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm B\:\: & \:\:\:\:\: & -\;\dfrac{\gamma \mathbf v_{0}}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v_{0}^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{03}\label{03}$$$$ with $$$$\rm B= \begin{bmatrix} \:\gamma & \:\:0\:\: & \:0\:\vphantom{\dfrac{a}{b}}\:\:\\ \:0 & \:\: 1\:\: & \:0 \: \vphantom{\dfrac{a}{b}}\:\:\\ \:0 & \:\: 0\:\: & \:1 \: \vphantom{\dfrac{a}{b}}\:\: \end{bmatrix} \,,\qquad \mathbf v_{0}= \begin{bmatrix} \:v \:\vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\dfrac{a}{b}}\: \end{bmatrix} \,,\qquad \mathbf v_{0}^{\boldsymbol{\top}}= \begin{bmatrix} \:v \: & \:0 \:& \:0 \:\vphantom{\dfrac{a}{b}}\:\: \end{bmatrix} \tag{04}\label{04}$$$$

In order to make the boost more general, that is not restricted to velocities parallel to the common axis $$\left(y'_1,y_1\right)$$, we make in the system $$\:Oy_1y_2y_3\:$$ a rotation $$\;\rm R\;$$ of the spatial coordinate system to map $$\:(y_1,y_2,y_3) \equiv \mathbf y\: \longrightarrow\: (x_1,x_2,x_3)\equiv\mathbf x \:$$ such that the velocity $$$$\mathbf v_{0}=(v,0,0)=v\,\mathbf e_1 \tag{05}\label{05}$$$$ of system $$\:O'y'_1y'_2y'_3\:$$ relatively to $$\:Oy_1y_2y_3\:$$ to be transformed to $$$$\mathbf{v}=(\upsilon_1,\upsilon_2,\upsilon_3) \tag{06}\label{06}$$$$ To keep the spatial coordinate system right orthonormal we choose any real orthogonal matrix $$\;\rm R\;$$ with positive unit determinant : $$$$\mathrm R = \begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\\ R_{21} & R_{22} & R_{23} \vphantom{\dfrac{a}{b}}\\ R_{31} & R_{32} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \,,\qquad \mathrm R^{-1} =\mathrm R^{\boldsymbol{\top}}= \begin{bmatrix} R_{11} & R_{21} & R_{31} \vphantom{\dfrac{a}{b}}\\ R_{12} & R_{22} & R_{32} \vphantom{\dfrac{a}{b}}\\ R_{13} & R_{23} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{07}\label{07}$$$$ So we must have $$$$\rm R\mathbf{v}_{0}=\mathbf{v} \tag{08}\label{08}$$$$ or $$$$\begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\:\\ R_{21} & R_{22} & R_{23} \vphantom{\dfrac{a}{b}}\:\\ R_{31} & R_{32} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \:v \:\vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\tfrac{a}{b}}\: \end{bmatrix} = \begin{bmatrix} \:v_1 \:\vphantom{\dfrac{a}{b}}\:\\ \:v_2 \: \vphantom{\dfrac{a}{b}}\:\\ \:v_3 \: \vphantom{\tfrac{a}{b}}\: \end{bmatrix} \tag{09}\label{09}$$$$ hence $$$$R_{11}=\dfrac{v_1}{v}\,,\qquad R_{21}=\dfrac{v_2}{v}\,,\qquad R_{31}=\dfrac{v_3}{v} \tag{10.1}\label{10.1}$$$$ Note that due to the property $$\:\rm R\rm R^{\boldsymbol{\top}}=\rm I\:$$ the three rows of $$\:\rm R\:$$ constitute an orthonormal system, so on one hand they are unit 3-vectors \begin{align} R^2_{11}+R^2_{12}+R^2_{13} & =1 \implies R^2_{12}+R^2_{13} =1-R^2_{11}=1-\dfrac{v^2_1}{v^2} \tag{10.2}\label{10.2}\\ R^2_{21}+R^2_{22}+R^2_{23} & =1 \implies R^2_{22}+R^2_{23} =1-R^2_{21}=1-\dfrac{v^2_2}{v^2} \tag{10.3}\label{10.3}\\ R^2_{31}+R^2_{32}+R^2_{33} & =1 \implies R^2_{32}+R^2_{33} =1-R^2_{31}=1-\dfrac{v^2_3}{v^2} \tag{10.4}\label{10.4} \end{align} and on the other hand they are normal to each other \begin{align} R_{11}R_{21}+R_{12}R_{22}+R_{13}R_{23} & =0 \implies R_{12}R_{22}+R_{13}R_{23} =-R_{11}R_{21}=-\dfrac{v_1v_2}{v^2} \tag{10.5}\label{10.5}\\ R_{11}R_{31}+R_{12}R_{32}+R_{13}R_{33} & =0 \implies R_{12}R_{32}+R_{13}R_{33} =-R_{11}R_{31}=-\dfrac{v_1v_3}{v^2} \tag{10.6}\label{10.6}\\ R_{21}R_{31}+R_{22}R_{32}+R_{23}R_{33} & =0 \implies R_{22}R_{32}+R_{23}R_{33} =-R_{21}R_{31}=-\dfrac{v_2v_3}{v^2} \tag{10.7}\label{10.7} \end{align} relations useful for what it follows.

In the $$3+1-$$dimensional Minkowski space the rotation $$\rm R$$ is represented by the following $$4\times 4-$$matrix $$$$\mathcal R= \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\:\: & \:\:\:\:\: & \hphantom{-}\boldsymbol 0 \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & \hphantom{--} \boldsymbol 0^{\boldsymbol{\top}} & & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \,,\qquad \mathcal R^{-1}=\mathcal R^{\boldsymbol{\top}}= \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R^{\boldsymbol{\top}}\:\: & \:\:\:\:\: & \hphantom{-}\boldsymbol 0 \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & \hphantom{-}\: \boldsymbol 0^{\boldsymbol{\top}} & & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{11}\label{11}$$$$ where $$$$\boldsymbol 0= \begin{bmatrix} \:0 \:\vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\dfrac{a}{b}}\:\\ \:0 \: \vphantom{\dfrac{a}{b}}\: \end{bmatrix} \,,\qquad \boldsymbol 0^{\boldsymbol{\top}}= \begin{bmatrix} \:0 \: & \:0 \:& \:0 \:\vphantom{\dfrac{a}{b}}\:\: \end{bmatrix} \tag{12}\label{12}$$$$ hence $$$$\mathbf X = \begin{bmatrix} \vphantom{\dfrac{a}{b}}\\ \mathbf x\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ c\,t\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} \hphantom{=}\rm R\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} y_1\vphantom{\dfrac{a}{b}}\\ y_2\vphantom{\dfrac{a}{b}}\\ y_3\vphantom{\dfrac{a}{b}}\\ y_4\vphantom{\dfrac{a}{b}} \end{bmatrix} =\mathcal R\,\mathbf Y \tag{13}\label{13}$$$$ Now, if in the primed system $$O'y'_1y'_2y'_3$$ exactly the same rotation $$\rm R$$ is used to map $$\:(y'_1,y'_2,y'_3) \equiv \mathbf y'\:\longrightarrow\: (x'_1,x'_2,x'_3)\equiv\mathbf x' \:$$ then $$$$\mathbf X' = \begin{bmatrix} \vphantom{\dfrac{a}{b}}\\ \mathbf x'\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ c\,t'\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} x'_1\vphantom{\dfrac{a}{b}}\\ x'_2\vphantom{\dfrac{a}{b}}\\ x'_3\vphantom{\dfrac{a}{b}}\\ x'_4\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} \hphantom{=}\rm R\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} y'_1\vphantom{\dfrac{a}{b}}\\ y'_2\vphantom{\dfrac{a}{b}}\\ y'_3\vphantom{\dfrac{a}{b}}\\ y'_4\vphantom{\dfrac{a}{b}} \end{bmatrix} =\mathcal R\,\mathbf Y' \tag{14}\label{14}$$$$ The transformation matrix $$\:\rm L\left(\mathbf v\right)$$ (1) for the Lorentz boost with velocity $$\:\mathbf v\:$$ of arbitrary direction is that connecting the 4-vectors $$\:\mathbf X\:$$ and $$\:\mathbf X'$$ $$$$\mathbf X' = \mathcal R\,\mathbf Y'=\mathcal R\,\mathcal B\,\mathbf Y=\mathcal R\,\mathcal B\,\mathcal R^{-1}\,\mathbf X=\mathcal R\,\mathcal B\,\mathcal R^{\boldsymbol{\top}}\,\mathbf X \tag{15}\label{15}$$$$ that is $$$$\mathbf X' = \rm L\left(\mathbf v\right)\mathbf X \tag{16}\label{16}$$$$ where $$$$\rm L\left(\mathbf v\right) = \mathcal R\,\mathcal B\,\mathcal R^{\boldsymbol{\top}} \tag{17}\label{17}$$$$ So \begin{align} \rm L\left(\mathbf v\right) & = \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\:\: & \:\:\:\:\: & \hphantom{-}\boldsymbol 0 \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & \hphantom{--} \boldsymbol 0^{\boldsymbol{\top}} & & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm B\:\: & \:\:\:\:\: & -\;\dfrac{\gamma \mathbf v_{0}}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v_{0}^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R^{\boldsymbol{\top}}\:\: & \:\:\:\:\: & \hphantom{-}\boldsymbol 0 \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & \hphantom{-}\: \boldsymbol 0^{\boldsymbol{\top}} & & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \nonumber\\ & =\begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\:\: & \:\:\:\:\: & \hphantom{-}\boldsymbol 0 \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & \hphantom{--} \boldsymbol 0^{\boldsymbol{\top}} & & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm B\rm R^{\boldsymbol{\top}}\:\: & \:\:\:\:\: & -\;\dfrac{\gamma \mathbf v_{0}}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v_{0}^{\boldsymbol{\top}}\rm R^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\rm B\rm R^{\boldsymbol{\top}}\:\: & \:\:\:\:\: & -\;\dfrac{\gamma \left(\rm R\mathbf v_{0}\right)}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \left(\rm R\mathbf v_{0}\right)^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\rm B\rm R^{\boldsymbol{\top}} \:\: & \:\:\:\:\: & -\;\dfrac{\gamma\mathbf v}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{18}\label{18} \end{align} that is $$$$\rm L\left(\mathbf v\right)= \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm R\rm B\rm R^{\boldsymbol{\top}} \:\: & \:\:\:\:\: & -\;\dfrac{\gamma\mathbf v}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{ccc|c} \:\:L_{11} & \:\:L_{12}\:\:\: & \:\: L_{13}\:\:\: & L_{14} \vphantom{\dfrac{a}{b}}\:\:\\ \:\:L_{21} & \:\:L_{22}\:\:\: & \:\: L_{23}\:\:\: & L_{24} \vphantom{\dfrac{a}{b}}\:\:\\ \:\:L_{31} & \:\:L_{32}\:\:\: & \:\: L_{33}\:\:\: & L_{34} \vphantom{\dfrac{a}{b}}\:\:\\ \hline \:\:L_{41} & \:\:L_{42}\:\:\: & \:\: L_{43}\:\:\: & L_{44} \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{19}\label{19}$$$$ Now \begin{align} \rm R\rm B\rm R^{\boldsymbol{\top}} & = \begin{bmatrix} L_{11} & L_{12} & L_{13} \vphantom{\dfrac{a}{b}}\\ L_{21} & L_{22} & L_{23} \vphantom{\dfrac{a}{b}}\\ L_{31} & L_{32} & L_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &= \begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\\ R_{21} & R_{22} & R_{23} \vphantom{\dfrac{a}{b}}\\ R_{31} & R_{32} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \:\gamma & \:\:0\:\: & \:0\:\vphantom{\dfrac{a}{b}}\:\:\\ \:0 & \:\: 1\:\: & \:0 \: \vphantom{\dfrac{a}{b}}\:\:\\ \:0 & \:\: 0\:\: & \:1 \: \vphantom{\dfrac{a}{b}}\:\: \end{bmatrix} \begin{bmatrix} R_{11} & R_{21} & R_{31} \vphantom{\dfrac{a}{b}}\\ R_{12} & R_{22} & R_{32} \vphantom{\dfrac{a}{b}}\\ R_{13} & R_{23} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &= \begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\\ R_{21} & R_{22} & R_{23} \vphantom{\dfrac{a}{b}}\\ R_{31} & R_{32} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \gamma R_{11} & \gamma R_{21} & \gamma R_{31} \vphantom{\dfrac{a}{b}}\\ R_{12} & R_{22} & R_{32} \vphantom{\dfrac{a}{b}}\\ R_{13} & R_{23} & R_{33} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{20}\label{20} \end{align} Since the matrix $$\:\rm R\rm B\rm R^{\boldsymbol{\top}}\:$$ is symmetric it's sufficient to determine 6 of its elements $$\:L_{ij}$$ \begin{align} L_{11} & = \gamma R^2_{11}+\overbrace{R^2_{12}+R^2_{13}}^{\eqref{10.2} =1-R^2_{11}}=1+\left(\gamma-1\right) R^2_{11}\stackrel{\eqref{10.1}}{=\!=\!=}1+\left(\gamma-1\right)\dfrac{v^2_1}{v^2}=1+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v^2_1 \tag{21.1}\label{21.1}\\ L_{22} & = \gamma R^2_{21}+\overbrace{R^2_{22}+R^2_{23}}^{\eqref{10.3} =1-R^2_{21}}=1+\left(\gamma-1\right) R^2_{21}\stackrel{\eqref{10.1}}{=\!=\!=}1+\left(\gamma-1\right)\dfrac{v^2_2}{v^2}=1+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v^2_2 \tag{21.2}\label{21.2}\\ L_{33} & = \gamma R^2_{31}+\overbrace{R^2_{32}+R^2_{33}}^{\eqref{10.4} =1-R^2_{31}}=1+\left(\gamma-1\right) R^2_{31}\stackrel{\eqref{10.1}}{=\!=\!=}1+\left(\gamma-1\right)\dfrac{v^2_3}{v^2}=1+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v^2_3 \tag{21.3}\label{21.3}\\ L_{12} & = \gamma R_{11}R_{21}+\overbrace{R_{12}R_{22}+R_{13}R_{23}}^{\eqref{10.5}=-R_{11}R_{21}}=\left(\gamma-1\right) R_{11}R_{21}\stackrel{\eqref{10.1}}{=\!=\!=}\left(\gamma-1\right)\dfrac{v_1v_2}{v^2}=\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v_1v_2 \tag{21.4}\label{21.4}\\ L_{13} & = \gamma R_{11}R_{31}+\overbrace{R_{12}R_{32}+R_{13}R_{33}}^{\eqref{10.6}=-R_{11}R_{31}}=\left(\gamma-1\right) R_{11}R_{31}\stackrel{\eqref{10.1}}{=\!=\!=}\left(\gamma-1\right)\dfrac{v_1v_3}{v^2}=\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v_1v_3 \tag{21.5}\label{21.5}\\ L_{23} & = \gamma R_{21}R_{31}+\overbrace{R_{22}R_{32}+R_{23}R_{33}}^{\eqref{10.7}=-R_{21}R_{31}}=\left(\gamma-1\right) R_{21}R_{31}\stackrel{\eqref{10.1}}{=\!=\!=}\left(\gamma-1\right)\dfrac{v_2v_3}{v^2}=\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}v_2v_3 \tag{21.6}\label{21.6}\\ L_{21} & = L_{12}\,,\qquad L_{31}=L_{13}\,,\qquad L_{32}=L_{23} \tag{21.7}\label{21.7} \end{align} The list of elements $$\:L_{ij}\:$$ of the symmetric matrix $$\:\rm L\left(\mathbf v\right) = \mathcal R\,\mathcal B\,\mathcal R^{\boldsymbol{\top}}\:$$ is completed with that of the fourth row, identical with those of the fourth column $$$$L_{41}=L_{14}=-\dfrac{\gamma v_1}{v}\,,\quad L_{42}=L_{24}=-\dfrac{\gamma v_2}{v}\,,\quad L_{43}=L_{34}=-\dfrac{\gamma v_3}{v}\,,\quad L_{44}=\gamma \tag{21.8}\label{21.8}$$$$ So finally $$$$\rm L\left(\mathbf v\right)= \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm I+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\mathbf v\mathbf v^{\boldsymbol{\top}} \:\: & \:\:\:\:\: & -\;\dfrac{\gamma\mathbf v}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & -\;\dfrac{\gamma \mathbf v^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{ccc|c} \:\:L_{11} & \:\:L_{12}\:\:\: & \:\: L_{13}\:\:\: & L_{14} \vphantom{\dfrac{a}{b}}\:\:\\ \:\:L_{21} & \:\:L_{22}\:\:\: & \:\: L_{23}\:\:\: & L_{24} \vphantom{\dfrac{a}{b}}\:\:\\ \:\:L_{31} & \:\:L_{32}\:\:\: & \:\: L_{33}\:\:\: & L_{34} \vphantom{\dfrac{a}{b}}\:\:\\ \hline \:\:L_{41} & \:\:L_{42}\:\:\: & \:\: L_{43}\:\:\: & L_{44} \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{22}\label{22}$$$$

By above equation the Lorentz boost transformation \eqref{16} is expressed explicitly as follows \begin{align} \mathbf x' & = \mathbf x+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\left(\mathbf v\boldsymbol{\cdot}\mathbf x\right)\mathbf v-\gamma\,\mathbf v\,t \tag{23a}\label{23a}\\ t' & = \gamma\biggl(t-\dfrac{\mathbf v\boldsymbol{\cdot}\mathbf x}{c^2}\biggr) \tag{23b}\label{23b}\\ \gamma & = \left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}} \tag{23c}\label{23c} \end{align} and in differential form \begin{align} \mathrm d\mathbf x' & = \mathrm d\mathbf x+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\left(\mathbf v\boldsymbol{\cdot}\mathrm d\mathbf x\right)\mathbf v-\gamma\,\mathbf v\,\mathrm dt \tag{24a}\label{24a}\\ \mathrm d t' & = \gamma\biggl(\mathrm d t-\dfrac{\mathbf v\boldsymbol{\cdot}\mathrm d \mathbf x}{c^2}\biggr) \tag{24b}\label{24b} \end{align} It's worth to mention at this point that from this differential form we could find the relativistic sum of two velocity 3-vectors not necessarily collinear. Dividing \eqref{24a},\eqref{24b} side by side we have $$$$\dfrac{\mathrm d\mathbf x'}{\mathrm d t'} = \dfrac{\dfrac{\mathrm d\mathbf x}{\mathrm d t}+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\left(\mathbf v\boldsymbol{\cdot}\dfrac{\mathrm d\mathbf x}{\mathrm d t}\right)\mathbf v-\gamma\,\mathbf v}{\gamma\left( \begin{matrix} 1-\dfrac{\mathbf v\boldsymbol{\cdot}\dfrac{\mathrm d\mathbf x}{\mathrm d t}}{c^2} \\ \hphantom{-} \end{matrix}\right)} \tag{25}\label{25}$$$$ Replacing \begin{align} \dfrac{\mathrm d\mathbf x'}{\mathrm d t'} & = \mathbf w'\!=\texttt{particle velocity with respect to }O'x'_1x'_2x'_3 \tag{26a}\label{26a}\\ \dfrac{\mathrm d\mathbf x}{\mathrm d t} & = \mathbf w=\texttt{particle velocity with respect to }\:Ox_1x_2x_3 \tag{26b}\label{26b} \end{align} we have $$$$\mathbf w' = \dfrac{ \mathbf w+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\left(\mathbf v\boldsymbol{\cdot} \mathbf w\right)\mathbf v-\gamma\,\mathbf v}{\gamma\biggl(1-\dfrac{\mathbf v\boldsymbol{\cdot} \mathbf w}{c^2}\biggr)} \tag{27}\label{27}$$$$ the relativistic sum of $$\:\mathbf w\:$$ and $$\boldsymbol{-}\mathbf v$$.

Note that for the inverse of the Lorentz boost matrix we have from \eqref{22} $$$$\rm L^{-1}=\rm L\left(-\mathbf v\right)= \begin{bmatrix} \begin{array}{ccc|c} \:\: & \:\:\:\:\: & \:\:\:\:\: & \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\hphantom{-}\rm I+\dfrac{\gamma^2}{c^2\left(\gamma+1\right)}\mathbf v\mathbf v^{\boldsymbol{\top}} \:\: & \:\:\:\:\: & +\;\dfrac{\gamma\mathbf v}{c} \vphantom{\dfrac{a}{b}}\:\:\\ \:\: & \:\:\:\: & \:\:\:\:\: & \hphantom{-} \vphantom{\dfrac{a}{b}}\:\:\\ \hline & +\;\dfrac{\gamma \mathbf v^{\boldsymbol{\top}}}{c} & & \hphantom{-} \gamma \vphantom{\dfrac{\dfrac{a}{b}}{b}}\:\: \end{array} \end{bmatrix} \tag{28}\label{28}$$$$

(1) I keep the symbol $$\rm L(\mathbf v)$$ for Lorentz boosts and $$\Lambda$$ for the proper homogeneous Lorentz transformations

This is just a formula for a boost. The most general Lorentz transformation includes combined boosts and rotations.

I assume that you know the usual formula $$\left[\matrix{ t'\cr x'}\right]= \left[\matrix{ \cosh s&-\sinh s\cr-\sinh s &\cosh s}\right]\left[\matrix{ t\cr x}\right]=$$ where $$s$$ is the rapidity with $$|{\bf v}|=\tanh s$$,and $$\gamma=\cosh s$$, etc.

I don't have a copy of Weinberg to hand, but if you want a formula similar to yours for a boost $${\bf v}$$ that preserves $$t^2-{\bf x}^2$$ then it follows from the above formula that $$\left[\matrix{ t'\cr {\bf x}'}\right]= \left[\matrix{0\cr (1-P){\bf x}}\right]+ \left[\matrix{ \cosh s&-\sinh s\cr-\sinh s &\cosh s}\right]\left[\matrix{ t\cr P{\bf x}}\right]=$$ where $$P$$ is the projection operator along $${\bf v}$$ with matrix elements $$P_{ij}=v^iv^j/|{\bf v}^2|$$. This is just taking the standard rapidity formula for an $$x,t$$ boost and only altering the components of $${\bf x}$$ parallel to $${\bf v}$$.

• Okay, fine! But how do you derive it? Oct 22, 2020 at 16:29