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{equation}
    \begin{gathered}
    v_k\Lambda^k_{\,i}=\gamma v_i
    \end{gathered}
\end{equation}
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.
 A: 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.
A: 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{equation}
    \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}
\end{equation}
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}
A: \begin{equation}
    \begin{gathered}
    \text{Assume }{\Lambda^k}_i = \delta^k _i +c^k _i   
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
    v_k\Lambda^k_{\,\,\,i}=\gamma v_i
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
    v_k\Lambda^k_{\,\,\,i}-\gamma v_i =0
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
    v^i v_k\Lambda^k_{\,\,\,i}-\gamma v^i v_i =0
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
    v^i v_k(\delta^k _i + c^k _i) -\gamma v^2 =0
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
    (v^2 + v^i v_k c^k _i) -\gamma v^2 =0
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
      v^i v_k c^k _i =\gamma v^2 -v^2 =v^2 (\gamma -1)
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
     v_i v^i v_k c^k _i = v_i v^2 (\gamma -1)
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
      v_k c^k _i = v_i  (\gamma -1)
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
      v^k v_k c^k _i = v^k v_i  (\gamma -1)
    \end{gathered}
\end{equation}
\begin{equation}
    \begin{gathered}
       c^k _i = \frac{v^k v_i}{v^2}  (\gamma -1)
    \end{gathered}
\end{equation}
A: 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 ${\bf v}$.
