As I understand it, the value of a 4-vector $x$ in another reference frame ($x'$) with the same orientation can be derived using the Lorentz boost matrix $\bf{\lambda}$ by $x'=\lambda x$. More explicitly, $$\begin{bmatrix} x'_0\\ x'_1\\ x'_2\\ x'_3\\ \end{bmatrix} = \begin{bmatrix} \lambda_{00}&\lambda_{01}&\lambda_{02}&\lambda_{03}\\ \lambda_{10}&\lambda_{11}&\lambda_{12}&\lambda_{13}\\ \lambda_{20}&\lambda_{21}&\lambda_{22}&\lambda_{23}\\ \lambda_{30}&\lambda_{31}&\lambda_{32}&\lambda_{33}\\ \end{bmatrix} \begin{bmatrix} x_0\\ x_1\\ x_2\\ x_3\\ \end{bmatrix} $$ I have seen examples of these components written in terms of $\beta$ and $\gamma$, which are defined as $$\beta=\frac{v}{c}$$ $$\gamma=\frac{1}{\sqrt{1-\beta\cdot\beta}}$$ where $v$ is the 3-velocity and $c$ is the speed of light. My question is this: How can the components of $\lambda$ be written in terms of the 4-velocity $U$ alone?

I know that $U_0=\gamma c$ and $U_i=\gamma v_i=\gamma c\beta_i$ for $i\in\{1,2,3\}$, but I'm having trouble deriving the components for $\lambda$ using the matrices based on $\beta$ and $\gamma$. An example of one of these matrices can be found at Wikipedia. How can I rewrite this matrix in terms of $U$ alone?

  • $\begingroup$ Not sure I understand your question. $U = \gamma c (e_t + \vec \beta)$. You substitute the appropriate components of $U$ in for $\beta, \gamma$, and the result follows. Is this substitution step where you're having trouble? $\endgroup$
    – Muphrid
    Commented Dec 12, 2013 at 5:43
  • $\begingroup$ Sorry for asking, but what do you mean by $e_t$? $\endgroup$
    – user76284
    Commented Dec 13, 2013 at 3:07
  • $\begingroup$ The unit vector in the time direction. $\endgroup$
    – Muphrid
    Commented Dec 13, 2013 at 6:43

1 Answer 1


Take units $c=1$.

You have $U_0^2-\vec U^2=1$, that is $\gamma^2(1-\beta^2)=1$. With some basic transformations, you will get : $\frac{\gamma - 1}{\beta^2}= \frac{\gamma^2}{\gamma + 1}$

Now, from your Wikipedia matrix, you have obvious term, $ U_0 =\gamma , U_i =\gamma \beta_i$

You have $(\gamma -1) \frac{\beta_i\beta_j}{\beta^2} = \frac{\gamma^2}{\gamma + 1}\beta_i\beta_j = \frac{U_i U_j}{U_0 + 1}$

Finally, $1 + (\gamma -1) \frac{\beta_i^2}{\beta^2} = 1 + \frac{U_i^2}{U_0 + 1}$

  • $\begingroup$ Thanks for the answer! Could the entire matrix be written in the form $$I+\frac{UU^T}{U_0+c}$$ including the time components? $\endgroup$
    – user76284
    Commented Dec 13, 2013 at 3:08
  • $\begingroup$ @user1667423 : This works only for the spatial part of the matrix, but it does not work for the entire matrix, you can check this with elements $M_{Oi}$ of the matrix which are not equal to $\delta_{0i} + \frac{U_0U_i}{U_0 +1}$ $\endgroup$
    – Trimok
    Commented Dec 13, 2013 at 9:46
  • $\begingroup$ Of course, I suppose those components can be written as $$\Lambda_{0i}=\Lambda_{i0}=-\frac{U_i}{c}$$ and $$\Lambda_{00}=\frac{U_0}{c}$$? $\endgroup$
    – user76284
    Commented Dec 13, 2013 at 22:32
  • $\begingroup$ Yes, obviously. $\endgroup$
    – Trimok
    Commented Dec 14, 2013 at 9:38

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