1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Dec 12 '13 at 5:43
  • $\begingroup$ Sorry for asking, but what do you mean by $e_t$? $\endgroup$ – user76284 Dec 13 '13 at 3:07
  • $\begingroup$ The unit vector in the time direction. $\endgroup$ – Muphrid Dec 13 '13 at 6:43
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Dec 13 '13 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 Dec 13 '13 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 Dec 13 '13 at 22:32
  • $\begingroup$ Yes, obviously. $\endgroup$ – Trimok Dec 14 '13 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.