General Relativity * – Page 3

$\mathbf{A}=\begin{bmatrix} A'^t\\ A'^x\\ A'^y\\A'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} A^t \\ A^x \\ A^y \\ A^z \end{bmatrix}$

$\mathbf{B}=\begin{bmatrix} B'^t\\ B'^x\\ B'^y\\B'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} B^t \\ B^x \\ B^y \\ B^z \end{bmatrix}$

are two four-vectors.

$\begin{aligned} A'^tB'^t & =(\gamma A^t-\gamma\beta A^x)(\gamma B^t-\gamma\beta B^x)\\ & = \gamma^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2\beta^2A^xB^x \end{aligned}$

$\begin{aligned} A'^xB'^x & = (-\gamma\beta A^t+\gamma A^x)(-\gamma\beta B^t+\gamma B^x)\\ & = \gamma^2\beta^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2A^xB^x\\ \end{aligned}$

$\begin{aligned} A'^yB'^y & = A^yB^y \end{aligned}$

$\begin{aligned} A'^zB'^z & = A^zB^z \end{aligned}$

$\begin{aligned} &\quad\enspace -A'^tB'^t+A'^xB'^x+A'^yB'^y+A'^zB'^z \\ & = (-\gamma^2+\gamma^2\beta^2)A^tB^t+(-\gamma^2\beta^2+\gamma^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\gamma^2(1-\beta^2)A^tB^t+\gamma^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^tB^t+\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -A^tB^t+A^xB^x+A^yB^y+A^zB^z \end{aligned}$

Scalar product of any two four-vectors is frame-independent.

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Category: General Relativity *

201902210102 Exercise 3.1.1