Leonhard
Jun17-11, 07:58 AM
Hi, I've been breaking my head on the matrix form of the lorentz transformation between one set of coordinates in one inertial frame (t,x^1,x^2,x^3) and what those coordinates will be in another inertial frame (t',x'^2,x'^2,x'^3).
Now I understand that if have a set of coordinates in one inertial frame, and we then those coordinates in an inertial frame with a boost along the x-axis, the transformation matrix between those two coordinates will be
L(\beta \hat{x}) = \left ( \begin{matrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right )
A boost along the y-axis and z-axis are given by
L(\beta \hat{y}) = \left ( \begin{matrix} \gamma & 0 & -\beta\gamma & 0 \\ 0 & 1 & 0 & 0 \\ -\beta\gamma & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right )
and
L(\beta \hat{z}) = \left ( \begin{matrix} \gamma & 0 & 0 & -\beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta\gamma & 0 & 0 & \gamma \end{matrix} \right )
Now I understand that, in general, a transformation from one set of coordinates to another is given by
\tilde{x}^\mu = \Lambda^\mu_\nu x^{\nu}
Where \Lambda is the lorentz transformation between the two frames of references, but I'm not sure how to derive it. I've been told that a general boost (\beta_x, \beta_y, \beta_z) is given by
L(\beta_x \hat{x} + \beta_y \hat{y} + \beta_z \hat{z}) = \left ( \begin{matrix} \gamma & -\beta_x\gamma & -\beta_y\gamma & -\beta_z\gamma \\ -\beta_x\gamma & 1 + (\gamma - 1)\frac{\beta^2_x}{\beta^2} & (\gamma - 1) \frac{\beta_x\beta_y}{\beta^2} & (\gamma -1)\frac{\beta_x \beta_z}{\beta^2} \\ -\beta_y\gamma & (\gamma - 1)\frac{\beta_y}{\beta_x} & 1 + (\gamma - 1)\frac{\beta^2_y}{\beta^2} & (\gamma - 1)\frac{\beta_y\beta_z}{\beta^2} \\ -\beta_z\gamma & (\gamma - 1)\frac{\beta_z\beta_x}{\beta^2} & (\gamma - 1)\frac{\beta_z\beta_y}{\beta^2} & 1 + (\gamma - 1)\frac{\beta^2_z}{\beta^2} \end{matrix} \right )
Is it simple derived by multiplying the transformation matrices?
L(\beta_x \hat{x} + \beta_y \hat{y} + \beta_z \hat{z}) = L(\beta_x \hat{x})L(\beta_y \hat{y})L(\beta_z \hat{z})
Now I understand that if have a set of coordinates in one inertial frame, and we then those coordinates in an inertial frame with a boost along the x-axis, the transformation matrix between those two coordinates will be
L(\beta \hat{x}) = \left ( \begin{matrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right )
A boost along the y-axis and z-axis are given by
L(\beta \hat{y}) = \left ( \begin{matrix} \gamma & 0 & -\beta\gamma & 0 \\ 0 & 1 & 0 & 0 \\ -\beta\gamma & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right )
and
L(\beta \hat{z}) = \left ( \begin{matrix} \gamma & 0 & 0 & -\beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta\gamma & 0 & 0 & \gamma \end{matrix} \right )
Now I understand that, in general, a transformation from one set of coordinates to another is given by
\tilde{x}^\mu = \Lambda^\mu_\nu x^{\nu}
Where \Lambda is the lorentz transformation between the two frames of references, but I'm not sure how to derive it. I've been told that a general boost (\beta_x, \beta_y, \beta_z) is given by
L(\beta_x \hat{x} + \beta_y \hat{y} + \beta_z \hat{z}) = \left ( \begin{matrix} \gamma & -\beta_x\gamma & -\beta_y\gamma & -\beta_z\gamma \\ -\beta_x\gamma & 1 + (\gamma - 1)\frac{\beta^2_x}{\beta^2} & (\gamma - 1) \frac{\beta_x\beta_y}{\beta^2} & (\gamma -1)\frac{\beta_x \beta_z}{\beta^2} \\ -\beta_y\gamma & (\gamma - 1)\frac{\beta_y}{\beta_x} & 1 + (\gamma - 1)\frac{\beta^2_y}{\beta^2} & (\gamma - 1)\frac{\beta_y\beta_z}{\beta^2} \\ -\beta_z\gamma & (\gamma - 1)\frac{\beta_z\beta_x}{\beta^2} & (\gamma - 1)\frac{\beta_z\beta_y}{\beta^2} & 1 + (\gamma - 1)\frac{\beta^2_z}{\beta^2} \end{matrix} \right )
Is it simple derived by multiplying the transformation matrices?
L(\beta_x \hat{x} + \beta_y \hat{y} + \beta_z \hat{z}) = L(\beta_x \hat{x})L(\beta_y \hat{y})L(\beta_z \hat{z})