Lorentz Transformations in general

[Leonhard]

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?

$$L(\beta_x \hat{x} + \beta_y \hat{y} + \beta_z \hat{z}) = L(\beta_x \hat{x})L(\beta_y \hat{y})L(\beta_z \hat{z})$$

 

[George Jones]

Is it simple derived by multiplying the transformation matrices?

No. See

https://www.physicsforums.com/showthread.php?t=503767.

 

[Bill_K]

Leonhard, You're going to shoot me, but it's obvious by inspection!

You have the transformations along the three axes. Just use the fact that L_tt is a scalar under a spatial rotation, L_ti and L_it are vectors under rotation, and the space-space part L_ij is a symmetric tensor. All you have to do is think of two vectors and a tensor that match the cases you are given. This will uniquely determine the solution.

Let β = (β_x, β_y, β_z) = β u where u is a unit vector. Clearly L_tt = γ and both vectors are L_ti = - β γ. The only part that takes some thought is the space-space part. It is L_ij = I + (γ - 1) u u. Write out the nine components of that tensor in terms of β_x, β_y and β_z, and you have the result.