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Lorentz Transformations in general

  1. Jun 17, 2011 #1
    Hi, I've been breaking my head on the matrix form of the lorentz transformation between one set of coordinates in one inertial frame [itex](t,x^1,x^2,x^3)[/itex] and what those coordinates will be in another inertial frame [itex](t',x'^2,x'^2,x'^3)[/itex].

    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

    [tex]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 )[/tex]

    A boost along the y-axis and z-axis are given by

    [tex]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 )[/tex]

    and

    [tex]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 )[/tex]

    Now I understand that, in general, a transformation from one set of coordinates to another is given by

    [tex]\tilde{x}^\mu = \Lambda^\mu_\nu x^{\nu}[/tex]

    Where [itex]\Lambda[/itex] 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 [itex](\beta_x, \beta_y, \beta_z)[/itex] is given by

    [tex]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 )[/tex]

    Is it simple derived by multiplying the transformation matrices?

    [tex]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})[/tex]
     
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  3. Jun 17, 2011 #2

    George Jones

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  4. Jun 17, 2011 #3

    Bill_K

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

    You have the transformations along the three axes. Just use the fact that Ltt is a scalar under a spatial rotation, Lti and Lit are vectors under rotation, and the space-space part Lij 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 = (βx, βy, βz) = β u where u is a unit vector. Clearly Ltt = γ and both vectors are Lti = - β γ. The only part that takes some thought is the space-space part. It is Lij = I + (γ - 1) u u. Write out the nine components of that tensor in terms of βx, βy and βz, and you have the result.
     
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