I know that the Lorentz transform, when two frames $\mathcal{S}$ and $\mathcal{S}'$ are in standard configuration (the axes are all parallel to their counterparts in the other inertial frame) is given by

$$L~=~\left( \begin{array}{cccc} \gamma & \beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

Is the definition of the Lorentz transform actually just

$$L^T\cdot \eta \cdot L=\eta$$

Where $\eta=\text{diag}(-1,1,1,1)$? I am trying to find a fundamental definition of the Lorentz transformation so I can show that some transformations are LTs, e.g. an ordinary rotation is obviously a Lorentz transformation but I'd like to show this explicitly and without any reference to the Lorentz group if possible. Is the defintion simply: a transformation in Minkowski space that preserves the length of 4-vectors under the Minkowski scalar product? If so then it would be easy.

  • 7
    $\begingroup$ You are completely correct. The equation you wrote above ($\Lambda^t \cdot \eta \cdot \Lambda = \eta$) is probably the most concise definition. Actually it defines the Lorentz group $SO(3,1)$, see en.wikipedia.org/wiki/Indefinite_orthogonal_group. $\endgroup$
    – Vibert
    May 11, 2013 at 11:10
  • 1
    $\begingroup$ This question inspires the shortest, complete and correct answer for this site: "Yes" :-) (Well OK, I suppose the shortest conceivable answer is "No") $\endgroup$
    – twistor59
    May 11, 2013 at 11:42
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    $\begingroup$ There are some variants. The most basic is what you have written - any orthogonal transform with respect to Minkowsky scalar product. But there are also some constraints that could be posed, usually leaving the orientation of the time axis, or the space, or both, the same, or having a positive determinant. That would select 2 or 1 of initially 4 connected components of an orthogonal group. They do not affect much formulas, though. $\endgroup$
    – firtree
    May 11, 2013 at 11:47
  • $\begingroup$ So is the definition of an inertial frame given in a similar way? It is a frame in which the co-ordinates transform the old frame the new one by a Lorentz transformation? $\endgroup$
    – shilov
    May 11, 2013 at 14:14
  • $\begingroup$ @shilov You should consider translations and Lorentz transformations (Poincaré group). I think you could define an inertial frame as a frame where the speed of light is always $c$. $\endgroup$
    – jinawee
    May 11, 2013 at 15:03


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