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I know that the Lorentz transformation, 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 general definition of the Lorentz transformation actually just

$$[L^T][\eta] [L]~=~[\eta]?$$

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.

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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. –  Vibert May 11 '13 at 11:10
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This question inspires the shortest, complete and correct answer for this site: "Yes" :-) (Well OK, I suppose the shortest conceivable answer is "No") –  twistor59 May 11 '13 at 11:42
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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. –  firtree May 11 '13 at 11:47
    
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? –  shilov May 11 '13 at 14:14
    
@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 $c$. –  jinawee May 11 '13 at 15:03
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