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If we take the Minkowski metric, $\eta_{\mu\nu}=(1,-1,-1,-1)$, instead of the usual $(-1,1,1,1)$, does this change the form of the Lorentz Transform? I think the standard Lorentz Transform looks like: $$ \begin{matrix} \gamma & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 &0 & 1 \end{matrix} $$

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The conventional way to indicate which one you are using is by writing $\mathrm{Tr}(g) = \pm 1$ or saying "using the trace equals (minus) one metric". –  dmckee Jan 13 '13 at 3:56

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An interesting question indeed :-) Yes, you can flip the overall sign of the Minkowski metric, and in fact a lot of physicists do this! The sign choice $\operatorname{diag}(-1, 1, 1, 1)$ is conventional in fundamental quantum field theory and in quantum gravity, if I remember correctly, whereas $\operatorname{diag}(1, -1, -1, -1)$ is conventional in particle physics.

This doesn't affect the Lorentz transform, though. If you apply the Lorentz transform to a metric tensor, it computes as $g'_{\alpha\beta} = \Lambda_\alpha^\mu \Lambda_\beta^\nu g_{\mu\nu}$, and so you will automatically come out with the same sign convention that you put in.

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Thanks very much for your help! –  Atreyu Jan 13 '13 at 4:01
However, four-vector norms change sign. A timelike one (inside the light-cone) has a negative norm when using (-1+1+1+1) and positive otherwise. Einstein used (-1 -1 -1 +1) in the Princeton Lectures, for GR. Some of the equations result also formally different. –  Eduardo Guerras Valera Jan 13 '13 at 7:44

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