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The relativity group of Minkowski spacetime is the subgroup $P < Aff(4,\mathbb{R})$ which preserves the proper time $c^2 (x_4-y_4)^2 - \|\mathbf{x}-\mathbf{y}\|^2$ between two events $X=(\mathbf{x},x_4,1)$ and $Y=(\mathbf{y},y_4,1)$, where $c$ is a positive constant called the speed of light.

Basically I need to find out what $P$ is. i tried taking a matrix $a$ in the affine group and finding out $a(X-Y)$, then trying to put the entries of this column vector into the proper time equation and choosing matrix entries so that it works, and after a lot of expanding brackets etc i came up with

$$P=\left\{\begin{pmatrix} A & 0 & u \\ 0 & 1 & x \\ 0 & 0 & 1 \end{pmatrix} \ : \ A\in O(3), u\in\mathbb{R}^3, x\in\mathbb{R}\right\}~$$

but then I found out this is wrong. but I'm not sure what is right and how to derive it.

I then need to determine its Lie algebra. which i could do if I knew $P$ i think.

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PS. if someone could just let me know the correct group P then i could try to derive it, so that would be very useful :) – blib Dec 1 '11 at 23:48
Well, the first obvious thing to try is to add in the constraint that it fixes the origin of spacetime. Compute the subgroup of these transformations, and then take the semidirect product with the subgroup of translations to recover the whole group. – Zhen Lin Dec 2 '11 at 0:12
thanks for responding. i don't really understand how to add in that constraint? and how do you know to take the semi-direct product? – blib Dec 2 '11 at 0:42
It is actually easier to determine the Lie algebra. Do you know how to determine the Lie algebra of the Euclidean group? This is quite analogous. – Qiaochu Yuan Dec 2 '11 at 1:50
@blib: Have you derived the generators of Poincare group in matrix form? – omehoque Feb 19 '14 at 21:02

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