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I am trying to deduce the Minkowski's dot product for two dimentional space: $$g=x^1y^1-c^2t_xt_y$$ If $f$ denote the Lorentz's transformation for two dimentional case: $$\begin{array}{rcll} f:&\mathbb{R}&\longrightarrow &\mathbb{R}\\ &x&\longmapsto &f(x)=\Lambda x \end{array}\qquad \text{ with }\ \Lambda = \gamma\left(\begin{array}{cc} 1 &-v\\ -v/c^2 & 1 \end{array}\right)$$ we have to find a matrix: $$G=\left(\begin{array}{cc} g_{11} & g_{12}\\ g_{21} & g_{22} \end{array}\right)$$ such that $\chi^tG\chi= (\Lambda \chi)^tG(\Lambda \chi)$ or equivalently: $$G=\Lambda ^tG\Lambda$$ we get the system: $$\begin{array}{rlr} \frac{v^2}{c^2}g_{11}-\frac{v}{c^2}g_{12}-\frac{v}{c^2}g_{21}+\frac{v^2}{c^4}g_{22}&=0&(1)\\ -vg_{11}+\frac{v^2}{c^2}g_{12}+\frac{v^2}{c^2}g_{21}-\frac{v}{c^2}g_{22}&=0& (2)\\ -vg_{11}+\frac{v^2}{c^2}g_{12}+\frac{v^2}{c^2}g_{21}-\frac{v}{c^2}g_{22}&=0& (3)\\ v^2g_{12}-vg_{12}-vg_{21}+\frac{v^2}{c^2}g_{22}&=0&(4) \end{array}$$ which is equivalent to: $$\begin{array}{rll} vg_{11}-g_{12}-g_{21}+\frac{v}{c^2}g_{22}&=0\\ c^2g_{11}-vg_{12}-vg_{21}+g_{22}&=0 \end{array}$$ if we denote $g_{21}=\lambda$ and $g_{22}=\mu$, it's easy to prove that: $$G=\left(\begin{array}{cc} -\mu/c^2& -\lambda\\ \lambda &\mu \end{array}\right)$$ As $G$ must be symmetric, $\lambda=0$; so then: $$G=\left(\begin{array}{cc} -\mu/c^2& 0\\ 0 &\mu \end{array}\right)$$ My question is:

How we can get that $\mu$ must be $-c^2$?

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    $\begingroup$ Minkowski's metric is the definition of Minkowski's metric... What do you want to "deduce" the definition from? The only derivation one can do is from the requirement that c is the same in every inertial system... which leads to the Lorentz transformations and the metric from a physical point of view. c is just a choice of units, anyway, so you might as well make it 1. $\endgroup$ – CuriousOne Jan 31 '16 at 21:58
  • $\begingroup$ Hi FUUNK1000. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Feb 1 '16 at 22:00
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You can't deduce this, because all metrics with different $\mu$ are invariant under Lorentz transformations.

The best you can do is to choose the units in which you measure the invariant interval such that $\mu$ is equal to whatever you like it to be.

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  • $\begingroup$ Actually it is possible to deduce this and is a standard exercise in these lecture notes, see pg.40 after eq. (3.41): users.physics.ox.ac.uk/~smithb/website/coursenotes/rel_A.pdf $\endgroup$ – udrv Feb 4 '16 at 7:34
  • $\begingroup$ @FUUNK1000 Consider the invariance relationship as applied to the space time-interval: for an event $(ct, x)$ it is known that $(ct, x)G(ct, x)^T = (ct)^2 - x^2$. $\endgroup$ – udrv Feb 4 '16 at 7:39
  • $\begingroup$ @udrv I dont understand your notation, but consider plugging $2g_{\mu \nu}$ instead of $g_{\mu \nu}$ in your formulas for the invariant interval. What would change exactly? $\endgroup$ – Prof. Legolasov Feb 10 '16 at 13:52
  • $\begingroup$ I had in mind matrix notation, with $G$ a $2\times 2$ matrix and $(ct, x)$ as it reads, a 2d row vector. This makes the transposed $(ct, x)^T$ the column vector. Explicitly, $$\left(\begin{array}{cc} ct & x \end{array}\right)\left( \begin{array}{cc} g_{11} & g_{12} \\ g_{21} & g_{22} \end{array}\right)\left(\begin{array} {c} ct \\ x \end{array}\right) = (ct)^2 - x^2$$ $\endgroup$ – udrv Feb 10 '16 at 15:39
  • $\begingroup$ Then using $g_{\mu\nu} \rightarrow 2g_{\mu\nu}$ or $G \rightarrow 2G$ should give $$2\left(\begin{array}{cc} ct & x \end{array}\right)\left( \begin{array}{cc} g_{11} & g_{12} \\ g_{21} & g_{22} \end{array}\right)\left(\begin{array} {c} ct \\ x \end{array}\right) = (ct)^2 - x^2$$ If you have in mind that $G = \Lambda^T G \Lambda$ is invariant under a rescaling of $G$, then yes, the space-time interval condition fixes the metric's scale. $\endgroup$ – udrv Feb 10 '16 at 15:39

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