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Let $K$ and $\bar K$ be two cartesian co-orditate systems in $\mathbb{R}^3$. The element: $$s^2=(\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2$$ is an invariant in all co-ordinate system. I want prove that the relations between de co-ordinates of $K$ and $K$ must be linear.

In the book: "The meaning of relativity, by A. Einstein" we can find the following reasoning:

Suppose that $\bar x^\nu=\bar x^\nu(x^1,x^2,x^3)$ and note that: $$\left\{\begin{array}{lll} (\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2= \text{const.}&(1)\\ (\Delta \bar x^1)^2+(\Delta \bar x^2)^2+(\Delta \bar x^3)^2 = \text{const.}&(2) \end{array}\right.$$ Also by Taylor's theorem: $$\Delta \bar x^\nu = \sum_\alpha \frac{\partial \bar x^\nu}{\partial x^\alpha}\Delta x^\alpha+\frac{1}{2}\sum_{\alpha,\beta} \frac{\partial^2\bar x^\nu}{\partial x^\alpha\partial x^\beta}\Delta x^\alpha\Delta x^\beta+\ldots\qquad (3)$$ Substituting $\Delta x^\nu$ given by $(2)$ to $(3)$ and compare this with $(*)$ then we deduce that $\bar x^\nu=\bar x^\nu(x^1,x^2,x^3,x^4)$ must be linear.

Can anybody help me to take $\Delta x^\nu$ given by $(2)$ and put into $(3)$ to deduce that the transformation must be linear?


marked as duplicate by Qmechanic Jun 2 '16 at 21:25

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  • 1
    $\begingroup$ Qmechanic gives the general argument in this answer (the thread contains several different methods for the linearity of the Lorentz transformations, most of the arguments apply equally well to the Euclidean/Cartesian case). It's really plugging in and noting that the higher order terms have to have properties inconsistent with them being non-zero, so they are zero. $\endgroup$ – ACuriousMind Jun 2 '16 at 18:42
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/253305/2451 $\endgroup$ – Qmechanic Jun 2 '16 at 21:23
  • $\begingroup$ Crossposted from math.stackexchange.com/q/1810027/11127 $\endgroup$ – Qmechanic Jun 2 '16 at 21:24