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Can someone show me how the Lorentz transformation can be derived from the metric equation: $s^2=c^2t^2-x^2$

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View the hyperbolic geometry derivation on the corresponding wikipedia entry: en.wikipedia.org/wiki/… –  neutrino Aug 2 '13 at 9:55
    
The derivation isn't hard, but it's somewhat tedious. If you Google for "derive lorentz transformation from metric" there's an article at physicsinsights.org/derive_lorentz.html that goes through it. –  John Rennie Aug 2 '13 at 10:09
    
Essentially a duplicate of physics.stackexchange.com/q/12664/2451 –  Qmechanic Aug 2 '13 at 16:37
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2 Answers

Look, I will give you the derivation the same as Einstein used. It is historical derivation and consist of redundant postulate about the invariance of speed of light. If you want, I can give to you other method, which uses group axioms and is more fundamentally.

If you don't want details, you can directly see the fourth step.

The first step.

The first step is common to the two methods. In the beginning we analyze two-dimensional transformations (which can be easy generalized on the case of 4-space-time). Let's have two transformations $$ x' = f(x, u, t),\quad t' = g(x, u, t), $$ where $u = const$ is the relative speed between two frames. Let's use the postulate of homogeneous space-time: according to it, the $x'$ differential (for t' - analogically) $$ dx' = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial t}dt $$ doesn't depend on point (x, t), in which we want to calculate it. So $$ \frac{\partial f}{\partial x} = A(u), \quad \frac{\partial f}{\partial t} = B(u), $$ and $$ x' = Ax + Bt, \quad t' = Cx + Dt \qquad (.1) $$

The second step.

Let's have transformations in 4-space-time: for $\mathbf v || Ox$, by using step 1, $$ x' = Ax + By + Cz + Dt, \quad y' = A_{1}x + B_{1}y + C_{1}z + D_{1}t, \quad z' = A_{2}x + B_{2}y + C_{2}z + D_{2}t, \quad t' = A_{3}x + B_{3}y + C_{3}z + D_{3}t. \qquad (.2) $$ Because $x'$-asis is always matches with $x$-asis, $y' = 0 \Rightarrow y = 0, \quad z' = 0 \Rightarrow z = 0$. So from $(.2)$ we can get $$ 0 = A_{1}x + C_{1}z + D_{1}t, \quad 0 = A_{2}x + B_{2}y + D_{2}t. $$ For a case of arbitrary $x, y, z, t$ from these explessions follows statement, that $A_{1} = C_{1} = D_{1} = A_{2} = B_{2} = D_{2} = 0$. Then, $$ y' = B_{1}(u)y, \quad z_{1} = C_{2}(u)z, $$ where $B_{1}, C_{2}$ are the even functions (in a reason of postulate of isotropic space).

If we use relativity principle, we can get $$ y = B_{1}(-u)y' = B_{1}(-u)B_{1}(u)y = B_{1}^{2}(u)y \Rightarrow B_{1}^{2} = 1 . $$ By using identity transform, $y'(u = 0) = y \Rightarrow B_{1} = 1$.

This result leads to the statement, that $y, z$ don't mixes with each other. This means that $B_{1} = C_{1} = B_{3} = C_{3} = 0$. So, finally, for $(.2)$ we can get that $$ x' = Ax + Bt, \quad y' = y, \quad z' = z, \quad t' = Cx + Dt. \qquad (.3) $$

The third step.

We can use principle of homogeneous space-time one more time for getting the relations of $A, B, C, D$. First, let's have system in which $x' = 0$. For this case $x = ut $, and we have from $(.3)$ $$ \begin{cases} 0 = Aut + Bt \Rightarrow B = -Au \qquad (.4) \\ t' = Cx + Dt \\ \end{cases}. $$ Let's have system in which $x = 0$. Then $x' = -ut'$, and $$ \begin{cases} -ut' = Bt \Rightarrow B = -\frac{ut'}{t} = -uD \qquad (.5) \\ t' = Dt \Rightarrow \frac{t'}{t} = D \\ \end{cases}. $$ From $(.4), (.5)$ follows that $A = D = \gamma (u)$, where $\gamma (u)$ is an unknown function of speed, $B = -Au$, and $C$ we can write as $D\frac{C}{D} = D\sigma (u)$. Finally, $$ x' = \gamma (x - ut), \quad t' = \gamma (t - \sigma x), \quad y' = y, \quad z' = z . \qquad (.6) $$

The fourth step.

Let's have the spherical wave equations in two frames, $$ x^{2} + y^{2} + z^{2} = c^{2}t^{2}, \quad x'^{2} + y'^{2} + z'^{2} = c^{2}t'^{2}. \qquad (.7) $$ By substitution $(.6)$ into $(.7)$ and we can get $$ \gamma^{2}x^{2} -2\gamma^{2}uxt + \gamma^{2}u^{2}t^{2} + y^{2} + z^{2} = c^{2}\gamma^{2}t^{2} - 2\sigma \gamma^{2}c^{2}xt + \gamma^{2}\sigma^{2}x^{2}c^{2}t^{2}. \qquad (.8) $$ For getting the sphere equation in the $S$ frame we must to identify $2\gamma^{2}uxt$ and $2\sigma \gamma^{2}c^{2}xt$. So $$ \sigma = \frac{u}{c^{2}}. $$ Then $(.8)$ will reduce to $$ \gamma^{2}\left( 1 - \frac{u^{2}}{c^{2}}\right)x^{2} + y^{2} + z^{2} = c^{2}t^{2}\gamma^{2}\left(1 - \frac{u^{2}}{c^{2}} \right) \Rightarrow \gamma^{2} = \frac{1}{1 - \frac{u^{2}}{c^{2}}}. $$ So, by using identity $x'(u = 0) = x \Rightarrow \gamma = \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}$, we finally can get for $(.6)$ $$ x' = \frac{x - ut}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}, \quad t' = \frac{t - \frac{u}{c^{2}}x}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}. $$

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For a boost in the $x$ direction, propose a linear transformation such that

$$t'=t'(x,t)=ax+bt \\ x'=x'(x,t)=cx+dt $$

Impose the constraint

$$c^2t^2-x^2=c^2(t')^2-(x')^2 $$

Solve for $a,b,c,d$

About the linearity of the Lorentz transformations, you may check Gravitation and Cosmology by Weinberg p.27.

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You need to explain why it should be linear –  Larry Harson Aug 2 '13 at 15:23
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@LarryHarson added reference about that matter –  Jorge Aug 2 '13 at 16:13
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