- #1

jk22

- 729

- 24

$$\left(\begin{array}{c} x'\\ict'\end{array}\right)=\left (\begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)\left(\begin{array}{c} x'\\ict'\end{array}\right)$$

Writing the first line gives $$x'=cos(\theta)(x-i tan(\theta)ct)$$

If I want this expression to be like a Lorentz transformation I should have the form $$x'=\gamma(v)(x-vt)$$ hence $$ic tan(\theta)=v \Rightarrow tan(\theta)=-i\frac{v}{c}$$

Then $$cos(\theta)=\frac{1}{\sqrt{1+tan(\theta)^2}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ which is the gamma factor.

This gives the usual Lorentz transformation.

The problem arises when I consider the 'isomorphism' $$i\equiv \left (\begin {array} {cc} 0 & -1\\ 1 & 0 \end {array}\right)$$

Indeed I can then write $$e^{i\theta}\equiv \left( \begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)$$

But inserting the value of theta gives : $$\left(\begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \begin {array} {cc} 1 & i\frac{v}{c} \\ -i\frac{v}{c} & 1 \end {array}\right)\equiv \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$

Which is the Bondi k factor.

Hence using this mapping of $$i$$ to real 2x2 matrices produce an error since we get then an homotethy instead of the Lorentz transform.

Is it because the application mapping of $$i$$ is injective and not bijective ?