Using hyperbolic geometry to derive Lorentz transforms

100 Views Asked by At

I want to know how to derive the below Lorentz transformation formula using hyperbolic geometry

$$ct' = ct \cosh\phi - x \sinh\phi$$ $$x' = x \cosh\phi - ct \sinh\phi$$

Here is a spacetime diagram

From the diagram, the line OH is $ct$ and the line OF is $x$. I think $x$ equals $AC\cosh\phi + OB\sinh\phi$ and $ct$ equals $CE\cosh\phi + OD\sinh\phi$.

It doesn't look right to me. I am not sure why the correct formula use $-\sinh$ not $\sinh$. Also I can't tell which line (OA or OB) equals to $ct'$ and which line (OE or OD) equals to $x'$.

I need someone to show me how to geometrically derive the correct Lorentz transform formulas.