# Obtaining Lorentz transformations in hyperbola geometry

In Tevian Dray's book The Geometry of Special Relativity, 1st edition, page 27, he writes:

5.3 Lorentz Transformations

We now relate Lorentz transformations, based on the physical postulates of special relativity, to hyperbola geometry. The Lorentz transformation between a frame $$(x,t)$$ at rest and a frame $$(x',t')$$ moving to the right at speed $$v$$ was derive in Chapter 2. The transformation from the moving frame to the rest frame is given by \begin{align} x &= \gamma (x' + v t'), \tag{5.1} \\ t &= \gamma \left(t' + \frac{v}{c^2} x'\right), \tag{5.2} \end{align} where, as before, $$\gamma = \frac{1}{1 - \frac{v^2}{c^2}}. \tag{5.3}$$ The key to converting this description to hyperbola geometry is to measure space and time in the same units by replacing $$t$$ by $$ct$$. The transformation from the moving frame, which we now denote by $$(x',ct')$$, to the frame at rest, now denoted $$(x,ct)$$, is given by \begin{align} x &= \gamma \left(x' + \frac{v}{c} ct'\right), \tag{5.4} \\ ct &= \gamma \left(ct' + \frac{v}{c} x'\right), \tag{5.5} \end{align} which makes the symmetry between these equations much more obvious.

I understand that he obtained 5.4 from 5.1 by changing $$t'$$ to $$ct'$$ and $$v$$ to $$v/c$$. But by substituting $$v$$ for $$v/c$$ in 5.2, wouldn't I obtain $$(v/c^3)x'$$ in 5.5? I'm sure I'm missing something very obvious and would appreciate some help.

• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. I edited your question to adhere to these standards. Commented Jul 17, 2022 at 22:42
• Thank you for editing and apologies for posting the image, I was unaware of that rule. Commented Jul 17, 2022 at 23:42
• Your $\gamma$ is missing the \sqrt . Commented Jul 18, 2022 at 17:15

You don't need to redefine variables. It is just a matter of doing nothing with Eq. (5.1) and multiplying Eq. (5.2) by $$c$$. Notice that $$vt' = \frac{v}{c} c t',$$ by pure algebra. Hence, Eq. (5.1) leads us to \begin{align} x &= \gamma (x' + v t'), \tag{5.1} \\ &= \gamma \left(x' + \frac{v}{c} ct'\right), \tag{5.4} \end{align}
Now, if we multiply Eq. (5.2) by $$c$$ we get \begin{align} t &= \gamma \left(t' + \frac{v}{c^2} x'\right), \tag{5.2} \\ ct &= \gamma \left(ct' + \frac{v}{c} x'\right). \tag{5.5} \end{align}