Reading "On the Electrodynamics of Moving Bodies", why do I get $\beta^2$ instead of $\beta$?

Einstein writes $$\xi=\phi(v)\frac{c^2}{c^2-v^2}x'$$ and $$x'=x-vt$$ in his paper "On the Electrodynamics of Moving Bodies". Additionally, he defines $$\beta=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$.

If I substitute the second equation into the first, I derive the following \begin{align*} \xi&=\phi(v)\frac{c^2}{c^2-v^2}x'\\ \xi&=\phi(v)\frac{c^2}{c^2-v^2}(x-vt)\\ \xi&=\phi(v)\frac{1}{1-\frac{v^2}{c^2}}(x-vt)\\ \xi&=\phi(v)\beta^2(x-vt) \end{align*} This result is incorrect, the correct result should be $$\xi=\phi(v)\beta(x-vt)$$, as is stated in Einstein's paper.

I am having similar difficulties writing the equations for $$\tau$$, $$\eta$$, and $$\zeta$$.

Where is my mistake and how would I derive the correct result? Unless I'm being really stupid, I don't think the error lies in the algebraic manipulation shown above. Perhaps I am misunderstanding the meanings of some of the symbols, as I have not found the author very clear in this respect.

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– rob
Commented Jan 2, 2019 at 0:17

The confusion is due to the definition of the unknown constant $$\,a$$. While in a previous paragraph it's written that $$^{\prime\prime}$$...where $$a$$ is a function $$\,\phi(\upsilon)\,$$ at present unknown...$$^{\prime\prime}$$(1) later on it's derived the equation $$$$\eta\boldsymbol{=}a\dfrac{c}{\sqrt{c^2\boldsymbol{-}\upsilon^2}}y \quad \text{and} \quad \zeta\boldsymbol{=}a\dfrac{c}{\sqrt{c^2\boldsymbol{-}\upsilon^2}}z. \tag{A}\label{A}$$$$ and then equates those $$\,\eta,\zeta\,$$ with $$$$\eta\boldsymbol{=}\phi(\upsilon)y \quad \text{and} \quad \zeta\boldsymbol{=}\phi(\upsilon)z \tag{B}\label{B}$$$$ This means that $$\,a\,$$ is not exactly the function $$\,\phi(\upsilon)\,$$ but $$$$a\dfrac{c}{\sqrt{c^2\boldsymbol{-}\upsilon^2}}\boldsymbol{=}\phi(\upsilon) \quad \boldsymbol{\Longrightarrow} \quad a\boldsymbol{=}\dfrac{\phi(\upsilon)}{\beta} \tag{C}\label{C}$$$$
(2) In the German paper: Zur Elektrodynamik bewegter K$$\ddot{\rm o}$$rper was written $$^{\prime\prime}$$...wobei $$a$$ eine vorl$$\ddot{\rm a}$$ufig unbekannte Funktion $$\,\phi(\upsilon)\,$$ ist...$$^{\prime\prime}$$
• Thank you, this makes sense. I had wrongly interpreted "...where $a$ is a function $\phi(v)$ at present unknown... to mean $a\equiv\phi(v)$. Commented Dec 31, 2018 at 18:53