Doubt in tranformations of perpendicular directions coordinates to the motion of frame in Lorentz transformation

Question

Consider a frame $S$ which is at rest. The frams $S'$ at $t=0$ coincides with $S$ and then start moving with velocity $v$ in the $+x$ direction.
In Galilean transformation we can easily see that $y'=y$
But while deriving the transformation formulas for Lorentz transformation, we take $y'=y$
If we consider $y'=y'(x,y,z,t)$ then definitely it is linear with respect to all its arguments so that Newton's law holds.
So, we can write $y'=a_1x+a_2y+a_3z+a_4t\tag{1}$
I am trying to prove $y'=y$ from $y'=y(x,y,z,t)$

If the particle is on $x-axis$ in $S$ then it remains on $x-axis$ on $S'$
So, $y'=y=0$ and $z'=z=0$
From (1),
$0=a_1x+a_4t$
As $x$ and $t$ are independent of each other.
We can conclude that $a_1=a_4=0$
Thus (1) becomes $y'=a_2y+a_3z \tag{2}$
If the particle is in $x-z$ plane in $S$ at $y=0$ the particle remains in $x-z$ plane at $y'=0$ otherwise we get a contradiction that $S'$ moves in $y$ direction also.
So, (2) becomes, $0=a_3z,\;\;\forall z$
$\implies a_3=0$

Thus (2) becomes, $y'=a_2y\tag{3}$

But I have trouble in proving $a_2=1$.
Some authors use the concept of symmetry, but I am not able to understand their argument completely.
Please help in understanding how $a_2=1$?
I am very confused.

Answer 1

I think we must use the symmetry concept.

So, for notation convenience consider that your frames $\,\rm S,S'\,$ are $\,\rm S_1,S_2\,$ respectively as in above Figure-01⁽¹⁾. Your equation (3) is then⁽²⁾ \begin{equation} y_2\boldsymbol{=}a_2\,y_1 \tag{A-01}\label{A-01} \end{equation} Below this configuration consider two frames $\,\rm S_3,S_4\,$ as follows : The system $\,\rm S_3\,$ is at rest with respect to $\,\rm S_2\,$ with reverse the $\,x-,y-\,$ axes so

\begin{align} x_3\boldsymbol{=-}x_2 \tag{A-02a}\label{A-02a}\\ y_3\boldsymbol{=+}y_2 \tag{A-02b}\label{A-02b}\\ z_3\boldsymbol{=-}z_2 \tag{A-02c}\label{A-02c} \end{align} The system $\,\rm S_4\,$ is at rest with respect to $\,\rm S_1\,$ with reverse the $\,x-,y-\,$ axes so

\begin{align} x_4\boldsymbol{=-}x_1 \tag{A-03a}\label{A-03a}\\ y_4\boldsymbol{=+}y_1 \tag{A-03b}\label{A-03b}\\ z_4\boldsymbol{=-}z_1 \tag{A-03c}\label{A-03c} \end{align}

The configuration of frames $\,\rm S_3,S_4\,$ is exactly that of $\,\rm S_1,S_2$ : that is the second frame of a pair is moving along the common $\,x-$axis with velocity $\,\boldsymbol{\upsilon}\,$ with respect to the first frame of the pair. So, corresponding to equation \eqref{A-01} we have \begin{equation} y_4\boldsymbol{=}a_2\,y_3 \tag{A-04}\label{A-04} \end{equation} which by equations \eqref{A-03b}, \eqref{A-02b} yields \begin{equation} y_1\boldsymbol{=}a_2\,y_2 \tag{A-05}\label{A-05} \end{equation} Combining equations \eqref{A-01}, \eqref{A-05} we have \begin{equation} y_1\boldsymbol{=}a^2_2\,y_1 \tag{A-06}\label{A-06} \end{equation} so \begin{equation} a^2_2\boldsymbol{=}1 \quad \boldsymbol{\Longrightarrow} \quad a_2\boldsymbol{=\pm}1 \tag{A-07}\label{A-07} \end{equation} We choose $a_2\boldsymbol{=+}1$ in order to exclude space inversion.

The Figure-02 below is a back view of Figure-01.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

^{(1) This Figure is extracted from my answers here Schutz's geometrical proof that spacetime interval is invariant and here Special Relativity - Reference Frames S and S′ relative velocity.}

^{(2) To be precise the coefficient $\,a_2\,$ must be considered as dependent on the velocity $\,\boldsymbol{\upsilon}\,$ and we must write $\,a_2(\boldsymbol{\upsilon})$. This fact doesn't modify the final conclusion.}

Answer 2

the lines element are

$$s^2=-c^2 t^2+x^2+y^2+z^2\\ {s'}^2=-{c'}^2 {t'}^2 + {x'}^2 + {y'}^2 + {z'}^2\tag 1$$

with this Ansatz:

$$c=c'\\ x'=A\,(x-v\,t)\\ t'=B\,x+C\,t\\ y'=a_2\,y\\ z'=z$$

and ${s'}^2-s^2=0$ you obtain:

$$\underbrace{\left(1-A^2 + c^2 B^2\right)}_{eq_1=0} x^2 + \underbrace{\left(2\,A^2 v + 2\,c^2 CB\right)}_{eq_2=0} t\,x + \\ \underbrace{\left(-c^2 - A^2 v^2 + c^2 C^2 \right)}_{eq_3=0} t^2 +\underbrace{(1-a_2^2)}_{eq_4=0}\,y^2= 0.$$

You have now four equations for the four unknowns $A~,B~,C~,a_2$.

The solution

$$A = \gamma\\B = -\frac{\beta}{c}\,\gamma\\ C = \gamma\\a_2=1$$

where

$$\gamma=\frac{1}{\sqrt{1-\beta^2}}\quad \beta = \frac{v}{c}$$ $\Rightarrow$

\begin{align*} \begin{bmatrix} t' \\ x' \\ y' \\ z' \\ \end{bmatrix} = &\left[ \begin {array}{cccc} {\gamma}&-{\frac {{\gamma}\,v}{{c}^{2}}}&0 &0\\ -{\gamma}\,v&{\gamma}&0&0\\ 0 &0&1&0\\ 0&0&0&1\end {array} \right]\, \begin{bmatrix} t \\ x \\ y \\ z \\ \end{bmatrix} \end{align*}