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

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. 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.

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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*}

• How you havewritten eq (1). I think $s^2=x^2+y^2+z^2$. Also how can we conclude that $s^2=s'^2$?
– Iti
May 10, 2021 at 16:37
• @Iti we are in Minkowski space and $~s^2=s'^2$ is requirement of the Lorentz transformation
– Eli
May 10, 2021 at 16:47
• thanks for the reply. But I have not studied about Minkowski space. I am an undergraduate student and have studied Newtonian mechanics. Can you please explain accordingly? Some authors lso used the concept of symmetry. But I have trouble understandng their arguments.
– Iti
May 10, 2021 at 16:54