Prove $y=y',z=z'$ in Lorentz transformation I am trying to prove the invariance of $ds^2$ in any 2 arbitrary inertial frames of reference, and I have reduced the proof to proving that when the relative velocity (of the 2 frames) is along the common $x$-axis, the coordinate transformation does not affect the $y$ and $z$ coordinates.
$$y=y'$$
$$z=z'.$$
This seems much simpler, yet i cannot arrive at this result.
These are the properties I tried to use for proving $y=y',z=z'$:

*

*principle of relativity of all Galileian frames


*constancy of the speed of light i.e $(ds)^2=0⇔(ds')^2=0$


*gravity is absent (since frames are Galileian)
Any help is appreciated.
 A: The Galilean transformation, where in this case you consider a boost along the x-axis only
$x ' = x - vt$
$y' =y$
$z' = z$
$t' =t$
If there is motion along the x-axis only, then surely $y' = y$ and $z' = z$, since there (once again) is no motion along the y or z-axes. So $dy = 0$ and $dz = 0$, meaning that the invariant interval,
$ds^2 = dx^2 + dy^2 + dz^2 = dx^2 + 0 + 0 = dx^2$
and
$ds^2 = ds'^2 = dx^2 =dx'^2$
You mentioned also that you wanted to use the principle of the constancy of the speed of light, yet you also mention that you are using Galilean or Newtonian relativity. Galilean assumes that velocities would add like
$\vec c' = \vec c + \vec v$
However, because of this principle, the speed of light is independent of the relative motion of the (inertial) reference frames, and
$\vec c + \vec v = \vec c$
in special (not Galilean) relativity. A consequence of this principle is the Lorentz transformations, which would not be applicable to your case. And yes, there is no gravity or any kind of accelerations considered in the case of special relativity. For this, you would need General Relativity. Your question is a bit "all over the place" and contradictory, and I'm not sure exactly what you are doing, but hopefully I have provided enough information to answer your question.
A: The transformation equations between cartesian coordinates of two inertial
frames $\Sigma (x, y, z, t)$ and $\Sigma' (x ', y', z ', t')$  must be linear:
$$ \left(
\begin{split}
x' = a_{11}x + a_{12}y + a_{13}z + a_{14}t \\
y' = a_{21}x + a_{22}y + a_{23}z + a_{24}t \\
z' = a_{31}x + a_{32}y + a_{33}z + a_{34}t \\
t' = a_{41}x + a_{42}y + a_{43}z + a_{44}t \\
\end{split}
\right. $$
If the reference frames $\Sigma (x, y, z, t)$ and $\Sigma' (x ', y', z ', t')$ are
in standard configuration we have two united coordinated plans:
$\forall x,z,t: y=0 \leftrightarrow y'=0
\qquad  \text{therefore: } y' = a_{22} y $
$\forall x,y,t: z=0 \leftrightarrow z'=0
\qquad  \text{therefore: } z' = a_{22} z $
Furthermore, the third coordinated plane must be the same at the initial instant:
$ \forall y,z: (t=0,x=0) \leftrightarrow (t'=0,x'=0) $
$ \qquad \text{therefore: } 
x' = a_{11}x + a_{14}t \qquad t' =a_{41}x +a_{44} t $
So by adoption of the standard configuration for $\Sigma$ and $\Sigma'$
we greatly simplify the transformations (erasing $10$ parameters):
$$ \left(
\begin{split}
& x' = a_{11}x + a_{14}t \\
& y' = a_{22}y  \\
& z' = a_{33}z  \\
& t' = a_{41}x + a_{44}t \\
\end{split}
\right.
$$
By reversing the arrow of time the two frames $\Sigma$ and $\Sigma'$ swap their roles:
therefore the inverse transformations is obtained from the
first by changing the sign of the time variables $t$ and $t'$
and leaving the sign of the spatial variables unchanged:
$$ \left(
\begin{split}
& x = +a_{11}x' - a_{14}t' \\
& y = a_{22}y'  \\
& z = a_{33}z'  \\
& t = -a_{41}x' + a_{44}t' \\
\end{split}
\right.
$$
The product of the direct and the inverse transformation must give the identity:
$$ \left(
\begin{split}
& x' = \ldots \\
& y' = a_{22}y = a_{22}^2 y' \qquad \to \quad a_{22}^2 =1 \\
& z' = a_{33}z = a_{33}^2 z' \qquad \to \quad a_{33}^2 =1 \\
& t' = \ldots \\
\end{split}
\right.
$$
The choise of the positive sign $ a_ {22} = + 1 $ and $ a_ {33} = + 1 $ corresponds
the same orientation for the axes $ y, y '$ and $ z, z' $.
If you develop this argument also for the variables $x$ and $t$
only one variable $a_{11}=a_{44}=\gamma$ remains unkown.
It's remarkable that such pre-formatted transformation equations
follow directly from the concept of inertial frame of reference itself!
With an absolute time you write $\gamma=1$ (Galileo transformations),
if you postulate the constancy of the speed of light you easily
obtain for $\gamma$ the well known Lorentz factor.
