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This question follows the same ideia of another question of mine: Doubts on derivation of Lorentz Transformations

So, the problem here is the following phrase:

The remain transformation equations for $x'$ and $t'$,namely,

\begin{cases}\tag{1} x' = a_{11}x+a_{12}y+a_{13}z+a_{14}t\\ t' = a_{41}x+a_{42}y+a_{43}z+a_{44}t\\ \end{cases}

Let us look first at $t'-equation$. For reasons of symmetry,we assume that $t'$ does not depend on $y$ and $z$. Other wise,clocks placed symmetrically in the $y-z$ plane, about the $x-axis$ would appear to disagree as observed from $S'$, which would contradict the isotropy of space. Hence $a_{42}=a_{43}=0$

We know that a point havind $x' = 0$ appears to move in the direction of the positive $x-axis$ with speed $v$ so the statemant $x'=0$ must to be identical to the statement $x=vt$

First of all, I didn't grasp quite well the arguments of the two bolded statements, I mean, the connection between geometry and physics. Principally when he said: "We know that a point havind $x' = 0$ appears to move..." and "clocks placed symmetrically in the y−z plane, about the x−axis would appear to disagree as observed from $S′$"

However, for the first bolded statement I think that means that for we have $a_{42}=a_{43}=0$ then $t'=0$ and $x=0$, $t=0$. So:

$$t' = a_{41}x+a_{42}y+a_{43}z+a_{44}t$$ become

$$0 = a_{41}0+a_{42}y+a_{43}z+a_{44}0 $$

then

$$0 = a_{42}y+a_{43}z $$

Therefore,

$$a_{42}=a_{43}=0$$

But,as well as I said, I didn't grasp the physical significance of this manipulation (if it is correct).

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The x direction is special because that's the direction of the (observable) velocity. The y and z directions are arbitrary: You can pick them to point in different directions. The underlying physical reality doesn't depend on how you pick arbitrary directions for your axes. So the equations shouldn't either.

Your derivation then works out the consequence of that statement.

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