For two coordinate systems $(x,y,z,t)$, $(x',y',z',t')$ the Lorentz transform is $$t'=\gamma \left( t-\frac{vx}{c^2} \right)$$ $$x'=\gamma (x-vt)$$ $$y'=y$$ $$z'=z$$ i have 2 questions about that, first i don't understand why $y$ and $z$ remain unchanged, when an observer moves in 3 dimensions relative to another observer , shouldn't $y$ and $z$ coordinates change? and second if, for example, $t$ or $x$ or both remain unchanged, how will the transform look like? What is a general rule to write the transform for any case?

  • $\begingroup$ Are you looking for physical or mechanical explanations, rather than substitutions in mathematical form ? $\endgroup$
    – Sean
    Commented Mar 27, 2018 at 21:49

2 Answers 2


The vector transformation for Lorentz boosts of arbitrary directions and velocities is given by

$$ \begin{align}t' & =\gamma\left(t-\frac{v\mathbf{n}\cdot\mathbf{r}}{c^{2}}\right)\\ \mathbf{r}' & =\mathbf{r}+(\gamma-1)(\mathbf{r}\cdot\mathbf{n})\mathbf{n}-\gamma tv\mathbf{n} \end{align} $$ where $\mathbf{n}=\frac{\mathbf{v}}{v}$.



The form of the transformations you have written assumes that the velocity is along the $x$ axis i.e. it has the form:

$$ \mathbf v = (v_x, 0, 0) $$

That's why we get $y'=y$ and $z'=z$, because the components of the velocity in the $y$ and $z$ directions are zero.

It is certainly possible to write the transformations for a velocity with an arbitrary direction - just use similar expressions for $y'$ and $z'$ with $v_y$ and $v_z$. However this is an unnecessary complication, which is why it is never done. We can simply rotate our axes to make the $x$ axis parallel to the velocity and shift our origin so the velocity vector passes through the origin.

  • $\begingroup$ So what is the transform if the velocity is $v=(v_x,v_y,v_z)$? Are $y'$ and $z'$ like this $$y'=\gamma (y-vt)$$ $$z'=\gamma (z-vt)$$ $\endgroup$
    – paradox
    Commented Mar 27, 2018 at 8:03
  • 1
    $\begingroup$ Yes, but use $v_y$ and $v_z$. $\endgroup$ Commented Mar 27, 2018 at 8:25

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