0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are you looking for physical or mechanical explanations, rather than substitutions in mathematical form ? $\endgroup$ – Sean Mar 27 '18 at 21:49
4
$\begingroup$

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 $\mathbf{n}=\frac{\mathbf{v}}{v}$.

https://en.wikipedia.org/wiki/Lorentz_transformation#Vector_transformations

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Mar 27 '18 at 8:03
  • 1
    $\begingroup$ Yes, but use $v_y$ and $v_z$. $\endgroup$ – John Rennie Mar 27 '18 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.