The transformation, as known, is the following:

$$\begin{align} x'&=\gamma(x-vt) \\ y'&=y \\ z'&=z \\ t'&=\gamma\left(t-\frac{vx}{c^2}\right) \end{align}$$

Is there a meaning to the direction of $v$?

If object A is at rest, and an object B is moving with speed $v$ towards him while an object C is moving with speed $v$ away from him. Will the transformation of time for example, be the same in both cases? And what about place ($x$)?

My misunderstanding comes from seeing that the direction DOES matter while calculating the Doppler effect.

Previous posts didn't help me, will the lorentz transformations in the case above be exactly the same?

  • $\begingroup$ You can use MathJax to typeset equations on this site. $\endgroup$
    – PM 2Ring
    Mar 3, 2020 at 14:27
  • 1
    $\begingroup$ Does this answer your question? Is the velocity a scalar or a vector in one dimensional Lorentz transformations? Please see the answer by G. Smith. $\endgroup$
    – PM 2Ring
    Mar 3, 2020 at 14:30
  • $\begingroup$ In case it's not completely clear from that linked question, yes the direction of the velocity does matter. The equations you list in your question are for the simplest case of the Lorentz transformation. The axes of the 2 frames are aligned with each other, they coincide at t=0, and the primed frame is moving in the +X direction with speed v relative to the unprimed frame. Wikipedia shows the transformations for more general cases. $\endgroup$
    – PM 2Ring
    Mar 3, 2020 at 14:39

1 Answer 1


I will answer your question for a Galilean transformation because it extends naturally to the Lorentz transformation and it will be less confusing this way (I hope).

Consider two observers: observer A (blue) and observer B (red)

enter image description here

If we consider A's frame then B is moving to the right with velocity $\vec v=(v,0,0)$. The coordinates of B are related to A's by \begin{align} x'&=x-v\,t\tag{1}\\ t'&=t \end{align} Where B has the primed coordinates. Remember that the blue gridlines are all the points which have a constant $x$ or $t$ coordinate for observer A and the red gridlines have constant $x$ or $t$ coordinates for observer B. For example the red gridline in the middle has $x'=0$ (I didn't draw the time gridlines for observer B).

So to answer your question: the direction of $\vec v$ does matter. Before $t=0$ observer B is always moving towards A and after $t=0$ B is always moving away from A. The sign of $v$ can switch in which direction B is moving though. If $v<0$ then B is moving to the left. When you are considering left moving observers you have to be careful. You can define the left moving transformation in two ways. Either define $v$ to always be positive, so the transformation for B moving to the left becomes \begin{align} x'&=x+v\,t\tag{2}\\ t'&=t \end{align} or allow $v$ to be negative, which means you can still use equation (1).

The sign in the doppler shift matters because it matters if the emitter is either moving towards or away from you. Which one you need depends on the convention you use so make sure your answer makes sense (something moving towards you will increase the frequency).

Note that for the inverse transformation you always get an additional minus sign because A will be moving in the direction $-\vec v$ according to B's frame. So \begin{align} x&=x'+v\,t'\tag{3}\\ t&=t' \end{align}


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