If we apply the Lorentz transformation


to the rest-frame coordinates of a light signal

$$x=ct$$ we have

$$x'=\gamma(ct-vt) = ct \sqrt{ \frac{c-v}{c+v}}$$

However, we can apply the principle of relativity and consider the situation in one (stationary) reference frame only but have instead the light source moving or not. The light speed postulate tells us in this case that the speed of light should be independent of the state of motion of the light source, so (using the primed coordinate for the case of the moving light source and the unprimed coordinate for the light source at rest)


(obviously, the clocks in the system can't be affected by the state of motion of the light source, so $t'=t$ here).

This however differs from the result obtained from the Lorentz transformation by the square root factor $\sqrt{(c-v)/(c+v)}$. How is this inconsistency resolved?

For clarification, I have produced a diagram for this

enter image description here


1 Answer 1


Note that the Lorentz transformation must be applied to both $x$ and $t$. So you'll end up with a new equation $x' = ct'$, and $c = \frac{x}{t} = \frac{x'}{t'}$. Both observers agree that the light travels at speed $c$, but in general they disagree on the distance it travels and the time it takes.

  • $\begingroup$ I have formulated $x'$ in terms of $t$ not $t'$, and if you apply the principle of Relativity and have the light light source instead of the observer moving, there is only the one observer time $t$, and the algebraic expression for $x'$ should be the same. $\endgroup$
    – Thomas
    Jun 15, 2021 at 17:35
  • $\begingroup$ Space and time coordinates have to go together. Either $(x, t)$ are the observer coordinates and $(x', t')$ are the light source coordinates, or vice-versa. It doesn't matter which one you consider to be at rest, but either way their times are different. There is no absolute time. $\endgroup$
    – Eric Smith
    Jun 15, 2021 at 23:45

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