Is it possible to derive Lorentz transformations just by assuming that if two spaceships are moving away from each other with a constant speed, it is impossible for them to tell who is moving, even if they can exchange light signals. (speed of light is assumed to be a constant). It seems to me that is shoud be possible, but to my surprise I have never seen such a derivation

  • 2
    $\begingroup$ it seems to you how? $\endgroup$ – Wolphram jonny Jan 21 at 18:18
  • $\begingroup$ This has been done by Albert Einstein himself, with the earth taking the place of one of the spaceships, and a fast-moving train taking the place of the other one, if I recall correctly. What derivations of the Lorentz transformations are you familiar with? $\endgroup$ – Cuspy Code Jan 21 at 19:18
  • $\begingroup$ Here is a reference to Einstein's derivation. marxists.org/reference/archive/einstein/works/1910s/relative $\endgroup$ – mmesser314 Jan 22 at 0:53

I will give a derivation of the Lorentz boosts requiring (what at least seem to be) minimal assumptions, and we will look at what assumptions we used, and see if some of them can be derived from each other, etc. Note that by "the Lorentz transformations", I mean the Lorentz transformation of spacetime position -- Lorentz transformations of other four-touples (i.e. proving that they are Lorentz vectors) would require other assumptions, of course. I've given a more full explanation of the derivation here.

(a) The first important fact you need to prove anything about the Lorentz transformations is that they are linear. Linearity is logically equivalent to the following conditions: (under the transformation),

  • all straight lines remain straight lines -- the physical interpretation of this is that if an object's velocity is constant in one inertial reference frame, it is constant in all inertial reference frames. This follows from the principle of relativity.
  • the origin remains fixed -- this is true by definition of the transformations we are considering -- boosts passing through the same origin.

With this, we know that we can use a matrix to write down the Lorentz transformations. Which matrix?

(b) The tilt/angle of the $t'$, $x'$ axes with respect to the $t$, $x$ axes. The tilt of the $t'$ axes follows from the definition of velocity as the gradient of the worldline. To prove the tilt of the $x'$ axis is equal to this tilt, we need to first define the $x'$ axis within the unprimed co-ordinate system.

This is possible by considering invariant features under a boost, i.e. from the principle of relativity -- the obvious invariant is as follows: if you had emitted a light ray $a$ seconds in the past, it reflects off some object and returns to you $a$ seconds in the future, it was on your x-axis at time 0.

enter image description here

By the principle of relativity, this should apply in the primed reference frame as well. By the invariance of the speed of light, the slope of the light ray is the same in the primed reference frame. Now figuring out the angle of tilt of the $x'$ axis becomes an exercise in geometry.

enter image description here

And it's easy to prove, by drawing an appropriate circle, that the two tilts are equal.

(c) We now know the lines the column vectors of our matrix land on -- they are multiples of $(1, v)$ and $(v, 1)$, but which vector on that line exactly? In other words, what's the scale on the axes? This requires one extra assumption: if you boost into the frame with velocity $v$, then boost $-v$ back, that's equivalent to not boosting at all, i.e. $L(v)L(-v)=I$. Then it's just computation:

\begin{gathered} \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \alpha &{\beta v} \\ {\alpha v}&\beta \end{array}} \right]\left[ {\begin{array}{*{20}{c}} \alpha &{ - \beta v} \\ { - \alpha v}&\beta \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\alpha ^2} - \alpha \beta {v^2}}&{{\beta ^2}v - \alpha \beta v} \\ {{\alpha ^2}v - \alpha \beta v}&{{\beta ^2} - \alpha \beta {v^2}} \end{array}} \right] \hfill \\ {\alpha ^2}v - \alpha \beta v = 0 = {\beta ^2}v - \alpha \beta v \Rightarrow {\alpha ^2} = \alpha \beta = {\beta ^2} \Rightarrow \alpha = \beta \hfill \\ {\alpha ^2} - \alpha \beta {v^2} = 1 = {\beta ^2} - \alpha \beta {v^2} \Rightarrow {\alpha ^2} = 1 + \alpha \beta {v^2} = {\beta ^2} \Rightarrow {\alpha ^2} = 1 + {\alpha ^2}{v^2} \hfill \\ \Rightarrow \alpha = \beta = \frac{1}{{\sqrt {1 - {v^2}} }} \hfill \\ \end{gathered}

Then the change of basis matrix is simply the inverse of this matrix, which is:

$$\Lambda=\gamma \left[ {\begin{array}{*{20}{c}} 1&-v \\ -v&1 \end{array}} \right]$$


\begin{gathered} x' = \gamma \left( {x - vt} \right) \\ t' = \gamma \left( {t - vx} \right) \\ \end{gathered}

(d) There's still one final step, however -- we need to verify that $y$ and $z$ aren't transformed under the Lorentz boost. To prove this, consider two twins with paintbrushes running towards each other, painting the wall at waist level -- if the orthogonal axis were transformed in any way, each twin would see his paint-streak as above the other's -- the fact that the paint-streaks' relative positioning can't be different can be seen, e.g. from supposing that the two paints cause an explosion in the mix. The fact that the presence of explosions (or any boolean quantity) is invariant under Lorentz transformations is a consequence of the principle of relativity.

We used three physical assumptions:

  • The principle of relativity
  • The invariance of the speed of light
  • $L(v)L(-v)=L(0)$, or "if I see you moving at $v$, you see me moving at $-v$"

The first two are the assumptions you wanted. As far as I can see, the last assumption can't really be proven from the other two -- it requires some sort of symmetry principle. But that's okay.

  • $\begingroup$ see Mermin, N. David. "An introduction to space–time diagrams." American Journal of Physics 65.6 (1997): 476-486 $\endgroup$ – ZeroTheHero Jan 22 at 4:24

The Lorentz transformations were originally derived by Lorentz and Fitzgerald as the symmetry group that leaves Maxwell’s equations invariant. (This isn’t easy to see in the way they’re written using vector notation as the symmetry is not manifest).

Einstein didn’t actually derive the Lorentz transformations; what he did was made classical mechanics consistent with this symmetry group, this is Special Relativity.


Your Answer

