# Why do we write the lengths in the following way? Question about Lorentz transformation

Yesterday we have studied the Lorentz transformation in school. So we have two frames of reference, $S$ and $S'$ . $S$ is stationary and $S'$. $S'$ has a constant velocity $v$, relative to the $S$ frame. $v$ is directed along the Ox axis. Ox is parallel to Ox' and Oy is parallel to Oy'.

If we apply the Galilran Transformations we get:

$x = x' + ut'$ $y = y'$ $z = z'$ $t = t'$

$x' = x - ut$ $y'=y$ $z'=z$ $t' = t$

Now, our physics teacher, assumed that:

$x=k(x'+ut')$ $x'=k(x-ut)$ with k being a constant.

Why did he do that? I didn't understand. I undrstood that the length of an object depends o the frame of reference and that the speed of light is the same in the two frames.

Assuming the above facts, we can derive the $k$ constant:$$\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$$

But why did we make that first assumptikn? I didn't get the logic. Could somebody explain, please?

• For the "derivation" we assume linearity of the transformation and that c is constant in all systems. Both, of course, are experimentally verified facts for flat spacetime (and neither holds, at least not naively, in the curved spactime of general relativity). – CuriousOne Sep 27 '14 at 16:58
• Why do e assume linearity of transformation?? It doesn't seem that obvious for me. – Bardo Sep 27 '14 at 17:06
• Linearity is experimentally tested with very high precision. I can't tell you off hand what the best evidence during Einstein's time was. Today it's probably optical experiments, the GPS system and high precision astronomy that set the limits for how well we know that distances add up in a linear fashion independently of the inertial system. – CuriousOne Sep 27 '14 at 18:32
• @CuriousOne: "Both, of course, are experimentally verified facts for flat spacetime." Where did they find this flat spacetime? – bright magus Oct 28 '14 at 20:58
• Are you asking me how physicists establish the numerical validity of their approximations (of which flatness is just one of many)? By looking at experimental data, of course. How else? – CuriousOne Dec 11 '14 at 23:22

The assumption of linearity is based on the observation that measurement results between reference frames depend only on their relative motion, not their absolute position (which doesn't exist anyway). The derivation is a bit involved, so I can understand why it was glossed over. Good on you for noticing the gloss.

In general, we start with $$x' = F(x,t; a)$$ where $F$ is an unknown functions of position ($x$), time ($t$), and some as yet unknown parameter ($a$). There's an entire other argument as to why there's only one parameter.

Let's look at how $x'$ changes with $x$ and $t$: $$dx'= \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial t}dt.$$ Here's where relativity comes in: the laws of physics and our measurements don't seem to depend on position in some absolute, universal coordinate system, nor on time as measured from some universal starting point. This leads to the conclusion that $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$ cannot depend on $x$ and $t$. Thus (with a minus sign for future convenience), $$\frac{\partial F}{\partial x} = H(a)$$ $$\frac{\partial F}{\partial t} = -K(a)$$ where $H$ and $K$ are two yet-to-be-determined functions of the unknown parameter $a$. Thus, $F$ is a linear function in $x$ and $t$. We can write this as $$x' = H(a)x - K(a)t$$ A little rearranging: $$x' = H(a)\left(x - \frac{K(a)}{H(a)}t\right)$$ Just looking at the units, $\frac{K(a)}{H(a)}$ must be a velocity. To find this velocity, let's put a particle at rest in the origin of the primed frame: $x' = 0$. $$0 = H(a)\left(x - \frac{K(a)}{H(a)}t\right)$$ $$x = \frac{K(a)}{H(a)}t$$ We also know that this particle is moving with velocity $v$ in the unprimed system: $$x = vt.$$ Thus, $$\frac{K(a)}{H(a)} = v.$$ We can also conclude (not definitively, but suggestively) that the unknown parameter $a$ is $v$.

With a suitable rewriting of variables ($H \rightarrow \gamma$, $a \rightarrow v$): $$x' = \gamma(v)(x-vt)$$

The derivation of $$\gamma(v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ continues from here.

The above is shamelessly stolen from Jean-Marc Lévy-Leblond's One More Derivation of the Lorentz Transformation. I like this paper because it doesn't assume the constancy of the speed of light. This fact is actually derivable from some very general and safe assumptions of how space and time work. In fact, in a universe with a massive photon (i.e., in which nothing could reach the now misnamed "speed of light"), this derivation would still work. It's a longer, subtler, and more involved derivation, so I can see why it's not used in classrooms. But, it's a bit wild to think that anyone from Newton on (or Galileo with some pseudo-calculus) could have discovered the Lorentz transform.

• I have given it a little thougt. Thank you for your response, is very accurate, although I haven't understand it compeltely. I have come to the next explanation: the Galilean transformation must be a particular case of the Lorentz transformation, when v<<c. Sp, if the relation is not linear, we can't reduce the Lorentz to the Galilean one. I don't have a rigurous proof, but I have the argument that.. – Bardo Oct 2 '14 at 19:23
• If F(x,t,a) would be something like $F(x,t,a) = x^kB + t^qA + a$ , then, this cannot br reduced to x-ut, because the rank of x and t would be higher than 1, i.e., the function would be a polynomial-like. It's more kf an intuitive argument, I know, but what do you tjink of it..? – Bardo Oct 2 '14 at 19:27
• @Bardo That does work as an argument that the exponents of $x$ and $t$ have to be 1 to keep the units correct, but you need other arguments to get to $(x-vt)$. One way is to say that the origin of $O'$ is moving in the $O$ frame at velocity $v$. Therefore, starting from $x' = H(a)(x-K(a)/H(a) t)$, when $x' = 0$, $x = vt$. So, $x' = H(a)(x-vt)$. – Mark H Oct 4 '14 at 3:18
• You may know that there is a long history of papers like the one you cite, beginning with Ignatowski in 1911. Even Einstein's 1905 paper mentions the group postulates. I like the approach and have had considerable success explaining things in these terms to intelligent teenagers who are driven mad by the question "what's light got to do with it anyway". One can see that $c$ is then simply a parameter that generalises Galileo's relativity, and SR is then simply what happens to Galileo's relativity when one relaxes the assumption of absolute .. – WetSavannaAnimal Dec 1 '14 at 12:44
• ... time. For what it's worth, here's my writeup of the same ideas: physics.stackexchange.com/a/143421/26076 Anyhow, +1, I hadn't seen the argument for a linear transformation before, but it is a good one, obvious when it is seen – WetSavannaAnimal Dec 1 '14 at 12:45