# Why does the Lorentz transformation in special relativity have to be like this?

Basically I think Albert Einstein (A.E.) was trying to find a transformation that:

1. Always transform a constant-velocity movement into a constant-velocity movement.
2. Always transform a light-speed movement into a light-speed movement.
3. If an object with a speed $v$ in frame $A$ is rest in frame $B$, then any rest object in $A$ has a speed $-v$ in $B$.

A.E. gave:

$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}},$$

$$t'=\frac{t-(u/c^2)x}{\sqrt{1-u^2/c^2}}.$$

But there are more than one transformation that can do this.

Multiply a factor, we can get another one:

$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}(1+u^2),$$

$$t'=\frac{t-(u/c^2)x}{\sqrt{1-u^2/c^2}}(1+u^2).$$

It satisfies all three postulates given. So why can't the latter be the Lorentz transformation?

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well, our experience tells us that for speeds we can experience conveniently the transformation law must absolutely reduce to galilean invariance. $t'=t$ and $x'=x-ut$. Your example doesn't. – luksen Sep 27 '11 at 16:22
@luksen: I think you have a good point. This should be the fourth postulation in my list. – trVoldemort Sep 27 '11 at 16:30
@luksen: 1. indicates that the transformation must be linear; 2. indicates that the eigenvectors of that transform matrix must be (c, 1) and (-c, 1); 3. reveals the matrix (1, -u; -u/c^2, 1), but with an unknown factor; and 4. given by you finally provides the factor 1/sqrt(1-u^2/c^2). Wonderful! – trVoldemort Sep 27 '11 at 16:39
feel free to answer your own question. – luksen Sep 27 '11 at 17:20
Just for completeness: the way the Lorentz transformation is derived in the literature is by posing that it preserves spacetime intervals $s^2 = (ct)^2 - \mathbf{x}^2$. This is a strong enough demand to completely determine the form of the Lorentz transformations. – Gerben Sep 27 '11 at 20:22

## 1 Answer

You have discovered the well-kept secret that in 2 dimensions, the transformations which keep light-rays fixed include conformal transformations, not just Lorentz transformations. You need to pick a subset of the conformal transformations which form a group, and which are compatible with reflections.

The ones you used are not the good ones, because if you use your transformation with speed u, and invert it, it is not the transformations with speed -u. The factor you get is not multiplicative, so if you compose two transformations with u and u', you don't get something in the group. If you keep transforming, your coordinates just get a bigger and bigger scale factor.

But there is another subgroup of the 1+1 d conformal group which is a group which obeys all of Einstein's speed-of-light postulates:

$$x' = e^{k\alpha} ( \cosh(\alpha) x - \sinh(\alpha) t) = ( \sqrt{1+v\over 1-v})^k{x-vt\over \sqrt{1-v^2}}$$ $$t' = e^{k\alpha} ( \sinh(\alpha) t - \cosh(\alpha) x )= (\sqrt{1-v\over 1+v})^k{t-vx\over\sqrt{1-v^2}}$$

This transformation scales by the rapidity (relativistic analog of 2d rotation angle) with the only 1d scale factor that forms a group. These transformations are the alternate Lorentz transformations you want. Their orbits are relativistic analogs of geometric spirals, not circles, and they form a one-dimensional group, and they reduce to Galilean transformations at low velocities ($c=1$ in the formulas above).

In his derivations of the Lorentz transformations, Einstein implicitly used reflection symmetry, by assuming that the transformation for -v will be the same as the transformation to speed v with just the sign on x reversed. This assumption allows you to kill this possibility, because it is asymmetric, the scale for positive velocity transformations is inverse to the scale with negative velocity transformations.

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You've thought about this before, right? – Physiks lover Jun 13 '14 at 11:16