1

Look at the wikipedia article (paragraph: Spherical wavefronts of light) about the derivation of Lorentz transformation between two reference frames $O(x,y,z,t)$ $O'(x',y',z',t')$ in standard position. At a certain point one reads :

" The relation between $x$ and $x'$ should be in linear form and in such a way that it should reduce to Galilean transformation at $v << c$. Therefore, such a relation can be written of the form: \begin{align} x' &= \gamma \left( x - v t \right)\\ \end{align} where γ, not necessarily a constant, is to be determined "

I don't understand why we are searching a transformation written in this form.

Thanks in advance.

Dubious
  • 13,350
  • 12
  • 53
  • 142

1 Answers1

3

It must be a linear form because it needs to preserve vector addition, which is precisely what linear transformations do. Otherwise, I can interpret vectors $a,b,c$ as force vectors and find that an inertial frame ($a+b+c=0$) transformed into a non-inertial frame. So linearity is a must.

To be honest I'm not too convinced by their second portion either. I go about this by

  1. requiring that the light vectors $[c,1]^T$ be eigenvectors of unspecified eigenvalues

  2. requiring that the transformation for a given boost send some vector $[v,1]^T$ to $x=0$ (with the $t$ value left unspecified)

  3. requiring that the inverse be found by simply making velocity negative. i.e. the same transformation takes $[0,1]^T$ to $[-vt,t]^T$ with some unspecified $t$ value.

After some simple algebra the Lorentz transforms emerge. Even though it's a little more tedious, I like this method because it invokes all of the major physical properties that the Lorentz transform needs to have through the derivation.

Robert Mastragostino
  • 15,727
  • If a transformation sends straight lines in straight lines, this not implies that the transformation is linear. This wrong reasoning is used in many textbooks. – Dubious Sep 09 '13 at 12:50
  • @Galoisfan I said that it preserves vector operations. These are precisely the linear transformations. I suppose if you interpret 'vector operations' in a freer way then you can include affine transformations, but the question asks about the Lorentz group and not the Poincare group, so I left this out. Affine maps are simply linear map+translation anyway, so that's trivial. – Robert Mastragostino Sep 10 '13 at 02:41