I know the hypothesis that the light speed is constant is retained by experiments. But is there any theory explaining why the light speed is constant no matter how an observer moves relative to light?
My question is, specifically: Suppose an observer $O$ launches a light and $O$ starts to move at the same time with a uniform velocity $v$ in the same direction that light points. Then why $c$ is still the light speed that $O$ will measure rather than $c-v$?
Answer
A personal point of view is that you may consider that Lorentz transformations apply primarily on momenta, and not primarily on (infinitesimal or not) space-time coordinates.
This is, of course, a "strong" postulate.
If you assume (some additional postulates are needed there) that transformations are linear, and that there is a rotation invariance, you are going to study "boost" transformations : $\begin{pmatrix} p'_z\\E'\end{pmatrix} = A(v) \begin{pmatrix} p_z\\E\end{pmatrix}$. You may show that, because $A(v)A(-v)=1$, $det A(v)=1$. Supposing a group structure, you finally are looking at the one-dimensional subgroups of $SL(2, \mathbb R)$, which are :
$$\begin{pmatrix} \lambda&\\&\lambda^{-1}\end{pmatrix}\quad \begin{pmatrix} 1&v\\&1\end{pmatrix}\quad \begin{pmatrix} 1&\\v&1\end{pmatrix}\quad \begin{pmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{pmatrix}\quad \begin{pmatrix} \cosh \theta&\sinh \theta\\\sinh \theta&\cosh \theta\end{pmatrix}$$
If you add additional postulates that, in a boost transformation, energy and momentum must change, that there exist a transformation which puts the momentum to zero , and that, if the energy is positive for an observer, energy will be positive for all observers, the first $4$ one-dimensional subgroups of $SL(2, \mathbb R)$ are excluded, and the last dimensional subgoup corresponds to a Lorentz transformation.
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