dx said:
It is possible to have well defined postulates from which the linearity of Lorentz transformations follows.
Let V be a four dimensional vector space. An inertial frame is a map ψ from the set of events into V which satisfies the following postulates:
1. The world lines of free particles are straight lines.
2. Clock rates are uniform, i.e. intervals measured by clocks agree with the linear structure of V.
I agree with your opening statement, but I would drop your second postulate and add stuff to the first. I'd choose V=\mathbb R^4, and change #1 to
1. Each transition function* corresponding to two inertial frames is a smooth** bijection that takes straight lines to straight lines.
*) See my posts earlier in this thread for a definition.
**) All its partial derivatives up to arbitrary order exist.
Let T be a transition function. The axiom guarantees that it can be Taylor expanded.
T(x)=T(0)+x^\mu\partial_\mu T(0)+\frac 1 2 x^\mu x^\nu\partial_\mu\partial_\nu T(0)+\cdots
Let's call a transition function with T(0)=0 a "Lorenz transformation". (This will be our definition of a Lorentz transformation for the rest of this post). Note that a Lorentz transformation defined this way takes straight lines through the origin to straight lines through the origin.
Now let T be a Lorentz transformation, and let x and y be two points on a straight line through the origin. We must have y=kx. Postulate #1 and our definition of a Lorentz transformation imply that we also have T(y)=k'T(x), but T(y)=T(kx), so we have
T(kx)=k'T(x)
for all x. Let's Taylor expand both sides.
kx^\mu\partial_\mu T(0)+\frac 1 2 k^2 x^\mu x^\nu\partial_\mu\partial_\nu T(0)+\cdots=k'\Big(x^\mu\partial_\mu T(0)+\frac 1 2 x^\mu x^\nu\partial_\mu\partial_\nu T(0)+\cdots\Big)
These two expressions must mach term by term, and that's only possible if k'=k and all the higher order terms are =0. So any "Lorentz transformation" must be linear.
I don't have any objections to this sort of argument, but one could point out that the axiom is extremely strong. I mean, we're assuming that transition functions take straight lines to straight lines, so it's not exactly a surprise that they turn out to be
linear. So one could argue that we might as well have started by requiring linearity. The counter argument to that is that this approach is more intuitive and "natural" than the abstract requirement of linearity. It only expresses the idea that any inertial observer should be able to describe any other inertial observer as moving with constant velocity.