Showing that Lorentz transformations are the only ones possible

[Fredrik]

This sort of thing is one reason why I prefer to start from inertial observers defined as those that feel no acceleration.

But that's what I do. That doesn't solve the problem. Now that I think about it, it makes things slightly worse than I understood when I wrote my previous post.

We are looking for theories in which there's a set K of curves in M (i.e. in spacetime) such that each member of K represents a possible motion of an accelerometer that measures 0. A global inertial coordinate system should be a bijection from M into ℝ⁴ that takes every one of those curves to a straight line. But we can't take this as the definition of a global inertial coordinate system, because we know that in SR, those curves are all timelike, and a global inertial coordinate system in SR also takes spacelike geodesics to straight lines.

I think we need to leave the term "global inertial coordinate system" partially undefined at this point. We can define it properly after we have found a group of inertial coordinate transformations.

The partial definition of "global inertial coordinate system" doesn't imply that inertial coordinate transformations take *all* straight lines to straight lines. It just implies that there's a set L of straight lines such that each member of L is taken to a member of L.

It does seem natural to also require that every inertial coordinate transformation takes all constant-velocity motions to constant-velocity motions, but this assumption doesn't pin down what an inertial coordinate transformation does to an infinite-speed straight line.

If one finds the maximal dynamical group applicable to the zero-acceleration equations of motion, the problematic case you mentioned can be handled by taking a limit afterwards.

After we have found the group? But we used the assumption that *all* straight lines are taken to straight lines to find the group. Also, limits require a topology. If we could do this they way I originally intended (as described in the long list a few posts back), we would find, without any assumptions about topology, that inertial coordinate transformations are affine. Since affine maps are continuous with respect to the Euclidean topology, this could even be thought of as justification for choosing the Euclidean topology later.

 

[Fredrik]

Why do you think relativity of simultaneity implies nonlinear transformations?

I don't. I said that there's no relativity of simultaneity if inertial coordinate transformations can't take infinite-speed straight lines to finite-speed straight lines. In other words, there's no relativity of simultaneity if inertial coordinate transformations can't change the slope of a horizontal line in a spacetime diagram. You don't need a non-linear transformation to change the slope of a horizontal line. A Lorentz transformation with non-zero velocity will do fine.

 

[TrickyDicky]

I don't. I said that there's no relativity of simultaneity if inertial coordinate transformations can't take infinite-speed straight lines to finite-speed straight lines.

Ok, so I don't know why you bring up this, there is no such thing as infinite speed in SR, thus the relativity of simultaneity, there's no transformation from spacelike vectors to timelike ones.

 

[Fredrik]

Ok, so I don't know why you bring up this, there is no such thing as infinite speed in SR, thus the relativity of simultaneity, there's no transformation from spacelike vectors to timelike ones.

Who said anything about spacelike to timelike?

I'm just saying that I haven't yet seen a satisfactory way to explain why inertial coordinate transformations should take *all* straight lines to straight lines. For example, why should the straight line with (t,x,y,z) coordinates t=0, x=0, y=z be taken to a straight line?

 

[TrickyDicky]

Who said anything about spacelike to timelike?

I'm just saying that I haven't yet seen a satisfactory way to explain why inertial coordinate transformations should take *all* straight lines to straight lines. For example, why should the straight line with (t,x,y,z) coordinates t=0, x=0, y=z be taken to a straight line?

Hmmm, I think I see what you mean, and I'm not sure there is a satisfactory way using your path.

But of course if you took as starting point a flat spacetime that would trivially come from the fact that all geodesic lines in such a space are straight lines by definition. Still, this begs the question why should one choose such a space for SR in the first place. And the only answer is the postulates which are arbitrary to some extent.

 

[Fredrik]

I think I have an argument that works. Consider the set of "vertical" lines (t variable, x,y,x constant) through a segment on that line. They would represent the motion of the component parts of a thin non-accelerating rod, in a comoving inertial coordinate system. Their union is the "world sheet" of the rod. We have already assumed that all the finite-speed lines in the world sheet, including the ones with arbitrarily high speeds, are taken to straight lines.

Suppose that the world sheet's intersection with the t=x=0 plane (i.e. the line I described), isn't taken to a straight line, then the images of the vertical lines under the inertial coordinate transformation have discontinuities at the t=x=0 plane, and a straight line with one point removed and put somewhere else isn't really a straight line. So this contradicts the assumption that finite-velocity straight lines are taken to straight lines, and this means that the line I described must be taken to a straight line.

My explanation may not be perfectly clear, but I have a spacetime diagram in my head that's seems clear enough, so I think this idea works even if I didn't explain it well enough.