I Compute Length Contraction: Is ##l=1## a Possible Model?

[PeterDonis]

I think this shows, that the derivation method of the OP does not require before a deduction, that length contraction is possible (in case of spacetime invariance), nor in this case, that it is not possible.

I didn't say "deduction", I said "assumption". Your argument has the same form as the OP, so whatever assumptions the OP is making, your argument is making as well.

One could, of course, also make an argument similar to the one I made in post #27, but using the Newtonian invariance property instead of the SR one. Whatever difference there is between my argument in post #27 and the OP, there would be the same difference between a Newtonian argument similar to post #27 and your argument in post #30.

 

[Sagittarius A-Star]

Your argument has the same form as the OP, so whatever assumptions the OP is making, your argument is making as well.

Newton's assumption of an absolute time enforces the length-"contraction" factor to be 1. Because the OP does not assume an absolute time, he must also drop this restiction of the length-contraction factor.

 

[PeterDonis]

Newton's assumption of an absolute time enforces the length-"contraction" factor to be 1. Because the OP does not assume an absolute time, he must also drop this restiction of the length-contraction factor.

Nothing you are saying here contradicts what I said.

 

[Sagittarius A-Star]

Summary:: Try to go straight from invariance of spacetime intervals to length contraction
...
Obviously the latter solution corresponds to actual special relativity. Does the former solution correspond to anything?

In your derivation, which is valid only for $v \neq 0$, an additional constraint is missing, that $l=1$ is only possible at $v=0$. You would need to derive from the spacetime invariance and additional equation, which contains $v$, for example the time dilation formula. Then you could derive the Lorentz transformations and get the missing constraint from the relativity of simultaneity.

But here is an easier derivation of the length contraction from spacetime invariance:

I define the length of the rod as $\Delta x' := 1$
I call the length of the rod in the unprimed frame $\Delta x = l$
The measurement events have the time difference $\Delta t = 0$.

I put this into the spacetime invariance
$\Delta t^2 - \Delta x^2 = \Delta t'^2 - \Delta x'^2$
and get
$0 - \Delta x^2 = \Delta t'^2 - \Delta x'^2$
$l^2/1 = \frac{\Delta x^2}{\Delta x'^2} = 1 - \frac{\Delta t'^2}{\Delta x'^2}$

Because of the symmetry between the temporal coordinates and spatial coordinates (in the direction of movement) in the spacetime invariance formula, it follows (in analogy to the time dilation formula):
$l^2 = 1 - v^2$

Visualization:
In an Minkowski diagram ($c=1$), the angle between the t'-axis and the t-axis is the same as the angle between the x'-axis and the x-axis.
$v=\frac{\Delta x}{\Delta t}$ if in the other frame $\Delta x' = 0\ \ \$ (relativity of "same location")
$v=\frac{\Delta t}{\Delta x}$ if in the other frame $\Delta t' = 0\ \ \ \ \$(relativity of simultaneity)