# How are spacetime diagrams compatible with moving observers measuring shorter lengths? [duplicate]

I understand that length contraction is a consequence of the relativity of simultaneity. I found this spacetime diagram online:

Here it's clear that what constitutes a rod for one observer, is a sequence of snapshots of this rod for another observer. Now I have always read the following about length contraction: an observer moving with respect to the rod will measure a shorter length. But according to this diagram, the moving observer who measures the length of the rod along his $$x'$$ axis, should actually measure a longer length?

• @BenCrowell Someone is fudging your identity. Commented Nov 13, 2019 at 18:57

Here is a visualization of three methods
to show that the "length of the rod" $$OM$$ observed by the moving observer is shorter than the rest length $$OL$$.

1. Draw concentric hyperbolas centered at the left end of the rod.
The hyperbola corresponding to the rest-length OL is larger than for any other length (such as OM).
So, $$OM < OL$$.

What is special about OL is that OL is Minkowski-perpendicular to the worldlines of the rod (that is, the tangent at L is parallel to the worldlines of the rod).
The tangent at M is not parallel to the worldlines of the rod.
2. Using rotated graph paper to help determine the area of "causal diamonds", observe that the causal diamond with spacelike diagonal OM has an area smaller
than that of the diamond with OL.
So, $$OM < OL$$.

3. If you construct the ticks along the axes (with the help of rotated graph paper, where all ticks have the same area), you'll see that $$OM < OL$$.

Short answer: When drawing diagrams like this, one must realize that the scale of the primed axes is not the same as the non-primed. That is, if you measure one centimeter along the $$x$$ axis and equated that to 1 m, then one centimeter along the $$x'$$ axis on your piece of paper is not 1 m. There is a hyperbolic transform, not a linear one.