How Does Length Contraction Work in Einstein's Relativity?

[nonequilibrium]

Hello,

I was reading Einstein's RELATIVITY (his non-mathematical 1916 (or so) book).

Having shown the Lorentz transformation formulas, i.e.
$$t' = \gamma( t - vx/c^2)$$
$$x' = \gamma(x - vt),$$
he then talks about length contraction. He takes a meter-rod fixed in a certain coordinate system S' so that the left end is at $$x_1' = 0$$ and the right end is at $$x_2' = 1$$ and he then tells us to view these two points from system S at a certain time for which he takes t = 0. We then get that in S $$x_1 = 0$$ and $$x_2 = 1/ \gamma$$ (we're ignoring units here). Thus we get that if we see a rod passing us at a speed v, the length at any given moment is $$\gamma$$ times smaller than the so-called eigenlength of S' where the stick is fixed. (this is, of course, the famous Lorentz contraction formula)

Now in this case it makes sense to say "the length of the stick in S is smaller than the length of the stick in S' ". But notice that in our previous case, $$t_1' \neq t_2'$$ (because $$t_1 = t_2$$ and $$x_1 \neq x_2$$). This was okay because the two points of the rod were fixed in S' anyway, so it didn't matter what "time it was" for a certain point of the rod (i.e. the position of the rod was independent of t'). But imagine the situation where we want to compare the length of a certain rod, moving relatively to us at c/2 and relatively to somebody else at c/3. It has no meaning to say the length I see is smaller/larger than what he sees; is this correct? By taking a certain time for both points in, say, S, then the two points in S' can't be at the same instant, but then you can't compare "two photographs" of each situation... (Then again, I just came to think of the fact that each of these systems can be compared with the eigenlength, thus offering a comparison after all by then cancelling the eigenlength out of those two equations... How come two of my reasonings lead to different statements? Where is there a mistake?)

Thank you.

 

[granpa]

the length of a rod in a given frame is the distance between the front and the back at one simultaneous moment in that frame.

so, yes.
Relativity of simultaneity does enter into it.

 

[JesseM]

In each frame you can pick two events, one on the worldline of the front and the other on the worldline of the back, that occur simultaneously in that frame, and consider the difference in position coordinates between the events as the object's "length" at that moment. As long as both front and back are moving inertially, each frame should see the length as constant with change (i.e. it won't matter what time we pick the two simultaneous events at).