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The length contraction means that an object is the longest in the frame in which it is at rest.

Lets assume i have a meter stick with length $\Delta x$ in my rest frame which is $x,ct$ and i want to know how long my meter stick seems to an observer moving with a frame $x',ct'$.

1st: I draw world lines of a meter stick in a rest frame and they are vertical (parallel to $ct$ axis) as meter stick is stationary in this frame.

2nd: If an observer in a moving frame $x',ct'$ wants to measure my meter stick he measures its edges at the same moment in his time, so i draw a tilted line (parallel to $x'$ axis).

3rd: If i mesure the length $\Delta x'$ which is a length of a meter stick as observer in frame $x',ct'$ sees it, it seems to me that he sees a longer distance than me.

This is not correct. Could anyone tell me what am i missing here?

enter image description here

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The first sentence is wrong, it is upside down. An object is longest in its rest frame. In all other frames - the moving ones - it's shorter. Shorter is why it's called "contraction". This was probably a typo. The solution to your puzzle is that your paper probably preserves the Pythagorean theorem, $dist^2 = x^2+y^2$. The right paper you should use is the spacetime paper which has $dist^2 = x^2 - c^2 t^2$, but this paper with a minus sign is hard to buy. The minus sign reverts the conclusion. –  Luboš Motl Oct 23 '12 at 17:44
    
It was a typo yes. –  71GA Oct 23 '12 at 18:03

3 Answers 3

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What you've missed is that the distance along the $x'$ axis is not the same as the distance along the $x$ axis. The locus of events that are 1 unit of proper distance from the origin is a hyperbola. This can be used to calibrate the $x'$ axis. See calibration hyperbola.

enter image description here

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Thank you this allso contributed to the solution and it is a really good anwser. In fact i did add my own anwser after studying your anwser. –  71GA Oct 23 '12 at 18:58

One picture is worth a 1000 words here... enter image description here

The important point is that we make a snapshot of the moving object in a time coordinate which is not its proper time. Indeed, if we drew just one image of the moving object in a cut given by $t' = \rm const.$, we would get a projection on the $x$ axis longer than $l_0$. But we must use $t = \rm const.$ instead if the measurement is done in the un-primed coordinate system. Thus, we measure the "front" of the object earlier (in terms of its proper time) than the "rear". The difference in $t'$ allows the latter to cover some extra distance, putting the two events closer in $x$. If the math is worked out, it indeed gives $l_1 < l_0$.

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Thank you. I was looking for a graphical demonstration and you provided an excelent picture! –  71GA Oct 23 '12 at 18:05

After thinking about aditional options to explain this it returned to me, that in frame $c',ct'$ marks on $x'$ axis are further appart than marks on $x$ axis (pay attention to $1$ and $1'$). So my picture was correct from the beginning, but i didn't draw the marks on $x'$ axis. Thank you Alfred Centauri.

Well now it is obvious why moving observer sees shorter distance. Take a look at the picture and give me some feedback please:

enter image description here

Q: But what if a meter stick is stationary in frame $c',ct'$? I guess then $\Delta x'$ is the proper length and i have to transfer its edge points to $x$ axis using lines parallel to $ct'$ axis?

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what did you use to draw the graphs? –  Physiks lover Oct 24 '12 at 17:00
    
I use inkscape. –  71GA Oct 24 '12 at 20:16

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