Deriving Length Contraction From Lorentz Transform

[Delzac]

Homework Statement

From Lorentz Transform,

<br /> x^{\prime} = \gamma (x - vt)<br />

From textbooks and wikipedia,

L_0 = x&#039;_2 - x&#039;_1 = \gamma (x_2 - x_1 )

Where x_1 and x_2 = L

Thus,

\L_0 = \gamma L

Question is this:
If i take the same method and us the Inverse Lorentz transform, i seem to get a different answer, namely:

\L = \gamma L_0

Which obviously is wrong. I suspect the problem is with where the observing is that is implicitly assume when one use either Lorentz or inverse Lorentz. But, i cannot be sure nor can i resolve this problem.

Any help will be appreciated.

 

[ideasrule]

From textbooks and wikipedia,

L_0 = x&#039;_2 - x&#039;_1 = \gamma (x_2 - x_1 )

Note that this is only true if t1=t2. To put it another way, (x1,t1) represents the event of measuring x1 while (x2,t2) represents the event of measuring x2. In the reference frame where the stick is moving, the two measurements have to be performed at the same time. They don't have to be performed at the same time in the rest frame--no matter when you measure x1' or x2', they're always going to be the same.

Question is this:
If i take the same method and us the Inverse Lorentz transform, i seem to get a different answer, namely:

\L = \gamma L_0

Here, you're assuming that t1=t2, where both are rest frame coordinates. (Otherwise, the right-hand side would not equal L_0.) However, if t1=t2, t1' does not equal t2' because of relativity of simultaneity! Unlike in the previous case, neither x1' nor x2' remain the same as time passes, so x2'-x1' does not equal L.

 

[Delzac]

Ah, i see. Thanks, got it.

 

[Delzac]

But, then how dose one use Inverse Lorentz Transform to get length contraction formula? Since t' is not the same, so we use the lorentz transform for t'?