View Single Post
Mentor
P: 17,308
 Quote by Piyu For length contraction, L_0 is the person who is can measure the distance at the same time meaning he is not moving with respect with the object. Is that correct?
No, this is not correct. Let me explain conceptually what we are doing in the length contraction and time dilation formulas and then walk you through the derivation.

In time dilation we have one worldline, the clock, and we pick two events on that worldline and calculate the time between those events in the frame where the clock is at rest and a frame where the clock is moving. That is relatively simple to understand, and the frame where it is at rest always records the minimum amount of time.

Length contraction is not so simple. In length contraction we have two parallel worldlines, the two ends of the rod, and each frame picks two events which are simultaneous (dt=0 and dt'=0) in that frame and calculates the distance between those different pairs of events. It is critically important in deriving length contraction that you realize that the measurements are simultaneous in both frames and that they are therefore different pairs of events in each frame.

So for the derivation let A be one end of the rod and B be the other end of the rod. Then, in the rest frame we can write:
$$x_A(t)=0$$
$$x_B(t)=L_0$$
$$x_B(0)-x_A(0)=L_0$$

Now the Lorentz transform gives us
$$x'_A(t)=\gamma t v$$
$$t'_A(t)=\gamma t$$
$$x'_B(t)=\gamma (L_0 + t v)$$
$$t'_B(t)=\gamma (t + L_0 v/c^2)$$

Here is where you are making your mistake. You are calculating
$$x'_B(0)-x'_A(0)=\gamma L_0$$
but note that
$$t'_B(0)-t'_A(0) \ne 0$$
therefore the calculated distance is not the length of the rod in the primed frame since the positions of the ends are not measured at the same time and the rod has moved between the two measurements. To calculate the length of the rod we must use simultaneous events in the primed frame. We do this by using each expression for t' to eliminating the t from the corresponding expressions for x'.
$$x'_A(t') = t' v$$
$$x'_B(t') = t' v + L_0/\gamma$$
$$x'_B(0')-x'_A(0')=L_0/\gamma$$

Hope that clears it up.