<p>We could also define the length as the product of the velocity of the moving rod by the time interval between the instant when one end point of the moving rod passes a fixed marker and the instand when the other end point passes the same marker. Show that this definition leads also to the Lorenz contraction formula for length. </p>
<p>How do I do this? I’ve tried several times and it’s not working for me…</p>
<p>First, set up the reference frames. Let the observer be in frame S and the rod be in frame S’. The rod is stationary in S’ and S’ is moving with velocity v, relative to S. Let the observer designate point x (in S) as the point, where the time will be recorded when front or back of the rod are at x. Notice that the time recorded will be the proper time.</p>
<p>When the observer sees the front of the rod at point x, a time, t1, is recorded. In S’, the position of the front of the rod is x1’=y<em>(x-v</em>t1), where y is gamma.</p>
<p>When the observer sees the end of the rod at point x, t2 is recorded. In S’, the position of the back of the rod is x2’=y<em>(x-v</em>t2).</p>
<p>x1’-x2’=y<em>(x-v</em>t1)-y<em>(x-v</em>t2)=y<em>v</em>(t2-t1)</p>
<p>Notice that x1’-x2’ is the proper length of the rod because the rod is at rest in S’. Thus x1’-x2’=Lp. Furthermore, recall our definition of length. L=v*(t2-t1).</p>
<p>Hence, Lp=y*L.</p>
<p>Hope this clears things up.</p>
<p>Hmm… I don’t quite see how that answers the question. Who is to say that the rod doesn’t have a different velocity than the coordinate system? And doesn’t that basically assume that the definition of the length is x1-x2 not u*(t2-t1) where u = speed of the rod?</p>
<p>The rod is moving relative to the observer in frame S. In the S’ frame, it is not moving. Also, I didn’t assume the definition of length, I assumed the definition of “proper length” which is the length of an object in a reference frame where the object is at rest (hence, this is not based on time or velocity).</p>