I have a doubt regarding the proof of Length Contraction. The proof of length contraction formula goes as follows: You choose a frame S (as shown in figure 1), suppose w.r.t. S, AB is moving with velocity v. Length of AB is L in AB frame and L1 in S frame. At t=0, a light pulse (shown with red color in figure) is left at A, it strikes B and reflects back to A. Figure 1: http://img166.imageshack.us/my.php?image=figure1fz0.png Time taken for first reflection t1 = L1/(c-v) and similarly for return t2 = L1/(c+v). So total time taken equals t = t1 + t2 = 2cL1/(c2-v2). Suppose time taken for these reflections in frame AB is t’, so t = t’γ (from Time Dilation). And in AB frame, t’ = 2L/c. Now eliminate t and t’ from these 3 equations to get the formula for Length Contraction: t = 2cL1/(c2-v2) and t = t’γ give t’γ = 2cL1/(c2-v2). Put t’ from t’ = 2L/c into this equation: 2Lγ/c = 2cL1/(c2-v2) L1 = L/γ, this is the formula for length contraction. Now, my doubt starts, consider figure 2, and suppose that we are not looking the second reflection. Figure 2: http://img525.imageshack.us/my.php?image=figure2ia2.png Then t = t1 = L1/(c-v) and t’ = L/c. And t = t’γ. Now eliminate t and t’ from these 3 equations: t’γ = L1/(c-v), put t’ = L/c to get Lγ/c = L1/(c-v) this gives: L1 = Lγ(1 - v/c) = L [(c-v)/(c+v)]1/2 and not L1 = L/γ. Is there any mistake if we choose to not to consider the second reflection? Further suppose that we only consider the second reflection, in that case t = t2 and hence finally we will get: L1 = L [(c+v)/(c-v)]1/2.