By P. W Bridgman

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However, it has a completely different meaning from the contraction introduced by Lorentz in connection with the Michelson experiment. Lorentz assumed that the longitudinal contraction appears only when an object is moving relative to the ether which serves as a universal system of reference. As a consequence, a segment that is stationary relative to the ether possesses the greatest length. In reality, the length contraction is observed for any object moving relative to any inertial system. And the segment has the greatest length in its own system of rest, which may be moving relative to a given inertial system at an arbitrary speed V .

Then the equation for the light pulse can be written as s212 ˆ 0 …54† The quantity s12 is called the four-dimensional interval (or just the interval) between events 1 and 2 [recall Eq. (31)]. We see that the interval for a light signal in vacuum is equal to zero. Suppose that the same process is being observed from another inertial reference frame K'. The space and time coordinates for the same two events (emission and absorption of the signal) in system K' will be (ct'1, x'1, y'1, z'1) and (ct'2, x'2, y'2, z'2), respectively.

In our case, when the velocity V is parallel to the x axis, the transverse coordinates must have the same values in both systems: y ˆ yH Y z ˆ zH …33† The reason is that the speed of relative motion of the systems K and K' is zero in the directions y and z. 6 The Lorentz transformations c t ˆ a00 ct H ‡ a01 x H x ˆ a10 ct H ‡ a11 x H y ˆ yH z ˆ zH i e y e t (34 b) However, the point O' moves at a speed V relative to K, so by the moment t its position in K is given by the coordinate x = Vt. Assuming x' = 0 and putting into Equation (34 b) x = Vt, we have ' c t ˆ a00 c t H V t ˆ a10 c t H (35) from which there follows a10 V ˆ c a00 …36† Thus we have the following matrix equation for the transformation of x and t:  ct x  H a00 ˆdV c a00 I  H e c tH x a11 a01 …37† To find the remaining three unknowns, we substitute the linear Equation (37) into the equation c2 t2 À x2 ˆ c2 t H2 À x H2 …38† obtaining …a00 ct H ‡ a01 x H †2 À …a00 Vt H À a11 x H †2 ˆ …c t H †2 À x H2 …39† Now we demand that the expression on the left might be reduced to the form c2 t'2 – x' .

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