## There is no length contraction

I have demonstrated that the correct "Lorentz transformation" are X' = V't', where V' = (V-v)/(1-Vv/c^2), and t' = t*sqrt(1-v^2/c^2). Those relations have been obtained 1) by considering two frames of reference, S and S', each in uniform translatory motion relative to the other, the velocity of S' relative to S being v. 2) by counting time from the instant at which the origins of the frames, O and O', momentarily coincide. At that instant, it is assumed that t = t' = 0. 3) by considering an object starting from the coincident origins at t = t' = 0, with a constant velocity V. See The Lorentz transformation (LT) are false According to the frame S, x=vt is the position of the origin of S' after a time interval t, and X=Vt the position of the object after the same interval t. But after such time interval, the clock of S' will read t', and the object will be at X' = V't' from the origin of S'. The existence of length contraction has been put forward by both Fitzgerald and Lorenz in order to explain the negative result of the Michelson and Morley experiment. They supposed that motion through the ether caused the material composing the interferometer to shorten by the factor sqrt(1-v^2/c^2) in a direction parallel to the motion. This contraction would equalize the light paths and prevent a displacement of the fringes. Indeed, let's call S the "ether" frame, and S' the interferometer frame. Both arms of the interferometer have the same length L. The arm of the interferometer, which is parallel to the x,x'-axis (the "parallel arm"), is limited by a mirror M2, whereas the arm parallel to the y-axis (the "perpendicular" arm) is limited by a mirror M1.Galilean analysis________________________ Let's call L' the length of the parallel arm. When the apparatus is at rest in the "ether", L' = L. As the interferometer is supposed to move at v relatively to the frame S, the light wave moving along the x,x'-axis has a velocity c-v on the outgoing trip, and c+v on the return trip. Hence, the total time required by this wave is t(x) = L'/(c-v) + L'/(c+v) = (2L'/c) * 1/(1-v^2/c^2). According to the frame S, the wave moving transversely travels a distance ct' along the hypotenuse of a right-angled triangle, whereas in the same time, the mirror M1 advances a distance vt'. Hence, (ct')^2 = (vt')^2 + L^2, and t' = (L/c) * 1/sqrt(1-v^2/c^2). This wave returns to the origin of the frame S' after a total time t(y) = 2t' = (2L/c) * 1/sqrt(1-v^2/c^2). If the length of the parallel arm is contracted by sqrt(1-v^2/c^2), one has have of course t(x) = t(y), hence no fringe shift can be observed. However, the Galilean analysis doesn't take the phenomenon of "time dilation" into account: If the interferometer is at rest in the "ether", i.e., if the frame S' is at rest relatively to the frame S, the light wave sent at t=t'=0 will be, according to S or S', at X=ct, and at Y=ct after the time interval t. Let's note that ct corresponds to the arms'length L. But if the frame S' moves at v relatively to the frame S, one gets X' = V't' = ct' for the wave following the x'-axis (which slides along the x-axis), and Y' = ct' for the wave following the y'-axis. As t' = t * sqrt(1-v^2/c^2), one obtains (1) X' = ct * sqrt(1-v^2/c^2) (2) Y' = ct * sqrt(1-v^2/c^2) on the outgoing trip, and also, for reasons of symmetry, on the return trip. The lengths of the two paths X' and Y' being equal, no fringe displacement will be observed. As X=ct, one could represent the relation (1) by L' = L * sqrt(1-v^2/c^2), and claim that the length L' of the arm moving along the x-axis is contacted by sqrt(1-v^2/c^2). But this would be false, as the apparent contraction is entirely due to the time "dilation". This can look paradoxical, but let's consider the following scenario: "Two cars run side by side on parallel lanes from A to B at a velocity V = 50 miles/h. According to the map, the distance AB = 100 miles. Arrived at B, one of the driver looked at his watch, and claimed that, since he has traveled during t = 2 hours at 50 miles/h, the distance AB = Vt is indeed 100 miles. The other driver claimed that, according to his own watch, he drived during t' = 1 hour 1/4, adding that, since t'is smaller than t, the distance AB is smaller than 100 miles, because he travelled only AB = Vt'= 87.5 miles. The first driver said: "You are wrong, because your watch must be 3/4 hours slow, compared to mine." The other drive replied: "No, I am right, because my lane contracted by the factor t'/t = 7/8." THe first driver then asked: "Isn't your last name Einstein?" See also the following excerpt from "Modern Physics" (Krane, K.S., Modern Physics, J. Wiley, New York, 1983, [23-25]): "It must be pointed out that time dilation is a real effect that applies not only to clocks based on light beams but to time itself. All clocks will run more slowly as observed from the moving frame of reference. ... The length measured by the moving observer is shorter. It must be emphasized that this is a real effect." Let's emphasize that this "real effect" doesn't imply a physical length contraction.SR analysis_______________ Let's suppose that a light signal, starting from the coincident origins of frames S and S' at t = t' = 0, travels toward positive x. After a time t, it will be at x = ct, and also at x' = ct', since *the speed of light is the same in all frames* . If the signal travels toward negative x, x = -ct and x' = -ct' Now, a light signal follow the y' axis. Relatively to S, it travels obliquely, for, while the signal goes a distance ct, the y'-axis advances a distance x = vt. Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t = ct * sqrt(1-v^2/c^2). But, also, y = y' = ct', etc... Iow, according to Einstein, x' = ct' and y' = ct', meaning that the two paths x' and y' are equal. Let's note that ct represents the rest length L of the arms. As x' = ct' and y' = ct' = ct * sqrt(1-v^2/c^2) = L * sqrt(1-v^2/c^2), x' = L * sqrt(1-v^2/c^2), thus L' = L * sqrt(1-v^2/c^2). We have seen above that this last relation represents an apparent length contraction, that is in fact explained by time "dilation". However, Einstein used his position LT x' = g(x - vt) to "prove" the existence of length contraction: Assuming that the length of an object at rest in S' corresponds to the difference Lo between the coordinates x2' and x1' of its ends, he substituted in Lo = x2' - x1' the values of x2' and x1' calculated from his position LT for a given value of t, thus obtaining Lo = g(x2 - x1) = gL, where L is the length of the object in S. He then concluded: 1) that any body measures shorter in terms of a frame relative to which it is moving with speed v than it does as measured in a frame relative to which it is at rest, the ratio of shortening being sqrt(1-v^2/c^2). 2) that, relative to a single frame, any physical body set into motion with speed v shortens in the direction of its motion, as was postulated by Fitzgerald and Lorentz, in the same ratio sqrt(1-v^2/c^2). By the way, length contraction has been variously explained as 1. true but not really true (guess who) 2. real 3. not real 4. apparent 5. the result of the relativity of simultaneity 6. determined by measurement 7. a perspective effect 8. mathematical. (See the chapter 2.1 "Tower of Babel", in the page "Sapere aude", by G. Walton.)A simple logical proof of the falseness of the LT, based on the MMX._______________________________________________________________________________________________ Let L be the rest length of the arms. When the apparatus is at rest in the ether, ct = L. When the apparatus is moving at v wrt the ether, according to the Galilean analysis of the MMX, the two-way trip time of light along the parallel arm is t(x) = 2L'/c(1 - v^2/c^2), where L' is the length of the parallel arm, whereas, for the perpendicular arm, one gets, by applying the Pythagorean theorem, t(y) = 2L/c*sqrt(1 - v^2/c^2). As no fringe shift is observed, one logically assumes that t(x) = t(y). This implies that L' = L * sqrt(1-v^2/c^2), iow, that the length of the parallel arm is contracted by sqrt(1-v^2/c^2). Now let's call S the "ether" frame and S' the interferometer's frame. S' moves at v wrt S. Let's remember that the parallel arm moves parallel to the x,x'-axis, whereas the perpendicular arm is situated along the y'-axis. According to SR, y' = ct'. As the result of the MMX is negative, the light paths must be equal, hence x' = y' = ct' = L'. Let's assume that t' = t * sqrt(1-v^2/c^2). Then, x' = ct * sqrt(1-v^2/c^2), to which corresponds the relation L' = L * sqrt(1-v^2/c^2). Let's remember that this relation doesn't imply a physical length contraction. By replacing L' by L * sqrt(1-v^2/c^2) in the Galilean relation t(x) = 2L'/c(1 - v^2/c^2), one gets t(x) = 2L/c*sqrt(1 - v^2/c^2) = t(y). The travel times of the light signal along the two arms being equal, the MMX must be negative. Now, instead of t' = t * sqrt(1-v^2/c^2), let's use the Einsteinian "time" transform t' = g(t-xv/c^2). Since x = ct, t' = t * sqrt[(1-v/c)/(1+v/c)], and x' = ct' = ct * sqrt[(1-v/c)/(1+v/c)]. The equivalent form being L' = L * sqrt[(1-v/c)/(1+v/c)], t(x) is no more equal to t(y) when using this value of L'. One has to conclude that the Einsteinian "time" LT is false, and that the true relation is t' = t * sqrt(1-v^2/c^2), as shown above. As x' = ct', the falseness of the "time" LT automatically implies the falseness of the "position" LT x' = g(x-vt).By the way, the LT being false, "events" which are simultaneous in one frame are simultaneous in all other frames. In other words, contrarily to Einstein's claim, simultaneity is not relative.Marcel Luttgens

May 23, 2003. muttgens@spamorange.fr Back Home Page