by Ottawa, Canada Wednesday, November 28, 2001
It can easily be shown that when the Lorentz transformation equations are applied — and even applied correctly — contradictions can arise. One such contradiction is brought to light below. Suppose that a primed inertial frame of reference is moving relative to an unprimed inertial frame of reference at velocity v in the x direction. Then according to the Lorentz transformations, the space and time coordinates t, x, y, z of an event in the unprimed frame are related to the space and time coordinates t', x', y', z' of the same event in the primed frame by the following equations (when the units are expressed in terms in which speed of light, c, is unity): x = gvt' + gx' y = y' z = z' (See for example <http://casa.colorado.edu/~ajsh/sr/construction.html>.) Now let's suppose we have two identical and accurate clocks C1' and C2' in the primed frame. Suppose they are both located in a very long spaceship: a spaceship of "proper length" 10 lightseconds. (This means it would take light 10 seconds, as measured by a clock carried on board the spaceship, to reach its tail end from its nose end.) Suppose that the spaceship is moving past a third clock C, in a direction parallel to the line between the two clocks C1' and C2', at a velocity v = 0.75^{0.5} c = 0.86602540378444 c. Suppose all the clocks are accurate to one part in a billion: that is, they gain or lose one second in a billion seconds — which, by the way, is almost 32 years. In other words, we're talking about quite a high level of accuracy here. And additionally, suppose that the readouts of the clocks are all the way up to the microsecond level. That is, the clocks read not only hours, minutes and seconds, but thousandths and even millionths of a second. Only if the two clocks both agree to the microsecond do they truly agree. Let's suppose that we perfectly synchronise both clocks C1' and C2' at the front end of the spaceship, and then we move clock C2' slowly but steadily away from clock C1' towards the rear end of the spaceship at a speed u equal to one hundredmillionth (i.e., 10^{8}) of the speed of light. Since the speed of light is 299,792,458 m/s, one hundredmillionth of that is just a touch less than 3 m/s, or about 10.8 km/h: in ordinary terms, not much faster than a jogging pace. At that speed it will take, of course, a billion seconds — almost 32 years — for clock C2' to reach the rear end of the spaceship, which as we said is a billion lightseconds away (as measured in the spaceship's frame). And let's suppose that once clock C2' gets to its destination, it comes to a stop relative to the spaceship. So now let's see whether clock C2' could have lost any time at all relative to clock C1' during this long journey, due to socalled Relativistic time dilation. The formula for time dilation is: Dt_{o }is the "proper time" indicated by C1': that is, the time interval which would be measured by an observer at rest with respect to C1', and Dt is the "coordinate time": that is the time interval which would measured by an observer at rest with respect to C2'. So g here would be 1/[1  (10^{8})^{2}] ^{0.5} which is 1/(1  10^{16})^{0.5}. Now of course (1  10^{16}) is 0.999 999 999 999 999 900 000 exactly, and so (1  10^{16})^{0.5} is 0.999 999 999 999 999 950 000 (accurate to 21 decimal places), and thus 1/(1  10^{16})^{0.5} is 1.000 000 000 000 000 050 000. So the difference between this and 1.00 exactly is only about 5 parts in 10^{17}. So, since Dt_{o}= a billion seconds = 10^{9} seconds, Dt = 10^{9 }(1  5*10^{17}) = (10^{9}  5*10^{8}) seconds. In other words, an observer at rest with respect to clock C2' would see clock C1' lose time by an amount equal to about five hundredmillionths of a second — i.e., five hundredths of a microsecond. Since each of the clocks are only accurate to one part in a billion, and it takes one billion seconds for clock C2' to reach the rear end of the spaceship from the front end, suffering much less than one whole second of time loss in the process, this basically means that time dilation would not occur at all within the limits of accuracy of the clocks. It wouldn't even register on them, in fact, since their readouts are only capable of displaying a microsecond difference. So now we have two perfectly synchronised clocks C1' and C2' in the spaceship, one at either end. Because of their great accuracy, they could read at most only +/ 1 second different from one another. Since the "proper length" of the spaceship is 10 lightseconds, the distance between the two clocks C1' and C2' in the frame of the spaceship — i.e., in the primed frame — is 10 lightseconds. In other words, if we were to assume clock C1' to be at primed coordinate x_{1}' = 0.00, then clock C2' would be at primed coordinate x_{2}' = 10 lightseconds (i.e., 2,997,924,580 m.) Now since the velocity v between clock C and the spaceship is 0.75^{0.5} c = 0.86602540378444 c, g = 1/(1  0.75/1^{2})^{0.5} = 1/0.25^{0.5} = 1/0.5 = 2.00. And according to the equations quoted above, the time coordinate t of an event in the unprimed frame should be related to the time coordinate t' of the same event in the primed frame by the equation t = gt' + gvx' Consider the set of events in which clock C is just a few centimetres away from clock C1' while the spaceship is whizzing by clock C at velocity v = 0.75^{0.5} c = 0.86602540378444 c at an instant when clock C1' is reading t'. In this set we have the following three events:
Now let's calculate, using the Lorentz transformation equations, what the time coordinate of clock C should be, using the time coordinates of each of the clocks C1' and C2' in turn. Applying the equation t = gt' + gvx' to calculate the time coordinate of clock C, using the figures from [1.] and [3.] above, we get:
… while
applying the same
equation to calculate the time coordinate of clock
C, using the
figures from [2.] and [3.] above, we get:
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