A Caveat on Length Contraction

G.R.Dixon, 11/9/2004

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G.R.Dixon, 11/9/2004

Three particles, P1, P2, and P3, are initially at rest in inertial frame K. Their positions are:

, (1a)

, (1b)

. (1c)

At time t(init) = 0 they are all given a common, constant acceleration of .95c/sec, and at time t(final) they come to rest in inertial frame K’, which moves in the positive x-direction of K at speed .95c. Assuming their acceleration returns to zero at time t(final), we have

. (2)

The instantaneous K positions at time t(final) are:



, (3b)

. (3c)

Thus according to K, there is no change in the particle separations.

The K’ space-time coordinates at which the particles come to rest are:


, (4b)

, (4c)

, (4d)

, (4e)

. (4f)

The particles do not all come to rest at the same moment, according to K’, because (according to K’) P3 began to decelerate before P2 did, etc. Of course eventually all 3 particles come to rest in K’, and their separations (according to K’) are:

, (5a),

. (5b)

As a check (once the particles have reached a speed of .95c relative to K), the K separations are:

. (6)

The fact that the particles are still separated by 1 meter (according to K), after they have come to rest in K’, may at first seem to contradict length contraction. For we should certainly expect a rigid rod, originally at rest and with a length of L in K, to eventually have a length of g-1-1L if it accelerates until it comes to rest in K’. The difference between the independent particles and the rigid rod lies in the fact that the atoms comprising the rod are bound to one another. Viewed from K’, the right end of the rod is subjected to a decelerating fractional force first. But when this fractional force is applied, this end of the rod pulls on parts to the left. The details depend upon the elasticity of the interatomic bonds and other factors. Note that it would be simplistic to suppose that, viewed from K, the left end of the rod accelerates faster than the right end, so that the length at all times t (where t(init) < t < t(final)) is just (1/g)L. For if L is large enough, this would require that the left end of the rod achieve speeds greater than c!

For those who would like to consider the details of an accelerated, rigid object, it may prove helpful to bear in mind that the spring "constant" is in fact dependent upon a spring’s speed and orientation in an inertial frame. For example, Figs. 1a and b depict 2 charges held at rest relative to each other by identical springs. The charges move to the right at a common, relativistic speed. The repulsive Lorentz force between the charges differs for the two orientations (and from the common electrostatic force in the charges’ rest frame). And the springs’ lengths and amounts of stretch depend upon the orientations, owing to length contraction.

Figure 1a

Perpendicular (to v) Orientation

Figure 1b

Parallel (t v) Orientation

Suffice it to say here that if all parts of an extended entity are in fact subject to the same acceleration (as viewed from a given inertial frame), then the entity will not change its dimensions as its speed varies. For example, in a previous article an oscillating spherical shell of charge was discussed. The very stipulation that all parts of the sphere experienced the same acceleration, a(t), assured that the charge would remain spherical throughout an oscillation.