Discussion: Two perpetual machine puzzles.

Answer: Each time a photon rebounds from one of the mirrors, it loses energy to the mirror. Its own loss of energy does not change its speed, for the vacuum speed of light is constant, but does change (lower) its frequency. (There's also a decrease of momentum of the photon, and an increase in the momentum of the mirror.) The mirrors can, therefore, only attain an energy equal to the initial energy of the photons (which was finite), and it will take forever (an infinite time) to attain that speed.

Relevant equations: The photon energy is E = hν = hc/λ. The photon momentum is h/λ, where c is the speed of light, h is Planck's constant, ν is the frequency associated with the photon, and λ is the wavelength associated with the photon.

The comment, "the beam never loses intensity" was intended to mislead, and was an outright lie. A beam reflected from a stationary mirror does not lose intensity. But if the mirror recoils, the mirror gains energy and momentum from the photons, and the photons lose energy and momentum.

2. This one may require some knowledge of field theory. Imagine two oppositely charged plates placed very close to each other. Some elementary analysis reveals that there is an electric field between the plates, and essentially no field just outside the plates. Drill a small hole in the center of the plates and drop in a charged particle. (An ordinary electron will do.) The particle will be accelerated by the electric field and pop out the hole in the other plate. Now set up a magnetic field outside the plates that causes the particle to move in a circle back to the first hole. It drops in as before, is accelerated again, and so on forever, eventually attaining arbitrary velocities.

Answer: This one has a lot of distractors. One might wonder about the fact that a charged particle moving in a circle must radiate, thereby losing kinetic energy and decreasing its orbital radius. Of course we could capture that radiated energy and consider it "output" of the machine. We might also suppose that that energy output won't be more than the energy gain in going between the charged plates, so there's still hope this might be an over-unity device. Vain hope, but before you get out your calculator to do the math, you'd better look again at the place where that energy gain was supposed to occur: the capacitor plates.

We misled you by noting that there's "essentially no field just outside the plates". While true, it's misleading. There's "fringing field" that extends all around the edges of the plates and the edges of the holes, and though it is weak in any small region (compared to the field strength between the plates) it extends over a very large region. The electric field is conservative, which means that a charged particle carried around any closed path will experience a net work of zero from the field. The fringing field is strongest near the holes and near the edges of the capacitor plates. The electron moves around the orbit and tries to re-enter the hole in the capacitor plate, where it encounters a fringing field that exerts a force to slow it down. It will reach its starting point with the same speed (zero) that it began with. No energy gain here.

Actually this answer did not require much knowledge of field theory.

All material in this musem is © 2002, 2003 by Donald E. Simanek, with the exception of text and materials indicated as from other sources.

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