Discussion: Two perpetual machine puzzles.
1. Two perfect mirrors float in empty space. A laser beam bounces
back and forth between them. Each time the beam reflects it imparts
a recoil momentum on one of the mirrors. Because the mirrors are
perfect the beam never loses intensity and so continues to bounce
back and forth forever, eventually accelerating the mirrors to
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 = hn = hc/l. The
photon momentum is h/l, where c is the
speed of light, h is Planck's constant, n
is the frequency associated with the
photon, and l is 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 distrators. 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 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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