nL;

Figure 6
owner of a perpetual engine! You
don't believe me? We1l, I'll prove it
mathematically.
Place two equally heated bodies
at points F, and Fr. Notice that any
ray lof light or heat) issuing from F,
must arrive atFzffig.6l. Aray propagated
along the same trajectory from
F, will arrive back at F, of course.
But look at a ray aimed frorn F, at
the vertical barrier. If there were no
partition, this ray would rebound
from the erased part of the ellipse
and hit Fr. But now it will be reflected
at some point R of the partition
in the direction it would have
had if it had propagated from F, to R
(fig. 6). Therefore, it will come back
to Fr! And, clearly,leaving point F,
along the same trajectory in the opposite
direction, we again come
back to Fr. So there's a considerable
number of rays that make "light
loops." In the cross section of our
"engine" shown in figure 5, these
are aLL the rays within the angle
AFrB and the symmetric angle
A',FzB'..
So, the body at F, will be heated,
while the body at F, willbe cooled.
We can even compute the eventual
temperatures of the bodies after the
process stabilizes.
Suppose the initial temperatures
were 7o K (or "kelvins," named after
the English physicist W. Thomson
mentioned above, on whom the
title of Lord Kelvin was con{erred for
his outstanding scientific achievements).
It's known that the rate of
energy loss by radiation (the luminosity)
is proportional to the fourth
power o{ the temperature (in
kelvins) of the radiating body.
Therefore, if the temperature at
Figure 7
point F, is 7r, and at Frit's Tr, then
the luminosity at F, and F, canbe
written as k( andkTf , respectively.
A11 the rays from F, after one reflection
arrive at F,b:ut only some of
the rays from F, arrive at Fr. How
many? Consider a small sphere centered
atFr. The rays that return to F,
cut a ring out of this sphere lfi4. Tl-
If its area is A, and the total area of
the sphere is A, than the amount of
radiation that comes back to F,
equals ArkTtlA, and the portion
that arrives at F, is (A - A)kftlA.
In the steady state, the energy radiated
from point F, equals the energy
that arrives at this point in the
same time interval. That is,
(o- +)U = krl.
\t/A
On the other hand, by the law of
conservation of energy, the loss of
energy atF, equals the gain of energy
atFz-thatis, C(7, -Tol: ClTo-721,
where C is the heat capacity of either
body,(we assume that the bodies
are absolutely alike). The last
equation gives 7, + Tr:2T0, so using
the previous relationship and
letting b = l(A - Ar) I A| la, andnoting
that b < 1, we finally get
T,
'
=T&.To, r+b
2T^
"= ri"o'
That's it! You can check the reasoning
from the very beginning and
be satisfied that it's perfectly correct.
So, did we really refute the second
law of thermodynamics? Unfortunately
(or fortunately), no. We've
made a mistake. But where? Think
about it yourself, and then compare
your answer with what's written
below. I have a feeling that not every
reader will be able to find the
correct answer.
You were on the wrong track if
you tried to find an error in the physics-for
instance, in the fact that we
ignored convection. Why? We can
cteate a vacuum inside the shell. But
what else is there in our little "engine"
except math and physics? The
error is hiding at the border, so to
speak, between these two sciencesin
the transition from the physical
process to its mathematical model.
Recall that we dealt with two
bodies placed at the foci. Thus, we
neglected the sizes of the bodies.
This is common in physics reasoning,
and the phrase "a body is 1ocated
at point M" doesn't cause
anyone to protest. In many situations,
this disregard of the sizes of
the bodies is justifiable. For instance
the motion of a body under given
forces applied to its center of mass
doesn't depend on its size and
shape. Or, as Newton showed, a
body consisting of a number of concentric
homogeneous spherical layers
attracts other bodies as the same
mass concentrated at the center of
this body. So to reptrace a body with
a point is indeed a customary operation
in physics.
But in our case this operation
leads to an error. Let's see whether
our reasoning remains valid if we
consider balls of nonzero radius centered
at the foci.
Consider three rays reflected from
the ellipse at the same point M in its
left half (fig. B): a ray issuing from the
foci F, and two rays, AM arrd BM,
with endpoints at a distancer fr.or;-i,Fr.
M
IB
Figure B
Y',
OUANTUllll/IIATUBE