On this page we'll be discussing some aspects of blackbody radiation in
terms. No math! (Or very little.)
We'll be presenting a brief description of what is meant by a
"blackbody" and what's meant by "blackbody radiation", and we'll be
briefly discussing the first two laws of thermodynamics, once again in
completely qualitative form. And then, by applying the second law
of thermodynamics to the behavior of blackbodies, we'll draw some
rather surprising conclusions about the real world (or, at any rate,
we'll be looking at some conclusions which I
In particular, we'll be looking at the reason telescopes can't make a
scene look brighter, what you can and can't produce in the way of
one-way glass, and what kind of radiation is produced when a polarizing
filter is heated. But first we will need to talk a bit about
Throughout this page, when we say radiation
we mean electromagnetic
Unfortunately, this page is rather long, so here's a table of contents:
The First and Second
Laws of Thermodynamics
We'll now present a very brief and rather casual description of the
first two laws. The third law doesn't concern us here and we
won't discuss it.
The form in which we will present the Second Law treats energy, rather
than entropy. It means the same thing, either way. For most
of us energy is a more intuitive concept than entropy, which is one
reason why we prefer to state the law in terms of energy. Perhaps more to
the point, in our discussion of blackbodies, we'll be concerned the
flow of heat, which is a form of energy.
- You can't win.
A "closed system" (one which can't obtain anything from the
outside world) contains a fixed amount of energy, and you can't create
more. You can change the existing stuff from one form to
another but at the end of the day, all you've got is what you started
In terms of heat, an isolated object can't simply get hot, all by
itself. If it warms up, the energy to warm it must come from
- You can't break even.
You can only actually make use of energy -- have
the energy "do work" -- by letting it "flow downhill"from
a place (or state) of high energy density to a place (or state) of low
density. Think of water in a pool on a mountain top. It
contains lots of potential energy, just sitting there. However,
want to get work out of it, you must let it flow downhill, operating a
turbine or mill wheel as it falls. But once it's fallen, you
can't just wish it back to the top -- you must pump it
back up, which takes more work ... and to do that, to 'do the
pushing it back up hill, you need to let some energy, someplace else,
"flow downhill"! No matter how you arrange things, once you've
energy to do work, the energy has moved to a lower energy density
place, and you can't use the same energy to do the same work over
again. In other words, you really can't "break even"; no matter
what you do, at the end of the day you have less usable energy
than you had to start with.
In terms of heat, what this means is heat flows from hot things to
cold things, never from cold things to hot things. To put it
another way, two things, isolated from everything else, and which
start at the same temperature, stay at the same temperature.
If you have two things at the same temperature (the inside of your
house and the outdoors, for example), and you want to move heat from
one to the other, you need to use energy to do it. It's
not going to happen on its own.
The third law is commonly stated as "you can't get out of the game" and
has to do with the impossibility of cooling something down to absolute
zero. However, it doesn't come into our discussion in any way so
I will say nothing further about it here.
Second Law Violations
and Perpetual Motion
Suppose we could violate the second law. Suppose we could so
arrange it that we could start with a collection of objects, all at the
same temperature, and seal them all inside a big box, which was also at
the same temperature as all the objects. Then suppose further
that, because of our clever arrangement of the objects within the box,
two of the objects would change temperature
-- object A (inside
our big outer box) would warm up while object B (also in our big box)
cooled off. Energy is conserved, but we have somehow contrived to
make heat "flow uphill" from a slightly cooler object to a slightly
Then we could run a heat pipe from each of objects A and B to a
Stirling motor and get mechanical energy out of the system -- we would
have a way to convert heat into mechanical energy, directly.
(Obviously, I'm assuming there's no
additional "energy source" inside
the box which is "doing work" to drive the temperature change -- you
didn't, for instance, sneak a battery and a Peltier module into the box
along with the other objects...) And, if the box is at "room
temperature", so that the heat we change to mechanical energy is
replenished from its surroundings, we could keep getting useful
(mechanical) energy out forever (or until the motor wore out), without
ever using any "fuel" of any sort.
That is "perpetual motion".
And that is exactly what the second law of thermodynamics says you cannot
Definition and Model of a
An "ideal blackbody" absorbs all energy which falls on it. The
thing coming away from a blackbody is energy it radiates. That
radiation depends only on its temperature; once you know its
temperature you know exactly what radiation it's producing (as we shall see later on this page
It might seem like "ideal blackbodies" would be few and far between,
but as we'll see, it's easy to construct something which comes quite
close to the ideal.
An Ideal Black Surface
An ideal black surface is one which absorbs all radiation which falls
on it. It may re-radiate it afterwards, but initially, any light
or other electromagnetic radiation which hits it is absorbed.
An Ideal Black Box in an Evacuated
Oven: Our Model Blackbody
|Figure 2.2a: "Black Box" with hole
Imagine an evacuated chamber, in which all the walls (and all parts of
every wall) of the chamber are at the same temperature, T
The walls are made of ideal black material, as described in the previous section
. That's our "oven".
Now, let's place a box inside the oven. The box is also made of
ideal black material. (We've drawn the box in gray
in figure 2.2a
just so you can see its shape. We'll talk about that hole in the
box in just a couple paragraphs.) To start with, all parts of the
box are the
same temperature as the walls of the chamber in which it's placed.
What can we conclude? Well, according to the second law, the box
the same temperature as the oven. That means
the box must be radiating
exactly as much energy as it's receiving
Since the box and oven are made of the same material, and they're the
same temperature, they must be radiating with identical intensities --
i.e., the total radiation per unit area is the same for box and oven
walls. Since no part
of the box is changing in
temperature, the radiation intensity it's receiving must also be
completely uniform across its entire surface, and equal to the rate at
which the chamber walls are radiating.
This is already a rather remarkable conclusion, because it doesn't
depend on the shape of the box or the oven
That also implies that the radiation within the oven is "isotropic" --
it's the same in all directions. If it weren't, we could
construct a weirdly shaped box which would take advantage of the
anisotropy to get hotter or cooler than the oven walls. Of
course, the same conclusion applies to the contained box as well, since
it's just like the oven: it has uniformly hot walls made of ideal
black material, so the radiation intensity within
the box must
also be completely uniform and isotropic. This implies something
about the walls of the box, and in turn it implies something about
anything which is black: Any black object, heated until it glows,
must radiate in all directions, uniformly. Otherwise, the
uniformity and anisotropy of the radiation intensity within the box
couldn't be guaranteed.
|Figure 2.2.b: Black Box with Hole, in Oven
Next, let's put a tiny hole
in one side of the inner box.
(We assume the walls of the box are relatively
thin.) We've shown 2-dimensional view of the box with a hole, in the oven, in figure 2.2b
. The radiation going in through the hole has the same
intensity as the radiation falling on all outer surfaces of the box, of
course. Consequently, the radiation coming out
the hole must have exactly the same intensity as the radiation produced
by all surfaces of the box. For, if it were stronger, the box
would start cooling off, as more heat came out than went in;
conversely, if it were weaker, the box would heat up, as more energy
went in through the hole than came out. Either way it would
violate the second law.
Finally, note that all radiation going into the hole is absorbed;
everything coming out must have been radiated by the interior walls of
the box. So, the radiation from the hole must be independent
of the surroundings
; it depends only on the temperature of (the interior of) the
box. The hole
is behaving as an "ideal
This box with a hole in one side is our "model blackbody".
The radiation coming out of the hole is what we mean by "blackbody
An Ordinary Black Box
in an Evacuated Oven: Also a Blackbody
Now let's replace the ideal box of the previous section with one made
of ordinary materials; this box is also at the same temperature as the
oven walls. What can we say about it?
To start with, the total energy flux
coming from the exterior
surface of the box must be equal to
the energy flux striking the exterior surface of the box.
Otherwise, if it were smaller, the box would warm up; if it were
larger, the box would cool off. However, any real material is not
fully absorptive at all frequencies; it's semitransparent and
semireflective. The total energy flux is consequently due to the sum
of radiation, reflection, and transmission of radiation from inside the
box. The sum must be the same as that striking the box, but in
general we can't say anything about the breakdown of how much each of
the three "sources" contributes; that depends on the material used.
Not only must the total
energy leaving the walls of the box be
the same as the total
energy impinging on them, but the distribution across the box surface
must be the same: the energy flux away from the walls is uniform
across the box, as we shall now show. To see why the flux must be
uniform, imagine that we "shrink wrap" the box in an ideal black
wrapper -- or, rather, we surround it with such a wrapper, such that
there is a tiny gap
between the "wrapper" and the box.
The wrapper is at the same temperature as everything else in the oven,
of course. The wrapper, being made of the same ideal black
material as the oven walls, is producing radiation of the same
intensity as that shining throughout the oven, so its presence doesn't
change the radiation falling on the surface of the box, which must
still be entirely uniform. But suppose the energy leaving the
surface of the box isn't totally uniform. Everyplace where it's
stronger, the wrapper will get warmer
, and where it's weaker
the wrapper will get cooler
-- and that can't happen, by the second law. So, the intensity of
the radiation leaving the box surface must be uniform -- and this is
true no matter what the box is made of, and no matter whether the
material of the box walls is itself uniform or a conglomerate.
Now consider the energy actually being radiated
by the walls of
the box. Let's assume the material of which the box is made is
uniform. Then, since the temperature of the box is uniform, the
radiation intensity from the box must be uniform, too.
Furthermore, since the radiation intensity impinging on the box is
uniform, if the material is uniform, the amount of reflected
radiation must be uniform across the box as well. And at last, by
simple arithmetic, the transmitted
radiation (which is the radiation "leaking out" through the walls of
the box) must be uniform too. So, the contributions from the
three "sources" of radiation from the box are uniform across the
surface of the box.
Let's put a tiny hole in the box, just as we did with our "ideal black box
". (figure 2.3a
, below.) Once again, the
radiation going in
through the hole will be the same as the radiation striking an
equivalent area of the side of the box. If the box is neither to
warm up nor cool down, as must be the case by the second law, then it
can't gain or lose heat, and the amount of energy coming out
the hole must equal the amount going in
Consequently, the radiation intensity coming out of the hole is equal
to the radiation intensity in the oven. This is identical to what
we saw with our "ideal black box": The hole
in the side
of the box acts (almost) as an "ideal" blackbody
despite the fact that the box itself is made of ordinary materials.
Since the placement of the hole was arbitrary, we can conclude that the
radiation intensity within the box must be perfectly uniform, no
matter what material the box is made from
Finally, we should say a few words about reflections and transparency,
and why we said "(almost)" up above. What we've said about the
behavior of the hole in the side of the box is true enough, as long as
the box is located within an oven which is as the same temperature as
the box. However, to the extent that the box is semitransparent
(as normal materials all are, at some frequencies), some of the
radiation from the hole is actually radiation which entered through the
(semitransparent) back wall of the box and came on through and out the
hole. When we remove the box from the oven, that part of the
spectrum will no longer be present. So, the box will behave as an
ideal blackbody only to the extent that it's opaque.
|Figure 2.3a: Opaque non-ideal box.
The hole still acts like a blackbody!
Reflections, on the other hand, are far less of a problem. For
most reasonably shaped boxes, unless the box is filled with corner
reflectors any radiation entering through a tiny
hole will need
to be reflected from the walls many times before it finally finds its
way back out through the hole. If the material is even moderately
absorbent, such that the probability of a photon being absorbed is
substantial each time it bounces from a wall, then nearly all entering
radiation will be absorbed before it can find its way back to the hole,
and we can treat the hole as being "fully absorptive" (save at
frequencies at which the box itself is transparent). Consequently, the hole in the box will appear black
-- far blacker than the surface of the box, if the box is made of any
ordinary material. And, since radiation going into the hole is
absorbed, the radiation coming out of the hole depends only on the
temperature of the inner surface of the box. Thus, whether the
box is in the oven or not, to the extent that it is opaque, the hole in
the box will act as a nearly ideal blackbody.
So, to construct a physical model of a blackbody, we could just build a
small box -- say, the size of a shoebox -- out of reasonably heavy
gauge aluminum. That's opaque at all common frequencies. We
then drill a small hole in the side of the box; a hole 1/16" in
diameter will probably do nicely. And that's all we need to
do. The box is shiny on the inside -- but that's OK! The
inner surface is not perfectly reflective; a fraction of the light
which hits it is absorbed. Because the hole is so small relative
to the size of the box, a photon entering the hole will be reflected
from the walls many times before it finds its way back to the hole --
and one one of those reflections, it's likely to be absorbed rather
than reflected. And so the hole will appear black
despite the fact that the box is shiny inside. Radio waves won't
fit through the hole, and gamma rays may go right through the sides of
the box, but most radiation in between will enter through the hole and
bounce around until it's absorbed.
If you want to find a premade
blackbody, look for an old
camera. With the lens stopped down and the shutter open (on 'B'
or 'T') the pupil in the lens is a pretty good approximation to a blackbody.
What does a Blackbody Look Like?
This question may sound absurd until we explain it a little.
Suppose we took a box with a small hole in the side, and heated it and
all its contents red-hot or orange-hot, so that it was glowing with
visible light. What would we see when we looked in through the
hole? We could try to deduce it, but we don't have to.
Here's a very striking description of what's seen in a real-world
example of such a thing, sent to a science discussion group by Horace Heffner
cavity and its contents heat up, everything in the cavity eventually
disappears from view through the peep-hole. I have personally sat
and watched through a gas forge observation port, which I kept open,
the cover lifted, as that gas forge, which was about 1' by 2' by 2',
heated up. Initially, I could clearly see the far walls of the forge
and things in it through the port. When the temperature rose to an
orange glow, suddenly nothing was visible inside the forge. There
was a pure orange glow coming from the observation port that had
nothing to do with the contents of the forge. One moment I could see
the other side of the forge, which had some hot spots and dark spots
on it, and the next it was replaced by flat orange glow. I could see
nothing at all inside the port. It was as if the hole surface itself
(which is not a physical thing) was radiating."
(Complete message from which this was excerpted may be viewed here
Testing a Material
We'll want to "test" a number of materials as we go along, to see how
they behave. To "test" them we'll allow radiation to impinge on
both sides, and see what can or can't happen, according to the second
Place a "Liner" in
|Figure 3.1a: A box with a "liner"
To start with, we'll place a liner
in the box. This is really a second box, just like the first box,
but just a little bit smaller. It starts out the same temperature
as the outer box, and it also has a hole in it, which exactly coincides
with the hole in the outer box (see figure 3.1a
, to right). So, the energy
coming out the hole in the box(es) is now all coming from the inner
box -- the "liner" -- rather than the outer box.
What can we say about the liner? By the second law, it must remain
the same temperature as the outer box. That means that the
radiation coming from the surface of the inner box must match the
radiation impinging on it, which is, of course, the radiation coming
from the inner surface of the outer box. And, as we already know,
the radiation intensity coming from the hole in the liner will be the
same as the radiation intensity from the hole in the box before we
added the liner.
3.1b: Box, liner, and insulator
Now one interesting thing we can do is place another layer
between the box and the liner. Call it an "insulating
layer (figure 3.1b
). If the liner is to remain at its original temperature, the
"insulating" layer must allow the same amount of radiation to fall on
the liner as was hitting it before we added the insulator. If the
is to remain at its original temperature, the
"insulator" must also allow the same level of radiation to reach the
inner surface of the box as was reaching it before. So, we
could say that, by the second law, our "insulator" must not
insulate at all
, no matter what it's made of.
"Test Sandwich": Parallel Plates and a Test Sheet Between
|Figure 3.2a: "Insulator sandwich"
now have a black box, with a liner of the same material, and an
"insulating layer" between the box and liner. We know that the
"insulator" must not impede the radiation flow between the inner surface of
the box and the liner, or one of the three layers will change temperature, which is
not allowed. So, at this point, we can dispense
boxes, and just look at a sandwich of three layers (figure 3.2a
We have two layers which we can assume are ideal black material, and a
third layer, in the middle, whose properties we're trying to
determine. All three layers are the same temperature. By
the second law, the radiation striking the two sides of the middle
layer must exactly match the radiation leaving the two sides of that
"ideal mirror" reflects all the radiation which impinges on it.
What can we say about an ideal mirror, when we place it in our test
Since all radiation is reflected, the reflected radiation equals the
incident radiation, and the mirror obviously absorbs nothing -- and it
also must radiate nothing
. For, if it radiated anything,
the sheets on either side of it would warm up
, as they'd be
receiving more radiation than they were radiating.
So, an ideal mirror does not radiate. In fact, the temperature
of an ideal mirror is irrelevant (as long as it's in an evacuated
Ideal Dichroic Mirrors
Dichroic mirrors are very cool gadgets, and what is more, they are not
just idealized things made of unobtanium; they're easily available, in
old enlargers, in some fancy ribbon for wrapping packages, and in
birds' wing feathers, to name three places you find them. A
dichroic mirror is semitransparent in a very special way: it
reflects only certain colors, and transmits the rest. It's called
because it has two colors: Its color by reflected
light is different from its color by transmitted light. So, for
instance, a dichroic mirror which reflects yellow light will look
yellow when it hangs on the wall, but blue if you hold it up to a
window and let daylight shine through it, because it reflects yellow
and transmits blue.
dichroic mirror is one which reflects 100% of exactly
the frequencies we want it to reflect and transmits 100% of all the
rest, without regard for what we can actually physically build using
All Blackbodies of
the Same Temperature have Identical Spectra
Of course, there is a formula which predicts the radiation spectrum of
a blackbody given its temperature, which certainly shows that they all
must have identical spectra. But this particular fact can be
shown far more easily than by deriving the formula for it.
Put two of our "model blackbodies
" (at the same temperature) back to
back, so that their holes meet. Call them A and B. Then all
the radiation flowing out of A goes into B, and all the radiation
flowing out of B goes into A.
Certainly both "blackbodies" must be emitting the same total amount
of radiation -- otherwise the one emitting more would cool off, and the
one emitting less would warm up. But how do we know their spectra
|Figure 4.0a: Splitting bright and dim with a mirror
Suppose "A" is emitting more red light than "B", and "B" is emitting
more green light than "A", but the total amount of energy they're each
emitting is equal. All seems well. But now, let's take a dichroic
which reflects all red light, and transmits all other colors
(including green). Place our dichroic mirror between
blackbodies (figure 4.0a
). Now, all the red light A emits is reflected right
back into A, along with
all the other colors -- including green
-- which are being emitted by B. Now A is receiving the extra
red light (reflected back, from itself), and the extra bright
light (transmitted from B), while B is receiving its own dim
red light and
green light -- and hey presto, we realize that A is now receiving more
total radiation than B, despite the fact that they are both emitting
the same amount. So, in short order, A will be getting warmer
The second law says that can't happen. So, the assumption that A
is emitting more strongly at one color than B must have been false -- A
and B must have identical spectra. Since they were arbitrary
blackbodies, the conclusion is that all
blackbodies of the same
temperature have identical spectra.
A very similar argument, in which we use a polarizing filter in place
of a dichroic mirror, shows that all blackbodies must emit nonpolarized
light, as well.
A Semisilvered Mirror
must be Symmetric
|Figure 5.0a: Semisilvered Mirror Sandwich
For this case, we'll use our test sandwich
which we described above
. We'll use a semisilvered mirror
for the middle layer.
Suppose the mirror reflects k%
of the radiation striking it
from the left
(while transmitting the remaining 100-k
and reflects m
% of the light striking it from the right
Must we have k
, and here is why (see figure 5.0a
Call the total light striking the mirror on the left 100%. Then
light leaving the mirror and going (back) to the
left is k
%. The transmitted
through the mirror from the right
and going away from the
mirror to the left, on the other hand, is (100-m)
%, since m
of the light coming from the right is reflected back to the right.
But the amount of light going away from the mirror to the left must
since the mirror is the middle layer in the sandwich, and that's true
no matter what material we use for that layer. So, (100-m
) + k
must be 100%
which can only be true if m
In short, if the mirror reflects more light on one side than the other,
the "brighter" side ends up with more total energy, and objects placed
on that side of the mirror will warm up, while objects on the "dimmer"
side of the mirror will cool off. The second law says that can't happen.
equal External Reflections
Consider a sheet of ordinary glass. When light enters it, not all
the light goes in -- some is reflected. Call that fraction k
%. Again, when the light exits the other side, some light is reflected inside
the glass, and doesn't get out; call that fraction m
Let's just consider light which strikes the glass perpendicularly (at a
90 degree angle), so we don't have to worry about the effects of
refraction. Can we say anything about the values of k
We can, if we can eliminate one surface from consideration -- and
that's something we can do. By careful choice of coatings on one
surface of the glass, we can (nearly) eliminate reflections, both
internal and external, from that surface. This is done
commercially with multicoated camera lenses; they reflect far less
light than ordinary (uncoated) glass. While real coatings only
affect certain frequencies, we'll assume for the moment we can have an ideal anti-reflective coating
which works on all frequencies at once. (In fact we only really
need to consider one frequency at a time for the following argument to
work, but it simplifies things to assume we can do them all at once.)
Now, what can we say about our single
remaining reflective surface? We can say ... we've already
analyzed it, in the previous section
! Because, as a single partially reflective surface
, it's identical in its action to a semi-silvered mirror. So, as we showed for such a mirror, we must have k
, and the strength of an internal
reflection must be exactly equal to the strength of an equivalent external
(Note, however, that as the incidence angle changes from perpendicular,
the behavior of internal (and external) reflections from glass becomes
more complex and this simple analysis is no longer adequate.)
|Figure 6.0a: Transmitting, reflecting, absorbing
Most real transparent materials actually absorb or reflect part of the
radiation passing through them. So, let's suppose we have a sheet
of material which reflects k
% of the radiation falling on it,
% of the radiation, and allows the rest to pass
through. We'll suppose further that it does so symmetrically
the same regardless of which side the radiation enters. What can
we say about it?
If the radiation impinging on it on each side totals 100%, then the
radiation leaving the sheet on each side must also total 100%.
Consider the radiation leaving on the left
The total of the reflected and transmitted parts is (k
which is equal to (100-m
)%, so for the grand total to reach
100%, the sheet must be radiating
an amount equal to m
But that is the same as the fraction of the light which the sheet absorbs
So, we can conclude that, according to the second law, a material which
% of the radiation falling on it must radiate m
as strongly as an ideal black material, or as a blackbody of the same
temperature. Whether the remaining 100-m
of the radiation
is transmitted or reflected makes no difference to the strength with
which the material radiates. In practical terms, this means that
if you have old steam radiators in your house, painting them white or
"aluminum" colored is counter productive, in that it cuts down the heat
they radiate: you should paint them black! (Of course the your
radiators aren't radiating visible light -- they're not nearly hot
enough for that! -- but they're radiating infrared, and the common
white pigment titanium dioxide also appears light in infrared light,
and carbon black, which is commonly used for black paint, is dark in
Suppose we have an ideal yellow filter. It absorbs all blue light
which falls on it, and transmits the rest. If it's heated until
it glows, what can we say about the color of the light it will radiate?
Place the filter in the center of a "sandwich" between sheets of ideal
black material, with all three at the same temperature. The light
coming from the filter will be made up of light passing through the
filter, which is yellow, plus light radiated by the filter; the total
of the two must be 100%. But we still can't say what color the
radiated light must be.
Now, let's add a dichroic mirror
to the sandwich, just to the right of the filter. Let's assume the mirror reflects all blue
light, and passes everything else through. Now, what's coming off
the right side of that mirror? There's the yellow light which was
radiated by the other black wall, and which passed through both the
filter and the mirror; and there's the blue light which was radiated by
wall and which was reflected by the mirror.
Everything's accounted for; we've got 100% intensity coming from the
mirror. But... what about the light radiated by the filter?
of it gets through the mirror, the intensity coming from
the mirror will be more than 100%, and that can't happen -- so, all the
light radiated by the filter must be reflected by the mirror.
Since the mirror only reflects blue light, the light radiated by the
filter must, therefore, be blue.
(Of course, we've ignored the "colors" outside the visible spectrum in this discussion.)
Of course, this argument didn't depend on the choice of
colors. From it, we can see that in general, the light radiated
by a hot filter must be the color the filter absorbs
A red filter will radiate cyan light, a blue filter will radiate yellow
light, a green filter will radiate magenta light, and so forth.
|Figure 6.2a -- Glass in
the oven at Corning
So-called lead crystal glass is extremely transparent. I don't
know the exact amount of light it absorbs, but even in thick sections,
it's certainly very little -- nearly all the light passes
through. Consequently, as we've just seen, it must radiate
far less than an ideal blackbody, at least in the visible spectrum.
However, when glass is heated very hot in an oven, it certainly does radiate, and does so quite visibly.
This isn't something one sees every day, but at Corning, New York, one
can watch Steuben glass animals being worked, and the glass is heated red-hot
to orange-hot when it's being formed -- and it glows orange. My
recollection from a trip to Corning about 25 years ago was that the
glass was actually glowing very brightly. What could we conclude
To the extent that the glass is glowing, it must not be transparent.
Though the glow of hot glass may be dimmer than the glow of equally hot
opaque material (such as steel) I none the less recalled it as being
quite bright -- far
brighter than one would expect if it were radiating only a tiny
fraction as much light as hot metal. If my memory was correct, we
would be forced to conclude that, according to the second law, glass,
when heated very hot, must turn opaque
, or at least it must
|Figure 6.2b -- Hot glass
The first question to ask, though, is just how bright that glow really
is. My memory being what it is, it seemed like a trip to the
closet was called for, where I pulled out my cartons of 35mm slides and dug around, looking
for the shots from the Corning trip. Rather to my surprise I actually found
the slides I was looking for, after only an hour or so of searching,
and I scanned the relevant pictures. We see the glass being
heated in the furnace (figure 6.2a
, above right), and then being rolled
out on a slick steel surface (right). The "double image" is a
reflection; the surface it's being rolled on is so smooth and polished
that it's acting as a mirror.
|Figure 6.2c -- Hot glass, blowup
A blowup of the glass knob on the end of the rod is shown in figure 6.2c
(again, the "double image" is a reflection from the mirrored surface on
which the glass is being shaped). Note that the glass
in direct contact with the environment of the oven, while the metal rod
was "protected" by the glass; at the same time, metal conducts heat
well, so the end of the rod is no doubt being cooled rather rapidly by
heat leaking down the rod. Consequently, we can probably assume
that the main bulk of the solid lump of glass is hotter
the end of the metal rod. The light wasn't great, and the blowup
is substantial, so the image isn't terribly sharp. None the less,
we can see pretty clearly that the end of the metal rod, which is
clearly visible through the transparent glass, is glowing far brighter
than the glass. The rod is glowing bright red; the glass is
glowing a relatively dim orange. In fact, if we look carefully,
we see that the part of the glass immediately next to the end of the
rod actually looks dark
; that part of the glass may be the
coolest part, though it's certainly at least as hot as the rod
end. The visible orange glow seems to be limited to the portion a
finger's width away from the end of the rod, where we might expect the
glass to be substantially hotter than the rod.
So, what can we say? The glass is apparently transparent, or
nearly so; it is glowing, but apparently quite a bit less strongly than
the metal rod embedded in it. The second law predicts that
transparent glass should glow less brightly than the opaque rod, and
that to the extent that the glass glows, it must not be fully
transparent. While these photos don't let us conclude anything
quantitative about either the brightness or transparency of the glass,
they seem to be consistent with what was predicted. Unfortunately
we also can't tell from these pictures whether the glass has become any
less transparent when heated to the point of glowing. This is an
unsatisfactory conclusion, and I may add to this section if I can
obtain some more quantitative results.
-- Shadow of candle, cast by flashlight
Candle flames glow yellow-orange. Must they, like yellow-orange glowing metal, be opaque?
Well, I hope not, because when I checked the shadow cast by a candle,
the flame cast no shadow -- only the wick was silhouetted (figure 6.3a,
to right). We can see a very faint shadow from the upper part of
the flame, but while that could indicate semitransparency of that
portion of the flame, it could also very well just be due to refraction
of the light by the hot and consequently less dense body of the
flame. In either case, the flame is not even close to being opaque
and the part of the flame near the bottom, which is supposedly hottest,
is apparently completely transparent.
Let's look at the colors in a candle flame (figure 6.3b
right). It's blue at the base,
orange a little higher up, and yellow in the middle. We might
suppose that the radiance of the flame is caused by its heat
that is to say, the gas in the flame is just radiating because it's
hot, like the glass in the previous section. What's more, if we
up candle flames, we'll find a fair amount of material which implies
exactly that, and which goes on to explain that the hottest part of the
flame can be recognized by its blue
color -- presumably because
it's "blue-hot" there, but only "yellow-hot" in the cooler part up
above. (But if this were correct, then we'd also expect the flame
to be opaque, and it's not.)
-- Candle Flame, showing colors of the flame
This simple interpretation of the flame color is clearly false,
however, as it leads directly to false conclusions about the flame
temperature. The temperature of a
candle flame, according to numerous references, is less than 2000 K
throughout. When one looks up color temperatures, on the other
hand, one finds a simple correlation between temperature of a radiating
body and its color; for example, see this
. The temperature of a body radiating blue light ("blue
as the base of the candle flame does, is about 9000
Kelvin. Forget the references on candle flame temperature for a
moment, and just consider what a flame at this temperature could do --
at 9000K, water splits into oxygen and hydrogen, iron melts -- rather,
iron boils --
aluminum vaporizes ... the flame is obviously
not burning at 9000K!
In fact, if we go by color, even the "cooler" orange and yellow parts
of the flame should be at about
2200K to 3000K, which is still quite a bit hotter than any part of a
real candle flame.
So, what's going on? In short, the flame is not
due to its temperature: that's not "blackbody radiation" we're
seeing. It's radiating as a result of the
combustion which is taking place within the flame
wax doesn't burn as a solid or liquid. Rather, the solid wax
melts, and is drawn up along the wick by capillary action; the liquid
wax then evaporates from the surface of the wick, and then the
burns as a it mixes with air in the rising current of hot gas
above the candle. The light is given off as a direct result of
the chemical reactions which constitute combustion, which releases
substantial energy. Some of the released energy is kinetic:
The molecules which are the reaction products fly apart rapidly.
But some of the energy is also released as radiation
light, in this case. And the color of that radiation is
characteristic of the material being burned; it is not
related to the temperature of the flame. So, for instance,
methane burns with a blue flame, and the flame on a typical gas range
burning natural gas is consequently blue -- but that in no way implies
that a kitchen gas range has flames burning at 9000 K
, which is
what their color temperature appears to be.
Above, we considered symmetric semitransparent materials. Is it
possible to have an assymetric
material -- one which absorbs k
of the radiation striking it on one side, and m
% of radiation
striking it on the other side, with k
Let's consider what happens when we place a sheet of this hypothetical
stuff between two ideal black surfaces. Radiation striking each
side of our semiabsorbing sheet totals 100%. The sheet absorbs k
of the radiation striking it on the left, passing 100-k
It absorbs m
% of the radiation striking it on the right,
So that the radiation leaving each side of the sheet totals 100%, it
must radiate an amount equal to m
% of the incident radiation on
, and an amount equal to k
% of the incident
radiation on the right
. In other words, the radiance of
one side of the sheet is equal to the absorbency of the other
side of the sheet. This is an extremely weird material!
While the simple arguments given here don't show that this is
completely impossible, I seriously doubt such a material exists.
Let's consider an ideal linear polarizer, which is oriented
"vertically". When unpolarized light strikes it, half goes
through, and half is absorbed; the half which goes through is
vertically polarized. If horizontally polarized light strikes the
filter, it will be entirely absorbed; if vertically polarized light
strikes the filter, 100% of it will pass through.
Let's place our polarizing filter between two ideal black sheets, with
all three at the same temperature. The polaroid material is
absorbing 50% of the light which hits it, so, like the semitransparent
material discussed above
, it must also radiate 50% as strongly as the ideal black surfaces. So far, there is nothing surprising here.
But now, let's place a second
polarizer to the right of the
first one, oriented identically. Call the left hand polarizer
"A", and the right hand polarizer "B". Radiation strikes the left
side of "A" with what we'll call 100%
intensity. Of that,
50% passes through, and 50% is absorbed; the 50% which passes through
is now vertically polarized. Polarizer "A" also radiates at 50%
the rate of an ideal black sheet. The sum of what "A" radiates,
with the radiation which passed through, is again 100%
, and that's what strikes polarizer "B".
The radiation which already passed through "A" is vertically polarized, so it will all
pass through "B". So, the radiation "passed through" both polarizers which emerges from "B" is at 50%
intensity. "B" is also radiating
at 50% intensity, just like "A". So, the sum of the radiation
which passed through both polarizers (and emerged from "B"), plus
the radiation produced by "B", is 100%.
But there's also radiation produced by "A" striking "B" -- what about that? It totals 50% intensity. If any
of it passes through "B", then the intensity leaving "B" will be more
than 100% -- and we know that can't happen. So, all the radiation
from "A" must be absorbed
by "B". Consequently, we see that the radiation produced by "A" must be horizontally polarized
So, we see that a polarizer must radiate exactly the kind of light it absorbs
-- the polarity of its radiation is the opposite
of the polarity of light which it allows through.
An identical argument works for circular polarizers, and leads us to
conclude that a circular polarizer's radiation must be circularly
polarized, but in the direction opposite to that of the radiation which
would pass through the polarizer.
In Arthur Clarke's Childhood's End
, an alien "ambassador" to
Earth is concealed behind a one-way mirror; the aliens will not allow
Earthlings to see them. The protagonist finally learns what the
ambassador looks like, however, by pressing a powerful flashlight
against the "one-way" mirror, and thus illuminating the chamber behind
it, which he can then see by looking through the mirror. Clarke
then explains that there is no such thing
as a "true" one-way mirror; all a one-way glass really is, is a semi-silvered mirror with a darkened room on one side.
From time to time over the years, I've thought of this, and wondered if
it was really correct. Can we do better than Clarke's
As we saw above
, a semi-silvered mirror is necessarily symmetric
-- if that's all that's used, then you can indeed see through it
equally easily from either side, and the only thing which makes it "one
way" is that the "observer" sits in a dark room, while the "subject" is
in a brightly lit chamber. And in that sense Clarke was indeed
correct. However, though we can't find a true "one-way" glass, we
can do a little better than a simple semi-silvered mirror, as we shall
First, though, I should point out that it appears to be impossible to
make the image of the "observer" which is seen by the "subject" any
dimmer than the image of the "subject" as seen by the "observer".
That is to say, if the glass transmits k
% of the light going one way, it's going to transmit k
% going the other way, too, and there's nothing we can do about it. I don't have an airtight proof (cf. Non-Symmetric Semi-Transparent Materials
, above), but I'm reasonably certain this is correct.
However, the thing which makes a one-way mirror hard to see through the "wrong way" is the reflection
of the "subject" which appears in it -- it's like hash marks on an envelope. And it's the reflection of the observer
which in turn makes it necessary for the observer's room to be
darkened; the darkness not only conceals the observer by dimming the
image seen by the subject, but with the room darkened, the reflection
on the observer's side is eliminated, so the observer can see the
It turns out that by clever mirror design, we can eliminate the
reflection of the observer without affecting the reflection of the
subject, and without the need to darken the room.
In the following two sections, we'll assume that the "subject" and
"observer" are equally brightly lit, and we'll see how much we can
improve the "seeing" of the observer relative to the "seeing" of the
Using a Simple Filter
The first thing we can do is place a simple filter on the observer's
side of the semisilvered mirror. If the mirror transmits a
of the light (where we've switched to using fractions of 1 rather than percentages), and the filter transmits a fraction m
of the light, then the brightness of the image of objects on the other side of the glass will be km
, for both subject and observer.
However, the brightness of the reflection
seen by the subject
will be (1-k
), while the brightness of the reflection
seen by the observer
will be m2
). The reason for the difference is that reflected light seen by the observer must pass through the filter twice
, while the transmitted light is filtered just once
, and the reflection seen by the subject is not filtered at all.
Suppose we use a mirror which transmits 1/2 of the light, while
reflecting 1/2; suppose further that our filter transmits 1/4 of the
light which strikes it, while absorbing 3/4. Then, overall, the subject
will see a reflected image of himself which has brightness 0.5, while
he'll see a transmitted image of the observer which has brightness
0.5·0.25 = 0.125. For the subject, the reflected image will be four times brighter
than the transmitted image. Even if the observer and subject are
equally brightly lit, the subject's view of the observer will be
largely obscured by the reflection.
On the observer's side, the brightness of the transmitted image of the
subject will also be 0.125, which is rather dim but still bright enough
to see (a brightness ratio of 1/8 is about like standing in bright sun
and looking at something in the shade). But the brightness of the
image will be 0.25·0.25·0.5 = 1/32 = 0.03125, which is very dim. So, for the observer, the transmitted
image of the subject will be four times brighter
than the reflected
image she sees of herself. Consequently the subject should be easily seen in spite of the (very dim) reflection.
But we can do better than this.
Using a Circular Polarizer
To see why this is going to be useful, we need a fact about mirrors which I can't prove from thermodynamics: A mirror reverses
circular polarization. Thus, if right-hand circularly polarized light strikes a mirror, it's reflected as left
hand circularly polarized light. To see why, think about the electric field vector of the light. It's rotating
, to the right -- that is to say, if we face the way the ray is going, the electric field vector is rotating clockwise
At the surface of a mirror, the electric field vector is reversed (head
and tail are swapped) but its orientation is unchanged. So, the
reflected ray will also have its electric field vector rotating to the
right. But the ray is going the other way
. So, if we face the way the ray is going -- which is out
of the mirror -- we'll see the electric field vector rotating counter clockwise
. So, the reflected ray is indeed left
hand polarized. A quicker way of saying this is to note that a
right-hand circularly polarized ray is like a screw with a right-hand
thread, and when we reflect a right-hand thread screw, we get a
left-hand thread screw.
What this means is that if we put a circular polarizer in front of a
mirror, the reflection of someone looking through the polarizer will be
completely absorbed -- it will appear black, because the light which
passed through the polarizer was reversed by the mirror, and won't go
back through the polarizer. However, if we use a semisilvered
mirror, the image from the other side will still be perfectly visible.
For example, if we put the circular polarizer on the observer's side,
and we use a semisilvered mirror which transmits 1/4 of the light, then
will see a reflection
of 0.75 intensity, and a transmitted
image of the observer with intensity 0.5·0.25 = 0.125. The
reflection will be 6 times brighter than the transmitted image of the
observer, and the observer will be difficult to see.
On the observer's side, the transmitted image will also be seen with
intensity = 1/8 = 0.125. However, the reflection will be missing
entirely, absorbed by the circular polarizer. Consequently the
subject will be clearly visible.
You might be wondering what this whole section is doing here, as it
doesn't seem to have anything to do with blackbodies. As it
happens, it was in attempting to understand the "one-way glass"
described in this section in terms of blackbody radiation that I first
realized that a glowing circular polarizer must radiate light which is
polarized the opposite way
from the light that it will pass. However, the analysis of polarizers I gave above
is simpler than the one I first stumbled across when thinking about
one-way glass, so I won't show the original diagrams here (it's
straightforward, involving nothing more than adding up percentages in
arrows, but there were a rather large number of arrows).
We showed, on our brightness of an image page
that you can't make a scene appear brighter just using combinations of
lenses. You can make it bigger, which is what binoculars,
telescopes, and microscopes all do, but you can't make it
brighter. On that page, we derived the result using a case by
case analysis of all combinations of positive and negative lenses.
In this section, we'll find that the result follows directly from the
properties of blackbodies. We'll start with two preliminary
You Can't Increase the Energy Flow through an Aperture
Of course we mean a uniformly lit
aperture; obviously in certain
cases, such as focusing an image of the Sun, you can increase the
energy flow in a particular spot. In that case, the "scene" is
not at all uniform: The bright sun is in a relatively dim sky.
But suppose, instead of that,
we had an optical system which would, given a uniformly bright
scene, increase the energy flow through an aperture beyond the flow due to the scene alone. Call this device a brightening scope
Let's take two blackbodies
, called "A"
and "B", and place them hole-to-hole; they're both within the same
oven, and everything's at the same temperature. The same amount
of energy flows out of each, and into each, and they each remain
at the same temperature -- as they must, by the second law. But
now, let's put our brightening scope into box "A", with the scope's
"output" placed at the hole in box A (figure 9.1b
, below). The inside of the box
itself provides the "uniformly lit scene", and so our brightening scope
must send more
energy out through the hole than was coming out
before we added it to the system. But that energy's going
straight into box "B" -- and box "B"'s output, which is not
increased, is all going into "A". So, now we have more energy
going from A to B than is coming back from B to A.
Consequently A must start to cool off, B must start to warm up -- and
that can't happen, by the second law. So, there is no such thing
as a "brightening scope".
9.1b: Two Black Boxes and a
The "brightening scope" brightens the
light coming out of box "A". As a result, more light flows from
"A" to "B" than from "B" to "A", and box "B" will get warmer. The
second law forbids this.
You Can't Selectively Increase the Intensity in a Cone of Rays
Suppose we have a device which doesn't increase the amount of energy
going through an aperture (which we now know we can't do), but it concentrates
it. Faced with a uniformly lit scene, instead of emitting its
light uniformly over a hemisphere, it emits it selectively, in a
"bright cone" of rays coming straight out the middle. Let's call
this device a concentrator
Let's consider one "black box" blackbody
, in a
uniformly hot oven. Place a sheet of ideal black material a short
distance from the hole in the black box; the sheet of black stuff is at
the same temperature as the rest of the oven. Now the light
shining on the black sheet must be completely uniform, as we know from earlier on this page
That light comes partly from the oven walls, partly from the outside
walls of the black box which constitutes our "blackbody", and partly from
the hole in the black box. Now, let's put our concentrator inside
the box, with its "output" placed exactly where the hole in the box is,
so the stuff coming out of the box is now the output of the
concentrator. The inside walls of the box provide the
concentrator with its "uniformly list scene", and so the light coming
out of the concentrator, though the same average intensity as the light
which was coming out the hole before, has a "hot spot" in the
middle. So, there is now a cone of rays
coming out the
middle of the aperture in the black box which is more intense than it
was before, and the light emitted outside that cone is a little weaker
than it was before.
What happens to the sheet of black material in front of the hole?
Where the cone of concentrated rays strikes it, it's now receiving more radiation
than it was receiving previously -- and more than it's radiating.
Meanwhile, the region around that spot, which is outside the cone of
more intense rays, is receiving less light than it was previously --
and less than it's radiating. The consequence is that the sheet
will develop a hot spot
-- where the "cone" of more intense
radiation strikes it, it will warm up, so that spot's temperature is
higher than the rest of the oven. And the second law says that
can't happen. So, it must be impossible to build a "concentrator".
Why You Can't Brighten a Scene
Let's suppose we have a telescope which makes a scene look brighter
to our eye.
Let's point the scope at the clear blue sky. Since it makes the
scene brighter, when we look through it, the blue light which falls on
our retina, at least in the central area, must be brighter than it was
when we looked at the sky without the scope.
But that, in turn, means that the light coming in through the pupil of our eye must be brighter
than it was when we looked at the sky directly. Either it must be
brighter overall, or, at least, the light entering at certain angles
must be brighter. Otherwise, looking at the uniform sky
with the telescope would produce an image on our retina which was no
brighter than what we saw without the scope.
But we've just seen, in the previous two sections, that the telescope
can't increase the total amount of light coming in through the pupil of
the eye (a "brightening scope
"), and can't increase the brightness of light coming from any particular direction (a "concentrator
If it can't do either of those things, it certainly can't increase the
intensity of the image of the sky formed on your retina -- and so it
can't actually brighten a scene.
But You Can Brighten a Scene!
As we mentioned earlier, we have shown elsewhere, using direct
arguments regarding lens behavior, that a telescope can't brighten a
uniformly lit scene. However, on that page, we also found a way around it
: If we place a lens in the middle of the eyeball
we can arrange to have the scene appear brighter. Why does that
work -- what's special about that lens which allows it to brighten the
image of the scene?
The answer is that the (imagined) lens in the middle of the eye is not
forming an image from a uniformly lit scene
which is what all our arguments have dealt with. Like the example
of a "burning glass" which forms a painfully hot image of the Sun, the
lens in the center of the eye is forming an image from a small brightly lit object
against a dark background. The "small brightly lit object" is the
pupil of the eye! From the point of view of a lens placed
in the middle of an eyeball, the front of the eye is a dark scene with
one bright spot, which is the pupil.
If we imagine the entire interior surface of the eye glowing with light
of the same intensity as that coming in through the pupil, the story is
different; in that case, adding a lens in the middle of the eye would
not make a difference to the brightness of the light falling on the
Images of Stars
It's worth taking a little side excursion to note that images of stars
are "special". Stars are so far away that the disk which would be
formed on the retina by a perfectly focused image of a star is far, far
smaller than the "blur disk" which results from imperfections in the
atmosphere, the lens of our eye, the retina itself, and, of course, the
diffraction limit of the pupil. Consequently, increasing the
magnification of the image doesn't increase the size of the disk formed
for each individual star in the image -- the same imperfections are
still dominating the star's disk size, and they don't change just
because the image of the star has more light in it. The result is
that, when we look at the night sky with a telescope, the images of the
individual stars are
brighter. The average brightness of
the image of the night sky isn't increased, as the distances between
the stars in the image are increased, but the individual points of
light which represent individual stars do indeed appear brighter (and
easier to see).
Finally, let's briefly consider reflections. We can learn a fair
amount about possible angles of reflection from the second law.
However, I will just mention a couple of results here.
Just as we can't build a "concentrator
using lenses, we can't build one from mirrors; it would violate the
second law either way. This means that a surface which reflects a
uniform fraction of all light which strikes it, from any angle, must,
if illuminated uniformly
(from all directions), reflect the
light uniformly, in all directions. On the other hand, a surface
which selectively reflects more light at particular angles from the
vertical must reflect that light at the same
angle from the
vertical. Consequently, a surface which reflects selectively at
certain angles will appear "shiny" -- which is, of course, exactly what
we observe with so-called "specular" reflections from painted surfaces.
Page first uploaded on 6 August 2009.
Minor corrections, and addition of some illustrations, on 18 August
2009. Two more illustrations, 22 September 2009.