The Zero Principle has been thoroughly verified by countless scientific experiments. It has a main impact on civilization and philosophy. Should there exist any object that would become colder or hotter than another object with whom it is in contact, there would be no need for any sort of fuel-consuming power plant, no need for slavery, perhaps even no need for food. We would just have to put a kettle on such a "warm" object, let the vapor stream out of the kettle and lead it towards a turbine to produce mechanical energy forever. This has been an Eldorado no scientist ever managed to reach. So it became thought of as a Law of Nature. Power plants are obliged to consume fuel in order to heat their kettle. At best they can rely on the nuclear fuel of the Sun by using the energy of the light coming from the Sun.
Let's focus on one of the many ways to let heat flow between objects : the thermal radiation. Here follows some common data about thermal radiation.
Thermal radiation are electromagnetic radiation that are continuously emitted by every object. Every point of the object emits thermal radiations in all directions. They are the same as radio waves, microwaves, infrared light, visible light, ultraviolet light, X rays and gamma rays. Electromagnetic radiation are emitted under the form of photons. Whether a photon is a radio wave or up to a gamma ray depends on its wavelength (its size, roughly speaking). Radio wave photons have a wavelength of thousand of kilometers up to a few millimeters. Visible light light photons have a size of about 600 nm (yellow) (that is 0. 000 000 600 meter). The bigger the size, the less energy a photon carries. That's why the most little amongst them are so dangerous : the gamma rays.
Every object emits a whole spectrum of electromagnetic radiation. From radio waves even up to gamma rays. This is due to the heat inside the object. The heat of the object transforms into photons and those photons travel away from the object. This tends to let the object cool down. It looses its heat by loosing photons. (To emit no radiation an object should be perfectly white, reflecting or transparent, or be at a temperature of 0 Kelvin.)
Yourself, for example, are currently emitting deep infrared light waves whose wavelength is mainly of about 3 µm (0. 000 003 meter). The power of the light waves emitted by your skin is of about 20 mW per square centimeter. Your whole skin emits about 200 W, that is the power emit by a strong lamp.
This is the formula to calculate the main wavelength l (in meters) of the photons emit by an object at a temperature T (in Kelvin):
The hotter an object, the littler the average wavelength of the emitted
photons will be. Thus the more energy each average photon will carry.
In order to emit mainly photons at 600 nm the temperature of an object should be of 1666 Kelvin, that is 1393 °C. That's why fire produces light and why the tungsten wire inside a light bulb is heat up to a high temperature by the electricity send through it.
This is the formula to calculate the power P (in watts) emit by a surface of S square meters, at a temperature T (in Kelvin) if that surface has a darkness of s (s is a number between 0 and 1: 0 for perfect white and 1 for perfect black):
The hotter an object, the more energy it will radiate. If it is 2 times
hotter, it will radiate 16 times more energy. If it is 3 times hotter, it will
radiate 81 times more energy. That's why the tungsten wire inside a light bulb
is so thin. Because it has a high temperature it only needs to have a very
little surface in order to radiate a tremendous quantity of visible and
invisible photons (100 Watt for example).
The flame of a candle emits a lot of light because the top of the flame is filled with unburned coal powder. That coal comes from the paraffin molecules that compose the candle body. The paraffin molecules don't burn completely because the center of the flame lacks oxygen. Their carbon skeletons remain and turn into a greasy black coal powder.
The flame of a gas stove produces very few light because the hot gasses of the flame are perfectly transparent. A good mixture of gas and oxygen was made before it was lit, so the combustion instantaneously produces (right when the gas mixture leaves the little holes) perfectly transparent water and CO2 gas molecules.
Every object (unless it would be perfectly white, reflecting or transparent) absorbs a part of the electromagnetic waves that hit its surface and makes another part bounce away. The part that is catch is turn into heat. Rather white or reflecting objects make most of the waves that hit bounce away. Rather transparent objects let a part of the photons travel through them unhampered and make another part bounce away. Rather black objects absorb most of the waves that hit. This is why fireman fire suits are white or reflecting. And this is why kettles warmed up by solar installations are painted black.
That's all why in winter you can feel the warmth of somebody's hand close to your cold cheek even without it touches you. The warm hand emits deep infrared photons. Those photons are partly absorbed by your skin, turn into heat and thus make your cheek skin increase temperature.
Suppose two theoretical huge and perfectly black walls placed closely facing each other (each has s = 1).
Every thermal photon emit by one wall will hit the other wall and will be catch/swallowed and turned into heat.
Suppose both walls have the same temperature. So they will emit each the same power/amount of thermal radiation. Each second, each wall looses the same amount of heat energy by radiation (which tends to make them cool down) and gets that same amount from the other wall (which tends to make them warm up). So both stay at the same temperature.
Suppose one wall is hotter than the other. The hot wall will emit more photons than the cold wall. This will make the cold wall warm up and the warm wall cool down. Till their temperatures become the same.
The Zero Principle implies one first corollary : a perfectly black surface emits a given amount of radiation power at a given temperature and there cannot exist any sort of other black surface that would emit even just a little more. Indeed, would there exist such a "super black" surface color, one could paint one of the walls with it. That wall would emit more radiation than the other wall, even while both walls have the same temperature. So the other wall would warm up and the super-black wall would cool down, till the radiation flux between the two equilibrates and they stabilize at a given temperature difference. This is why the sigma factor inside the formula goes from 0 to 1. There cannot exist any matter or color with a sigma above 1. Otherwise the Zero Principle would not be followed.
Now what if one of the walls is painted a perfect white or reflecting color (this is impossible, but let's suppose). In that case the difference of temperature between the two walls does not matter. The perfect white wall will emit no thermal radiation and thus cannot cool down. It will neither warm up because any photons that hits bounces back. Every photon coming from the other black wall will bounce back on the white or reflecting surface and will go back towards the black wall. The black wall will receive exactly as much radiation energy it emits; it receives every photon back. Neither wall will change temperature. That's why thermos bottles have a reflecting inside wall. In order to keep cool if they are cool and to keep warm if they are warm (a thermos bottle is made of two inside walls facing each other, they are both reflecting and there is a vacuum of air between them so the heat cannot flow even by conduction or convection through the air).
What if one of the walls is painted gray ? Say this gray surface absorbs 30% of the photons it receives and makes the other 70% bounce back. The Zero Principle implies this surface must emit 30% of the thermal radiation a black surface would. Indeed, we know that in order to keep the same temperature the other black wall must receive exactly as much photons it has emit. So if the gray surface swallows 30% of the photons coming from the black surface, it must emit an amount of radiation that will exactly compensate. That is an amount of 30% of the amount of a black surface. 30% + 70% = 100%. It has an s of 0.3. The Zero Principle thus implies any surface has an s factor equal to the ratio of incoming radiation that surface swallows.
So, whatever the gray shade of the wall the black wall is facing, if both walls have the same temperature the black wall will receive an amount of radiation exactly the same as the amount of radiation it emits. If it is facing a perfect black surface it will receive that given amount of radiation from the facing surface. If it is facing a gray surface it will receive partly its own photons bouncing back on the gray surface and partly photons emit by the gray surface, the sum of the two yielding exactly the given amount.
More generally, we can state that if all inert objects inside a box have the same temperature, the radiation power at any point inside the box is exactly the same, whatever direction you are looking. Whatever the sort of paint you are using, whatever the shape of the objects, whatever the presence of lenses or mirrors. Indeed, should there be a little more radiation power at any given point inside the box, you could place a little black object at that point and it would heat towards a temperature above the box temperature. Should there be a little less radiation at that point, the object would become a little cooler.
Whatever the objects inside the box, the radiation inside the box will be like if the box was empty and painted perfectly black. Whatever the objects you are looking at, it will be like if you were looking to a perfect black wall.
This can be generalized to colors and to surfaces with directional reflectivity. A red surface is a surface that swallows green and bleu photons and makes red photons bounce back. It is "white" for red photons and its is "black" for green and bleu photons. So if that surface is heat it will emit green and bleu photons and no red photons. It will emit a cyan color (green + bleu). A surface that tends to reflect photons towards a direction A and swallows the photons that would have gone towards direction B, that surface will radiate few thermal photons in direction A and will radiate a lot of photons in direction B. (The radiation behavior of a given object can be very complicated, since it can be for example transparent for radio waves, reflecting for microwaves, transparent for infrared light, "white" for red and green light, "black" for bleu light and varnish-like for ultraviolet light...)
This is too why photons reflect partly or completely when they hit a glass surface sideways. Otherwise a glass lens could be used to focus the thermal radiation towards a little black object and make it heat. The formula of the reflection of photons that hit a surface with a given refraction indices can be calculated from the Zero Principle.
The thermal radiation inside a box is perfectly isotropic. It has exactly the same intensity and spectrum wherever you measure and in whatever direction you are looking. You cannot use it to "see" any object inside the box.
This implies whatever the orientation of a surface you look at you will receive exactly the same radiation flux:
This in turn implies the radiation power emit by a given amount of surface
depends on the angle you are looking at that surface. If the surface is facing
you, you will receive the maximum amount of radiation (angle of 90°). The more
the surface turns (towards an angle of 0°) the less radiation you will
receive:
This is the formula. The power P (in watts) emit by a given amount of
surface under an
angle a (in degrees or radiants) will be proportional to the sinus
of that angle:
This is a polar plot of the radiation intensity at all angles 180° around the
piece of surface:
Now, let's make a calculated experiment. Let's build a virtual 2D world.
There is one big black wall on the left and one small wall on the right. These
"walls" are just lines because they are in a two dimensional world:
We enclose those two walls within two mirrors. That way any photon emitted by
one of the walls will necessarily end on one of the two walls:
At first hand we could expect this will make every photon emit by the big
wall be guided towards the little wall and every photon emit by the little
wall be guided towards the big wall. That way the little wall would become
hotter than the big wall, because it receives the whole bunch of photons
radiated by the big wall while it emits just a little amount of photons. But
there is one phenomenon that hampers this: while every photon radiated by the
little surface necessarily hits the big surface, not every photon radiated by
the big surface will hit the little surface. Some of them will bounce back
towards the big surface, due to the angle of the mirrors:
We let a computer compute the trajectory of a huge amount of photons, emitted
by every point of the two surfaces in virtually every possible direction. We
take into account a given part of a surface emits less photons sideways to the
surface than perpendicular to the surface. That's our formula P ~ sin(a). This
is the result :
The whiteness of the lines shows the amount of photon hits. The whiter, the
more hits. Alas, the little surface, on the right, gets more photon hits than
the big
surface. This means the little surface is supposed to warm up and the big
surface is supposed to cool down!
This is contradictory to the Zero Principle! How comes?
Simply because there is an error inside my computer program. The two lines should have been the same.
So let's build another calculated experiment, this time in a virtual 3D world, with two square "real" surfaces facing each other and with a program written the right way:
We enclose them between four mirrors:
This time we get the right result. The two surfaces will be hit by exactly
the same amount of photons (the little square sometimes seems
a little brighter, but that's an illusion because of its size,
you can load the picture inside a paint program and use the color picker
to check):
Don't care for the faded hits around the two squares.
they are there because I used two big squares and relied
on the mirrors to "cut" a little square out of one
of the big squares. The photons coming out of that
big square, around the little square, bounced away on
the mirors and hit around the squares.
Note every part of each surface is hit by exactly the same amount of photons. This is quite impressing: the photons coming from the little surface will tend to hit the center of the big surface and the photons emitted by the big surface and bouncing back towards the big surface due to the oblique mirrors will tend to hit the sides of the big surface, yet the sum of the two will compensate exactly and yield a perfect uniform amounts of hits on the big surface.
We could also have expected the mirrors to make the photons coming out of the big surface focus more on the center or on the side of the little surface. This is neither the case : the amount of hits is perfectly uniform too. Whatever the position of the surfaces and the shape of the mirrors. A perfectly isotropic radiation flux keeps being isotropic in a still world. That's what we stated above for the inside of a box. It's a rule of thermodynamics too, yet applied to photons: entropy cannot decrease.
To get the programs and source codes I used to calculate the photon exchanges in the 2D and 3D virtual worlds click here. They are C++ code for Linux using the Qt graphic library.