A much-simplified alternative explanation:

In quantum field theory, every type of fundamental particle has its own quantum field ; fundamental particles are merely the spatially localized form of the smallest amount of a change in the quantum field… a quanta.

The universe is permeated by waves (although they’re not actually waves, they’re spirals, a consequence of the photon’s electronic and magnetic fields oscillating in quadrature, forming a circle. A circle translated through space-time is a spiral.

The image above shows the real (cosine… labeled ‘Re’ in the image) and imaginary (sine… labeled ‘Im’ in the image) components of an electromagnetic ‘wave’. When viewed in line with its direction of travel, it will appear to be a circle, and when viewed orthogonal to its direction of travel, it will appear to be a sinusoid, when in reality it’s a spiral.

This is because a sinusoid is a circular function.

You’ll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You’ll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.

Thus the magnetic field and electric field (oscillating in quadrature) of a photon is a circle geometrically transformed into a spiral by the photon’s movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame, which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis… no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

All atoms and molecules (all comprised of fundamental particles, which are all comprised of quanta of waves) are harmonic oscillators.(with anharmonic force constants complicating the wave equation in the case of molecules).

This is why all matter has absorption peaks and valleys… wave resonance. Think in terms of everything in the universe composed of tiny antennae and waves.

Blackbodies absorb and emit at a wide range of wavelengths exemplified by the Planck curve, whereas a gas absorbs and emits at specific spectral lines. As a first approximation, atmospheric gases on Earth cannot and do not emit as a blackbody. The dominant source of blackbody radiation is transient oscillating dipoles induced by inter-molecular thermal vibrations within a material. Solids, liquids, plasma of sufficient density and gases of sufficient density can emit blackbody radiation, but our atmosphere cannot simply because the gas molecules spend the majority of their time relatively distant from each other, and thus they cannot sustain the inter-molecular oscillations necessary for blackbody radiation. As gas density increases, blackbody radiation production increases and eventually dominates the discrete emission spectra. Similarly, as gas density increases, blackbody absorption increases and eventually dominates the discrete absorption spectra.

So the universe is permeated by constantly-shifting energetic waves, with regions of higher and lower flux dependent upon the energetic processes occurring in that region of space-time.

Inherent to all invariant mass is a surrounding field (virtual photons) caused by Larmor radiation due to the bound electrons undergoing angular acceleration. This field has inherent to it field radiation pressure.

Photons, too, have field radiation pressure when they are incident upon invariant-mass matter… the angular momentum of the electronic and magnetic fields of a photon oscillating in quadrature is geometrically transformed to linear momentum upon incidence.

This field radiation pressure, for instance, is what prevents the sun from gravitationally collapsing. The process by which that field radiation pressure is generated also gives the sun its magnetic field.

Photons must be of sufficient energy (frequency) to overcome (electromagnetically cancel) this field radiation pressure in order to be emitted into the field or absorbed by an object.

As an analogy, think in terms of a bathtub filled with a fluid with special properties which will not allow any waves to form unless those waves are of greater energy than the energy of the bulk of the fluid. The waves, once formed, frictionlessly ‘ride along’ the surface of the fluid, but cannot be subsumed back into the bulk of the fluid until a region of the bulk fluid reduces its energy sufficiently to allow that wave back into the bulk fluid.

As two objects are coming into thermodynamic equilibrium, there will be photons emitted which are in transit between the two objects as the thermodynamic equilibrium state is passed through. Those photons are of insufficient energy to cancel the field radiation pressure of either object, and hence must be reflected, since they cannot be absorbed.

Simplistically, at exact thermodynamic equilibrium, the standing wave would preclude emission toward or absorption from either object due to the potential of each object being equal to the field radiation pressure of that standing wave.

Simplistically, if Object A is of higher temperature than Object B, Object A’s field radiation pressure is higher, thus the wave travels from A to B. And vice versa.

Of course, exact thermodynamic equilibrium can never really be achieved. It can be momentarily passed through, but never maintained. Heisenberg’s Uncertainty Principle (Fourier’s theorem for fields) ensures that there will always be at least a few quantum states which can be excited by absorption of (a few absorption resonances which will accept) the incident radiation.

We can move up one level to the fundamental particle level, with a higher abstraction of the concept.

We can move up another level to the atomic / molecular level, with yet more abstraction. This is the level you are discussing, in which it appears that radiation incident upon an object emitted by a lower potential object will be absorbed, and objects at a finite temperature emit under all circumstances.

The closer we get to the actual underlying physical reality, the less abstraction we deal with, and the more accurately we can describe energetic interaction on a per-interaction basis.

]]>Nuclear fusion processes in the sun decrease entropy for the sun…. check.

Radiative emission from the sun increases entropy for the sun… check.

Radiative emission from the earth is “negative entropy for the Earth (a gain by ‘out there’).”, therefore radiative emission from the earth decreases Earth’s entropy?

Sun radiative emission increases entropy of the sun.

Earth radiative emission decreases entropy of the Earth?

Something’s not adding up.

]]>John Millet,

The surface for gases is arbitrary, you pick it to define the thickness of the layer of gas of interest. You need a sufficient number of molecules in the layer so that temperature is defined and the concept of local thermodynamic equilibrium is applicable. It’s the radiation *through* the plane that’s important. For calculating atmospheric absorption/emission for any given total path length, you use enough layers that increasing the number of layers doesn’t change the result significantly. You need to do this in the atmosphere because it isn’t isotropic and isothermal. For monochromatic radiation in an isotropic, isothermal system, a single calculation is all that’s necessary

Thanks for the link which included:

To find the radiated power per unit area from a surface at this temperature, multiply the energy density by c/4.

The question remains: where in a gas body is the radiating surface? The only tangible matter in a gas body are molecules, a small proportion of which are radiators.

]]>John Millet,

W/m² of electromagnetic radiation from (or to) a planar surface with an area of one square meter integrated over a wavelength range and averaged over all angles, like the Stefan-Boltzmann equation is the integration of the Planck equation. See http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html and scroll down to the Planck Radiation Formula box.

]]>By my comment I wanted to tell that sentences like

The trigonometric integrations which give irradiance presumably apply to all 6 “surfaces”.

do not make sense at all except for a real cube opaque to the radiation.

The plane parallel approximation can be formulated precisely and is a very useful approach, but I don’t think that anything like that can be done with “all 6 surfaces”.

]]>Frank,

This discussion has gone into the question of minimal assumptions needed to derive certain results. Answering reliably such questions is difficult, because few textbooks, if any, discuss these issues carefully. It’s much more common to introduce a whole standard set of postulates in combination with the most common way of doing the calculations. That has often led to the erroneous idea that this is the only correct approach. With quantum theory that is not a valid conclusion. Everything can be calculated in several ways, which may appear very different. A particular approach may be clearly the best for a specific problem, but probably never the only one.

After this introduction I tell that my educated guess is that the formula E = hv can be derived from the combination of Maxwell’s equations and quantum mechanical description of matter. QM is certainly needed, but probably only for matter. I’m not certain of this, but I think that this is the case. In such a description the formula is not about photons as photons are not part of the theory, but the matter still behaves in the same way as in a theory that includes the concept of photon. it’s easier to use the concept of photons, but perhaps not absolutely required.

This answer should be qualified also by the comment that certain essential inconsistencies of the theory have not been resolved without full QED, and not totally satisfactorily even with it. (Renormalization works very well in practice and a lot is understood about it, but open questions do remain at the level of fundamentals.)

]]>Pekka,

It’s not the cosine law that puzzles me, it’s the Wm-2 in DeWitt’s formula – Watts per square metre of what?

]]>DeWitt and Pekka: I fully understand that electrons don’t orbit the nucleus. However, we do know that matter consists mostly of empty space with a positively charge nucleus surrounded by electrons. Most or all of these electrons remain attached to one atom. Given classical mechanics and Maxwell’s equations, is there any other solution besides orbiting that permits an electron to remain near a nucleus indefinitely? Isn’t an orbiting electron required to give off electromagnetic radiation and spiral into the nucleus?

Since classical theory and observation disagree, something is wrong with the classical picture. I now see that DeWitt is correct when insisting that the failure could (and does) point to a problem with classical mechanics rather than with Maxwell’s equations. We can’t tell if Maxwell’s equations are right or wrong on the atomic scale, if we can’t precisely specify where the electron is and how fast it is moving. In fact, we probably don’t know whether electrons move at all inside atoms and molecules. (Moving electrons are sometimes invoked to explain Van der Waals force, but that is fairly indirect evidence of any motion.)

After some reading, I vaguely understand that in quantum field theory, Maxwell’s equations describe the electric and magnetic fields associated with photons, the particles that carry forces between charged particles. The solutions to Maxwell’s equations in a vacuum apparently demand motion at the speed of light and permit sinusoidal solutions with a frequency/wavelength.

Can E = hv can’t be derived from Maxwell’s equations?

]]>