# Cavity Radiation, why should only standing waves persist in the cavity?

"Because when the box is in thermal equilibrium, there can never be any electric field at the walls for it would shake the charge in the wall around, changing the temperature, contradicting the claim that the box is in thermal equilibrium."

This argument just doesn't sit right with me. Thermal equilibrium does not mean that it's absolutely forbidden for the temperature in a system to locally fluctuate. Its like saying, when the box with gas is in thermal equilibrium, there are no collisions between the molecules anymore... Thermal equilibrium merely means that every degree of freedom is on average giving away as much energy as it is absorbing.

The book by Robert Eisberg and Robert Resnick, "Quantum Physics" has this to say:

"We assume for simplicity that the metallic-walled cavity filled with electromagnetic radiation is in the form of a cube of edge length a, as shown in Figure 1-3. Then the radiation reflecting back and forth between the walls can be analyzed into three components along the three mutually perpendicular directions defined by the edges of the cavity. Since the opposing walls are parallel to each other, the three components of the radiation do not mix, and we may treat them separately. Consider first the x component and the metallic wall at x = O. All the radiation of this component which is incident upon the wall is reflected by it, and the incident and reflected waves combine to form a standing wave. Now, since electromagnetic radiation is a transverse vibration with the electric field vector E perpendicular to the propagation direction, and since the propagation direction for this component is perpendicular to the wall in question, its electric field vector E is parallel to the wall. A metallic wall cannot, however, support an electric field parallel to the surface, since charges can always flow in such a way as to neutralize the electric field. Therefore, E for this component must always be zero at the wall. That is, the standing wave associated with the x-component of the radiation must have a node (zero amplitude) at x = O. The standing wave must also have a node at x = a because there can be no parallel electric field in the corresponding wall. Furthermore, similar conditions apply to the other two components; the standing wave associated with the y component must have nodes at y = 0 and y = a, and the standing wave associated with the z component must have nodes at z = 0 and z = a. These conditions put a limitation on the possible wavelengths, and therefore on the possible frequencies, of the electromagnetic radia- tion in the cavity."

Particularly this part: "A metallic wall cannot, however, support an electric field parallel to the surface, since charges can always flow in such a way as to neutralize the electric field. Therefore, E for this component must always be zero at the wall."

Well, then how can it emit any radiation? Clearly, metal objects outside of a box can also emit thermal radiation so doesn't that contradict the idea that "E for this component must always be zero at the wall"?

What should happen is that the emitted wave, once reflected, is reflected again from the surface that originally emitted it. This, however, will never cancel the resulting standing wave completely, even ad infinitum.

Maybe what is happening is that although the frequencies that don't create standing waves with nodes at the edges never die down, they also do not grow bigger and bigger after each reflection, whereas the frequencies that do create standing waves with nodes at the edges have their amplitudes increased, due to constructive interference, as a result of every reflection, overshadowing the other frequencies in amplitude.

This leads to my second question though: the intensities that we'd observe coming from this box are highly amplified due to these reflections and are therefore different from those emitted from normal thermally radiating surfaces.

This final point could be easily resolved if somebody could confirm that the spectra of the box and that of a regular surface only differ by a constant factor that takes into account how often the waves are reflected and with what ratios. I have seen no mention of such constant anywhere however.

All these questions about this whole experiment have been plaguing me for over two years now. I hope that somebody can help me finally make this intuitive and straightforward. Physics is not satisfactory to me when I'm not really trying to step into the shoes of the people that first did the experiments and the calculations to think it through to a point where I can see why they did everything the way they did. Any insights would be much appreciated!

• I'm not sure, but you may be confusing the steady-state solution (standing waves only) with an initial settling period? All the noncoherent power ends up getting cancelled (or split into vectors and "joining" a standing wave pattern) ; the usual analogy is to study a plucked guitar string. // or is it because the theoretical box has perfect conductivity (which is why nothing lost to 'heat')? Oct 29, 2020 at 12:56
• @Darl Witthoft A sinusoid bouncing back and forth between two walls never gets cancelled completely. I would also argue that a guitar string is different because the waves don't start as coming out of one end and being reflected at the other in the same way. If you fix one end of the guitar string, it can not be the point where the waves are generated. Like I said, it's different here. It is irrefutable though, that the decoherent power can never build up the way te coherent power ca. So maybe the decoherent power is just much smaller and negligent in proportion. Oct 29, 2020 at 13:13
• Imagine an ideal LC resonator whose capacitor is magically somehow initially charged but then is separated from its environment and upon connecting it to the inductor the E and M energies will fluctuate henceforth forever between the L and and the C. The wall is the "L" and the "C" is the inside of a very small cavity. Now model the ideal cavity as an infinite set of resonators of all different frequencies and distribute the E and M energies per Planck. Oct 30, 2020 at 10:30

This gets a little long to explain the situation. See the bottom for answer to your questions.

Many textbooks do a bad job explaining the theory of blackbody radiation. In particular, often they assume perfectly reflecting cavity without explaining why they do so. It seems to be often just copied from textbook to textbook.

I don't know who and where introduced the concept of perfectly reflecting walls (for EM radiation) into theory of blackbody radiation. It does not seem to be stressed by Rayleigh or Jeans in their papers. However, Planck uses this concept in his book The theory of heat radiation, where in section 50 he writes

Hence the radiation of a medium completely enclosed by absolutely reflecting walls is, when thermodynamic equilibrium has been established for all colors for which the medium has a finite coefficient of absorption, always the stable radiation corresponding to the temperature of the medium such as is represented by the emission of a black body. Hence this is briefly called “black” radiation.

and then in section 51 he writes

Hence in a vacuum bounded by totally reflecting walls any state of radiation may persist. But as soon as an arbitrarily small quantity of matter is introduced into the vacuum, a stationary state of radiation is gradually established.

In different words, in order to experimentally achieve a region of space where radiation can change into state of thermodynamic equilibrium, we must prevent leakage of radiation through the walls and any interaction of outside radiating objects with the stuff inside the cavity. Such interaction would prevent establishment of thermodynamic equilibrium inside.

Perfectly reflecting walls make the system inside the cavity isolated. They play the same role the perfectly solid immovable walls of gas container do in molecular theory of gas. Energy can't go out, because the perfectly reflecting walls are (in theory) keeping all radiation energy inside. If the walls weren't perfectly reflecting, some radiation would go through the walls to the outside, and some outside radiation would penetrate inside.

The assumption of perfect reflection has several interesting consequences. Perfectly reflecting walls means the walls have infinite electric conductivity. This implies electric field component tangential to the wall is always zero both on the metal and on the vacuum side of the boundary. This allows us to pair the terms in the Fourier series giving any electric field inside the cavity, leading to the concept of standing waves.

All that said, the standing waves or perfectly reflecting cavity aren't some necessary additional assumption for the usual modern derivation of Rayleigh-Jeans or Planck's formula for radiation spectrum, they are rather incidental (a consequence of the idea of practically isolating radiation systems to achieve thermodynamic equilibrium). All that is really needed for the derivation of those spectral formulae is that the radiation is in thermodynamic equilibrium. One can do all the usual steps of the modern presentation of the Rayleigh-Jeans derivation for imaginary cube in a vacuum containing EM radiation in thermodynamic equilibrium, ignoring any material walls. Electric field inside any cuboid can be expressed as Fourier series in space and time (with all the Fourier waves, not just the standing ones). We then arrive at the same familiar formulae for blackbody radiation spectrum.

Particularly this part: "A metallic wall cannot, however, support an electric field parallel to the surface, since charges can always flow in such a way as to neutralize the electric field. Therefore, E for this component must always be zero at the wall."

Well, then how can it emit any radiation? Clearly, metal objects outside of a box can also emit thermal radiation so doesn't that contradict the idea that "E for this component must always be zero at the wall"?

No, there is no contradiction. What the authors say is that there can't be net tangential electric field inside a perfect conductor, or on its surface. That's why perfect conductor reflects everything. Its thermal emission is zero, it just reflects. Metal objects outside of a box can and do have thermal emission radiation, but only if they are made of real metal (not perfect conductor).

• My question isn't: "Under what circumstances is thermal equilibrium achieved?" But: why should thermal equilibrium imply standing waves? If like you say: "This implies electric field component tangential to the wall is always zero both on the metal and on the vacuum side of the boundary" then how can any radiation be emitted as this radiation clearly has to pass the vacuum side of the boundary? I think my explanation that the radiation bouncing back and forth only interferes constructively in the case of the length of the box being an integer multiple of half the wavelength makes more sense. Oct 29, 2020 at 13:41
• If the walls can support an electric field because in reality they're not perfectly conducting then it becomes even harder to explain why only standing waves should persist, which is what is really my question. Oct 29, 2020 at 14:09
• I'm getting more and more certain that my idea is correct: for something capable of absorbing energy to really be considered a full degree of freedom that participates as an energy reservoir, it must be capable of absorbing an arbitrary amount of energy. That's what's important and what brings these particular "length of the box being an integer multiple of half the wavelength" frequencies to the forefront. We don't need to make up all sort of unphysical illogical reason to magically explain away all the other frequencies as being entirely nonexistent. Oct 29, 2020 at 14:11
• > "If like you say: "This implies electric field component tangential to the wall is always zero both on the metal and on the vacuum side of the boundary" then how can any radiation be emitted as this radiation clearly has to pass the vacuum side of the boundary?" -- I think I explained this in the last part of my text. Maybe you have some misconception about radiation that is causing your confusion. The perfectly reflecting cavity does not emit its own thermal radiation. Coefficient of thermal emission of a perfect conductor is 0. Oct 29, 2020 at 14:54
• The radiation inside the perfectly reflecting cavity that is assumed when deriving Rayleigh-Jeans' or Planck's formula doesn't originate as thermal emission of that cavity. It is there either as initial condition of the field, or it is due to thermal emission of real emitter inside the cavity. Or the cavity has a hole in the wall and the radiation is introduced from outside. Oct 29, 2020 at 14:55