Why Rayleigh-Jeans Law Assumes E Field in Cavity is Zero

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In summary: On the other hand, if you have a very thin wall of a very good conductor, you will not be able to produce any thermal radiation at all.In summary, the transverse component of the E field parallel to the walls of a cubical cavity is taken as zero because it simplifies the analysis and allows for the assumption of infinite conductivity for the walls. This assumption is necessary for the walls to act as perfect reflectors and for the cavity to behave as a blackbody. However, in reality, actual blackbodies are not 100% efficient and have finite conductivity. To account for this, a small hole is used as the actual blackbody source, with the walls acting as a perfect absorber for the radiation emitted through the
  • #1
Regarding the classical Rayleigh-Jeans Law for blackbody radiation, why is the transverse component of the E field parallel to the walls of an assumed cubical cavity taken as zero?

For non-zero conductivity σ of the walls,they are opaque to the incident radiation at least partly.However that does not mean an infinite attenuation such that E becomes zero instantaneously.The only way to reconcile this is to assume infinite conductivity for the walls.That means then that they are perfect reflectors.

But for a blackbody produced in this particular fashion(heated body with a hole to the outside) it was my understanding that the walls should be at least partly absorbing(i.e. finite conductivity) to the incident radiation so that it is eventually absorbed by the walls to fulfill the blackbody's prime condition of perfect absorptivity.

Where am i going wrong?Any help is appreciated.

Edit:Later in the proof I also realize that the average energy of the standing wave can not be kT unless the walls are partly absorbing as the standing waves need to be in equilibrium.Although I think this bit of info is immaterial because it was eventually proved wrong.
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  • #2
Short answer:
It makes the analysis much simpler. Of course, we never observe a 100-% ideal black body in nature, so we can attribute that to our simplified model.

Slightly more satisfactory answer:
Even though we tend to speak of "black-body cavities", you have to remember that it is the hole, and not the box, that is the actual black body. So for a small enough hole and large enough cavity, a ray of light that enters the hole will have a very small chance of ever finding it's way back to exit the hole (even after ~infinite reflections). In that sense, the hole is a perfect absorber of radiation. We can then assume that any radiation that leaves the hole must be due to thermal emission.
  • #3
Thanks.I had thought of that infinite reflections thing earlier.Wasn't too satisfying.

For walls having nonzero conductivity, the transverse electric field strength at the walls is E=0 because modes with E=0 at the walls are lossy. Only those standing waves with E=0 at the walls will persist after some time lag t>>a/c.

Link:http://www.cv.nrao.edu/course/astr534/BlackBodyRad.html" [Broken]
Will this view work too?
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  • #4
I am not sure whether to start a new thread for this but here is another unrelated query on cavity radiation:

In counting the number of waves in frequency interval ν to ν+dν for a standing 3-d wave we explicitly count it by showing

Then we plot nx,ny and nz on three axes which gives us that every point of the lattice(i.e. points corresponding to integers nx,ny and nz)represent an allowed frequency.Reasoning that the number density of the lattice points is unity(obviously) the volume of the spherical shell gives the number of allowed frequencies in the required interval.

So far so good.But in a rather unrelated but similar procedure we count the number of energy states in an energy interval ε+dε indirectly by counting the number of momentum states in the interval and reasoning that since a particular energy corresponds to a particular momentum their numbers of state are equal.
To do this a momentum space is constructed the axes labelled px,py and pz.Then each point in the space corresponds to an energy(momentum) state.Here the referred text says that the number of states in the interval is proportional to the volume of a shell in the momentum plane.However they go on to relate this proportionality by a constant ie.

Shouldn't the number density be taken into account here?If the number density is constant it's fine but it may change with energy and hence momentum space.And by the way what would be the number density here?This is,unlike the previous example where we had a lattice,a continuous distribution where there are infinite points corresponding to allowed states.
Any help is highly appreciated.

1.Concepts of Modern Physics-Arthur Beiser
2.Quantum Physics of Atoms,Molecules,Solids,Nuclei and Particles-R.Resnick,R.Eisberg.
  • #5
I've thought of the thermal radiation little bit as well, so maybe this is a good opportunity to share some ideas.

To your first question: I agree with you that if you had walls made of ideal conductor, you could not get thermal radiation. For this reason E=0 condition cannot be true for rapidly varying Fourier components of the field.

There are also modifications of this procedure that do not require any walls, and are just counting various configurations of arbitrary free field (assuming periodical conditions on some cube of empty space instead) and deriving the spectrum from the entropy of such configurations (the modern quantum field theroy derivations). But this is ridiculous as well since the thermal radiation is in reality always connected to matter particles. If you try to incorporate them into the standard calculation, you end up with nonsense, since you will got divergent integrals of E^2 + B^2.

I think more appropriate picture of the thermal radiation is due to dipole oscillators behaving in a certain peculiar way to maintain such radiation. They produce and are acted back by the radiation, but by no means they are reflective to all of it, because we know that every hot cavity from whatever material will cool down in a while.

1. What is the Rayleigh-Jeans law?

The Rayleigh-Jeans law is a mathematical formula that describes the spectral energy density of blackbody radiation in terms of the frequency and temperature of the radiation.

2. Why does the Rayleigh-Jeans law assume that the electric field in a cavity is zero?

The Rayleigh-Jeans law assumes that the electric field in a cavity is zero because it was based on classical physics, which did not take into account the quantum nature of electromagnetic radiation. In classical physics, it was believed that the energy of a blackbody radiation was evenly distributed between the electric and magnetic fields. However, this assumption was found to be incorrect when scientists discovered the photoelectric effect and the quantization of energy.

3. What happens if the electric field in a cavity is not zero?

If the electric field in a cavity is not zero, then the Rayleigh-Jeans law will not accurately describe the spectral energy density of blackbody radiation. This is because the energy of the radiation is not evenly distributed between the electric and magnetic fields, as assumed in classical physics. Instead, the energy is quantized and can only exist in discrete packets known as photons.

4. Why is the assumption of a zero electric field in a cavity not valid in modern physics?

In modern physics, we understand that the energy of electromagnetic radiation is quantized and can only exist in discrete packets known as photons. This means that the energy of the radiation cannot be evenly distributed between the electric and magnetic fields, and thus the assumption of a zero electric field in a cavity is not valid.

5. Are there any situations where the Rayleigh-Jeans law is still applicable?

Yes, the Rayleigh-Jeans law is still applicable in certain situations where the energy of the radiation is much lower than the energy of a single photon. This is known as the classical limit and is often used in cases where the temperature of the radiation is large and the frequency is small. However, in most cases, the Rayleigh-Jeans law is not accurate and has been replaced by more advanced theories such as Planck's law and the quantum theory of radiation.

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