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

[aim1732]

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.

 

[cmos]

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.

 

[aim1732]

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"
Will this view work too?

 

[aim1732]

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
ν=c/2a*(n_x²+n_y²+n_z²)

Then we plot n_x,n_y and n_z on three axes which gives us that every point of the lattice(i.e. points corresponding to integers n_x,n_y and n_z)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 p_x,p_y and p_z.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.
n(p)dp=(Constant)*p²dp

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.

References:
1.Concepts of Modern Physics-Arthur Beiser
2.Quantum Physics of Atoms,Molecules,Solids,Nuclei and Particles-R.Resnick,R.Eisberg.

 

[Jano L.]

Hello,
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.