Derivation of Rayleigh-Jeans law

In summary, the conversation discusses different approaches to deriving the Rayleigh-Jeans 'Law' for calculating the spectral energy density of thermal radiation. One approach involves using a finite-sized cavity and expanding the fields into Fourier series, while another approach involves considering the whole space outside of the blackbody and using a continuous Fourier integral. The latter approach raises difficulties in assigning energy to continuous modes and ultimately cannot be used to derive the Rayleigh-Jeans law. It is also mentioned that the Rayleigh-Jeans method may not be applicable for high frequencies.
  • #1
dEdt
288
2
All the derivations of the Rayleigh-Jeans 'Law' I've seen assume that the electromagnetic radiation is enclosed in a cube. I'm trying to derive the law using less arbitrary circumstances. That is, by starting with the equation [tex]U=\int \left[ \frac{\epsilon_0}{2}E^2 + \frac{1}{2\mu_0}B^2 \right]dV,[/tex]
then taking the Fourier transform of the electric and magnetic fields, appling Parseval's theorem, and finally using the equipartition theorem I hope to calculate the spectral energy density. Unfortunately I'm having trouble filling in the details, and would appreciate some help. Thanks.
 
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  • #2
Fourier transform is an integral over the whole real axis. Do you really want to consider thermal radiation in the whole space? Usually, cavity is assumed. If cavity is cuboid, one can use standard Fourier series with sines. If cavity has more complicated shape, one can use generalized Fourier series with eigenfunctions of the cavity (these are not sine functions of position, but they are still sine functions of time.)
 
  • #3
Jano L. said:
Do you really want to consider thermal radiation in the whole space?

Yes. Or at least, the whole space outside of the blackbody in question. I know that the cavity approach is standard (and easier), but I'd like to be able to do it this way.
 
  • #4
It's impossible to help you unless you post the details of the calculations you have thus far. Incidentally I'm having trouble seeing how you would use Parseval's identity when you don't have countably many normal modes of the electromagnetic field to work with since you aren't imposing periodic box boundary conditions. It may be that there is a generalized form of Parseval's identity; the one I know of is only for the relationship between Fourier transforms and sums.
 
  • #5
WannabeNewton said:
It's impossible to help you unless you post the details of the calculations you have thus far. Incidentally I'm having trouble seeing how you would use Parseval's identity when you don't have countably many normal modes of the electromagnetic field to work with since you aren't imposing periodic box boundary conditions. It may be that there is a generalized form of Parseval's identity; the one I know of is only for the relationship between Fourier transforms and sums.

Yes, it generalizes to $$\int_{-\infty}^\infty | x(t) |^2 \, dt = \int_{-\infty}^\infty | X(f) |^2 \, df, $$ where X(f) is the Fourier transform of x(t).
 
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  • #6
WannabeNewton said:
It's impossible to help you unless you post the details of the calculations you have thus far. Incidentally I'm having trouble seeing how you would use Parseval's identity when you don't have countably many normal modes of the electromagnetic field to work with since you aren't imposing periodic box boundary conditions. It may be that there is a generalized form of Parseval's identity; the one I know of is only for the relationship between Fourier transforms and sums.

My approach is as follows: as Dauto mentioned, [tex]\int (\phi(\mathbf{x}))^2 dV = \int |\tilde{\phi}(\mathbf{k})|^2 d^3 k,[/tex] where the tilde denotes the Fourier transform.

Applying this equation to the electromagnetic energy formula, we get that
[tex]U=\int \left[ \frac{\epsilon_0}{2} |\tilde{\mathbf{E}}(\mathbf{k})|^2 + \frac{1}{2\mu_0} |\tilde{\mathbf{B}}(\mathbf{k})|^2 \right] d^3 k.[/tex]

If I could rearrange this integral into something like [tex]U=\int \left[ \mbox{something that looks like a harmonic oscillator} \right] d\nu,[/tex] then I would be a happy man because I could apply the equipartition theorem (or at least some 'continuous extension' of it) to derive the Rayeigh-Jeans law. But I'm having trouble with that last step.
 
  • #7
I don't have the derivation on hand so I do not know if this is simply repeating the typical textbook derivation, but why not simply assume a cavity of finite size and take its definition via the Fourier series. Then take the volume of the cavity to infinity at a suitable point. You would have to introduce an infinitesimal loss when doing so to remove the incoming wave solution in the cavity mode. In essence, you are simply finding the vacuum modes. This is something that I have done for derivations like the Casimir force.
 
  • #8
I think you need to write the fields in terms of the potentials A and ø.
 
  • #9
dEdt said:
Yes. Or at least, the whole space outside of the blackbody in question. I know that the cavity approach is standard (and easier), but I'd like to be able to do it this way.

Equilibrium radiation spectral formula gives radiation energy per unit frequency interval. One of the assumptions behind its derivation is

(reasonable for low but wrong for high frequencies)
1) that the radiation is enclosed in a perfectly reflecting cavity of finite volume

or

2) unphysical but mathematically similar condition that the field in space repeats the pattern of the field in a finite-sized cuboid

Both cases allow for expansion of the field into Fourier series and lead to finite number of modes (oscillators) per unit frequency interval. Multiplicating by finite average energy of such oscillator, this in turn leads to finite spectral density of Poynting energy density (R-J or Planck function) - just divide the energy of oscillators per unit frequency interval by the volume of the cavity/cuboid.

However, if the region where the field is considered is infinite right from the beginning and no periodic conditions are imposed, the field cannot be expanded into Fourier series. It may be expandable into Fourier integral, but then the number of independent modes per unit frequency interval is infinite.

There is no obvious way to ascribe energy to such continuous mode or interval of modes, except for reverse-engineering the desired spectral function or returning to finite volume. In case of the Rayleigh-Jeans or Planck spectral function, one continuous mode has to be ascribed zero energy. Derivation of the spectral function in line with the Rayleigh-Jeans procedure does not work for infinite volume.

Physically, this is not much of a problem since there is little reason to think radiation in the whole space is equilibrium radiation. And I would like to say that the calculation method of Rayleigh-Jeans is of limited value even for finite cavities, as it does not (and should not) work for high frequencies.
 

1. What is the Rayleigh-Jeans law?

The Rayleigh-Jeans law is a mathematical formula that describes the spectral energy density of blackbody radiation at a specific temperature. It was developed in the late 19th century by Lord Rayleigh and Sir James Jeans, and was later found to be inaccurate at high frequencies, leading to the development of the Planck's law.

2. How is the Rayleigh-Jeans law derived?

The Rayleigh-Jeans law can be derived from classical electromagnetic theory, specifically from the equipartition theorem which states that the energy of a system is evenly distributed among all of its degrees of freedom. By applying this principle to the electromagnetic field inside a cavity, the Rayleigh-Jeans law can be derived.

3. What are the limitations of the Rayleigh-Jeans law?

The Rayleigh-Jeans law is only accurate at low frequencies and breaks down at high frequencies, a problem known as the "ultraviolet catastrophe". This led to the development of quantum mechanics and the eventual discovery of the Planck's law, which accurately describes blackbody radiation at all frequencies.

4. What is the significance of the Rayleigh-Jeans law?

The Rayleigh-Jeans law was an important step in understanding the behavior of blackbody radiation and the principles of classical electromagnetism. It also helped to highlight the limitations of classical physics and paved the way for the development of quantum mechanics.

5. How does the Rayleigh-Jeans law compare to other laws describing blackbody radiation?

The Rayleigh-Jeans law is only accurate at low frequencies, while the Planck's law accurately describes blackbody radiation at all frequencies. Additionally, the Wien's displacement law relates the peak wavelength of blackbody radiation to its temperature. The Rayleigh-Jeans law and Wien's displacement law can be derived from the Planck's law in certain limits.

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