Discrete vs Continuous Spectra in Blackbody Radiation?

Click For Summary
SUMMARY

The discussion centers on the nature of blackbody radiation spectra, specifically the distinction between discrete and continuous spectra. It highlights that standing waves in a cavity can only occur at specific, discrete wavelengths, suggesting that under certain conditions, a blackbody may emit a discrete spectrum. The conversation also touches on the differences between Planck's law and the Rayleigh-Jeans law, emphasizing that Planck's model accounts for quantum mechanics, while the classical Boltzmann law applies in thermodynamic equilibrium at temperature T.

PREREQUISITES
  • Understanding of blackbody radiation principles
  • Familiarity with Planck's law and Rayleigh-Jeans law
  • Knowledge of thermodynamic equilibrium concepts
  • Basic grasp of quantum mechanics and classical physics
NEXT STEPS
  • Research the implications of Planck's law on blackbody radiation
  • Explore the concept of standing waves in thermodynamic systems
  • Investigate the differences between quantum and classical interpretations of energy distribution
  • Learn about the equipartition theorem in statistical mechanics
USEFUL FOR

Astronomers, physicists, and students studying thermodynamics and quantum mechanics will benefit from this discussion, particularly those interested in the theoretical models of blackbody radiation.

greswd
Messages
764
Reaction score
20
I was reading this article which talks about the theoretical model behind blackbody spectra:
http://www.cv.nrao.edu/course/astr534/BlackBodyRad.html

At the start, it mentions standing waves in a cavity. Standing waves in this model consist of an integer number of wavelengths. The standing waves can only occur at specific, discrete wavelengths.

Therefore, could a blackbody emit in a discrete spectrum under certain conditions?
 
Physics news on Phys.org
If you want to discuss Planck vs Rayleigh-Jeans, take note that Planck's difference only comes in at this section:

James J. Condon said:
In thermodynamic equilibrium at temperature T, equipartition of energy implies that each mode has average energy <E>=kT according to the classical Boltzmann law (but not according to quantum mechanics).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
5K