I Questions about Jean-Rayleigh's derivation of Ultraviolet Catastrophe

  • I
  • Thread starter Thread starter LightPhoton
  • Start date Start date
AI Thread Summary
The discussion centers on Jean-Rayleigh's derivation of the Ultraviolet Catastrophe, specifically addressing the properties of a blackbody and the assumptions made in the derivation. It emphasizes that a perfect absorber, like Jean's cube, must also emit radiation with a blackbody spectrum due to thermal equilibrium. Questions arise about the validity of breaking electromagnetic waves into three independent components, with a request for a mathematical proof to support this assumption. Additionally, the use of the Equipartition theorem is debated, highlighting its application to the Hamiltonian of electromagnetic fields despite concerns about kinetic energy in EM waves. The conversation underscores the complexity of these concepts and the need for deeper mathematical insights.
LightPhoton
Messages
42
Reaction score
3
TL;DR Summary
Questions about Jean-Rayleigh's derivation of Ultraviolet Catastrophe related to black body, EM waves in cavity and use of equipartition theorem
I am following this video and; Eisberg and Resnick's Book for this derivation, for I cannot find other sources that go as in-depth as they do.


$$\Large\text{Question 1)} $$

Jean's cube, or the metallic cube, is assumed to be a perfect absorber. On this fact alone, authors state


Now assume that the walls of the cavity are uniformly heated to temperature T. Then the walls will emit thermal radiation which will fill the cavity. The small fraction of this radiation incident from the inside upon the hole will pass through the hole. Thus the hole will act as an emitter of thermal radiation. Since the hole must have the properties of the surface of a blackbody, the radiation emitted by the hole must have a blackbody spectrum; but since the hole is merely sampling the thermal radiation present inside the cavity, it is clear that the radiation in the cavity must also have a blackbody spectrum. In fact, it will have a blackbody spectrum characteristic of the temperature T on the walls, since this is the only temperature defined for the system.


Now, I am not sure how having the absorption properties of a black body implies that it must also have emission properties of the black body as well. The closest answer to this was in the comments of this question's answer, which is basically "that's what experiments tell us."



$$\Large\text{Question 2)} $$


In deriving the spectrum of EM waves inside the cube, we assume that the waves can be broken up into three independent components. This seems logical, but given the complexity of Maxwell's equations, I have a hard time buying this. In the video @19:42 the diagram helps in clearing this up and is pretty satisfactory, but a mathematical proof would be better.



$$\Large\text{Question 3)} $$



Why did it seem reasonable at the time to use the Equipartition theorem when it clearly only adds to the kinetic energy of the system? There is no sensible way of talking about the kinetic energy of EM waves, even if it has a quadratic form ##(\epsilon_0E^2/2)##
 
Physics news on Phys.org
LightPhoton said:
Now, I am not sure how having the absorption properties of a black body implies that it must also have emission properties of the black body as well. The closest answer to this was in the comments of this question's answer, which is basically "that's what experiments tell us."
We assume thermal equilibrium. In equilibrium, absorption is in equilibrium with emission.

LightPhoton said:
In deriving the spectrum of EM waves inside the cube, we assume that the waves can be broken up into three independent components. This seems logical, but given the complexity of Maxwell's equations, I have a hard time buying this. In the video @19:42 the diagram helps in clearing this up and is pretty satisfactory, but a mathematical proof would be better.
Why do you say three? EM waves have two polarizations. There are 3 directions in space so naively one might expect 3 polarizations, but the longitudinal polarization (the one in the direction of motion) is missing due to the fact EM field is massless. I don't know a simple explanation of this, but if you are ready for a complicated one it can be found in many places.

LightPhoton said:
Why did it seem reasonable at the time to use the Equipartition theorem when it clearly only adds to the kinetic energy of the system? There is no sensible way of talking about the kinetic energy of EM waves, even if it has a quadratic form ##(\epsilon_0E^2/2)##
The equipartition theorem is valid whenever the Hamiltonian is quadratic in canonical coordinates and momenta, i.e. something of the form ##H \sim ap^2+bx^2##, where ##a## and ##b## are constants. The EM Hamiltonian is exactly of this form, ##H \sim aE^2+bB^2##, where ##E## is proportional to canonical momentum and ##B## to canonical coordinates. To see this, one has to write the EM theory in terms of gauge potentials, which are the fundamental canonical variables.
 
  • Like
Likes LightPhoton
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top