Deriving the density of states for waves in a blackbody

In summary, the conversation discusses the calculation of the density of states for waves inside a blackbody box. The notes provided by the professor mention the need for the position of the walls to be a node for each dimension and how to calculate the number of waves that can fit inside the box. The conversation also includes a discussion of the derivation of the density of states and the need to account for 2 independent polarizations for each mode. The conversation concludes with a clarification of what is meant by the term "polarization" in this context.
  • #1
daselocution
23
0

Homework Statement


Hey all,

I am having trouble following some of the notes that my professor posted with regards to waves inside a blackbody; here is what he posted: (the part in bold is what I am just not understanding)

"Inside the blackbody box, we need for the position of the walls of the box to be a node for each dimension. For a one-dimensional cavity, e.g., we would choose x = 0 at one wall and then require the other end to be 2a/λ = n (where a is the length of of a side of the cube, which he defined as the shape of this particular blackbody). How many such waves can we fit inside a box of dimensions a^3? We can figure out how to calculate this geometrically by looking at a cube of points in a 3-dimensional space where each point represents a coordinate defined by (nx, ny, nz) and nx = 1, 2, 3, ... and the same for ny and nz. This is a graph in which only positive integer values for nx, ny, and nz are allowed and we want to know the size of the cube (i.e. the number of points defined by the coordinates we just defined) that fit inside of the 1/8 of a sphere whose radius is r = (2a/λ) with shell thickness givenby dr = (2a/λ^2)dλ. The net result of the density of states calculation is

N(λ)dλ= (2πa^3)/(cλ^2) dλ


I have absolutely no idea how he made the jump to that result, and he didn't indicate any other calculations in the note page.


The Attempt at a Solution



Since there was a dλ in the density of states he arrived at, I thought he might have taken the dV/dr of a sphere such that dv=4πr^2 dr

As he noted in the paragraph above, we should only be looking in 1/8 of this region, so I multiplied this dv*1/8 (which is the same effect as just multiplying the original volume by 1/8...) and got

dv=(πr^2)/2 dr

I then substituted dr=(2a/λ^2) as indicated in the above paragraph and got:

dV=aπr2a/λ2
I then substituted in r2=4a22...

What I did is clearly wrong. I have no idea how he arrived at having a 2 in the numerator rather than a 4, or how he got a c in the denominator. Any help any of you could give me would be HUGELY appreciated.

Thanks
 
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  • #2
It looks to me that your derivation is essentially correct! The instructor's result is clearly incorrect as N(λ)dλ should be dimensionless. You still need to multiply your result by 2 to take into account the 2 independent polarizations for each mode.

More often, the derivation of the density of states is carried out in terms of frequency (##\nu##) or wavenumber (k) instead of wavelength. But you can then easily express these results in terms of wavelength. For example, you can compare your result with that given http://www.phys-astro.sonoma.edu/people/faculty/tenn/p314/BlackbodyRadiation.pdf .
 
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  • #3
Ah thank you very much. I am glad to see that I am not quite as lost as I thought I was, your input is much appreciated!
 
  • #4
TSny said:
You still need to multiply your result by 2 to take into account the 2 independent polarizations for each mode.

Could you please explain to me what you mean by the two independent polarizations for each mode? Does this mean that the standing wave could be sin or -sin? I guess I never really understood this when my professor mentioned it...
 
  • #5
For a standing wave on a vibrating string, the string could oscillate in a vertical plane. Then you would have vertical polarization. Or, the string could oscillate in a horizontal plane. Other directions of polarization could be obtained by superposition of the vertical and horizontal modes. So, you can take the horizontal and vertical modes as 2 independent polarizations from which all other polarizations can be constructed by superposition. Likewise for electromagnetic waves where the polarization is determined by the direction of the electric field.
 

1. What is a blackbody?

A blackbody is an idealized object that is capable of absorbing and emitting all wavelengths of electromagnetic radiation with 100% efficiency. It does not reflect or transmit any radiation, and its temperature determines the distribution of wavelengths it emits.

2. How does the density of states relate to waves in a blackbody?

The density of states is a measure of the number of possible states that a system can have. In the case of waves in a blackbody, the density of states represents the number of possible modes or wavelengths of electromagnetic radiation that can exist in the blackbody at a given temperature.

3. What is the derivation process for the density of states in a blackbody?

The derivation process involves using statistical mechanics and quantum mechanics principles to calculate the number of modes or wavelengths that can exist in a blackbody at a given temperature. This involves considering the energy levels of the electromagnetic radiation and the number of possible states for each energy level.

4. Why is the density of states important for understanding blackbody radiation?

The density of states is important because it allows us to calculate the energy distribution of electromagnetic radiation from a blackbody at a given temperature. This is important for understanding the properties of blackbodies and their role in various physical phenomena, such as thermal radiation.

5. How does the density of states change with temperature for waves in a blackbody?

The density of states increases with temperature for waves in a blackbody. This is because an increase in temperature leads to an increase in the number of possible states or modes that can exist in the blackbody, resulting in a higher density of states.

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