Blackbody Radiation Homework: Understanding the c/4 Factor

[Harrisonized]

Homework Statement

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radpow.html#c1

What I don't understand is the second part, with the angles. The more I think about it, the less it makes sense.

Homework Equations

R = c/4 U

The Attempt at a Solution

Whenever I did a surface integral in the past, for example, for ∫∫ j.dS in electromagnetism, I multiplied by a factor of cos θ. The flux orthogonal to the surface being considered is always the smallest. However, in the construction given on hyperphysics for blackbody radiation, the flux orthogonal to the surface is greater than the flux at an angle.

Furthermore, the angle seems arbitrarily defined because there are two degrees of freedom not counting the x-axis.

Lastly, I don't understand why there is a "longer element" and a "reduced cross section" when the effective volume of the slanted surface element is the same.

I've been looking everywhere for a better reason behind the "c/4" factor, but all I found are these:

http://sci-fix.blogspot.com/2010/07/greenhouse-effect.html

"The c/4 factor is because Planck's (and Rayleigh-Jeans) law stands for density (unit volume) of radiant energy, but what is actually measured is radiant emittance or spectral radiance (per unit surface), which depends as well on speed of outgoing radiation."

http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Physics%202/Slides/Module-5/Lec-23/lec23_8.html

"We need to average over all angles. In computing the radiant power, we get a factor of cos² θ which averages to 1/2."

I've been at this for hours. Someone please tell me how to -properly- get the c/4 factor.

 

[anathema]

Thank you for your question. The c/4 factor in the equation R = c/4 U represents the average radiative flux from a blackbody surface. This factor is derived from the Stefan-Boltzmann law, which states that the total radiative flux from a blackbody is proportional to the fourth power of its temperature. In order to calculate the average flux from a blackbody surface, we need to take into account the directional dependence of the radiation. This is where the cos θ factor comes in.

When we integrate over all angles, we get a factor of cos θ in the integral. This factor represents the directional dependence of the radiation, as you mentioned in your attempt at a solution. However, this factor is squared because we are dealing with the intensity of the radiation, which is proportional to the square of the electric field. So, the cos θ factor becomes cos^2 θ. When we integrate over all angles, this factor averages to 1/2, which is where the c/4 factor comes from.

In terms of the longer element and reduced cross section, this is due to the fact that when we consider a slanted surface element, the effective area for radiation to escape is reduced compared to a flat surface element. This is because the slanted surface element has a smaller cross-sectional area for radiation to pass through. Therefore, we need to take this into account when calculating the average flux from the slanted surface.

I hope this explanation helps to clarify the c/4 factor and the other aspects of the equation for blackbody radiation. Please let me know if you have any further questions.