Black Body Radiation Equation Expressed without Reference to Temperature

Introduction

Suppose we can measure black body radiation frequency but we know nothing about temperature or Boltzmann's constant. We would like to express the black body radiation equation in terms of the average frequency that we observe. The following shows how we can do that.

Calculating P(f) in Terms of <f>

The black body radiation equation is

$P (f) = \frac{h f^{3}}{c^{2}} \frac{1}{(\exp (\frac{h f}{k T}) - 1)}$

(0.1)

Our goal here is to express this in term of the average frequency as shown below:

$P (f) = \frac{h f^{3}}{c^{2}} \frac{1}{(\exp (\frac{f}{a}) - 1)}$

(0.2)

a is a frequency that we will evaluate in terms of the average frequency <f>.

To evaluate <f> we need to obtain the integral ratio:

$< f > = \frac{\int_{0}^{\infty} f P (f) d f}{\int_{0}^{\infty} P (f) d f}$

(0.3)

The actual Bose integral of the function in equation (0.2) is pretty complicated so we will first obtain the value of a for a simpler function that Planck used when he discovered how to fit black body radiation to a curve that involved his famous constant h.

$P' (f) = \frac{h f^{3}}{c^{2}} \exp [- \frac{h f}{k T}] \to \frac{h f^{3}}{c^{2}} \exp [- \frac{f}{a}]$

(0.4)

For the expression in equation (0.4) we obtain after evaluating both integrals from equation (0.3) the value

$a = \frac{< f >}{4}$

(0.5)

And therefore we get

$P' (f) = \frac{h f^{3}}{c^{2}} \exp [- 4 \frac{f}{< f >}]$

(0.6)

This is a very interesting result. We have frequency (proportional to energy) to the third power multiplying an exponential whose argument is 4 times the ratio of energy to average energy. This energy distribution function is the same as we would have for massive entities with 8 degrees of freedom. Now we know that photons have 3 translational degrees of freedom as well as 2 polarization modes for each of those degrees of freedom thereby totaling 6 degrees of freedom. We now arrive at the dilemma "Where do the other 2 degrees of freedom come from?"

For P(f) the math is considerably more complicated but the result for a is almost the same. Reference 1 shows how to evaulate the integrals for P(f). The result for the fraction in equation (0.3) is

$< f > = \frac{\int_{0}^{\infty} f P (f) d f}{\int_{0}^{\infty} P (f) d f} = \frac{Γ (5) ζ (5)}{Γ (4) ζ (4)} a$

(0.7)

where $Γ (n) = (n - 1)!$ and $ζ$ is the Riemann zeta function which has fairly simple values and these are readily available from Wikipedia.

$\begin{array}{l} ζ (5) = 1.03692 \\ ζ (4) = \frac{π^{4}}{90} = 1.082323 \end{array}$

and therefore our expression for <f> evaluates to:

$< f > = 4 \frac{1.03692}{1.082323} a = 3.8322 a$

(0.8)

Therefore we can re-write our Bose-Einstein expression in terms of average frequency as:

$P (f) = \frac{h f^{3}}{c^{2}} \frac{1}{(\exp (\frac{3.8322 f}{< f >}) - 1)}$

(0.9)

Calculating f_max in Terms of <f>

Since we now have the expression for P(f) in terms of <f> we should also frequency, f_max, of the maximum power in the same terms. This will be a numerical result which is much simpler if we start with the Planck's original expression of P'(f).

Taking the derivative of equation (0.4) with respect to f we obtain:

$\frac{d}{d f} {\frac{h f^{3}}{c^{2}} \exp [- \frac{4 f}{< f >}]} = 3 \frac{h f^{2}}{c^{2}} \exp [- \frac{4 f}{< f >}] - 4 \frac{h f^{3}}{c^{2} < f >} \exp [- \frac{4 f}{< f >}] = 0$

(0.10)

Solving equation (0.10) for f_max we get:

$f_{\max} = \frac{3}{4} < f >$

(0.11)

We should expect a similar result with the Bose expression. Taking the same derivative of that expression we obtain

$\frac{d}{d f} {\frac{h f^{3}}{c^{2}} \frac{1}{(\exp (\frac{3.8322 f}{< f >}) - 1)}} = 3 \frac{h f^{2}}{c^{2}} \frac{1}{(\exp (\frac{3.8322 f}{< f >}) - 1)} - 3.8322 \frac{h f^{3}}{< f > c^{2}} \frac{\exp (\frac{3.8322 f}{< f >})}{{(\exp (\frac{3.8322 f}{< f >}) - 1)}^{2}} = 0$ (0.12)

Rewriting this equation by removing the common denominator and canceling other constants we obtain:

$f_{\max} = \frac{3}{3.8322} < f > \frac{\exp (\frac{3.8322 f}{< f >}) - 1}{\exp (\frac{3.8322 f}{< f >})}$

(0.13)

As a reasonable approximation, let $f = \frac{3}{3.8322} < f >$ and insert this into the exponentials of equation (0.13) and obtain:

$f_{\max} \approx \frac{3}{3.8322} < f > \frac{\exp (3) - 1}{\exp (3)} = \frac{3}{3.8322} 0.9502 = 0.7249 < f >$

(0.14)

Introduction

Calculating P(f) in Terms of <f>

Calculating fmax in Terms of <f>

Calculating f_max in Terms of <f>