# Planck's formula for Blackbody Radiation

1. Oct 6, 2008

### raghav

1. The problem statement, all variables and given/known data
Suppose that a blackbody spectrum is specified by Spectral Radiancy $$R_{T} (\nu) d\nu$$ and Energy Density $$\rho_{T} (\nu) d\nu$$ then show that
$$R_{T} (\nu) d\nu = \frac {c}{4}\cdot \rho_{T} (\nu) d\nu$$

2. Relevant equations

$$\rho_{T} (\nu) d\nu = \frac{8 \pi h \nu^{3} d\nu}{c^{3}(e^{\frac{h\nu}{kT}}-1)}$$

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 7, 2008

### gabbagabbahey

What do you know about spectral radiancy for a blackbody? What have you tried?

3. Oct 7, 2008

### raghav

Spectral Radiancy $$R(\nu)$$ is defined such that $$R_{T}(\nu)d\nu$$ gives the energy absorbed per unit area per unit time when radiation has frequency between $$\nu$$ and $$\nu + d\nu$$. It is essentially power

4. Oct 7, 2008

5. Oct 7, 2008

### gabbagabbahey

If you integrate $R_{T}(\nu)d\nu$ over all possible frequencies you should get the total radiated power per unit area:

$$\Rightarrow \int_0^{\infty} R_{T}(\nu)d\nu=\sigma T ^4$$

What quantity do you get when you integrate $\rho_{T} (\nu) d\nu$ over all possible frequencies?

6. Oct 7, 2008

### raghav

Since the latter quantity denotes energy density, the integral over all frequencies should give us total radiant energy contained in the cavity, which is $$kT$$ if we assume that $$T$$ is the absolute temperatre of the cavity.
Am I right sir?

7. Oct 7, 2008

### gabbagabbahey

No, the total radiated energy will not be kT; try carrying out the integration with the expression you provided for the energy density.

8. Oct 7, 2008

### raghav

hmm yes sir I did that, and got something like this:
$$\frac{8(kT)^{4}\pi^{5}}{(hc)^{3}\cdot 15}$$

Ok so now in order that i prove my problem All i do is to divide Stefan's Law by the above right? keeping of course in mind stefans constant in terms of $$h, c , k$$.
Is the relation true only between the integrals over $$0 to \infty [\tex] or is it to even for spectral Radiancy? Last edited: Oct 7, 2008 9. Oct 7, 2008 ### raghav I fail to understand why that sentence of mine is appearing latexified 10. Oct 7, 2008 ### Rake-MC backslash instead of forward slash? 11. Oct 7, 2008 ### raghav oh ok sir 12. Oct 7, 2008 ### gabbagabbahey Substitute [tex]\sigma= \frac{2 \pi ^5 k^4}{15h^3c^2}$$

into the expression, and notice that you get:

$$\int_0^{\infty} \rho_{T} (\nu) d\nu=\frac{4 \sigma T^4}{c}$$

$$\Rightarrow \sigma T^4=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu$$

What does that mean $R_{T}(\nu)d\nu$ is?

13. Oct 7, 2008

### raghav

Hence it means that $$R_{T}(\nu}d\nu$$ is the power absorbed, per unit area by radiation with frequency lying in $$\nu$$ and $$\nu + d\nu$$ . But I dont quite get the physical interpretation of the result. Do you mind throwing some light on that sir?

14. Oct 7, 2008

### gabbagabbahey

Actually. you already knew that $R_{T}(\nu)d\nu$ was the power radiated per unit area between $\nu$ and $nu +dnu$. You used that to show that

$$\int_0^{\infty} R_{T}(\nu)d\nu=\sigma T ^4$$

You then used your expression for the energy density to show

$$\sigma T^4=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu$$

Putting the two together you get:

$$\Rightarrow \int_0^{\infty} R_{T}(\nu)d\nu=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu$$

$$\Rightarrow R_{T}(\nu)d\nu=\frac{c}{4} \rho_{T} (\nu) d\nu$$

doesn't it?

The physical interpretation of this result is that the radiated power is proportional to the energy density.

15. Aug 7, 2010

### reckon

An old thread, but hope this posting will give something to anyone confused on this topic and ended up here (just like me).

I tried to derive this on my own but didn't get the factor 1/4. I got the proportionality value only c, which I thought physically sensible because the energy inside the cavity of a blackbody was carried away by EM waves whose speed is c. But my answer was not in agreement with the book I was reading, c/4.

After searching for sometime, I finally found this:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c5

the factor of 1/4 was there because of two facts. Roughly, the two factors are:
1. in thermal equilibrium, the radiated power is the same as the absorbed power. This account for 1/2 if we are going to calculate only the radiated power.
2. the radiation going through the hole comes from every direction, and thus, it needs to be multiplied by cos^2 (comes from dot products of area and velocity of radiation A.v) which then averaged and account for another 1/2

Hope this help anyone who ended-up here when searching the meaning for the factor 1/4.