Deriving the Total Energy from Blackbody Radiation: A Mathematical Approach

In summary, the conversation discusses the integration of an equation for the energy density of radiation from a black body. The person asking the question is confused about the steps in the integration process and is seeking clarification. The expert explains that the geometric series representation is used and then substituted into the integral, which is evaluated using integration by parts. The last line in the derivation is also explained. The person asking the question thanks the expert and the conversation ends.
  • #1
Xyius
508
4
This might be more of a mathematical question than a physical one. But I am taking a Quantum Mechanics course and the book starts out by introducing the equation for the energy density of radiation from a black body. They then integrate this expression over infinity to find the total energy per unit volume.

http://img256.imageshack.us/img256/5189/blackbody.jpg [Broken]

My question is, how did they do the integral? It looks like they turned [itex]\frac{1}{e^{x}-1}[/itex] into its geometric series representation. That part I understand. But what do they do in the step after that? Where does the geometric series go? And where does the [itex]\frac{1}{(n+1)^4}[/itex] come from? And for that matter, the last line in the derivation?

I know its not an incredibly crucial question in understanding the Physics, but it bugs me a lot when I cannot follow the mathematics.
 
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  • #2
Well when you pull the sum out to the front of the integral, you have [tex]x^3e^{-(1+n)x}[/tex]. Then when you substitute y = (n+1)x, you have to use x^3 dx= y^3dy / (n+1)^4. The integral can then be evaluated, presumably by parts, to get 6. Evaluating the sum is a bit tricky. If I was working through the derivation, I'd just be satisfied with looking up the answer. This page gives a few clever ways of doing it: http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90
 
  • #3
Thank you!
 

1. What is blackbody radiation energy?

Blackbody radiation energy, also known as thermal radiation, is the electromagnetic radiation emitted by a perfect blackbody at a given temperature. It is a form of heat transfer that occurs when an object's temperature is above absolute zero.

2. What is the significance of blackbody radiation energy?

Blackbody radiation energy is significant because it helps us understand the relationship between temperature and the amount of radiation emitted by an object. It also plays a crucial role in various fields such as astronomy, thermodynamics, and quantum mechanics.

3. How is blackbody radiation energy related to temperature?

The intensity and wavelength of blackbody radiation energy are directly related to the temperature of the object. As the temperature increases, the intensity and peak wavelength of the emitted radiation also increase.

4. Can blackbody radiation energy be observed in real life?

Yes, blackbody radiation energy can be observed in real life. For example, the glow of a hot metal object or the heat emitted by the sun are both examples of blackbody radiation energy.

5. How is blackbody radiation energy calculated?

The amount of blackbody radiation energy emitted by an object can be calculated using the Stefan-Boltzmann Law, which states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its absolute temperature. This law is expressed as E=σT^4, where E is the energy, σ is the Stefan-Boltzmann constant, and T is the temperature in Kelvin.

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