Power for a blackbody radiation

In summary, the power emitted by a blackbody in equilibrium with the surrounding electromagnetic field is related to the energy density of the field by the equation P=\frac{cu}{4}. Despite attempts to derive this equation, it is still considered a challenging task. However, the temperature dependence of the blackbody power spectrum is included in the expression for the energy density of the photon gas in the cavity. The standard derivation of this equation can be found in texts such as Reif's book, which provides a clear explanation. As the temperature of a blackbody radiator increases, the overall radiated energy also increases and the frequency of the peak shifts to higher frequencies.
  • #1
dEdt
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If a blackbody is in equilibrium with the surrounding electromagnetic field, the power emitted by the surface of the blackbody will be related to the energy density of the electromagnetic field by [itex]P=\frac{cu}{4}[/itex]. Try as I might, I haven't found a good derivation for this equation (the Hyperphysics one has several problems). I thought that it shouldn't be too hard to derive, but playing with Poynting's theorem led me nowhere, so now I humbly turn to physicsforums for help.
 
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  • #2
Wouldn't it also depend upon the temperature?
 
  • #3
dEdt said:
Try as I might, I haven't found a good derivation for this equation (the Hyperphysics one has several problems).

What's wrong with the standard derivation in texts? See e.g. pp. 385-388 of Reif.

UltrafastPED said:
Wouldn't it also depend upon the temperature?

The usual temperature dependence of the blackbody power spectrum is contained in the expression for the energy density of the photon gas in the cavity.
 
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  • #4
WannabeNewton said:
What's wrong with the standard derivation in texts? See e.g. pp. 385-388 of Reif.

I just checked Reif's book and found his derivation to be very clear. Thanks.
 
  • #5
dEdt said:
I just checked Reif's book and found his derivation to be very clear. Thanks.

No problem! It's a brilliant gem of a book.
 
  • #6
When the temperature of a blackbody radiator increases, the overall radiated energy increases
 

1. What is blackbody radiation?

Blackbody radiation, also known as thermal radiation, is the electromagnetic radiation emitted by a perfect blackbody at a specific temperature. A perfect blackbody is an object that absorbs all of the radiation that falls on it and emits radiation at all wavelengths.

2. What is the power of blackbody radiation dependent on?

The power of blackbody radiation is dependent on the temperature of the blackbody. It follows the Stefan-Boltzmann law, which states that the total power emitted by a blackbody is proportional to the fourth power of its temperature.

3. How does the power of blackbody radiation change with increasing temperature?

As the temperature of a blackbody increases, the power of its radiation increases at a much faster rate. This is because of the fourth power relationship between temperature and power in the Stefan-Boltzmann law. Therefore, a small increase in temperature can result in a significant increase in the power of blackbody radiation.

4. Can the power of blackbody radiation be calculated?

Yes, the power of blackbody radiation can be calculated using the Stefan-Boltzmann law. The equation is P = σAT⁴, where P is the power emitted, σ is the Stefan-Boltzmann constant, A is the surface area of the blackbody, and T is the temperature in Kelvin.

5. How is blackbody radiation used in science and technology?

Blackbody radiation is used in various scientific and technological applications, such as in thermal imaging cameras, solar panels, and in the study of the cosmic microwave background radiation. It is also an important concept in understanding the behavior of stars and other celestial bodies.

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