# Black Body Radiation & how it works?

• KuroAlchemist
In summary, the conversation discusses the concept of the Black Body curve and the UV catastrophe. The perfect black body absorbs all electromagnetic radiation and emits the same amount of energy from wall to wall. At a constant temperature, the black body is in thermal equilibrium and any decrease in temperature will result in a decrease in emission until equilibrium is restored. The amount of UV radiation emitted by a black body depends on the temperature and increases with higher temperatures. Planck's Law explains the relationship between temperature and emission of photons.
KuroAlchemist
This is just for grasping the concept of the Black Body curve and the UV catastrophe. I don't understand how ideal Black Bodies work? or how they absorb and re-emit the energy which they absorbed? Also, why does radiation bounce around the walls of a cavity of a Black Body? shouldn't the black-coated walls absorb the incoming radiation and not let it bounce?

Thank you anyone who helps! :)

By bounce they mean that it is continually absorbed and emitted. If the black body is heated and the inside is a hollow cavity with no way for light to escape, radiation passes back and forth between the walls continuously.

A "perfect" black body is one that absorbs ALL EM radiation that passes through it. In reality there are no perfect black bodies. It turns out that ALL objects will emit EM radiation if they are at a temperature above absolute zero. This radiation is emitted by the charged particles in the atoms that make up the object. It is effectively the reverse of the object absorbing radiation. In the cavity above, if all sides of the black body are equal in temperature, the radiation emitted by AND absorbed by each wall is equal in energy.

You've given a ton of info that was looking to verify what I thought.

Anyways, I understand that in reality there is no "perfect" black body and that it is a theoretical but ideal make for phsyics. However, I'm having trouble understanding how this black body is able to absorb and emit equal amounts of energy from wall to wall, are the walls in thermal equilibrium with each other? And what happens to the energy in the black body as it cools down? is the equilibrium distrubed?

I'm sorry but thank you for your reply!

Well, let's assume a few things here. Let's assume that the entire black body is in thermal equilibrium. Let's also assume that no heat is lost on the outside as well. Now, if all the walls are equal in temperature then they emit, on average, the same amount of radiation over time as they absorb. This is because each wall emits the same amount as every other wall. And assuming that it is a perfect black body, then each wall absorbs all the radiation that hits it. So, each wall is the same temperature on average, it emits the same amount of energy and absorbs the same as every other wall.

If we cool down one side of this black body, then the corrosponding wall inside the black body will emit less energy thanks to its lower temperature until everything equalizes again.

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Oh, it is in thermal equilibrium because the temperature of the walls are constant throughout, hence, causing them to emit, on average, the same amount of radiation and because it is a perfect black body, it will absorb all the radiation falling on the walls. This absorption and emission is done at a constant rate between the atom's electrons (as they lose and gain energy) and the energy avliable inside the hollow cavity?

So, if we drill a hole inside that cavity and cause radiation to sip out, the energy circulation inside the cavity is decreasing, so would that mean that the electrons are also losing energy and converting it to the pool of energy bouncing around the hollow cavity to adjust the thermal equilibrium, thus cooling down?

That was just my train of thought.

Yep. That is correct.

Oh sweet!
Thank you for solving my problems! I appreciate the help I got!

KA.

I'm sorry, could I ask one more thing?
In the Black Body Radiation curve that Planck theoretical proved, why is only a small amount of UV radiation or high frequency radiation given out of the ideal balck body? If we keep increasing the temperature high enough, above the temperatures we are able to produce on earth, would we see more UV radiation given out?

Is the amount of UV photons given out up to the probablity of the electrons falling back to their ground state? Is it more probable to be giving out photons of light than of UV or higher?

Yes the higher the temperature the higher the energy of the most probable photon. Hot enough and the most probable photon will be 100 Gev gamma rays.

The full version requires a description of Plancks Law, which I am not familiar enough to give accurately. See here for more though: http://en.wikipedia.org/wiki/Planck's_law

But I can say that yes, when you increase the temperature you have more energy available to make more photons with higher energies.

hmm, okay.
I'll look into it but thank you both for you're help!

## 1. What is black body radiation?

Black body radiation is the electromagnetic radiation emitted by an idealized object that absorbs all radiation that falls on it. It is a theoretical concept used to describe the emission of radiation from an object at a given temperature.

## 2. How does black body radiation work?

Black body radiation works by the absorption and subsequent emission of electromagnetic radiation by an object. The object absorbs all wavelengths of radiation that fall on it and then re-emits this energy in the form of electromagnetic radiation.

## 3. What is the relationship between temperature and black body radiation?

The intensity of black body radiation increases with temperature. As the temperature of an object increases, the average energy of its particles also increases, resulting in a greater amount of radiation being emitted.

## 4. What is Wien's displacement law and how does it relate to black body radiation?

Wien's displacement law states that the peak wavelength of black body radiation is inversely proportional to the temperature of the object. This means that as the temperature of an object increases, the peak wavelength of its radiation shifts to shorter wavelengths.

## 5. How does black body radiation play a role in the study of astrophysics?

Black body radiation is an important concept in astrophysics as it helps us understand the properties of stars and other celestial objects. By studying the electromagnetic radiation emitted by these objects, we can determine their temperature, composition, and other important characteristics.

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