Confusion With Blackbody Radiation

In summary: This happens for temperatures above $\sim 10^10 \, K$.In summary, a blackbody is a theoretical object that perfectly absorbs all light that falls on it, but does not exist in reality. Other objects are close to being blackbodies but do not absorb 100% of light. Some articles refer to blackbodies as "perfect" or "ideal", but the defining property is that they absorb all light. Blackbodies are also good emitters of light, as seen with the sun emitting energy back into its surroundings. This is necessary due to the second law of thermodynamics. All objects above absolute zero emit radiation, and if their spectrum matches that of a blackbody, it is called blackbody radiation. Examples
  • #1
Amanda5455
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A blackbody is a theoretical object that perfectly absorbs all the light that falls on it. From what I understand this is an ideal situation and does not actually exist in reality. Certain objects are close to being a blackbody but they do not absorb 100% of the light that hits it (i.e. some gets reflected).

I have seen a few articles refer to a blackbody as a "perfect blackbody" or an "ideal blackbody". Isn't it redundant to describe a blackbody as being perfect or ideal? The defining property of a blackbody is that it completely absorbs all light that hits it.

So this is how I understand it: a blackbody is a perfect absorber and all other object that aren't perfect absorbers of light are close-blackbodies or near-blackbodies. I found the following quote online and the terminology contradicts what I have read and understood from other sources. If anyone knows which one is actually correct can you please clarify it for me.

"A perfect blackbody absorbs all wavelengths of incoming electromagnetic radiation perfectly; none of it is reflected. A perfect blackbody at a given temperature emits radiation in accordance with Planck's law. Any real blackbody is imperfect in both of these respects; it does not absorb all incoming radiation, and it does not have an emission spectrum that perfectly matches Planck's law."​

A blackbody is also a good emitter of light. If this was not true, water placed next to the sun would freeze instead of evaporating away. Logically, this makes sense because the sun absorbs the energy around it including energy stored in the water. This isn't what we observe in real life so energy must be emitted from the sun back into its surroundings. I just don't understand why this has to be the case. Theoretically, why couldn't the sun just absorb the surrounding energy without emitting it back? In a similar fashion to black holes (this idea could be totally wrong, I don't really know much about black holes). Ultimately my questions is why do good absorbers have to be good emitters too?

As a side note, would it be correct to call this system (the sun absorbing energy from its surroundings) endothermic?

All objects above absolute zero emit radiation. The amount of radiation is dependent on the object's temperature. Do we call this radiation blackbody radiation? This is where I am really confused. I thought blackbody radiation was electromagnetic radiation emitted by a blackbody a theoretically perfect absorber and emitter.

What would be some examples of objects found in nature that are very close to being a blackbody? So far I have stars/our sun and carbon black.
 
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  • #2
Amanda5455 said:
Theoretically, why couldn't the sun just absorb the surrounding energy without emitting it back?
It would violate the second law of thermodynamics.
In a more microscopical way: a large emission means there are available energy transitions - those can be used to emit and absorb light in the same way. Or, alternatively: the laws of the universe are time-symmetric (at least to our current knowledge, and neglecting some details of particle physics not relevant here). A time-reversed emission is an absorption.

Black holes do radiate (Hawking radiation) - they are extremely cold objects, but their spectrum should be the best blackbody spectrum we have in the universe. In practice we don't see it because the radiated power is incredibly tiny and everything around them is much hotter.
Amanda5455 said:
Do we call this radiation blackbody radiation?
If its spectrum is the same as the spectrum of a blackbody.

Amanda5455 said:
I thought blackbody radiation was electromagnetic radiation emitted by a blackbody a theoretically perfect absorber and emitter.
For very hot objects, massive particles (like neutrinos, electrons and positrons) join the emission spectrum.
 

What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect blackbody, which is an object that absorbs all radiation that falls on it and does not reflect or transmit any of it. This radiation is emitted at all wavelengths and is dependent only on the temperature of the object.

Why is blackbody radiation important?

Blackbody radiation is important because it provides a theoretical basis for understanding the emission of radiation from objects, such as stars and planets. It also plays a crucial role in fields such as thermal and quantum physics.

What is the relationship between temperature and blackbody radiation?

The relationship between temperature and blackbody radiation is described by the Planck's law, which states that the amount of radiation emitted by a blackbody is directly proportional to the fourth power of its temperature. This means that as the temperature of a blackbody increases, the amount of radiation it emits also increases significantly.

What is the blackbody radiation spectrum?

The blackbody radiation spectrum is the distribution of the intensity of radiation emitted by a blackbody at different wavelengths. This spectrum follows a specific shape, known as the Planck curve, which peaks at a specific wavelength that is dependent on the temperature of the blackbody.

How does blackbody radiation relate to the color of objects?

The color of an object is determined by the wavelengths of light that it reflects, absorbs, and transmits. Blackbody radiation plays a role in this because the temperature of an object determines the wavelengths of light that it emits, and these emissions can contribute to the overall color of the object.

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