# Why must black bodies in thermal equilibrium emit the amount of radiation absorbed?

I'm studying black bodies for IB Physics and I'm having trouble understanding the full concept of black body radiation. I understand that a black body in thermal equilibrium with its surroundings will emit all radiation that is incident upon it. My question is, why must this radiation be emitted at all? Why can't the object continue to heat up from the absorbed radiation? If energy is transferred to the black body, why can't it's temperature increase rather than the energy being emitted? I'm barely starting to cover quantum mechanics, so my knowledge on the subject is bare which is probably the source of my confusion. Any help would be greatly appreciated. Thank you in advance.

• I guess maybe the electron can't contain the excitement for long... – user6760 Aug 13 '16 at 7:29
• Because if they don't then they aren't in equilibrium. That's what thermal equilibrium is. – immibis Oct 3 '16 at 9:35

The definition of thermal equilibrium is that at equilibrium there is no net transfer of heat, so it is this requirement that forces the blackbody to emit as much energy as it absorbs.

In general an object can heat up (or cool down) by absorbing (respectively emitting) radiation, but then the object is not in thermal equilibrium and will have a lower (respectively higher) temperature than its surroundings.

• So is my intuition backwards? I thought a black body emitted the same amount of radiation absorbed because it was in thermal equilibrium, but is it actually the case that a black body is in thermal equilibrium because it absorbs the same amount of radiation it emits? – B. Sandoval Aug 13 '16 at 7:31
• Its simpler than that. Saying "a blackbody is in thermal equilibrium" is just a fancy way of saying "its emitting the same amount of heat as it is absorbing". there is no real difference between the 2 statements. – By Symmetry Aug 13 '16 at 7:35
• I see. I'm used to thinking of heat transfer as merely convection and conduction, so I hadn't thought of it that way. – B. Sandoval Aug 13 '16 at 7:39

As long as a body has temperature it will radiate energy. All the way down to absolute zero. If the electrons are excited (accelerated) they will radiate photons. If energy goes in faster than it can radiate then the body will heat up. The hotter it gets the higher the frequency of the photons. But more to your question if an electron is accelerated it will radiate energy.

I understand that a black body in thermal equilibrium with its surroundings will emit all radiation that is incident upon it.

This is true when interpreted correctly, but it is not the best thing to begin with when trying to understand what is going on.

All bodies reflect some part of radiation that is directed at them and absorb the rest. They also emit their own radiation. All at the same time.

In thermodynamic equilibrium, the energy of body is constant, so if there are above processes going on, they must balance each other:

energy of ingoing radiation = energy of reflected radiation + energy of emitted radiation.


The defining property of black body is that it does not reflect any part of the ingoing radiation. If any foreign radiation is directed at it, it will absorb it all.

However, black body is still allowed to emit radiation. If black body is to be in thermal equilibrium with surroundings, energy of this emitted radiation must be equal to energy of the ingoing radiation.

My question is, why must this radiation be emitted at all? Why can't the object continue to heat up from the absorbed radiation? If energy is transferred to the black body, why can't it's temperature increase rather than the energy being emitted?

If the ingoing radiation is not in equilibrium with the body, the body can heat up or cool down, depending on the quality of the ingoing radiation. For example, if concentrated ray of light (say, due to contribution of LED or laser light) is directed at a body, the body will heat up. Black body is no different in this respect.

The speed of heating up will be highest for black body though, as it reflects nothing. But this kind of process immediately destroys thermodynamic equilibrium. When the ingoing radiation disappears, the system begins to change and gets closer to another state of thermodynamic equilibrium, only this time, with higher temperature.