In the answer of an exam said that a black body in thermal equilibrium with it's surroundings won't emit any energy, but I don't really understand why.

My logic is that every object emits electromagnetic radiation because of it's temperature, which is a form of transmission of energy, this happens microscopically because or molecules have a random motion, that is the temperature, so an accelerated charged particle emits electromagnetic radiation as a consequence of the varying $\vec{B}$ and $\vec{E}$, so independently the black body is in thermal equilibrium it will always emit radiation as a consequence of the temperature. The energy radiated minus the energy received is the heat flow, so in thermal equilibrium the radiation emitted must equal the received and then the black body and surroundings wouldn't spontaneously change their temperature.

So my idea was that the same it gives the same as it receives and so it wouldn't be correct to say it doesn't emit anything. Can someone explain what was the failure in my approach.


At thermal equilibrium, a blackbody will have no net emission of energy, but that's not the same thing as no emission at all. Obviously, a blackbody will emit blackbody radiation, but at equilibrium it will absorb exactly the same amount of energy. If it emits more than it absorbs, its temperature will fall; if it absorbs more than it emits its temperature will rise. In neither case will it be at thermal equilibrium.

I suspect whoever wrote the exam was being very careless.

  • $\begingroup$ (Assuming no other energy transportation mechanisms) $\endgroup$ – Rob Jeffries Oct 4 '17 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.