I'm having my first lectures on Quantum Physiscs. We just discussed the black body and thermal radiation. However, why does a heated body emit light? How it is that a rise in temperature produce a change in the electromagnetic field?

Feel free to use equations to answer.


3 Answers 3


IchVerloren, welcome to Physics SE!

The thing isn't temperature changing the electromagnetic field. In its most fundamental form, temperature $T$ is the relationship between lack of information (entropy $S$) and total energy $U$:

\begin{equation} \frac{1}{T}=\frac{\partial S}{\partial U}. \end{equation}

It is a statistical parameter that measures the spread of your energy over the degrees of freedom of the system. Loosely speaking, that equation means that the higher the temperature, the less new degrees of freedom have to be excited for me to increase the system's energy in a thermodynamically consistent way (with lower energy states being occupied first, in a proportion known as the Boltzmann distribution).

Now, consider the electromagnetic field as our system. It can have excitations, and therefore, have energy. We can define a temperature for anything with energy. The degrees of freedom of the EM field are precisely its modes of vibration. As such, the temperature is related to how many of these modes are occupied. Since lower energy modes need to be more excited at lower temperatures, and experiment tells us that lower frequencies dominate at low temperatures, we can postulate a direct relationship between energy resolution (energy of a single excitation) $E$ and mode frequency $\nu$ as


which is quite ad-hoc but can be made quite rigorous with in depth thermodynamic arguments. For the electromagnetic field to have temperature, it needs blackbody radiation.

But maybe you were interested in where this radiation comes from. Well, roughly speaking, electrons in materials also have statistical thermal distributions, and to maintain thermal equilibrium they are constantly being excited and dropping energy levels. This energy shifting is done through EM radiation (seeing electrons are charged particles) and has the blackbody spectrum.


You are here- und deshalb bist du nicht verloren! Here is why:

The surface of a hot body can be thought of as consisting of a huge number of tiny electromagnetic oscillators which can be excited by incoming photons. They can absorb the energy in those photons, and then by oscillating they can radiate that energy away again. In so doing, the oscillators strive towards an equilibrium between the radiation in the space next to the surface.


From my understanding, it's to stay in thermal equilibrium. Let's say that your object sit on a table and the room's lights are on. Then the object consistently absorbs photons from the light source. For the object to not have a net gain of energy, it needs to emit light in the infrared spectrum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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