# What causes hot things to glow, and at what temperature?

I have an electric stove, and when I turn it on and turn off the lights, I notice the stove glowing.

However, as I turn down the temperature, it eventually goes away completely. Is there a cut-off point for glowing?

What actually is giving off the light? Does the heat itself give off the light, or the metal?

It is due to thermal radiation. Bodies with temperature above absolute zero emits radiation. If frequency of the radiation is in the visual range the body "glows".

When the electrons in the atom are excited, for example by being heated, the additional energy pushes the electrons to higher energy orbits. When the electrons fall back down and leave the excited state, energy is re-emitted in the form of a photon. -wikipedia(emission spectrum)

This explains that heat itself is not giving off the light.

$u(\lambda,T) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}}-1}$

If you integrate $u(\lambda,T)$ wrt $\lambda$ from $\lambda = 380nm$ to $\lambda = 750nm$ you will find that there is always some radiation in visual range. But it is very very small at room temperature(T = 300K) and hence undetectable to naked eye.

To get cut off temperature($T_c$) for glow. You need to solve

$\displaystyle \int_0^{T_c} \int_{\lambda = 380nm}^{\lambda=780nm} u(\lambda,T)d\lambda dT$ = minimum required to be detected by human eye.

• Good answer. I think the most important point to understand is the one you make in the first paragraph. To put it another way: Cold stuff is emitting radiation too, just not at a high enough frequency to be visible to the human eye. – Tim Goodman Nov 11 '10 at 19:08
• This is a great answer :) Btw, is this process the same way by which a lightbulb shines, or a halogen lamp glows? – Justin L. Nov 11 '10 at 21:14
• @Justin L.: I think so! – Pratik Deoghare Nov 11 '10 at 21:17
• @Justin L: It is. Halogen lights are brighter because they're hotter. They can be hotter because the halogen gases reduces the decay of the tungsten filament. In turn, this makes them slightly more efficient, but still they produce lots of heat and little light. – MSalters Nov 12 '10 at 12:15
• This is not such a good answer. The "heat itself" is what makes the light. You will make thermal radiation even when the atoms are not excited, just by collisions of atoms. It is actually more difficult to understand how atoms with sharp spectral lines can stay in equilibrium with radiation, which is what led to the A,B coefficients and spontaneous emission prediction by Einstein. – Ron Maimon Dec 24 '11 at 14:49

I would like to emphasize a point mentioned in the first answer.

The classical black body radiation has to do with a gas of photons in thermal equilibrium. The photons interact with each other and that interaction gives the usual Planck spectrum for their energies.

Inside stars, there are photons produced which interact with the stellar matter and they both come to a thermal equilibrium. When the photons escape the star, they have a black body spectrum.

When you take some amount of matter (an electric stove) and heat it, there is the question of where do the photons come from. In low temperatures, you don't have enough thermal energy to excite electrons in atoms. So, where do the photons come from? The answer is that they must come from the excitation of the background electromagnetic field. Even if you tried to isolate the heated matter from every electromagnetic field, you could never block out the vacuum fluctuations.

If you do the approximation of a black body (and it is not a bad approximation), you can measure the temperature of a body with the help of the spectrum of an object. The key is the Wien's displacement law:

http://en.wikipedia.org/wiki/Wiens_displacement_law

This is the way a lot of thermometers work.

So the light given off from an incandecent light bulb is the same type (photonic) as the light given off from an LED (just a different frequency makup).

• The light given off by an LED or florescent is non-thermal, i.e. it is the electrical excitation of some mode which decays giving off a visible light photon. I think for florescents, the photons are actually in the UV, and the coating on the inside of the bulb converts the UV to visible light. – Omega Centauri Dec 13 '10 at 1:15