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Why are emitters colored black better emitters than other colors? Why is white a worse emitter?

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Black is the best emitter because if something is black it means it's strongly coupled to the electromagnetic field.

Thermal radiation

Physical objects are surrounded by electromagnetic radiation. One source of radiation is stars such as the sun. Stars emit visible light, infrared radiation, and in fact a whole range of wavelengths. You can see a plot of the spectrum of sunlight on Wikipedia.$^{[a]}$ It also turns out that matter itself emits radiation. Because matter is at a nonzero temperature, the thermal "motion" of the atoms/molecules randomly causes them to jump to excited states, and when they jump down they can emit photons of electromagnetic radiation. This is called thermal radiation.

Different kinds of matter radiate their thermal energy away as electromagnetic radiation with different levels of efficiency. This is roughly because the electron (or other) transitions in various materials are more or less strongly coupled to the electromagnetic field, and because different materials have more or less available transitions at each wave length.$^{[c]}$ Therefore, for a given temperature, some materials emit their radiation faster than others.

Good emitters are good absorbers

Now here's the important part: materials which are good at emitting their thermal radiation are also good at absorbing incoming radiation. This is actually not surprising: the process of absorbing an incoming photon is exactly the reverse of the process of emitting one, so if a material has lots of available transitions inside of it or transitions which are more strongly coupled to the electromagnetic field, then those transitions are available for both emission and absorption.

And now we answer the question

A material which is a very good absorber looks black because it absorbs all incoming light. Well that sounds convincing but if you think about it you would point out that I told you that good absorbers are also good emitters, so a good absorber should give off light too. It shouldn't look black! The key here is that it doesn't have to emit the photons at the same wavelength as it absorbs them. What generally happens is that after an e.g. visible photon comes in and gets absorbed, the energy is converted to heat, and it is later emitted by a thermal emission as described above. This thermal emission may be in the infra-red range (for practical temperatures), which you can't see with your eyes. That's why an object which is both a good absorber and a good emitter looks black when it's at thermal equilibrium near room temperature.$^{[d]}$ Note that this means that black objects are actually radiating more power than white ones, you just can't easily tell because that power is in a wavelength you don't see.

More information

A hypothetical object which perfectly absorbs all incoming radiation is called a black body. Using quantum mechanics and statistical mechanics you can compute the amount of radiation power a black body at a given temperature should emit at each wavelength.$^{[b]}$ Interestingly, if you look at the plot of the sun's emission, you see that it's pretty close to an ideal black body.

Ever notice that thermal blankets are shiny? Shiny materials are weakly coupled to the electromagnetic field. They're shiny precisely because they reflect incoming radiation instead of absorbing it. Since bad absorbers are bad emitters, this also means that when you wrap yourself in a shiny blanket you will radiate away your body heat more slowly, keeping you warmer in a cold environment. Of course, wearing a shiny blanket would also prevent you from warming up in the sun light. The way to think about it is that the shiny thing isolates you from the surroundings: it prevents you from getting warmed by incoming radiation, and it keeps you from getting cold from outgoing radiation.

This is also why thermos bottles are shiny.

$[a]$: Note that the peak radiation power from the sun is at around $500\,\text{nm}$ wavelength, which is right in the middle of the the visible spectrum. Coincidence?

$[b]$: In fact, it was in trying to calculate the black body emission spectrum that folks realized that classical physics had problems. In classical physics the black body would radiate an infinite amount of power. Planck discovered that he could fix this by assuming the energy was quantized. By his own admission it was a total un-motivated hack at the time but it worked and was a major jumping-off point for quantum mechanics.

$[c]$: For details simply refer to Fermi's Golden Rule.

$[d]$: Of course, hotter objects look non-black colors. For example, if you heat up a piece of metal in the fire it may glow red.

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    $\begingroup$ +1, but could you make 'strongly coupled to the electromagnetic field' precise for me? $\endgroup$ – Danu Jan 17 '15 at 9:21
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    $\begingroup$ @Danu: Simple case: if an atom has a resonance at a particular frequency it is a very good absorber/emitter at that frequency. $\endgroup$ – DanielSank Jan 17 '15 at 9:28
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    $\begingroup$ Its a very thorough answer and I've upvoted it, but you could point out that the best emitters can be all sorts of colours if you change their temperature (e.g. the photosphere of the Sun). They are only black below about 2000K. $\endgroup$ – Rob Jeffries Jan 17 '15 at 9:30
  • $\begingroup$ @RobJeffries perhaps a link to previous black body radiation questions is more fitting, e.g. physics.stackexchange.com/questions/89477/…. I only know about this one since I answered it, but there are bound to be some similar ones. $\endgroup$ – Danu Jan 17 '15 at 9:31
  • $\begingroup$ Could you elaborate on the comment about the peak radiation power being in the green range? Isn't that the only color that leaves don't absorb (or rather absorb the least) -- i.e. the opposite of what I would expect based on your comment? $\endgroup$ – marcianx Mar 23 at 17:08

protected by ACuriousMind Mar 29 '15 at 2:08

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