3
$\begingroup$

When I look at an object in the sunlight, I'm looking at photons reflected back from the object. But if I wear infrared goggles and look at an object in the dark, am I looking at photons that are emitted by the object itself, rather than reflected? This is explained briefly in Tales of the Quantum. As I understand, the warmer the object is, the more the atoms vibrate. Atoms are made of +protons and -electrons. These charged particle vibrations create waves in the electromagnetic field which are photons of IR light.

Edit: From Wikipedia on Thermography: "All objects above the absolute zero temperature emit infrared radiation. Hence, an excellent way to measure thermal variations is to use an infrared vision device..." So it specifically says "emit infrared radiation" Is this emitted from the vibration of charged protons and electrons? And would increased vibration of these particles, caused by increased heat, thereby increase the IR emissions?

$\endgroup$

1 Answer 1

4
$\begingroup$

"Infrared" covers a wide band of wavelengths, everything from about 0.8 um at the long-wavelength limit of human vision to the 1-mm limit of conventional radio-frequencies. An object at 310 K (typical human body temperature) has its peak emission at about 10 um. However, detectors for this wavelength are much more expensive than those for shorter "near-IR" wavelengths closer to the visible band.

Typically infrared night-vision goggles contain an LED light source (essentially, an IR flashlight), and you are seeing photons from this source reflected off the object you are viewing.

Other night vision goggles don't contain a light source; they simply use a very sensitive detector and amplify the received signal to allow imaging in dim conditions. This type of goggles won't work in complete darkness. Again they would be detecting visible and IR light reflected off the objects from other sources.

It is also possible to design an imaging system that detects the longer wavelengths (9-15 um) emitted by objects at typical temperatures themselves. This is usually called thermal imaging or thermography rather than night vision.

$\endgroup$
5
  • $\begingroup$ From your final paragraph, does that mean that my OP assumption is correct, that the heat caused vibrations of protons and electrons leads to waves in the electromagnetic field that are the photons I see? I understand how electrons can produce photons, but how do + protons produce EM waves that are photons? $\endgroup$ Jan 7, 2021 at 16:34
  • $\begingroup$ @foolishmuse, protons can produce EM waves the same way electrons do. They have charge, therefore they can interact with the EM field. The question of whether a body composed of only neutral particles (if such a thing were possible) could radiate, I'm not clear on. The explanations I know for black-body radiation don't address this point. For example, the Wikipedia article on Planck's Law. $\endgroup$
    – The Photon
    Jan 7, 2021 at 16:44
  • $\begingroup$ Looking further, neutrons have no charge but they do have spin, therefore can interact with the EM field through their magnetic moment. So ultimately it looks like all the massive particles that make up everyday matter can interact with the EM field and radiate/absorb EM radiation. $\endgroup$
    – The Photon
    Jan 7, 2021 at 16:51
  • $\begingroup$ I'd like to research this further. Can you think of a reference or a concept that I should follow down the rabbit hole? $\endgroup$ Jan 7, 2021 at 17:09
  • $\begingroup$ @foolishmuse, I was looking at the Wiki articles on Black-body radiation, Planck's Law, Neutron, and Neutron star as I answered your comments. Probably there are textbooks out there covering this material but I don't know of specific ones beyond the level of introductory physics (i.e. Giancoli for example) $\endgroup$
    – The Photon
    Jan 7, 2021 at 17:12

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.